mm-486
===
Subject: Re: [No math] web2news down?
> For a short while, I was using web2news.com to read and post
to this ng,
> but now I can never connect thereto. Anyone else having this
trouble?
yup, cannot seem to get there myself either. I'd give it a day
or two
before I'd suspect that the site was dead. May just be a big
network/hardware fault for them.
===
Subject: Re: [No math] web2news down?
> For a short while, I was using web2news.com to read and post
to this ng,
> but now I can never connect thereto. Anyone else having this
trouble?
Yup. Shame, isn't it?
- EM
===
Subject: Re: Amusing Terminology Trivia Question
|What is the noun form of (non)abelian? I am hoping not to
have to
|settle for (non)commutativity.:-)
I'm pretty sure it's abelianness. A Google ([tm] I assume)
search suggests that commutativity and noncommutativity are
more popular than abelianness and nonabelianness by a wide
margin, though. There were only 20 hits for nonabelianness.
Google's dictionary service has this to say:
abelian
A*beli*an, Abelite Abel*ite, AbelonianA`bel*oni*an, n.
(Eccl.
Hist.) One of a sect in Africa (4th century), mentioned by St.
Augustine,
who states that they married, but lived in continence, after
the manner,
as
they pretended, of Abel.
Source: Webster's Revised Unabridged Dictionary,  1996, 1998
MICRA, Inc.
Keith Ramsay
===
Subject: Re: Amusing Terminology Trivia Question
> I'm pretty sure it's abelianness. A Google ([tm] I assume)
> search suggests that commutativity and noncommutativity are
> more popular than abelianness and nonabelianness by a wide
> margin, though. There were only 20 hits for nonabelianness.
A Google search for nonabelianity yields 2 hits. I haven't yet
tried
abelianity or (non)abelianism.
---- David
===
Subject: Re: Ann on evolution of ideas
If you really are what you think you are, it naturally follows
that
visiting
a shrink would be no big deal.
But I know that you haven't got the guts to do it.
Imagine if he said you were crazy!
Haakon
> Jeez, Herc - I have never doubted that you are smart.
> But I seriously doubt your sanity!
> http://tinyurl.com/iky5
> http://tinyurl.com/iky8
> http://tinyurl.com/iky9
> http://tinyurl.com/iky4
> then why do people from around here admit i'm the truman?
> and how did I know Jim Carrey would costar Laurie Holden?
> and why did my post in rec.org.mensa PROOF OF GOD 0202 2002
> in the subject happen one year to the day before the shuttle
disaster, it
> even had the word preminition in it.
> and why do I have hundred of replies like this one
>
--------------------------------------------------------------
------------
------
> Randi will test you when you properly apply to be tested.
Sign up here:
> http://www.randi.org/research/challenge.html
> -----------------
> Rich Shewmaker
> CNote
> Wanda
> Rust
> James Randi is rather famous because he gets people onto a
stage
> and sets them up to do magic shows, literally. He even
offers to pay a
> million dollars if they succeed in disproving classical
science. He is
> a [rich show-maker]. Now 2 people from sci.math after days
of coersion
> have already guessed who the author of this post is. Is it
really a one
in 4
> coincidence? Could you guess it from 1000 names? And when I
beat 1000
> to one odds every 10 posts to me, then I'm not selecting
them from a
large
sample.
> You're all working back from the conclusion that no person
is unique, so
> claims of that are wrong somehow. I'm the truman dood, its
not that
unplausable,
> 100,000 people know it, media just prefers to stage it for
you. When I
> go out tomorrow 50 people will say along the lines of
there's the
truman
> or there's adam. then I come back onto the internet and
everyone is
docile.
> Herc
===
Subject: Re: Ann on evolution of ideas
> If you really are what you think you are, it naturally
follows that
visiting
> a shrink would be no big deal.
> But I know that you haven't got the guts to do it.
> Imagine if he said you were crazy!
Well OF COURSE he's going to say that. There's one in a billion
anyone is who i say i am, and theres one in 5 billion its me.
Not
going to affect his treatment result stats to make a
conclusion there.
Everyone in Townsville knows who I am though.
A peasant wearing Kings clothes is by definition crazy. It
doesn't mean
with absolute certainty he's not who he says.
you even looked at these?
> http://tinyurl.com/iky5
> http://tinyurl.com/iky8
> http://tinyurl.com/iky9
> http://tinyurl.com/iky4
> then why do people from around here admit i'm the truman?
I predicted Jim Carrey's costar of Majestic here before your
eyes, and
there's 130,000 witnesses who all tell me every day there's
the Truman,
I saw Adam
Exhibit A: http://tinyurl.com/fuf8 she looks exactly like
Laurie
Holden
Exhibit B: http://tinyurl.com/fuf2 government has spied on me
so long
Look I don't get why this goes past everybody this would be
admissible
in court against Warner Brothers as far as I see it.
they called the show TRUEman about a man on government allowed
camera.
It came out in 1998.
In 2002 Jim Carrey stars in a romance (obviously a follow on
from
Truman show) costarring a small screen actor for the first
time on big
screen Laurie Holden.
One, not only am I the only person in the world to claim this
show is
based on me
http://tinyurl.com/fufb 1998 : The Truman Show starring Jim
Carrey
http://tinyurl.com/fpg4 2002 : The Majestic Jim Carrey costars
Laurie
Holden
Here's the release dates of the movie.
when I said them. It's one thing to say look im the truman the
girl is
like laurie
holden to a bunch of people and then later they see it happen!
but to have the 2 posts in different newsgroups, and noone
here heard me
say directly the trumans affection is like laurie holden
before laurie
was big screen,
means you have to LOOK at the DATES.
And don't be indistinguishable from millions on usenet who
don't care
can't think
don't speak out
close minded
jump to conclusions from assumtions
believe only what their tv tells them
wont lift a finger to research
can't interpret time stamped evidence
can't reason IF there's a trueman, this is circumstial evidence
I'm going to the bank today, 50 people are going to call out to
me yelling things like I saw the truman. NO BULL
then I have to come back to usenet to try and put some sense
into the world
and wonder why no ing one can even consider the possibilty
its TRUE.
NIC CAGE does prison movies for s sake your all so ing
ignorant
THE TRUMAN THE ING TRUMAN work it out
Its the 5 billion minus the 100,000 in Townsville that are all
in fantasy
land
Herc
===
Subject: another very very simple question
X-URL:
http://mygate.mailgate.org/mynews/sci/sci.math/
d2455a26a63c5bc7e62a8c03f50132
02.35661%40mygate.mailgate.org
Definition:
Let I = {(a, b] | a,b are elements of R} We define the
collection of
borel sets in R as the smallest collection of subsets of R
satisfying
i). I is a subset of B
ii). if B(italic) is a subset of B, then R B(italic) is an
element of
B
iii). B is closed under countable additions.
Okay, what I'd like to know is whether the number of Borel
sets is
infinite in this case.
I think there must be an infinite number of subsets of R which
have
(a,b] as a subset, e.g. (a - delta, b], (a - 2*delta, b], (a -
3*delta]
etc.
So is it right that there is an infinite number of Borel sets?
Sorry, I
know this is child's play..........
David.
--
===
Subject: Re: another very very simple question
> Definition:
> Let I = {(a, b] | a,b are elements of R} We define the
collection of
> borel sets in R as the smallest collection of subsets of R
satisfying
> i). I is a subset of B
> ii). if B(italic) is a subset of B, then R B(italic) is an
element of
> iii). B is closed under countable additions.
> Okay, what I'd like to know is whether the number of Borel
sets is
> infinite in this case.
> I think there must be an infinite number of subsets of R
which have
> (a,b] as a subset, e.g. (a - delta, b], (a - 2*delta, b], (a
- 3*delta]
> etc.
> So is it right that there is an infinite number of Borel
sets? Sorry, I
> know this is child's play..........
Yes it is. There are already uncountably many inteverals (a,b].
They are all Borel sets.
--
Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html
His mind has been corrupted by colours, sounds and shapes.
The League of Gentlemen
===
Subject: Re: A Problem on probability
>* narendranath.ts@in.bosch.com
> I pick two numbers, randomly, and tell you one of them.
You are
supposed
> to guess whether this is the lower or higher one of the
two numbers I
> picked. Can you come up with a method of guessing that
does better
than
> picking the response low or high randomly
>If I know the distribution you pick numbers from, it is
easy. Your
>distribution, randomly, is not a distribution to calculate
from, so
>I could ask, What _is_ your distribution?
>Anyway, as far as I know, there _is_ a strategy to guess
better than a
>fifty/fifty one given you really use a distribution to pick
numbers
>from although it is a secret. However, I don't know it
because I
>never really understood it. Anyone care to elaborate?
>My question is: Given the optimal strategy, can you rethink
_your_
>strategy to fool me?
> There is a strategy which will do what is asked for,
> regardless of the distribution. There is no optimal
> strategy; for any method of guaranteeing better than .5,
> there is a strategy which will reduce the probability to
> as close to .5 as is desired.
There are responses worse than 0.5, but surely a random
response is always
0.5.
Herc
===
Subject: Re: A Problem on probability
> I pick two numbers, randomly, and tell you one of them. You
are
supposed
> to guess whether this is the lower or higher one of the two
numbers I
> picked. Can you come up with a method of guessing that does
better than
> picking the response low or high randomly
> Thaanx in Advance.
I don't think this is the context of the question but you
could use
feedback,
we are told ONE of them, but we don't know whether the one we
are
told is random from the pair, it could always be the lowest.
This is
assuming the
experiment is repeated.
Herc
===
Subject: Axiom of PC
An axiom of predicate calculus is VxVy(x=y->(p->p')), where V
stands
for
for all and p' is obtained from p after a substitution of some
x in p
with
y (not necessarily all).
What about Vy(x=y->(p->p')) ?? Is it true without Vx ??
I think it is, but I am not sure. That's because
Vy(x=y->(p->p')) == ~Ey(x=y & (p & ~p')).
Am I right?
===
Subject: Re: Axiom of PC
> An axiom of predicate calculus is VxVy(x=y->(p->p')), where
V stands
for
> for all and p' is obtained from p after a substitution of
some x in p
with
> y (not necessarily all).
> What about Vy(x=y->(p->p')) ?? Is it true without Vx ??
There is a convention that if a formula has unbound individual
variables
then it is to be read as if they were all bound by a prefix of
universal
quantifiers. According to that convention
Vy(x=y->(p->p'))
is to be read as
VxVy(x=y->(p->p')).
But whether your predicate calculus uses that convention, I
don't
know.
It isn't all together true to say An axiom of predicate
calculus is
VxVy(x=y->(p->p')), .... In some formulations it might be and
in
others it might not.
> I think it is, but I am not sure. That's because
> Vy(x=y->(p->p')) == ~Ey(x=y & (p & ~p')).
> Am I right?
--
G.C.
===
Subject: Re: Axiom of PC
> An axiom of predicate calculus is VxVy(x=y->(p->p')),
where V
stands
for
> for all and p' is obtained from p after a substitution of
some x in
p
with
> y (not necessarily all).
> What about Vy(x=y->(p->p')) ?? Is it true without Vx ??
> There is a convention that if a formula has unbound
individual variables
> then it is to be read as if they were all bound by a prefix
of universal
> quantifiers. According to that convention
> Vy(x=y->(p->p'))
> is to be read as
> VxVy(x=y->(p->p')).
> But whether your predicate calculus uses that convention, I
don't
> know.
Never heard of such convention for the moment.
> It isn't all together true to say An axiom of predicate
calculus is
> VxVy(x=y->(p->p')), .... In some formulations it might be
and in
> others it might not.
Can you show me an example? I thought VxVy(x=y->(p->p')) (with
p(x,a_1,...,a_n), p'(x,y,a_1,...,a_n) was invariant for
interpretation.
===
Subject: Re: Axiom of PC
 An axiom of predicate calculus is VxVy(x=y->(p->p')),
where V
stands
> for
> for all and p' is obtained from p after a substitution
of some x
in p
> with
> y (not necessarily all).
> What about Vy(x=y->(p->p')) ?? Is it true without Vx ??
> There is a convention that if a formula has unbound
individual
variables
> then it is to be read as if they were all bound by a
prefix of
universal
> quantifiers. According to that convention
> Vy(x=y->(p->p'))
> is to be read as
> VxVy(x=y->(p->p')).
> But whether your predicate calculus uses that convention,
I don't
> know.
> Never heard of such convention for the moment.
> It isn't all together true to say An axiom of predicate
calculus is
> VxVy(x=y->(p->p')), .... In some formulations it might be
and in
> others it might not.
> Can you show me an example? I thought VxVy(x=y->(p->p'))
(with
> p(x,a_1,...,a_n), p'(x,y,a_1,...,a_n) was invariant for
interpretation.
It will be a _theorem_ but not necessarily an axiom.
--
G.C.
===
Subject: Re: Axiom of PC
First, usually PC is a abbreviation for _propositional
calculus_.
A common abbreviation for predicate calculus/logic is FOPL
(first-order
predicate logic).
> An axiom of predicate calculus is VxVy(x=y->(p->p')), where
V stands
for
> for all and p' is obtained from p after a substitution of
some x in p
with
> y (not necessarily all).
Actually (if so) this is from an axiom of FOPL (predicate
calculus)
_with identity_.
> What about Vy(x=y->(p->p'))?? Is it true without Vx ??
Sure, if 'x' is replaced with a constant. (Depending an the
logical
system you are working with: there are several variants of
FOPL with
identity; some of them utilize constants others not.) Leading
to:
Ay(a = y -> (F(a) -> F(y)).
That's quite obvious, no? For any y that is a, we have F(y) if
F(a).
> I think it is, but I am not sure. That's because...
...there's a rule of derivation in predicate calculus that
allows for
dropping the universal quantifier(s). (If we have AxFx then
certainly we
also have Fa for a specific _a_.)
F.
===
Subject: Re: Axiom of PC
> There is a convention that if a formula has unbound
individual
variables
> then it is to be read as if they were all bound by a
prefix of
universal
> quantifiers.
> Never heard of such convention for the moment.
Actually, it's a convention quite popular in _mathematics_, I
guess.
> It isn't all together true to say An axiom of predicate
calculus is
> VxVy(x=y->(p->p')), .... In some formulations it might be
and in
> others it might not.
As I told you before: usually it [or a similar formula] is an
axiom of
_FOPL with identity_, but not of _FOPL_ (only).
F.
===
Subject: Re: Cantor's disproof Mk II
> You still do not seem to realize that you seem to
> be confusing the notions of a_n = x for some n and for
any
epsilon
> 0, there is an n such that |a_n - x| < epsilon. These
are very
> different notions. Take some topology. Learn some
mathematical rigor.
> Stop thinking so sloppily.
 but the list itself contains 2/3 just as it contains 1/3.
it got up to
0.22 already
> the essential_contradiction is cyclic, given a finite
space represent an
infinite field.
> this is not demonstrably the distinction to reals.
> What in the world are you trying to say here?!?!?! Speak
more clearly,
> for crying out loud!
> Herc
> Again I reiterate. 2/3 has no terminating representation in
base 4.
> Every number on your list does. Therefore 2/3 is not on your
list. You
> point out to me that this list contains .2, .22, .222...,
but what you
> are really doing is exhibiting an infinite subsequence which
converges
> to 2/3, not a member of the list which is 2/3. If 2/3 is on
your list,
> then tell me what value of n (where I call the nth number on
your list
> a_n) yields a_n=2/3. Get it into your head that what _we_
mean by 'the
> list contains the real number x' is 'a_n=x for some n', not
'I can get
> arbitarily close to x by choosing appropriate values of n'
already!
> What you mean by 'the list contains 2/3' is different from
what we
> mean (and what Cantor meant) by 'the list contains 2/3'! How
long will
> it take you to realize this?
about 3 years ago. I knew the original post was faulty I was
just seeing
if
you could distinguish between represented and identified,
since the opening
assumption on the 1/3 list is valid. Does this sound like I
don't
understand
diagonalization?
> I assert, then, that Cantor's principals are based
erroneously upon an
> axiom which is very subtle and ghostlike, so much so in fact
that even
> the most learned of the mathematical community are blind to
it, though
> it is right under their noses. To wit, I refer to that
ever-assumed
> hypothesis that we can construct a set whose members are
uncountable
> in the first place. But wait!, cry the critics, What of the
set of
> real numbers? But do not laugh at them, for these are very
subtle
> matters and their lack of understanding is worthy of our
sympathy.
> Let us, then, address their concern. For, indeed, we may say
Let R
> then be all such numbers which cannot be represented as the
quotient
> of two integers, and a unique well-defined set named R does
thus
> arise, there is no arguing about this. However, how do we
know R is
> uncountable? Because Cantor asserts that it is. But now I am
the one
> to cry But wait!. Cantor's proof subtly assumes that, even
if a
> transfinitude does exist (which we will for the moment
allow, just to
> give him the benefit of the doubt), that it is then possible
to
> construct a set of such magnitude. So when I assert that this
> constructability is unproven, and you offer R as a
counterexample with
> Cantor's assertian that R possesses such magnitude, it is
clear as day
> that you are using circular reasoning. But I do not hold
this against
Actually, Cantor assumes R are *countable* by putting them in
a list.
I have another refutation that he only makes minimal
assumptions here:
> all the reals arranged in some kind of (dis)order, then
constructing a
new
> real by taking the main diagonal of this array and
changing every
digit.
> This number is by construction different from every
member of what is
> already an infinite set (since each real can be tagged
with an index
number,
> which is an integer, and there is an infinity of
integers), therefore
it is
> not in the set, therefore the count of reals is a
greater infinity
than
> the count of integers.
 Correct me if I've got any of the above wrong.
 But does that not *assume* multiple infinities to start
with? If we
use my
> pet subject of the uniqueness of infinity, the modified
diagonal
number does
> find a place in the set of reals - at infinity, and no
fixup of
infinity is
> needed.
> it assumes a single infinity type, that reals can be
counted then it
> finds a contradiction. the difference is you can tell me
any rational
number
> (integer over integer) and I can count to it, but you
can't necessarily
count
> to a given real.
> Notice my comment about (dis)order in the original post.
Every account I
> have read about Cantor's method has illustrated the point
with a totally
> unordered (partial!) list of reals. Conceivably you could
find a real
number
> given its index number - even if there was no seek other
than to look
it
> up in an infinite book - but you could not go the other way.
I presume
> that's what you mean by can't necessarily count to a given
real.
it doesn't matter that the cantor list is random, he is trying
to see if
they can
be ordered and makes the assumption that they must be listed
somehow,
he arrives at a contradiction well before he has to establish
any
properties
of the possible ordering. can't necessarily count to a given
real i had
to
think how to word it correctly, another way is can't count to
every
real,
though that has a second meaning of all of them. basically you
can't
count to real numbers, but if you had specific real numbers
you could rig
a count to them, rationals are a subset of reals. the look up
process is
symmetric, 'from the integer index give the number' is what
the ordering
is, 'from the number what is the index' is just a process of
counting
through
each number until you find it. integers have definite
properties and sets
that map to integers must attain the properties.
> [unsnip Terry Wilder]
> There's a theorem that every set , let alone that of real
numbers, can
be
> well ordered, by reductio ad absurdum,
> using the axiom of choice.
> This was the original thought that started me off. Once a
systematic
> ordering scheme is applied to the reals one *can* go the
other way - to
find
> the index number of a real. A scheme that was discussed here
a while ago
was
> to reverse the order of the decimal digits, e.g. the real
0.5 will map to
> the index number 5, the real 0.00032759 will map to the
index number
> 95,723,000, etc. Conversely there is a unique real
corresponding to any
> integer one cares to name, to Graham's number and beyond,
and it is
equally
> easy to find. We *can* now put the reals in one-to-one
correspondence
with
> the integers.
> OK, so all nonterminating decimals, recurring or not, will
map to
infinity.
yes, your examples are all rationals, they can be counted.
> No problem, the one-to-one correspondence with the infinity
of integers
is
> still valid, there's enough room at infinity to accommodate
all the
> successive Godel-diagonalisations of the original list of
reals: each
step
> generates one more infinitely-long real and one more
infinite index
integer.
Neat, could be constuctively worded as longer rather than
infinitely long,
then you have a computable system that implies infinitely long
because the
index is infinite. I posted up Penrose version of this idea a
month ago.
#9. WE KNOW WE LEGITIMATELY CREATED THIS NEW Y & N PATTERN,
IE: IT IS TRUE.
YET NONE OF THE EXISTING AXIOM STATEMENTS PRODUCE THIS
DIAGONAL STATEMENT.
A
NEW AXIOM IS NEEDED TO EXPRESS THE DIAGONAL.
10. IF WE WRITE A NEW STATEMENT (CALL IT R) THAT INCLUDES A
PROCEDURE FOR
MAKING THIS DIAGONAL , AT SPACE R/R A NEW DIAGONAL LETTER WILL
APPEAR AND
WE
WILL HAVE TO ADD STATEMENT S TO REPRESENT THIS NEW SEQUENCE.
BUT AT S/S A
NEW DIAGONAL NUMBER WILL APPEAR, REQUIRING A STATEMENT T AND
SO ON,
INFINITELY.
the distintion is the diagonal you create every second row is
only
the diagonal of the preceding rows, not the entire set. put it
this way,
if that is acceptable then we can indeed make a well defined
order and make
our list of real numbers. what integer represents pi? can you
count to
it?
at any moment you index the list you only get a rational
number.
Cantors proof is valid because it gives a distinction between
classes of
numbers.
In essense it says infinitely long numbers don't all map to a
sequence. if
I
give you the number 0.3333 recurring, a consisely defined
number you can
say that is number 26 on my list. if I tell you square root of
2, another
consisely
defined number, it doesn't appear on the same list.
The missing diagonal number was only one number, say the list
of reals
starts
out with infinitely long random numbers. The axiom of choice
is the nth
digit
of the nth number. Not only can we add 1 to each digit, we can
add 2, 3,..
making 9 new diagonal numbers. We can add 1 to the first digit
making it
different to the first number, add 2 to the second digit
making it
different
to the second number, we can add any number to any digit of
the diagonal
and as long as every digit is changed we have a new number, so
you
have to add infinite diagonally constructed numbers, not just
one.
Try to count them out and it starts to look like a fractal
snowflake or
coastline, finite area but infinite perimeter, you try to
traverse but
never
move forward.
Herc
===
Subject: Closed form , improper integral
,, e denotes Napier's constant and let [.]
be the integral part. Prove or disprove that
Integral_{t=0 to t=infty}e^{-t}(t-1)^{n} dt = [ n!/e + 1/2] .
===
Subject: Re: comparing two lists
3QLpj-NoP*NzsIC,boYU]bQ]H'<P%JD/,.J>y<#4ga3$21:
> It's not C++, but in a relational database you'd just say:
> select count(*) from
> (select n.string from
> nlist n left join mlist m on n.string=m.string
> group by n.string)
> I wonder if there's some equally concise phrasing available
in
> something like Python.
I think the way to do it in Python is the same way people have
been
suggesting in C++: with a dictionary (hash table). The
following takes
as input two lists (or other iterable objects such as strings)
and
outputs a list of the number of occurrences in B for each item
in A.
def countMatches(A,B):
counts = {}
for x in B:
counts[x] = counts.get(x,0) + 1
return [counts.get(x,0) for x in A]
> countMatches(aeoiuy, Now is the time for all good men to
come to
the aid of their party.)
[3, 6, 8, 4, 0, 1]
--
David Eppstein http://www.ics.uci.edu/~eppstein/
Univ. of California, Irvine, School of Information & Computer
Science
===
Subject: Re: comparing two lists
...
>I've got two lists of objects (strings): the first list
contains N
*unique*
>objects, whereas the second list has M objects taken from
the first list
>(replicates and missing objects are allowed). Usually N is
much bigger
than
>M. I need to count how many objects from the list first are
contained in
>the second.
> Mildly unusual to find a problem where this N > M.
>I guess there must be something more efficien than just
comparing each
>single element from the first list to all the elements of
the second
one.
>Any suggestions?
...
> It's not C++, but in a relational database you'd just say:
> select count(*) from
> (select n.string from
> nlist n left join mlist m on n.string=m.string
> group by n.string)
> I wonder if there's some equally concise phrasing available
in
> something like Python.
...
I don't know for Python, but this is wrong in SQL as it will
return number
of records in table nlist (it's equivalent to select count(*)
from nlist).
Correct expression would be:
select count(*) from
(select n.string from
nlist n INNER join mlist m on n.string=m.string
group by n.string)
or :
select count(*) from NList where exists(select 1 from MList
where NList.str
= MList.str)
or:
select count(*) from (select distinct n.str from nlist n inner
join mlist
m
on n.str=m.str)
Goran
===
Subject: Re: comparing two lists
> It's not C++, but in a relational database you'd just say:
> select count(*) from
> (select n.string from
> nlist n left join mlist m on n.string=m.string
> group by n.string)
> I wonder if there's some equally concise phrasing available
in
> something like Python.
>I think the way to do it in Python is the same way people
have been
>suggesting in C++: with a dictionary (hash table). The
following takes
>as input two lists (or other iterable objects such as
strings) and
>outputs a list of the number of occurrences in B for each
item in A.
>def countMatches(A,B):
> counts = {}
> for x in B:
> counts[x] = counts.get(x,0) + 1
> return [counts.get(x,0) for x in A]
> countMatches(aeoiuy, Now is the time for all good men to
come to
>the aid of their party.)
>[3, 6, 8, 4, 0, 1]
Well if _that's_ what we want to do then that's the way to do
it all
right. But (confessing that I haven't paid any attention to
most of
the thread, didn't seem interesting until the word Python
appeared)
how did we decide that that's what we wanted to do given then
OP I need to count how many objects from the list first are
contained in the second. ? Seems to me like if the first list
is
aeoiuy and the second list is Now is the time for all good men
to come to the aid of their party. then how many objects from
the
list first are contained in the second is 5, not [3, 6, 8, 4,
0, 1].
I guess I gotta learn some English.
************************
David C. Ullrich
===
Subject: Re: comparing two lists
3QLpj-NoP*NzsIC,boYU]bQ]H'<P%JD/,.J>y<#4ga3$21:
>def countMatches(A,B):
> counts = {}
> for x in B:
> counts[x] = counts.get(x,0) + 1
> return [counts.get(x,0) for x in A]
> countMatches(aeoiuy, Now is the time for all good men
to come
to
>the aid of their party.)
>[3, 6, 8, 4, 0, 1]
> Well if _that's_ what we want to do then that's the way to
do it all
> right. But (confessing that I haven't paid any attention to
most of
> the thread, didn't seem interesting until the word Python
appeared)
> how did we decide that that's what we wanted to do given then
> OP I need to count how many objects from the list first are
> contained in the second. ? Seems to me like if the first
list is
> aeoiuy and the second list is Now is the time for all good
men
> to come to the aid of their party. then how many objects
from the
> list first are contained in the second is 5, not [3, 6, 8,
4, 0, 1].
Looking back at the original post, it seems you are correct.
So:
def intersectionLength(A,B):
A = dict([(x,None) for x in A])
B = dict([(x,None) for x in B])
return len([x for x in A if x in B])
> intersectionLength(aeoiuy, Now is the time for all good
men to
come to the aid of their party.)
5
In Python 2.3 this might be cleaner using sets instead of
dicts.
--
David Eppstein http://www.ics.uci.edu/~eppstein/
Univ. of California, Irvine, School of Information & Computer
Science
===
Subject: Re: comparing two lists
>I don't know for Python, but this is wrong in SQL as it will
return number
>of records in table nlist (it's equivalent to select count(*)
from nlist).
>Correct expression would be:
> select count(*) from
> (select n.string from
> nlist n INNER join mlist m on n.string=m.string
> group by n.string)
oops you're right, it's an inner join.
J.
===
Subject: Re: comparing two lists
quite the cautionary tale!
Good luck answering posts such as What is the point of rigor?
there,
guys.
--
Phlip
http://www.c2.com/cgi/wiki?TestFirstUserInterfaces
===
Subject: Re: comparing two lists
>def countMatches(A,B):
> counts = {}
> for x in B:
> counts[x] = counts.get(x,0) + 1
> return [counts.get(x,0) for x in A]
> countMatches(aeoiuy, Now is the time for all good men
to come
to
>the aid of their party.)
>[3, 6, 8, 4, 0, 1]
> Well if _that's_ what we want to do then that's the way to
do it all
> right.
Actually I think I lied there -
def countMatches(A,B):
counts = {}
for x in A:
counts[x] = 0
for x in B:
counts[x] = counts[x] + 1
return counts
seems better; Counts['a'] is a more obvious way to retrieve the
information we want than Counts[0]. (never mind...)
> But (confessing that I haven't paid any attention to most of
> the thread, didn't seem interesting until the word Python
appeared)
> how did we decide that that's what we wanted to do given
then
> OP I need to count how many objects from the list first are
> contained in the second. ? Seems to me like if the first
list is
> aeoiuy and the second list is Now is the time for all good
men
> to come to the aid of their party. then how many objects
from the
> list first are contained in the second is 5, not [3, 6, 8,
4, 0, 1].
>Looking back at the original post, it seems you are correct.
>So:
>def intersectionLength(A,B):
> A = dict([(x,None) for x in A])
> B = dict([(x,None) for x in B])
> return len([x for x in A if x in B])
Well, I should just break down and get a current version
instead
of asking, but I can't imagine why the dicts would be needed
there;
although I didn't know the exact syntax I conjectured
elsewhere that
a one-line
def intersectionLength(A,B):
return len([x for x in A if x in B])
would work. It doesn't? (Note we were given that A has no
repetitions.)
> intersectionLength(aeoiuy, Now is the time for all good
men to
>come to the aid of their party.)
>In Python 2.3 this might be cleaner using sets instead of
dicts.
I _hate_ installing new software when the old version still
does everything I want, but you may have talked me into it
here.
************************
David C. Ullrich
===
Subject: Re: comparing two lists
3QLpj-NoP*NzsIC,boYU]bQ]H'<P%JD/,.J>y<#4ga3$21:
> Actually I think I lied there -
> def countMatches(A,B):
> counts = {}
> for x in A:
> counts[x] = 0
> for x in B:
> counts[x] = counts[x] + 1
> return counts
> seems better; Counts['a'] is a more obvious way to retrieve
the
> information we want than Counts[0]. (never mind...)
First rule of posting Python code: always run a test case.
> countMatches(aeoiuy, Now is the time for all good men to
come to
the aid of their party.)
Traceback (most recent call last):
File <stdin>, line 1, in ?
File <stdin>, line 6, in countMatches
KeyError: N
That doesn't look like an improvement to me...
>def intersectionLength(A,B):
> A = dict([(x,None) for x in A])
> B = dict([(x,None) for x in B])
> return len([x for x in A if x in B])
> Well, I should just break down and get a current version
instead
> of asking, but I can't imagine why the dicts would be needed
there;
> although I didn't know the exact syntax I conjectured
elsewhere that
> a one-line
The dicts are not needed. You will get correct results with
just the
last line, if A and B are lists. The dicts will make it run in
near-linear time instead of quadratic time, though.
--
David Eppstein http://www.ics.uci.edu/~eppstein/
Univ. of California, Irvine, School of Information & Computer
Science
===
Subject: Re: comparing two lists
> Actually I think I lied there -
> def countMatches(A,B):
> counts = {}
> for x in A:
> counts[x] = 0
> for x in B:
> counts[x] = counts[x] + 1
> return counts
> seems better; Counts['a'] is a more obvious way to retrieve
the
> information we want than Counts[0]. (never mind...)
>First rule of posting Python code: always run a test case.
> countMatches(aeoiuy, Now is the time for all good men to
come to
>the aid of their party.)
>Traceback (most recent call last):
> File <stdin>, line 1, in ?
> File <stdin>, line 6, in countMatches
>KeyError: N
Aargh. Good point. What I shoulda just said was that if we're
going to calculate that histogram I'd rather have a dict than
a list, but that of course figuring out how to do that
properly was
beyond my feeble capabilities. Aargh.
>That doesn't look like an improvement to me...
>def intersectionLength(A,B):
> A = dict([(x,None) for x in A])
> B = dict([(x,None) for x in B])
> return len([x for x in A if x in B])
> Well, I should just break down and get a current version
instead
> of asking, but I can't imagine why the dicts would be
needed there;
> although I didn't know the exact syntax I conjectured
elsewhere that
> a one-line
>The dicts are not needed. You will get correct results with
just the
>last line, if A and B are lists. The dicts will make it run in
>near-linear time instead of quadratic time, though.
Ok, that makes sense. If I hadn't decided to shut up after the
humiliation above I'd suggest that the one-liner nonetheless
gives a cooler answer to the original conjecture about a
concise Python version.
Oops.
************************
David C. Ullrich
===
Subject: Re: COMPLEX MATHEMATICS - FAKE????
> A remarkable outcome of Gamma Function research is that
there SHOULD
> be a negation spiral in the Gamma Function.
> However, if you put one in, the function falls to pieces.
> This SHOWS that complex maths is incompatible with the Gamma
Function.
> That cannot be disputed.
> However, is complex mathematics incompatible with ALL
functions - in
> HIDDEN WAYS?
> The process of logic can be followed by studying
> Then you either click on the link at the bottom of the page,
or go to
> http://www.wehner.org/euler/solution.htm to see what is not
actually
> the solution, but the reason that the riddle is INSOLUBLE.
> Are there researchers out there who have come across Nature
rejecting
> the complex mathematics in relation to OTHER functions?
> It is all very odd.....
> Charles Douglas Wehner
Perhaps you should take up another line of work. You might
consider
numerology or astrology, for example.
--
There are two things you must never attempt to prove: the
unprovable --
and the obvious.
--
Democracy: The triumph of popularity over principle.
--
http://www.crbond.com
===
Subject: Re: COMPLEX MATHEMATICS - FAKE????
didn't get as far as your critique o'the Gamma func., but
anyone who bashes Stevie STTNG Pantheon Hawking gets my vote;
plus, it's a nice exercise.
> Then you either click on the link at the bottom of the page,
or go to
> http://www.wehner.org/euler/solution.htm to see what is not
actually
--UN HYDROGEN (sic; Methanex (TM) reformanteurs) ECONOMIE?...
La Troi Phases d'Exploitation de la Protocols des Grises de
Kyoto:
(FOSSILISATION [McCainanites?] (TM/sic))/
BORE/GUSH/NADIR @ http://www.tarpley.net/aobook.htm.
Http://www.tarpley.net/bushb.htm (content partiale, below):
17 -- L'ATTEMPTER de COUP D'ETAT, 3/30/81
===
Subject: Re: COMPLEX MATHEMATICS - FAKE????
X-Cise: tanbanso@iinet.net.au
X-CompuServe-Customer: Yes
X-Coriate: admin@interspeed.co.nz
X-Ecrate: tanandtanlawyers.com
X-Punge: Micro$oft
X-Sanguinate: themvsguy@email.com
X-Terminate: SPA(GIS)
X-Tinguish: Mark Griffith <markgriffith@rocketmail.com>
X-Treme: C&C,DWS
said:
>This SHOWS that complex maths is incompatible with the Gamma
>Function. That cannot be disputed.
You're a few fries short of a happy meal. This cannot be
disputed.
--
Shmuel (Seymour J.) Metz, SysProg and JOAT
Any unsolicited bulk E-mail will be subject to legal action. I
reserve the
right to publicly post or ridicule any abusive E-mail.
Reply to domain Patriot dot net user shmuel+news to contact
me. Do not
reply
to spamtrap@library.lspace.org
===
Subject: Re: Concise Proof of Goldbach's Conjecture
> So the guy posted a really _bad_ 'theorem'; did it merit 7
replies?
> Or can you respondents not see each others' posts?
Using Google, I could not see the two replies before mine when
I posted.
--
J K Haugland
http://www.neutreeko.com
===
Subject: Re: Congruent Multiple-Sums (harmonic-number related)
Oops, you did define H(n, m).Sorry about that.
> Let H(0,m) = 1/m, for m = all positive integers;
> and let, for all n = positive integers,
> H(n,m) = sum{k=1 to m} H(n-1,k).
> (So, H(1,m) = H(m) = the m_th harmonic-number.)
> (Also, H(n,m) = binomial(m+n-1,n-1) (H(m+n-1) -H(n-1)).)
> (I have posted on this sum-of-sum-of... many times before.)
> Now, let
> G(0,n,m) = H(n,m) *m;
> and let, for all positive integers r,
> G(r,n,m) = sum{k=1 to m} G(r-1,n,k).
> (G(r,n,m)
> = binomial(r+n+m-1,r+n) (n*H(r+n+m-1) -n*H(r+n) +1), I
think.)
> And, finally, let,
> for n+1 >= m,
> F(0,n,m) = H(n+1-m,m) *m *(-1)^m;
> and let, for all positive integers q,
> F(q,n,m) = sum{k=1 to m} F(q-1,n,k).
> (I have not come up with a closed-form for F as of
currently.)
> Then:
> (m-1)! G(r,n,m) is always congruent to
> (m-1)! F(q,n,m) (-1)^m (mod{m+q+r})
> Leroy
> Quet
> Leroy,
> I'm a bit lost with what you're saying (It's probably my
fault). You
clearly
> defined H(0,m) for m= all positive integers. Then you go on
to talk about
> H( n, m).Now this is where I get lost. In case n /= 0 I
think that you
did
> not define H( n , m ).
===
Subject: Re: constructing an ovoid from N given points
>... So you've got five degrees of freedom, which means you
can expect
> to pass a conic through most any set of five points, and
that conic is
> unique; if it looks like the points lie on an ellipse, they
probably
> do.
> Or, solving for y in terms of x in
> a x^2+ 2 h x y + b y^2 + 2 f x + 2 g y + 1 = 0 one obtains
> y=A +/- sqrt[A^2- B ] where A= - (g+h x)/b and B=(a x^2+2 f
x+1)/b
> At point of vertical tangency, there is a repeated root ,
radical
> disappears.
> Question is, when 5 points are given, can we solve out for
the
> coefficients a,h,b,f and g using any of the two equations?
By what
> method? Is the method of resultants indicated by Robert
Israel for
> polynomials applicable here ? Or Eliminate command of
Mathematica or
> Maple? Are least squares methods possible?
>That raises a question of how the conic section originated.
>Given : a x^2+ 2 h x y + b y^2 + 2 f x + 2 g y + 1 = 0 (5
constants)
>can we find the semi-vertical angle of cone and, the angle
between
>cone axis to the cutting plane in terms of these 5 constants
?
> This suggestion for cone angle and inclination was intended
for use on
> mechanical softwares like IDEAS, MDT or 3DStudio which
enables one to
> visualize the intersection between a cone and plane.
> Hope my queries supplement the OP ...
Take a look at 100 Great Problems of Elementary Mathematics
by Heinrich Dorrie, Dover Publications ($13).
It has, among its 100 problems (all with solutions):
Find a parabola from either 4 points or 4 tangents.
Draw a conic section from 5 elements, any combination of points
and tangents.
Draw the points of intersection of a given straight line
with a conic section of which 5 elements - points or tangents -
are given.
Draw the tangents from a given point
to a conic section of which 5 elements - points or tangents -
are given.
I find this book quite interesting a worth rereading.
Martin Cohen
===
Subject: Definitions and Theorems
I'm an undergrad in math and after reading some books I have a
question
about the uniqueness of definitions and theorems. Ie. Let's
say we're
referring to a standard theorem T (the T theorem) where it's
written in one
book as
If A then if B then C
and another as
If A and B then C.
Both of these theorems would be referred to as the T theorem
as if there
was
a unique theorem. But these are not the same sentences. Is the
T
theorem
supposed to refer to one of the similar ways of writing the
theorem?
Leon
===
Subject: Re: Definitions and Theorems
Visiting Assistant Professor at the University of Montana.
>I'm an undergrad in math and after reading some books I have
a question
>about the uniqueness of definitions and theorems. Ie. Let's
say we're
>referring to a standard theorem T (the T theorem) where it's
written in
one
>book as
>If A then if B then C
>and another as
>If A and B then C.
>Both of these theorems would be referred to as the T theorem
as if there
was
>a unique theorem. But these are not the same sentences. Is
the T
theorem
>supposed to refer to one of the similar ways of writing the
theorem?
Both statements are logically equivalent. This can be checked
with
truth tables, if you want. Remember that the truth table of an
implication X->Y is
X Y X->Y
0 0 1
0 1 1
1 0 0
1 1 1
That is, If X then Y is false if and only if X is true and Y is
false.
The truth table of the conjunction and is:
X Y X and Y
0 0 0
0 1 0
1 0 0
1 1 1
That is, X and Y is true if and only if both X and Y are true.
Now, If A then if B then C is A->(B->C). If A and B then C is
(A & B) -> C.
Now we can compare the truth tables of both A->(B->C) and
(A&B)->C and
see if they are the same:
A B C A&B (A&B)->C A->(B->C) (B->C)
0 0 0 0 1 1 1
0 0 1 0 1 1 1
0 1 0 0 1 1 0
0 1 1 0 1 1 1
1 0 0 0 1 1 1
1 0 1 0 1 1 1
1 1 0 1 0 0 0
1 1 1 1 1 1 1
As you can see, the only way in which either If A and B then C
and
If A then if B then C could be false is in the case where both
A and
B are true and C is false. So the two statements are logically
equivalent. There difference is as logically insignificant as
the
difference between the statement in English and its
translation into
Spanish.
It's not denial. I'm just very selective about
what I accept as reality.
--- Calvin (Calvin and Hobbes)
Arturo Magidin
magidin@math.berkeley.edu
===
Subject: Re: Definitions and Theorems
> I'm an undergrad in math and after reading some books I have
a question
> about the uniqueness of definitions and theorems. Ie. Let's
say we're
> referring to a standard theorem T (the T theorem) where it's
written in
one
> book as
> If A then if B then C
> and another as
> If A and B then C.
> Both of these theorems would be referred to as the T theorem
as if there
was
> a unique theorem. But these are not the same sentences. Is
the T
theorem
> supposed to refer to one of the similar ways of writing the
theorem?
Definitions and theorems generally have minor differences from
book to
book. As a common example, some books defined countable as
either
finite or countably infinite. Other books define countable as
countably infinite.
Depending on the definition being used, some theorems will
need slight
adjustment. In some books the statement, a countable set can
be put
into 1-1 correspondence with the natural numbers, is true. In
other
books it's false.
Moral of the story is to always make sure you understand the
particular
definitions your book is using.
===
Subject: Re: Definitions and Theorems
>I'm an undergrad in math and after reading some books I
have a question
>about the uniqueness of definitions and theorems. Ie. Let's
say we're
>referring to a standard theorem T (the T theorem) where
it's written in
one
>book as
>If A then if B then C
>and another as
>If A and B then C.
>Both of these theorems would be referred to as the T
theorem as if there
was
>a unique theorem. But these are not the same sentences. Is
the T
theorem
>supposed to refer to one of the similar ways of writing the
theorem?
> Both statements are logically equivalent. ...
If I recall correctly the two halves of the equivalence even
have names.
The inference from (A & B) -> C to A -> (B -> C) is called
exportation;
and the inference from A -> (B -> C) to (A & B) -> C is called
importation.
Or vice versa.
--
G.C.
===
Subject: Re: Definitions and Theorems
>I'm an undergrad in math and after reading some books I have
a question
>about the uniqueness of definitions and theorems. Ie. Let's
say we're
>referring to a standard theorem T (the T theorem) where it's
written in
one
>book as
>If A then if B then C
>and another as
>If A and B then C.
>Both of these theorems would be referred to as the T theorem
as if there
was
>a unique theorem. But these are not the same sentences.
But they obviously mean the same thing.
>Is the T theorem
>supposed to refer to one of the similar ways of writing the
theorem?
Both and neither: The T theorem is the _fact_ that's expressed
by
both sentences, not the precise wording.
>Leon
************************
David C. Ullrich
===
Subject: Re: Definitions and Theorems
> I'm an undergrad in math and after reading some books I have
a question
> about the uniqueness of definitions and theorems. Ie. Let's
say we're
> referring to a standard theorem T (the T theorem) where it's
written in
one
> book as
> If A then if B then C
> and another as
> If A and B then C.
> Both of these theorems would be referred to as the T theorem
as if there
was
> a unique theorem. But these are not the same sentences.
Right. Two different sentences, but both express the same
mathematical
fact (or state of affairs) if you like.
If A then if B then C is true if and only if If A and B then C
is
true. (Where A,B,C may be replaced with a n y declarative
sentences.)
F.
===
Subject: Re: Definitions and Theorems
> I'm an undergrad in math and after reading some books I have
a question
> about the uniqueness of definitions and theorems. Ie. Let's
say we're
> referring to a standard theorem T (the T theorem) where it's
written in
one
> book as
> If A then if B then C
> and another as
> If A and B then C.
> Both of these theorems would be referred to as the T theorem
as if there
was
> a unique theorem. But these are not the same sentences. Is
the T
theorem
> supposed to refer to one of the similar ways of writing the
theorem?
They're both logically equivalent.
Recall: If A then B has the value - A / B by definition.
Hence: If A then if B then C -> - A /(-B / C)
If A and B then C -> -(A / B)/ C
= (- A / -B)/ C by De Morgan's law
= - A /(-B / C) by associativity of
/
-Bill Dubuque
===
Subject: Re: Definitions and Theorems
Leon <Leon@nospam.net> escreveu na mensagem
> If A then if B then C
> and another as
> If A and B then C.
> Both of these theorems would be referred to as the T theorem
as if there
> was a unique theorem. But these are not the same sentences.
They are the same sentences (this is, they have the same
meaning), just
written in different words. More precisely, they are
equivalent.
Using logical notation to write the above sentences, let us
show that
A => (B => C) l.e. A / B => C (*)
where / means and and l.e. means logical equivalent.
Recall the
laws
P => Q l.e. ~P / Q (1)
(P / Q) / R l.e. P / (Q / R) (2)
~P / ~Q l.e. ~(P / Q) (3)
where / means or and ~ means not.
We have
A => (B => C) l.e. (apply (1))
A => (~B / C) l.e. (apply (1))
~A / (~B / C) l.e. (apply (2))
(~A / ~B) / C l.e. (apply (3))
~(A / B) / C l.e. (apply (1))
A / B => C
therefore (*) is true.
In the example that you presented, the sentences have the same
meaning. But
sometimes you can find in different books the name theorem T
attributed
to
different results (this is, results that do not have the same
meaning). Not
everyone calls the same names to the same things. I think that
the best
that
we can do is be careful with names and do not assume that
another person
use
the names with the same meaning that we do. So, we should
check if we are
talking about the same things.
I hope I have helped.
Jaime Gaspar
______________________________
Homepage: www.jaimegaspar.com
E-mail: e-mail@jaimegaspar.com
===
Subject: Re: Delta patterns related to the primes.
> Take the difference sequence of the primes, and iteratively
take the
> difference sequence of the previous sequence, but make sure
you take
> all differences in absolute value.
2 3 5 7 11 13 15 17 19 . . . .
1 2 2 4 2 2 2 2 . . . .
1 0 2 2 0 0 0 . . . .
1 2 0 2 0 0 . . . .
1 2 2 2 0 . . . .
> The claim is that the leftmost diagonal is always 1 (except
the initial
2).
> This seems to be some sort of delicate statement
Not all that delicate! As we shall see. It's almost worthless.
> about the distribution of primes. Has anyone seen this claim?
Yes, in some popularization or other; and I didn't think much
of it then,
and don't now. It struck me as having been made by someone
with little
real number sense. (But in that, it is similar to Goldbach's.)
I don't want to rain on anyone's parade, (well actually I
don't really
mind doing that as you probably know), but this is a parade
that should
have been rained out before it left the marshalling yard.
The conjecture will apply to almost any sequence of numbers
all of
the same parity, except for a first one of different parity,
that doesn't
grow too fast. And fast is loose - basically, size doubling is
the limit.
In a sense, the algorithm and initial conditions have been
chosen quite
cunningly; the absolute value of the differences ensures that
any local
variations (as in the primes) get smoothed out fairly quickly,
and the
parity conditions ensure that there will always be a one at
the very
left, provided there isn't any sudden fast growth or decay
anywhere.
As examples, here is a sequence beginning with primes 2,3,5
and continuing
to grow as fast as possible consistent with the conjecture.
2 3 5 9 17 33 65 129 ... essentially just doublings.
1 2 4 8 16 32 64 ...
1 2 4 8 16 32 ... not really very exciting, is it?
And if some of the terms are lowered, it has no effect so long
as
the subsequent terms are lowered accordingly (within VERY easy
limits!),
as the absolute differences will smooth out local variations,
as I say.
As a second example, we go to the other extreme from primes.
(Contrary to what someone was hoping about composite numbers.)
Here is a sequence of very round numbers according to a
criterion
which I have devised but will keep hidden for now.
1 2 4 8 12 20 24 36 48 60 90 120 180 200 240 300 ...
1 2 4 4 8 4 12 12 12 30 30 60 20 40 60 ...
1 2 0 4 4 8 0 0 18 0 30 40 20 20 ...
1 2 4 0 4 8 0 18 18 30 10 20 0 ...
1 2 4 4 4 8 18 0 12 20 10 20 ...
1 2 0 0 4 10 18 12 8 10 10 ...
1 2 0 2 6 8 6 4 2 0
1 2 2 4 2 2 2 2 2
1 0 2 2 0 0 0 0
1 2 2 2 0 0 2
1 2 0 2 0 2
As you can see, they still satisfy the conjecture.
(OK I admit I had to work a little to make it so, but not a
lot.)
Finally, I was going to make an example with the primes, but
each one
plus-or-minus one, more or less at random. But it was far too
easy.
So here is an example that shows the slow growth of the primes
isn't
necessary either; each term is 1.5 times the previous one,
rounded
to the nearest odd number. It grows geometrically fast, though
not
close to the doubling limit.
2 3 5 7 11 17 25 37 55 83 125 187 ...
1 2 2 4 6 8 12 18 28 42 62 ...
1 0 2 2 2 4 6 10 14 20 ...
1 2 0 0 2 2 4 4 6 ...
Again, you can plus-or-minus two to all the terms to introduce
more
local randomness, but provided you don't damage the very low
ones
too badly, it won't make any difference.
Well, ICBW, but AFAICS the conjecture is a crock.
--------------------------------------------------------------
--------------
--
Bill Taylor W.Taylor@math.canterbury.ac.nz
--------------------------------------------------------------
--------------
--
If we knew what we were doing it wouldn't be called research
--------------------------------------------------------------
--------------
--
===
Subject: Re: Delta patterns related to the primes.
Everything Bill says below is true, but it still doesn't give
a proof
of the conjecture.
> Take the difference sequence of the primes, and
iteratively take the
> difference sequence of the previous sequence, but make
sure you take
> all differences in absolute value.
> 2 3 5 7 11 13 15 17 19 . . . .
> 1 2 2 4 2 2 2 2 . . . .
> 1 0 2 2 0 0 0 . . . .
> 1 2 0 2 0 0 . . . .
> 1 2 2 2 0 . . . .
> The claim is that the leftmost diagonal is always 1
(except the initial
2).
> This seems to be some sort of delicate statement
> Not all that delicate! As we shall see. It's almost
worthless.
> about the distribution of primes. Has anyone seen this
claim?
> Yes, in some popularization or other; and I didn't think
much of it then,
> and don't now. It struck me as having been made by someone
with little
> real number sense. (But in that, it is similar to
Goldbach's.)
> I don't want to rain on anyone's parade, (well actually I
don't really
> mind doing that as you probably know), but this is a parade
that should
> have been rained out before it left the marshalling yard.
> The conjecture will apply to almost any sequence of numbers
all of
> the same parity, except for a first one of different parity,
that doesn't
> grow too fast. And fast is loose - basically, size doubling
is the
limit.
> In a sense, the algorithm and initial conditions have been
chosen quite
> cunningly; the absolute value of the differences ensures
that any local
> variations (as in the primes) get smoothed out fairly
quickly, and the
> parity conditions ensure that there will always be a one at
the very
> left, provided there isn't any sudden fast growth or decay
anywhere.
> As examples, here is a sequence beginning with primes 2,3,5
and
continuing
> to grow as fast as possible consistent with the conjecture.
> 2 3 5 9 17 33 65 129 ... essentially just doublings.
> 1 2 4 8 16 32 64 ...
> 1 2 4 8 16 32 ... not really very exciting, is it?
> And if some of the terms are lowered, it has no effect so
long as
> the subsequent terms are lowered accordingly (within VERY
easy limits!),
> as the absolute differences will smooth out local
variations, as I say.
> As a second example, we go to the other extreme from primes.
> (Contrary to what someone was hoping about composite
numbers.)
> Here is a sequence of very round numbers according to a
criterion
> which I have devised but will keep hidden for now.
> 1 2 4 8 12 20 24 36 48 60 90 120 180 200 240 300 ...
> 1 2 4 4 8 4 12 12 12 30 30 60 20 40 60 ...
> 1 2 0 4 4 8 0 0 18 0 30 40 20 20 ...
> 1 2 4 0 4 8 0 18 18 30 10 20 0 ...
> 1 2 4 4 4 8 18 0 12 20 10 20 ...
> 1 2 0 0 4 10 18 12 8 10 10 ...
> 1 2 0 2 6 8 6 4 2 0
> 1 2 2 4 2 2 2 2 2
> 1 0 2 2 0 0 0 0
> 1 2 2 2 0 0 2
> 1 2 0 2 0 2
> As you can see, they still satisfy the conjecture.
> (OK I admit I had to work a little to make it so, but not a
lot.)
> Finally, I was going to make an example with the primes, but
each one
> plus-or-minus one, more or less at random. But it was far
too easy.
> So here is an example that shows the slow growth of the
primes isn't
> necessary either; each term is 1.5 times the previous one,
rounded
> to the nearest odd number. It grows geometrically fast,
though not
> close to the doubling limit.
> 2 3 5 7 11 17 25 37 55 83 125 187 ...
> 1 2 2 4 6 8 12 18 28 42 62 ...
> 1 0 2 2 2 4 6 10 14 20 ...
> 1 2 0 0 2 2 4 4 6 ...
> Again, you can plus-or-minus two to all the terms to
introduce more
> local randomness, but provided you don't damage the very low
ones
> too badly, it won't make any difference.
> Well, ICBW, but AFAICS the conjecture is a crock.
>
--------------------------------------------------------------
---------------
-
> Bill Taylor W.Taylor@math.canterbury.ac.nz
>
--------------------------------------------------------------
---------------
-
> If we knew what we were doing it wouldn't be called research
>
--------------------------------------------------------------
---------------
-
--
Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email)
===
Subject: Re: Delta patterns related to the primes.
>Take the difference sequence of the primes, and iteratively
take the
>difference sequence of the previous sequence, but make sure
you take
>all differences in absolute value.
>2 3 5 7 11 13 15 17 19 . . . .
> 15?
> Robert Israel israel@math.ubc.ca
> Department of Mathematics http://www.math.ubc.ca/~israel
> University of British Columbia
> Vancouver, BC, Canada V6T 1Z2
Gulp...
===
Subject: Re: Delta patterns related to the primes.
> Take the difference sequence of the primes, and
iteratively take the
> difference sequence of the previous sequence, but make
sure you take
> all differences in absolute value.
Re-writing history here:
> 2 3 5 7 11 13 * 17 19 . . . .
> 1 2 2 4 2 4 * 2 . . . .
> 1 0 2 2 2 2 * . . . .
> 1 2 0 0 0 . . . .
> 1 2 2 2 . . . .
> The claim is that the leftmost diagonal is always 1
(except the initial
2).
> This seems to be some sort of delicate statement
> Not all that delicate! As we shall see. It's almost
worthless.
> about the distribution of primes. Has anyone seen this
claim?
> Yes, in some popularization or other; and I didn't think
much of it then,
> and don't now. It struck me as having been made by someone
with little
> real number sense. (But in that, it is similar to
Goldbach's.)
> I don't want to rain on anyone's parade,
Well, it certainly isn't _my_ parade, so don't worry about it.
I was
just reminded by this looking at the OP's statement, and
thought I'd
try to find out where it stands.
> The conjecture will apply to almost any sequence of numbers
all of
> the same parity, except for a first one of different parity,
that doesn't
> grow too fast. And fast is loose - basically, size doubling
is the
limit.
I absolutely agree that the conjecture has more to do with the
growth
rate than the arithmetic properties of the elements, but what
does it
say exactly? Does it say more, or less, than Bertrand's
Postulate?
- EM
===
Subject: Re: Delta patterns related to the primes.
> Take the difference sequence of the primes, and
iteratively take the
> difference sequence of the previous sequence, but make
sure you take
> all differences in absolute value.
> 2 3 5 7 11 13 15 17 19 . . . .
> 1 2 2 4 2 2 2 2 . . . .
> 1 0 2 2 0 0 0 . . . .
> 1 2 0 2 0 0 . . . .
> 1 2 2 2 0 . . . .
> The claim is that the leftmost diagonal is always 1
(except the initial
2).
> This seems to be some sort of delicate statement
> Not all that delicate! As we shall see. It's almost
worthless.
> about the distribution of primes. Has anyone seen this
claim?
> Yes, in some popularization or other; and I didn't think
much of it then,
> and don't now. It struck me as having been made by someone
with little
> real number sense. (But in that, it is similar to
Goldbach's.)
> I don't want to rain on anyone's parade, (well actually I
don't really
> mind doing that as you probably know), but this is a parade
that should
> have been rained out before it left the marshalling yard.
> The conjecture will apply to almost any sequence of numbers
all of
> the same parity, except for a first one of different parity,
that doesn't
> grow too fast. And fast is loose - basically, size doubling
is the
limit.
> In a sense, the algorithm and initial conditions have been
chosen quite
> cunningly; the absolute value of the differences ensures
that any local
> variations (as in the primes) get smoothed out fairly
quickly, and the
> parity conditions ensure that there will always be a one at
the very
> left, provided there isn't any sudden fast growth or decay
anywhere.
> As examples, here is a sequence beginning with primes 2,3,5
and
continuing
> to grow as fast as possible consistent with the conjecture.
> 2 3 5 9 17 33 65 129 ... essentially just doublings.
> 1 2 4 8 16 32 64 ...
> 1 2 4 8 16 32 ... not really very exciting, is it?
> And if some of the terms are lowered, it has no effect so
long as
> the subsequent terms are lowered accordingly (within VERY
easy limits!),
> as the absolute differences will smooth out local
variations, as I say.
> As a second example, we go to the other extreme from primes.
> (Contrary to what someone was hoping about composite
numbers.)
> Here is a sequence of very round numbers according to a
criterion
> which I have devised but will keep hidden for now.
> 1 2 4 8 12 20 24 36 48 60 90 120 180 200 240 300 ...
> 1 2 4 4 8 4 12 12 12 30 30 60 20 40 60 ...
> 1 2 0 4 4 8 0 0 18 0 30 40 20 20 ...
> 1 2 4 0 4 8 0 18 18 30 10 20 0 ...
> 1 2 4 4 4 8 18 0 12 20 10 20 ...
> 1 2 0 0 4 10 18 12 8 10 10 ...
> 1 2 0 2 6 8 6 4 2 0
> 1 2 2 4 2 2 2 2 2
> 1 0 2 2 0 0 0 0
> 1 2 2 2 0 0 2
> 1 2 0 2 0 2
> As you can see, they still satisfy the conjecture.
> (OK I admit I had to work a little to make it so, but not a
lot.)
> Finally, I was going to make an example with the primes, but
each one
> plus-or-minus one, more or less at random. But it was far
too easy.
> So here is an example that shows the slow growth of the
primes isn't
> necessary either; each term is 1.5 times the previous one,
rounded
> to the nearest odd number. It grows geometrically fast,
though not
> close to the doubling limit.
> 2 3 5 7 11 17 25 37 55 83 125 187 ...
> 1 2 2 4 6 8 12 18 28 42 62 ...
> 1 0 2 2 2 4 6 10 14 20 ...
> 1 2 0 0 2 2 4 4 6 ...
> Again, you can plus-or-minus two to all the terms to
introduce more
> local randomness, but provided you don't damage the very low
ones
> too badly, it won't make any difference.
> Well, ICBW, but AFAICS the conjecture is a crock.
>
--------------------------------------------------------------
---------------
-
> Bill Taylor W.Taylor@math.canterbury.ac.nz
>
--------------------------------------------------------------
---------------
-
> If we knew what we were doing it wouldn't be called research
>
--------------------------------------------------------------
---------------
-
Hi Bill,
BTW: I enjoy your art from other posts.
Your argument against the conjecture of the prime sequence and
its
deltas appears valid, but getting back to the original post.
Were my
sequence has a different order and each prime is represented
twice.
If you are also referring to my sequence (OP) then I propose
this
argument ---
When I first started to construct this sequence and I
discovered the
first 4 left diagonals of the deltas developed into a pattern
that
appeared to continue, I proceeded to add many more primes to
the
sequence without testing. I then tested to see if the pattern
continued and then discovered after 103 delta rows the pattern
broke
and larger numbers randomly appeared in the first 4 diagonals
and
other diagonals. The strange thing then, which is the crux of
my
argument, is after about 25 more delta rows with no patterns
in the 4
diagonals which then changed back to the original 4 patterns.
Why?
I then proceeded to find what I thought could be an error but
could
not find anything. I kept adding to the sequence and the
patterns
still persisted before and after the error, where the error
was within
about 25 delta rows.
I then finally found the error and corrected it. It was a
prime number
short of the needed prime count between the same two primes
that
represent the count. It then ran correctly where the 4 pattern
are
consistent throughout.
Granted, you can plant a similar error in another location in
the
sequence and nothing happens to the patterns in these 4
diagonals, but
if you can explain why the above happened with this particular
error
where from about 104 - 129 delta rows the patterns do not
exist. Then
after about 25 delta rows that were in error the 4 same
patterns
resumed!
Is this just a coincidence?
I don't believe so!
I believe that something deeper is going on with this sequence
that
right now no one can explain.
The error sequence is listed below with the location of the
error in
the sequence.
2,3,5,2,7,3,11,13,5,17,19,23,7,29,31,37,41,43,11,47,53,13,59,61
,67,71,73,
17,79,83,19,89,97,101,103,23,107,109,113,127,131,137,139,29,149
,151,31,
157,163,167,173,179,181,37,191,193,197,199,41,211,223,43,227,22
9,233,239,
241,47,251,257,263,269,271, (this spot should have had the
next prime
277 but instead 53 was placed here where 53 should have come
right
after
277),
277,281,283,293,307,311,313,317,59,331,337,61,347,349,353,359,
367,373,67,379,383,389,397,71,401,409,73,419,421,431,433,439,44
3,449,79,
457,461,463,467,83,479,487,491,499,503,509,521,89,523,541,547,5
57,563,569,
571,577,97,587,593,599,601,101,607,613,103,617,619,631,641,643,
107,647,
653,109,659,661,673,677,113,683,691,701,709,719,727,733,739,743
,751,757...
You need to run this amount of terms to observe the return of
the
original 4 delta diagonal patterns.
If you run this error sequence, make sure prime 53 is placed
before
prime 277.
If you do not think my argument is strong enough, please point
out the
weaknesses.
I hope there are no typos in my sequence because that would
mean one
error on top of another. :-)
Dan
===
Subject: Re: Delta patterns related to the primes.
> The conjecture will apply to almost any sequence of numbers
all of
> the same parity, except for a first one of different
parity, that
doesn't
> grow too fast. And fast is loose - basically, size doubling
is the
limit.
>I absolutely agree that the conjecture has more to do with
the growth
>rate than the arithmetic properties of the elements, but what
does it
>say exactly? Does it say more, or less, than Bertrand's
Postulate?
More, I think. There are sequences that satisfy a Bertrand's
Postulate
that won't satisfy the conjecture, e.g, I think, this one:
4 5 9 11 19 23 39 47 79
1 4 2 8 4 16 8 32
3 2 6 4 12 8 24
1 4 2 8 4 16
3 2 6 4 12
1 4 2 8
3 2 6
1 4
3
(where the second line seems to be EIS sequence A076736)
But I suspect the conjecture may imply Bertrand's postulate,
i.e. if
A_{1,1} >= 2, A_{1,j} = 1 otherwise, A_{i,j+1} = |A_{i,j} -
A_{i+1,j}|,
and A_{i,1} is strictly increasing, then A_{i+1,1} < 2 A_{i,1}
Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia
Vancouver, BC, Canada V6T 1Z2
===
Subject: Re: Delta patterns related to the primes.
Sniped ----
Sorry Bill, and other readers, I don't normally top post I
sniped
above just to shorten message which was not quite relivent
here.
This thread kind of strayed from the original post sequence.
Which is
ok but just trying to get back on track. See part way down and
bottom
for new message.
> Hi Bill,
> BTW: I enjoy your art from other posts.
> Your argument against the conjecture of the prime sequence
and its
> deltas appears valid, but getting back to the original post.
Were my
> sequence has a different order and each prime is represented
twice.
> If you are also referring to my sequence (OP) then I propose
this
> argument ---
> When I first started to construct this sequence and I
discovered the
> first 4 left diagonals of the deltas developed into a
pattern that
> appeared to continue, I proceeded to add many more primes to
the
> sequence without testing. I then tested to see if the pattern
> continued and then discovered after 103 delta rows the
pattern broke
> and larger numbers randomly appeared in the first 4
diagonals and
> other diagonals. The strange thing then, which is the crux
of my
> argument, is after about 25 more delta rows with no patterns
in the 4
> diagonals which then changed back to the original 4
patterns. Why?
> I then proceeded to find what I thought could be an error
but could
> not find anything. I kept adding to the sequence and the
patterns
> still persisted before and after the error, where the error
was within
> about 25 delta rows.
> I then finally found the error and corrected it. It was a
prime number
> short of the needed prime count between the same two primes
that
> represent the count. It then ran correctly where the 4
pattern are
> consistent throughout.
 Granted, you can plant a similar error in another location
in the
> sequence and nothing happens to the patterns in these 4
diagonals, but
> if you can explain why the above happened with this
particular error
> where from about 104 - 129 delta rows the patterns do not
exist. Then
> after about 25 delta rows that were in error the 4 same
patterns
> resumed!
See more of my tests I just conducted below and it will
clarify and
add more info. to the above paragraph.
> Is this just a coincidence?
> I don't believe so!
> I believe that something deeper is going on with this
sequence that
> right now no one can explain.
> The error sequence is listed below with the location of the
error in
> the sequence.
>
2,3,5,2,7,3,11,13,5,17,19,23,7,29,31,37,41,43,11,47,53,13,59,61
,67,71,73,
>
17,79,83,19,89,97,101,103,23,107,109,113,127,131,137,139,29,149
,151,31,
>
157,163,167,173,179,181,37,191,193,197,199,41,211,223,43,227,22
9,233,239,
> 241,47,251,257,263,269,271, (this spot should have had the
next prime
> 277 but instead 53 was placed here where 53 should have come
right
> after
277),
277,281,283,293,307,311,313,317,59,331,337,61,347,349,353,359,
>
367,373,67,379,383,389,397,71,401,409,73,419,421,431,433,439,44
3,449,79,
>
457,461,463,467,83,479,487,491,499,503,509,521,89,523,541,547,5
57,563,569,
>
571,577,97,587,593,599,601,101,607,613,103,617,619,631,641,643,
107,647,
>
653,109,659,661,673,677,113,683,691,701,709,719,727,733,739,743
,751,757...
> You need to run this amount of terms to observe the return
of the
> original 4 delta diagonal patterns.
> If you run this error sequence, make sure prime 53 is placed
before
> prime 277.
> If you do not think my argument is strong enough, please
point out the
> weaknesses.
> I hope there are no typos in my sequence because that would
mean one
> error on top of another. :-)
> Dan
Here is more which I have just tested.
Pursuing this further I tried forcing errors to prove my point
in the
above post.
I found that switching certain primes that are together in the
sequence, trying one switch at a time, either had no effect on
the 4
patterns or had a similar effect like the 277,53 switch. Where
prime
53 is placed just before 277 to create a temporary end to the
4 left
diagonal delta patterns.
In this column reverse In this column doing the
same
the smaller number to be first reversal will not have any
effect
to create a temporary end to on the 4 left diagonal delta
patterns.
the 4 left diagonal delta patterns.
[5,2] reversal drops 1st diagonal pattern. [7,3] reversal
change
has
[13,5] like (277,53) switch. no effect on
patterns
[23,7] ditto [43,11] ditto
[53,13] ditto
[73,17] ditto [83,19] ditto
[103,23] ditto
[139,29] ditto
[151,31] ditto
[181,37] ditto
[199,41] ditto [223,43] ditto
[241,47] ditto
[277,53] ditto
Etc.
The column on the left above are those certain adjoining
primes in the
sequence when switched produce a temporary end to the 4 left
diagonal
delta patterns then after a number of delta rows will return
to those
same 4 patterns and continue. Whereas the above column on the
right
the pairs when swiched never have any affect on the 4
patterns. Always
shift the smaller left prime to the right and the larger right
prime
to the left to create a possible error.
I believe this to be an important discovery because this
sequence
should be in a certain order and if some certain pairs of
primes in
the sequence are reversed the 4 left diagonal delta patterns
start out
ok but then are temporarily ended when more delta rows are
generated.
Then with more terms iterated creating more delta rows the 4
left
delta diagonal patterns eventually reappears and the same 4
patterns
continue on.
When a known forced error in a certain position caused by
switching
two different primes that are next to each other in the
sequence will
end the 4 patterns at a certain point. Then at a certain point
of more
delta rows will resume this pattern. How can these 4 patterns
begin
and then end for a short duration and then restart again and
continue
with the same 4 patterns?
What is going on here?
Can someone duplicate this with another sequence if you only
reverse a
certain two consecutive terms and at one point end the 4 left
diagonal
patterns then restart them again after more iterations and
delta rows?
It will help in the understanding of this sequence if the
delta rows
are created to show the 4 patterns and to create errors to see
first
hand what happens.
I hope I explained clearly of what is going on!
If not,any questions or replies welcome.
Dan
===
Subject: Re: determine the area bounded by the intersection of
five lines?
> This is not homework.
> Given five lines given by two pairs of (x, y) coordinates
each
> (not that it matters, it could be converted to y=mx+b form)
> how do I find the smallest area bounded by all five lines? If
> one of the lines runs through the middle of the area bounded
> by the other four, then I don't count the fifth line... I
only
> want the smallest area.
> Where can I look for this or does someone have an equation
the
> can point me to?
> Mike
First: one can understand the question in more than one way:
There are, for generic configurations, (5 choose 2) = 10 points
of intersection, and 16 regions (bounded or unbounded, but
certainly convex).
(1) Do you mean the area of the convex hull of those 10 points?
(2) Do you mean the smallest area of the 16 regions bounded by
parts of the 5 lines?
(3) Yet another possibility?
In all cases, this can be calculated by finite search - which
is
a legitimate mathematical, or programming, method. Methods for
determining the convex hull of a finite set of points can be
found (key phrase: convex hull), as well as areas of polygons.
===
Subject: Determining whether an array is a k-cycle
I am iterating a certain function, and I keep all the
resultant values
in an array: e:=array(0..499);
I need a local proc(e,n) which takes as input the array of
entries, its
actual size (actual size may be less than 500) and
heuristically tries
to determine if the entries are close to a k-cycle.
I tried to figure out a quick and dirty solution to this, but
after a
while it became obvious that the problem is rather difficult.
For starters one needs to make several assumptions. One needs
to have an
eps, which determines how close the entries need to be to be
considered
a match for the cycle and I need the proc to tell the user in
case
there is no match.
There are other difficulties also: Cases where there are
accidental
matches, but where the rest of the entries do not comprise a
cycle.
It looks to me that in the worst case, the algorithm seems to
be O(n^2),
since I need to check all entries against all previous ones
for matches.
I'd appreciate if anyone can help a bit with this.
--
Ioannis
http://users.forthnet.gr/ath/jgal/
___________________________________________
Eventually, _everything_ is understandable.
===
Subject: Re: Determining whether an array is a k-cycle
>I am iterating a certain function, and I keep all the
resultant values
>in an array: e:=array(0..499);
>I need a local proc(e,n) which takes as input the array of
entries, its
>actual size (actual size may be less than 500) and
heuristically tries
>to determine if the entries are close to a k-cycle.
>I tried to figure out a quick and dirty solution to this, but
after a
>while it became obvious that the problem is rather difficult.
>For starters one needs to make several assumptions. One needs
to have an
>eps, which determines how close the entries need to be to be
considered
>a match for the cycle and I need the proc to tell the user in
case
>there is no match.
>There are other difficulties also: Cases where there are
accidental
>matches, but where the rest of the entries do not comprise a
cycle.
>It looks to me that in the worst case, the algorithm seems to
be O(n^2),
>since I need to check all entries against all previous ones
for matches.
You can do it in O(n log(n)), as long as at most O(n log(n))
pairs are
within eps of each other.
myproc:= proc(e, n, eps)
# e is a numeric array with index 0 to n-1
local L,i,j,k,T,delta;
L:= sort([$0..n-1],(i,j) -> (e[i] < e[j]));
T:= Array(1..n);
for i from 1 to n-1 do
for j from i+1 to n while e[L[j]] < e[L[i]]+eps do
delta:= abs(L[j]-L[i]);
T[delta]:= T[delta]+1
od
od;
for k from 1 to n do
if T[k] = n-k then
printf(Found period of %dn,k);
return
fi
od;
printf(No period found, maybe increase eps);
end;
Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia
Vancouver, BC, Canada V6T 1Z2
===
Subject: Re: Distibutive inequalities in lattices
===
Subject: Distibutive inequalities in lattices
>I've been searching for a proof of the distributive
inequalities of
>lattice theory, but there don't seem to be any other than well
>obviously kind. Ultimately I'm looking to show what
properties are
>required for distributive lattices. Total order relations
imply
>distributivity, but there are certainly other examples where
the
>lattice is distributive without requiring a total order (e.g.
sets).
>Any help would be much appriciated.
Here's from my notes where a*b = inf a,b; a+b = sup a,b
Is this what you're looking for? The last part,
lattice cancellation implies distributivity, is complicated.
equivalent axioms for distributivity of a lattice
a(b+c) = ab + ac; a + bc = (a+b)(a+c)
ab + bc + ca = (a+b)(b+c)(c+a)
lattice cancellation: ax = ay, a+x = a+y ==> x = y
lattice cancellation equivalent distributivity.
----
===
Subject: Re: DNA Profile statistics
X-URL:
http://mygate.mailgate.org/mynews/sci/sci.math/
9efab30e9ead4b8677d7b2ffc96a20
d3.111419%40mygate.mailgate.org
> What I call the medians for each of these 10 loci is the
allele which
> has the most frequent occurance within a racial sub-group
- here UK
> caucasians.
> That's the mode, not the median. But other than this
> error in terminology, your question is a valid look at the
> question of allele independence. Based on those population
> statistics you can assign a probability to each event
> allele for marker X is one of these.
> The question then is, are these events independent?
> But that's only a single data point. It's not enough to
> draw a conclusion on.
> - Randy
Generally speaking the measurements are reproducible.
14 will always read 14. Part of the validation procedure
is to repeat analyses and also machine calibration
checks with standard samples.
I don't consider myself to be especially representative of
all uk caucasians - there is a bit of Irish and French and
width and breadth English geography in my ancestry.
Yet despite this I am within 2 alleles of the 'mode' for each
of the 10 loci. If I was particularly ichthonous ie many
generations
coming from one geographic locality then it would be more
understandable.
At the same time combined probability would say I'm 1 in 300
to 1 in
1000.
e mail nutteing2@quickfindit.....com (just one dot)
--
===
Subject: Re: Factorial/Exponential Identity, Infinity
I think it's interesting that there is an expression for
(5n/4)! in
terms of (n!), (n/4), and n^(n/4).
(5n/4)! = n! (n/4) n^(n/4)
In calculating the value of the expression, espcially as n
gets large
it may well be that it would cost less to calculate n!, (n/4),
n^(n/4), and their product than (5n/4)!. Well, maybe not. It
would
take n extended precision multiplications to calculate n!, one
to
calculate n/4, and n/4 to calculate n^(n/4), and two to
multiply the
three values together, yielding 5n/4 + 3. To calculate (5n/4)!
it
would take 5n/4 calculations, three less. Yet, it might take
less
time and resources. So where x is divisible by five, then 4x/5
could
be calculated as n and then that used to generate the exact
value of
(5n/4)! without multiplying (5n/4)(5n/4-1)(5n/4-2)...(n+1).
Is it so that (5n/4)! / n! (n/4) n^(n/4) = 1?
One thing here is that n is divisible by 4 or conversely x is
divisible by 5. If that is known then it takes a division or
right-shift by two bits to get n/4 or a division and two
multiplications or one and a left-shift to get x/5 and 4x/5.
Let's see, n=4, n/4=1, n!=24, n^(n/4)=4. The product n! *
(n/4) *
(n^(n/4) = 96. That is not equal to 120, 5!, 120/96= 5/4, that
is not
equal to one.
Let's see , n^(n/4) = 2^(n/4) 2^(n/4) (n/4)^(n/4).
I guess I should reevaluate the expression I derived from the
factorial definition of the gamma function.
Gamma(z) = lim n->oo n! n^z / ((z+1)(z+2)...(z+n))
If x = n/2 and z = n/4 =x/2, then:
Gamma(x/2) = limit n->oo (2x)! (2x)^(x/2) / ((5x/2)! /(x/2)!)
(x/2-1)! = (2x)! (2x)^(x/2) (x/2)! / (5x/2)!
(5x/2)! (x/2-1)! = (2x)! (2x)^(x/2) (x/2)!
(5x/2)! = (2x)! (2x)^(x/2) (x/2)
(5n/4)! = n! n^(n/4) (n/4)
One issue is that the Gamma function has two variables, n and
z, I
relate them by saying z = n/4. In the definition of the
function, it
is as the limit is taken as n diverges to infinity. Both sides
of the
above relation would diverge, but due to their expression of
equality
in the limit then their ratio would equal one.
lim n->oo (5n/4)! / ( n! n^(n/4) (n/4) ) = 1
Is this true or false? Why or why not?
Say y is n/8 instead of n/4.
Gamma(n/8)= limit n->oo n! n^(n/8) / ( (9n/8)! / (n/8)! )
(9n/8)! = n! n^(n/8) (n/8)
For integer x, this expression would show a similar relation:
((1+1/x)n)! = n! n^(n/x) (n/x)
lim n->oo n! n^(n/x) (n/x) / ((1+1/x)n)! = 1
This is just a reformulation of the Gamma function and is not
directly
related to the Factorial/Exponential Identity.
If that is so then it isn't immediately apparent from the
calculation
of f(n) = (5n/4)! / (n! n^(n/4) n/4) ):
f(4)=
1.250000000000000000000000000000000000000000000000000000000000
f(8)=
0.703125000000000000000000000000000000000000000000000000000000
f(16)=
0.443572998046875000000000000000000000000000000000000000000000
f(32)=
0.352519793887040577828884124755859375000000000000000000000000
f(64)=
0.444950048827587832031906868766203160007403471370990288846769
f(128)=
1.417186935509602084562734755564139602885006156205776690581632
f(256)=
28.747764530480100745279297021710678182560318273161584010488280
f(512)=
23656.230392256541850405423894512179641023942805494295940168978
839
f(1024)=
32035924072.135932556176618190547111645083296343519866651520060
543116994
f(2048)=
117500598137002400541538.03147002291573580137578978370871464143
63197832406934
33287440
f(4096)=
3161334239219732890548900496905568901855436766817.5591434477534
79442386466660
331677310351829378702512827084827
f(16384)=
191850593047417368739023783470251511308207492418128205723751863
61782239983313
438378190905194904438405630125326542726807509450428857025444551
24048319808869
1281269116834341160857218280248285816531957071631.9121440496871
49936446060645
956473757627640835673986634036602
f(32768)=
674220418048043976976300676749250732148049671918036997984568986
01755390099138
138097607964655239362756056018647146457182781858821516246731855
38769412401403
653969441531663405519210222760885327143826333678749651380494657
03759289950257
911889185477409102946662759777043295472466283207676469068185468
35331560898782
163362515559969279776451401329991541321683655246658575387480362
83204141037467
51372
72043545644920554.290075136553133247138361866911422321755093685
0919109897186
65
f(65536)=
166536362244043885954716941074705526807663435293445248718163478
23059702141365
068200957332945301967048772389116016263701195106563023002245626
91645353826931
272731256550249339108824177104141642692050218030300772966078015
04113833645740
752390331750949185147956719741896320444104917689939300843596433
47781096639452
482924540033222438356266287092998522715202537870779772931786406
50800436944699
77930
852888665145098335742935747361006784241139904270398725105954337
3336962984726
826686393395308356936594995013221857407406543218291232052632715
33244178654867
411205207198843717006073376882524205282978417711195481392965834
75657602471462
394942798653864184605690844265564199334076430484343665545736220
25627905522569
723767693087846914161049243717063400710054982467929795910901517
03203241143434
8954361152475576
6609412751467435655225345918.9086265128017987050560454951718476
3255484901033
4776618238475
Please explain your thoughts on this matter.
Ross
===
Subject: Re: Factorial/Exponential Identity, Infinity
> I think it's interesting that there is an expression for
(5n/4)! in
> terms of (n!), (n/4), and n^(n/4).
> (5n/4)! = n! (n/4) n^(n/4)
This is trivially false for n = 4,
Left hand side = (5n/4)! = (5*4/4)! = 5! = 120,
Right hand side = n! (n/4) n^(n/4) = 4! (1) 4^1 = 24 * 4 = 96
It is less trivial, but still false for n = 8
Is there any n for which it is true?
===
Subject: Re: Factorial/Exponential Identity, Infinity
<Pine.LNX.4.56.0308030801540.25542@cpw.math.columbia.edu>
<Pine.LNX.4.56.0308030826420.25542@cpw.math.columbia.edu>
<Pine.LNX.4.56.0308041520130.2473@cpw.math.columbia.edu>
===
:Subject: Re: Factorial/Exponential Identity, Infinity
:
:
:> I think it's interesting that there is an expression for
(5n/4)! in
:> terms of (n!), (n/4), and n^(n/4).
:> (5n/4)! = n! (n/4) n^(n/4)
:
:This is trivially false for n = 4,
:Left hand side = (5n/4)! = (5*4/4)! = 5! = 120,
:Right hand side = n! (n/4) n^(n/4) = 4! (1) 4^1 = 24 * 4 = 96
:
:It is less trivial, but still false for n = 8
:
:Is there any n for which it is true?
:
Fairly trivial for n=8 too. Here's a sort of strategy:
for some N, for any n>=N there is a prime p with n<p<=5n/4.
Then if 4 divides m and m>=n p will divide (5m/4)! but not m!.
So there can be finitely many n for which it is true.
You can probably attack the other side (especially if you can
figure out
N) as you can raise the lower bound fairly quickly.
If n=4r, then 5 must divide r, r=5t. Then you get 77 divides t
etc.
===
Subject: Re: Factorial/Exponential Identity, Infinity
> I think it's interesting that there is an expression for
(5n/4)! in
> terms of (n!), (n/4), and n^(n/4).
> (5n/4)! = n! (n/4) n^(n/4)
What's the point? Have you ever posted anything anywhere
anytime
that wasn't total bollocks?
--
Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html
His mind has been corrupted by colours, sounds and shapes.
The League of Gentlemen
===
Subject: Re: Factorial/Exponential Identity, Infinity
<Pine.LNX.4.56.0308080352370.5461@cpw.math.columbia.edu>,
===
> :Subject: Re: Factorial/Exponential Identity, Infinity
> :> I think it's interesting that there is an expression for
(5n/4)! in
> :> terms of (n!), (n/4), and n^(n/4).
> : :> (5n/4)! = n! (n/4) n^(n/4)
> :This is trivially false for n = 4,
> :Left hand side = (5n/4)! = (5*4/4)! = 5! = 120,
> :Right hand side = n! (n/4) n^(n/4) = 4! (1) 4^1 = 24 * 4 =
96
> :It is less trivial, but still false for n = 8
> :Is there any n for which it is true?
> Fairly trivial for n=8 too. Here's a sort of strategy:
> for some N, for any n>=N there is a prime p with n<p<=5n/4.
> Then if 4 divides m and m>=n p will divide (5m/4)! but not
m!.
> So there can be finitely many n for which it is true.
> You can probably attack the other side (especially if you
can figure out
> N) as you can raise the lower bound fairly quickly.
> If n=4r, then 5 must divide r, r=5t. Then you get 77 divides
t etc.
It is also trivially false for n being any integer multiple of
4 and
having a prime between n and 5n/4, since that prime must
divide the
left hand side but not the right hand side of the equation.
For example, when n is 100, there are 4 primes betweem 100 and
125,
they are 101.103.107.109 and 113.
Thus there are a large number of n, possibly even infinitely
many,
for which it is trivial to prove that the equation is false.
And, as yet, no known values for which it is true.
Not a vary good batting average.
===
Subject: Re: Factorial/Exponential Identity, Infinity
<Pine.LNX.4.56.0308030801540.25542@cpw.math.columbia.edu>
<Pine.LNX.4.56.0308030826420.25542@cpw.math.columbia.edu>
<Pine.LNX.4.56.0308041520130.2473@cpw.math.columbia.edu>
<Pine.LNX.4.56.0308080352370.5461@cpw.math.columbia.edu>
===
:Subject: Re: Factorial/Exponential Identity, Infinity
:
:<Pine.LNX.4.56.0308080352370.5461@cpw.math.columbia.edu>,
:
===
:> :Subject: Re: Factorial/Exponential Identity, Infinity
:> :
:> :
:> :> I think it's interesting that there is an expression for
(5n/4)! in
:> :> terms of (n!), (n/4), and n^(n/4).
:> :>
:> :> (5n/4)! = n! (n/4) n^(n/4)
:> :
:> :This is trivially false for n = 4,
:> :Left hand side = (5n/4)! = (5*4/4)! = 5! = 120,
:> :Right hand side = n! (n/4) n^(n/4) = 4! (1) 4^1 = 24 * 4 =
96
:> :
:> :It is less trivial, but still false for n = 8
:> :
:> :Is there any n for which it is true?
:> :
:> Fairly trivial for n=8 too. Here's a sort of strategy:
:> for some N, for any n>=N there is a prime p with n<p<=5n/4.
:> Then if 4 divides m and m>=n p will divide (5m/4)! but not
m!.
:> So there can be finitely many n for which it is true.
:> You can probably attack the other side (especially if you
can figure out
:> N) as you can raise the lower bound fairly quickly.
:> If n=4r, then 5 must divide r, r=5t. Then you get 77
divides t etc.
:
:It is also trivially false for n being any integer multiple
of 4 and
:having a prime between n and 5n/4, since that prime must
divide the
:left hand side but not the right hand side of the equation.
That was the point I made just after the word strategy above.
:
:For example, when n is 100, there are 4 primes betweem 100
and 125,
:they are 101.103.107.109 and 113.
:
:Thus there are a large number of n, possibly even infinitely
many,
:for which it is trivial to prove that the equation is false.
I also commented about this (rather more in fact that the set
of
counterexamples is not only infinite but cofinite and, quite
likely,
empty).
:
:And, as yet, no known values for which it is true.
:
:Not a vary good batting average.
:
===
Subject: Re: Factorial/Exponential Identity, Infinity
I know (5/4)! / (n! (n/4) n^(n/4)) does not equal one for any
finite
integer n. What I have surmised is that the limit as n
diverges of
that expression is equal to one. It is implicit in the context
of the
discussion about the equation there that the limit case is
that that
expression may be true where I already pointed out that 120
does not
equal 96, and provided values of the result of the expression
for
various larger values of n.
You might notice that that expression can be expressed in
terms of the
Gamma function instead of factorial, with a domain of r in RZ-
instead of n in Z+, and that between 4 and 8 the value of the
expression goes from higher to one to less than one, and
between 64
and 128 it goes back, with it being continuous, thus that for
some
finite, probably irrational, value r that it is true for the
Gamma
expression instead of the factorial expression.
The point is that the factorial expression is true in the
limit case,
or it is not, that is the point of discussion here. I
presented a
rationale why to consider it to be true, and want to clarify
here that
the prime in [n, (1+1/x)n] argument does not affect the
rationale of
the limit case argument.
This is different than lim n->oo sqrt(n Pi/2) n! / ( (n/2)!^2
2^n ) =
1. I am told that Maple evaluates that expression, yet I'm
unclear as
to how, although I have an idea. I think it uses the doubling
formula
for Gamma, and Gamma(1/2), and other than that I can't say how
it's
derived, only that I expressed it, and observed it to
correspond with
truth.
Given a random uniform distribution over all infinite binary
sequences the probability of a random sequence having equal
numbers
of ones and zeros is sqrt(n Pi/2), which equals sqrt(n Pi/2)
2^n /
2^n, in terms of n, an infinite variable.
So, the factorial expression doesn't hold true for a finite
integer
value n, I understand, although it does hold true for various
real
values in the gamma expression. I say that the expression is
true in
the limit case, do you understand?
Ross
===
Subject: Re: Factorial/Exponential Identity, Infinity
> I think it's interesting that there is an expression for
(5n/4)! in
> terms of (n!), (n/4), and n^(n/4).
 (5n/4)! = n! (n/4) n^(n/4)
 What's the point? Have you ever posted anything anywhere
anytime
> that wasn't total bollocks?
Hi Robin,
Why do you care?
Ross
===
Subject: Re: Finite subgroups of GL(n,Z)
> Anyway, my point is that in the States, the normal form of
address for
> someone in your position would be professor,
> I'm not in the States.
> so it's natural people are
> going to keep doing this.
> Why should common practice in the States be natural
> for anyone outside?
I'm not saying that. I am saying that there will be plenty of
sci.math
folks who follow that custom. So it seems futile to correct
them. On the
other hand, thinking about it now, perhaps it isn't futile.
> Are you going to keep correcting everybody who
> addresses you that way?
> No, sometimes I may not reply to them at all.
===
Subject: Generating function , degree of polynomials.
Suppose that [.] denotes integral part ,
p is a positive integer ,
a_p,a_{p+1},... are real numbers with a_p=/= 0 ,
H(t) is formal power series
H(t)=a_p*t^p+a_{p+1}*t^{p+1}+ ... .
Consider that (P_n(x))_{n>=0} is the sequence of
polynomials generated by
exp(x*H(t))= Sum_{k=0 to k=infty} P_k(x)t^k/k! .
Prove or disprove that degree(P_n)=[n/p] .
===
Subject: Generating Primes from Nonprimes
Well, I am already beyond the third time is a charm. Maybe the
fifth
try will actually work (reasonably well). I have revised this
idea
yet again: http://www.tln.net/~reriker/prime.html
Granted, it may be too unwieldly to be of any practical use.
===
Subject: Re: Geometry: The Implications of Homogeneity and
Isotropy
> It is often said that the Lorentz transformation can be
derived from
> the homogeneity and isotropy of space alone. I'm looking
for a
> concrete counterexample to this claim.
> The Galilean transformation is consistent with homogeneity
and isotropy,
> so these cannot imply the Lorentz transformation.
The Galilean transformation is understood to be a special case
of the
Lorentz transformation.
See: http://www.everythingimportant.org/relativity/
===
Subject: Re: Geometry: The Implications of Homogeneity and
Isotropy
> I'm wondering if it's mathematically permissible, if space is
> homogeneous and isotropic, for a moving rod to experience a
uniform
> expansion or contraction during the time it's not in its
stationary
> frame of reference. What's preventing a moving rod from
returning to
> its point of origin smaller or larger?
I doubt it is possible to invent a self-consistent theory with
that
property.
The analogy to the twin scenario would be that when the
traveling twin returns, she is not only younger than the
stay-at-home twin, but that she CONTINUES to age more
slowly than her sister while they both remain at home.
Real objects in our world do not behave that way....
> The expanding or shrinking
> effect could go like f(v)exp(kt) for as long as the rod
maintains a
> constant velocity v. Conceivably, this might be made to work
in 3
> spatial dimensions. Every observer could say that every
other frame
> is shrinking uniformly in time and, akin to the twin
paradox, all
> returning twins could end up YOUNGER and smaller.
> Can you prove that homogeneity and isotropy alone disallows
this
> possibility?
I doubt it.
One of the hidden assumptions of SR is that clocks and rulers
have no
memory. That is, when a meterstick is at rest in your frame it
has a
length of 1 meter regardless of its history in coming to rest
in your
frame. This agrees with our common experience, and with the
notion that
inter-atomic bonds of the meterstick will arrange to do that
(unless its
history includes violent actions that break it -- a broken
meterstick is
no longer a meterstick).
Tom Roberts tjroberts@lucent.com
===
Subject: Re: Godel numbering
>Can anyone please explain to me how Godel numbering works.
On an
intuitive
>level, PM seems too complex for such a numbering to exist.
Anyways,
since
>intuition does not always lead to the right conclusions,
how does Godel
show
>that there exists a one-to-one correspondence between all
statements in
PM
>and natural numbers?
> There is no need to use the numbering Godel used, nor is it
> necessary that there be a one-to-one correspondences. All
> that matters is that each statement correspond to an integer,
> with different statements corresponding to different
integers.
> One simple way to do this is to use ASCII or some other means
> to code the characters, and use the resulting number.
The crucial is not to find a numbering; that is a triviality.
What is
crucial is to find one that allows one to model formal
inferences as
arithmetic operations in such a way that correct proofs
correspond to
certain arithmetic operations involving only addition,
multiplication,
and attempted subtraction and division.
===
Subject: Re: Godel numbering
intuitively that what Godel does is a numbering of the
chararcters. What I
don't understand is how is this a one-to-one mapping? How is
it proved that
each statement correpronds to an integer?
Minos
> There is no need to use the numbering Godel used, nor is it
> necessary that there be a one-to-one correspondences. All
> that matters is that each statement correspond to an integer,
> with different statements corresponding to different
integers.
> One simple way to do this is to use ASCII or some other means
> to code the characters, and use the resulting number.
> --
> This address is for information only. I do not claim that
these views
> are those of the Statistics Department or of Purdue
University.
> Herman Rubin, Department of Statistics, Purdue University
> hrubin@stat.purdue.edu Phone: (765)494-6054 FAX:
(765)494-0558
===
Subject: Re: Godel numbering
> intuitively that what Godel does is a numbering of the
chararcters. What
I
> don't understand is how is this a one-to-one mapping? How is
it proved
that
> each statement correpronds to an integer?
Be careful with what you mean by one-to-one. Goedel's method
is a
method for setting up a one-to-one correspondence between the
expressions in the calculus and a _certain subset_ GC of the
integers.
In particular, not every integer is a Goedel number. Consider
the number
100 = 2^2*5^2. Looking at this prime factorization it becomes
immediatelly obvious that the number 100 cannot be a Goedel
codon, since
powers of the prime 3 are skipped, TOGETHER with the fact that
powers of
5 appear. This is by definition not allowed.
On the other hand, given G in GC, you can easily decompose G
into its
prime number factorization and assuming the decomposition
follows proper
codification rules (i.e. no numbers like 100) therefrom derive
a unique
sequence of exponents of the first n primes, which will lead
you back to
a sequence of fundamental codons, which can be used to recover
the
calculus expression uniquely.
The above paragraph can be used slightly modified to prove
what you
want.
> Minos
--
Ioannis
http://users.forthnet.gr/ath/jgal/
___________________________________________
Eventually, _everything_ is understandable.
===
Subject: Re: Godel numbering
> Be careful with what you mean by one-to-one. Goedel's method
is a
> method for setting up a one-to-one correspondence between the
> expressions in the calculus and a _certain subset_ GC of the
integers.
> In particular, not every integer is a Goedel number.
Consider the number
> 100 = 2^2*5^2. Looking at this prime factorization it becomes
> immediatelly obvious that the number 100 cannot be a Goedel
codon, since
> powers of the prime 3 are skipped, TOGETHER with the fact
that powers of
> 5 appear. This is by definition not allowed.
> On the other hand, given G in GC, you can easily decompose G
into its
> prime number factorization and assuming the decomposition
follows proper
> codification rules (i.e. no numbers like 100) therefrom
derive a unique
> sequence of exponents of the first n primes, which will lead
you back to
> a sequence of fundamental codons, which can be used to
recover the
> calculus expression uniquely.
> The above paragraph can be used slightly modified to prove
what you
> want.
number we get to a unique sequence of exponents and, given a
codification,
to a statement in PM. My question is: how is it proved that
there exists a
complete codification between statements in PM and these
exponent
(basically natural numbers). In other word how does Godel
prove that all
statements in PM can be mapped into the natural numbers?
Intuitively what
is
the Godel number of the class of all classes?
===
Subject: Re: Group generators
> Is there an algorithm to determine the minumum number of
generators for
a
> finite group ? Clearly, for a cyclic group it is 1, for a
symmetric
> permutation group or a dihedral group it is 2 but what
about any
finite
> group ?
> I asked John Conway the inverse question - Can every group
be
>generated from a set of generators whose orders multiply to
give the
>order of the group? and he assured me that they could, at
least up to
>the first Janko group. This puts an upper limit on the
number of
>generators needed.
> Yes, but this upper limit would be much too large in general.
>S4 needs all 4 {3x2x2x2=24).
> I am not sure what you mean by that. S4 is a 2-generator
group.
I was thinking of generation by string concatenation, where
a^3=b^2=c^2=d^2=1 and five re-write rules are needed to create
S4. Two
permutations, two tau matrices, or two gamma matrices also
create S4.
You say A5 needs 3 generators, but only two permutations
{{1,2,4,5,3},{3,4,5,1,2}} generate it. This is in example 25 in
GroupLoopDemo.nb in
http://library.wolfram.com/infocenter/MathSource/4894/
>GAP can provide a set
>of generators via
RelatorsOfFpGroup(Image(IsomorphismFpGroup(SmallGroup(m,n)))),
>but this may not be minimal.
> But the question was to find a minimal generating set. If
you just want
> a generating set, take the whole of G.
> Derek Holt.
As there may not be a general answer to the question, I was
setting an
upper bound on the problem.
Roger Beresford.
Make things a simple as possible, but no simpler. (A. Einstein)
===
Subject: Re: Group generators
> Is there an algorithm to determine the minumum number of
generators
for a
> finite group ? Clearly, for a cyclic group it is 1, for
a symmetric
> permutation group or a dihedral group it is 2 but what
about any
finite
> group ?
> I asked John Conway the inverse question - Can every
group be
>generated from a set of generators whose orders multiply
to give the
>order of the group? and he assured me that they could, at
least up to
>the first Janko group. This puts an upper limit on the
number of
>generators needed.
> Yes, but this upper limit would be much too large in
general.
>S4 needs all 4 {3x2x2x2=24).
> I am not sure what you mean by that. S4 is a 2-generator
group.
>I was thinking of generation by string concatenation, where
>a^3=b^2=c^2=d^2=1 and five re-write rules are needed to
create S4. Two
>permutations, two tau matrices, or two gamma matrices also
create S4.
>You say A5 needs 3 generators,
I didn't say that A5 needs 3 generators. I said that A5^20
(the direct
product
of 20 copies of A5) needs 3 generators. A5^19 is still a
2-generator group.
Derek Holt.
> but only two permutations
>{{1,2,4,5,3},{3,4,5,1,2}} generate it. This is in example 25
in
>GroupLoopDemo.nb in
http://library.wolfram.com/infocenter/MathSource/4894/
>GAP can provide a set
>of generators via
RelatorsOfFpGroup(Image(IsomorphismFpGroup(SmallGroup(m,n)))),
>but this may not be minimal.
> But the question was to find a minimal generating set. If
you just want
> a generating set, take the whole of G.
> Derek Holt.
>As there may not be a general answer to the question, I was
setting an
>upper bound on the problem.
>Roger Beresford.
>Make things a simple as possible, but no simpler. (A.
Einstein)
===
Subject: Re: Help finding a Carroll Problem
> I am looking for a problem referenced by Martin Gardner,
Chapter 11 in
> his collection entitled The 2nd Scientific American Book of
> Mathematical Puzzles & Diversions, Simon and Schuster,
1961.
> Gardner references Lewis Carroll telling that Carroll was
fond of
> inventing quaint and enormously complicated problems of
this sort.
> Eight are to be found in the appendix of his Symbolic
Logic.
> Gardner continues speaking of a large problem about judges
not smoking
> tobacco which was solved in 60's by Kemeny.
> Can you give me some help on where tofind the text of the
problem?
> Symbolic Logic volume 1, Appendix Addressed to Teachers.
> <http://www.amazon.com/exec/obidos/tg/detail/-/0486204928>
There is an on-line version of one of the editions (page
images):
http://durendal.org/lcsl/index.html
and also a text transcription of it:
http://durendal.org/lcsl/SymbolicLogic.txt
...but I am unable to find any problem about judges not
smoking tobacco.
Perhaps it's not the right edition (or I'm just overlooking
it).
Reinhold
===
Subject: Re: Help finding a Carroll Problem
> I am looking for a problem referenced by Martin Gardner,
Chapter 11 in
his
> collection
> entitled The 2nd Scientific American Book of Mathematical
Puzzles &
> Diversions,
> Simon and Schuster, 1961.
> Gardner references Lewis Carroll telling that Carroll was
fond of
inventing
> quaint and
> enormously complicated problems of this sort. Eight are to
be found in
the
> appendix of his Symbolic Logic.
> Gardner continues speaking of a large problem about judges
not smoking
> tobacco which was solved in 60's by Kemeny.
> Can you give me some help on where tofind the text of the
problem?
I'm afraid not, but I can give you a little information.
First, the
problem
Gardner refers to deals with magistrates not taking snuff
rather than
judges
not smoking tobacco. Similar concept, but that might help you
with
searches.
Second, none of the eight complicated problems in the appendix
of Symbolic
Logic involves either magistrates or snuff.
I skimmed through the Dover edition of Symbolic Logic and The
Game of
Logic,
and didn't find it in either. I looked at a few likely places
in The
Complete Works of Lewis Carroll (Modern Library edition, 1936,
which
doesn't
include all of his math work), and didn't find it in the
obvious places --
but I don't claim that search was exhaustive.
--
Jim Gillogly
12.19.10.8.14, 10 Ix 2 Yaxkin, Third Lord of Night
===
Subject: Re: Help! Help with gonio problem
> Could anyone maybe shed some light on the problem that i
sketched in
> http://www.rfjvanlinden171.freeler.nl in the file
PROBLEM.JPG.
> It is not that i am trying to have you do my homework or so.
I really
> puzzled with this problem for more than a week and i simply
dont see how
it
> can be solved. Either i am missing some clue or it is
reallly complex
after
> all.
> The angles alfa, beta and gamma add up to pi/2. The problem
is to find a
> simple expression for op in terms of oq and os in the circle
with radius
R.
> p, q and s are orthogonal projections.
We have:
Os=R*cos(alpha)
Op=R*cos(beta)
Oq=R*cos(gamma)
alpha+beta+gamma=pi/2
So,
cos(alpha)=Os/R
cos(gamma)=Oq/R
sin(alpha)=sqrt(R^2-Os^2)/R
sin(gamma)=sqrt(R^2-Oq^2)/R
Thus,
Op=R*cos(beta)
=R*sin(pi/2-beta)
=R*sin(alpha+gamma)
=R*(sin(alpha)*cos(gamma)+sin(gamma)*cos(alpha))
=R*((sqrt(R^2-Os^2)/R)*Oq/R + (sqrt(R^2-Oq^2)/R)*Os/R)
=(sqrt(R^2-Os^2)*Oq + sqrt(R^2-Oq^2)*Os)/R
I hope this helps.
--
Clive Tooth
http://www.clivetooth.dk
===
Subject: Re: Help! Help with gonio problem
with the sin and cosine functions while you simply express
them in the line
segments and R first before further calculating. Sometimes i
simply seem to
get blind for this kind of obvious solutions.
Rob
The Last Danish Pastry <TheLastDanishPastry@yahoo.com> schreef
in
bericht
> Could anyone maybe shed some light on the problem that i
sketched in
> http://www.rfjvanlinden171.freeler.nl in the file
PROBLEM.JPG.
> It is not that i am trying to have you do my homework or
so. I really
> puzzled with this problem for more than a week and i
simply dont see
how
> it
> can be solved. Either i am missing some clue or it is
reallly complex
> after
> all.
> The angles alfa, beta and gamma add up to pi/2. The
problem is to find
a
> simple expression for op in terms of oq and os in the
circle with
radius
> R.
> p, q and s are orthogonal projections.
> We have:
> Os=R*cos(alpha)
> Op=R*cos(beta)
> Oq=R*cos(gamma)
> alpha+beta+gamma=pi/2
> So,
> cos(alpha)=Os/R
> cos(gamma)=Oq/R
> sin(alpha)=sqrt(R^2-Os^2)/R
> sin(gamma)=sqrt(R^2-Oq^2)/R
> Thus,
> Op=R*cos(beta)
> =R*sin(pi/2-beta)
> =R*sin(alpha+gamma)
> =R*(sin(alpha)*cos(gamma)+sin(gamma)*cos(alpha))
> =R*((sqrt(R^2-Os^2)/R)*Oq/R + (sqrt(R^2-Oq^2)/R)*Os/R)
> =(sqrt(R^2-Os^2)*Oq + sqrt(R^2-Oq^2)*Os)/R
> I hope this helps.
> --
> Clive Tooth
> http://www.clivetooth.dk
===
Subject: Re: How I know, linchpin of my FLT proof
Visiting Assistant Professor at the University of Montana.
[.snip.]
>Well I did find a problem with the definition of the object
ring that
>I'd given, and I've updated it.
There have been at least two changes in recent memory, one
sort of
announced, one done in silence. And then there was another
change in
the past 36 hours, presumably what you are refering to here.
It is interesting that you DO NOT SAY what the problem you
supposedly
found was. What was it? Why was it a problem? Did you find it
yourself, or is it the problem I've been pointing out for many
months
that you had so diligently ignored?
The first change occurred back in January:
Your definition at the time was that units in object rings had
to have
complex norm equal to 1. You then modified the definition to:
Objects are members of commutative rings where any unit and its
multiplicative inverse are units in all possible commutative
rings in
which either are members, where no member is a factor
of an object for which it is not a factor in all possible
commutative
rings in which it and that object are members.
and then changed it (without acknowledging any problems or
announcing
the change) to:
Objects are members of commutative rings where any unit and its
multiplicative inverse are units in all possible commutative
rings in
which either and all integers are members, where no member is a
factor of an object for which it is not a factor in all
possible
commutative
rings that include all integers in which it and that object are
members.
throwing in the integers. You can even see that this was
hasty, since
you simply threw in the phrase that include all integers in the
middle of a sentence to end up with an ungrammatical one.
I see now that you have a ->further<- change, done in the past
36
hours; your webpage now reads:
The Object Ring is the set of all numbers where any member
that is a
unit, i.e. factor of 1, and its multiplicative inverse are
units in
all possible commutative rings in which either and all
integers are
members, and where no non-unit member a is a factor of any two
integers that are coprime.
You are still imprecise (it should most certainly NOT be all
numbers
and all possible commutative rings; under that definition,
NOTHING
is an object). Presumably you mean all COMPLEX numbers and all
subrings of the complex numbers.
I note, however, that you have not defined any operations on
this set,
nor proven that it is a ring under those operations.
This definition is certainly more inclusive than your previous
one
(under which the ONLY objects were the integers). But your
phrasing is
still too convoluted and unnecessary. First, you must specify
if the
integers are coprime in the ring of integers, or in your object
ring.
Assuming the latter, then all complex numbers are the Object
Ring,
since in that ring every two nonzero elements are coprime, and
any
element is a unit.
So I assume you mean two integers that are coprime in the ring
of
integers. But in that case, the second clause is vacuous:
Suppose R is ANY subring of the complex numbers (hence, must
contain
the integers), and let r be an element of R. Assume further
that a and
b are two integers that are coprime (in the ring of integers),
and
that r divides both a and b. The claim is that r is
necessarily a
unit. This follows because if r divides both a and b, then it
must
divide their integer gcd; but their gcd is 1, hence r divides
1, hence
r is a unit.
So your definition is reduced to:
The Object Ring is the [subring of the complex numbers]
satisfying
the condition that if u in the Object Ring is a unit in the
Object
Ring, then it is a unit in Z[u], the smallest subring
containing all
the integers and u.
Unfortunately, you have not proved that there is only one
object ring,
nor have you proven that it is well defined. Also, it
represents a
return to your attempted definition from some time ago. At the
time, I
posted several proofs showing that if such a ring contains all
algebraic integers, and all its elements are algebraic, then
it cannot
be anything OTHER than the ring of all algebraic integers.
Here are
the proofs again:
THEOREM. Let R be any subring of C, and assume that all
elements of R
are algebraic integers. If u in R is a unit, then u is a unit
in Z[u],
the smallest subring of C containing the integers and u.
Proof. Since u is a unit in R, and all elements of R are
algebraic
integers, it follows that 1/u is an algebraic integer. Let
f(x) be the
minimal polynomial of 1/u. Then
f(x) = x^n + a_{n-1}x^{n-1} + ... + a_1*x + a_0. Evaluating at
u^{-1},
we have:
0 = f(u^{-1})
= u^{-n} + a_{n-1}u^{1-n} + ... + a_1u^{-1} + a_0.
Multiplying through by u^{n-1} we have
0 = u^{-1} + a_{n-1} + a_{n-1}u + ... + a_1*u^{n-2} +
a_0*u^{n-1}.
Therefore,
u^{-1} = -(a_{n-1} + a_{n-1}u + ... + a_1*u^{n-2} +
a_0*u^{n-1}),
and so lies in Z[u], as claimed. QED
THEOREM. Let R be a subring of the algebraic numbers, and
assume that
R contains all algebraic integers. Assume further that if u is
a unit
in R, then u is a unit in Z[u]. Then R is exactly equal to the
ring of
all algebraic integers.
Proof. Assume not. Then R contains an algebraic number which
is not an
algebraic integers; call it r. We may write r as a quotient of
coprime
algebraic integers, since the ring of all algebraic integers
is a
Bezout domain: r = a/b. Moreover, there exist algebraic
integers x and
y such that ax+by = 1, again because the ring of all algebraic
integers is a Bezout domain, and a and b are coprime. And
since r is
NOT an algebraic integer, we know that b is not a unit in the
ring of
all algebraic integers.
Since R contains all algebraic integers, it contains x*r. It
also
contains y. Therefore, it contains x*r + y. But
x*r + y = (xa)/b + y = (xa+yb)/b = 1/b.
Therefore, R contains 1/b, so b is a unit in R.
However, b is not a unit in Z[b], since Z[b] is a subring of
the ring
of all algebraic integers, and we know that b is not a unit in
that
larger ring. This contradicts the property of R, so we may
conclude
that R does NOT contain any algebraic number which is not an
algebraic
integers.
QED
Now, what about transcendental complex numbers? Then you run
into some
serious problems. For example, the smallest subring of C
containing
all algebraic integers and pi satisfies your conditions,
because it is
isomorphic to the ring of polynomials with coefficients in the
algebraic integers. It's only units are the algebraic integers
which
are units. But the same is true of the smallest subring of C
containing all algebraic integers and 1/pi. That would make
both pi
and 1/pi into objects, but no subring of C may contain BOTH pi
and
1/pi, and be an object ring! So, presumably, your definition is
meant to be applicable only to algebraic numbers in any case.
So, your definition is still unclear as written, and if some
sense is
attempted to give to it, then the only conclusion that can
follow is
that you have not defined anything new, contrary to your
claims. You
assert (without proof) that object is a generalization of the
algebraic integers, yet if the notion includes all algebraic
integers,
it cannot include anything else. As such, it is a poor
generalization.
Why do you take so much trouble to expose such a reasoner as
Mr. Smith? I answer as a deceased friend of mine used to answer
on like occasions - A man's capacity is no measure of his power
to do mischief. Mr. Smith has untiring energy, which does
something; self-evident honesty of conviction, which does more;
and a long purse, which does most of all. He has made at least
ten publications, full of figures few readers can critize. A
great
many people are staggered to this extend, that they imagine
there
must be the indefinite something in the mysterious all this.
They are brought to the point of suspicion that the
mathematicians
ought not to treat all this with such undisguised contempt,
at least.
-- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan
Arturo Magidin
magidin@math.berkeley.edu
===
Subject: Re: How I know, linchpin of my FLT proof
I just figured something out. If you add all the bandwidth in
asserting
Short Proof of Fermat's Last Theorem, it becomes another
Long Proof of Fermat's Last Theorem.
Herc
===
Subject: Re: How I know, linchpin of my FLT proof
<deletedGiven acceptance of the intriguing little result that
given a factor g
>of a polynomial P(x), g=r+c, where c is a factor of the
constant term
>P(0), given by c=g at x=0, and r=g-c, the proof follows
easily enough.
>And yet I keep failing to follow it, and when I explained
why you keep
>failing to explain it, even though I've repeatedly pointed
out the post
>where I stated where I don't follow the proof.
> Yet, in your previous reply you said:
 No one is attacking this statement. Just what comes
*after* it.
> Yes, I am attacking the proof that comes after it, as I said
above.
Hmmm...reading back I can see where I read you wrong there. So
then
it's settled that you accept the nifty statement that given a
factor g
f a polynomial P(x), g=r+c, where c is a factor of the
constant term
P(0), given by c=g at x=0, and r=g-c.
> Consider
 P(m)/f^2 = (m^3 f^4 - 3m^2 f^2 + 3m) x^3 -
 3(-1+mf^2 )x u^2 + u^3 f.
> Now using b_1, b_2, b_3, w_1, w_2, and w_3, I have the
factorization
 P(m)/f^2 = (b_1 x + u w_1)(b_2 x + u w_2)(b_3 x + u w_3)
 where w_1 w_2 w_3 = f, and
 b_1 b_2 b_3 = (m^3 f^4 - 3m^2 f^2 + 3m),
 and at m=0
 P(0)/f^2 = 3xu^2 + u^3 f = u^2(3x + uf),
 so two of the b's must equal 0, which means
 P(0)/f^2 = w_1 w_2 u^2 (b_3 x + u w_3)
 which is
 P(0)/f^2 = u^2 (b_3 w_1 w_2 x + u f) = u^2(3x + uf)
 proving that w_1 w_2 must equal 1, as f is coprime to 3
from before,
> which leaves b_3 = 3.
> Notice that this depends on f coprime to three.
Yup.
 Essentially objections now come down to claiming that the
w's are
> dependent on m, but consider that w_1 w_2 = 1, when m=0,
here where f
> is coprime to 3.
 But that was an arbitrary choice *I* made, so let f=3.
> And now you look at the behavior when it isn't.
Well, f is a symbol, which represents a number, and typically
I say
it's a prime integer that is coprime to 3.
However, I found that some people like yourself apparently
oddly
believed that a constant factor f^2 could divide off as a
function, so
I consider what would happen if that is the case, as then
you'd get a
contradiction at f=3.
Imagine that w_1 and w_2 are functions of m.
Then from the above, you have
w_1(0)w_2(0) = 1.
So you might have something like w_1 = h(m)+1, where h(0) = 0.
But,
it turns out that if f=3, you have instead w_1(0) = 3^{1/3}.
So then, if w_1 is a function of m, it only is one if f does
not equal
3 or have any non unit factors in common with 3, which is
nonsensical.
And readers, I find it odd to need to remind the poster that an
equation doesn't just disappear because a variable is no longer
coprime to a prime integer. Debating that it does is
fascinating
behavior on the part of the poster.
 Now w_1 w_2 = 3^{2/3} as long as m is coprime to 3,
WITHOUT REGARD TO
> m.
> You say that w_1 w_2 has a certain value as long as m is
coprime to 3.
> That is stating a dependence on m.
Oh yeah, I made a mistake. If f=3, then 3^3 divides off, so in
fact
there is no variation as m varies, each of the w's just has a
factor
that is 3^{1/3}.
It's possibly confusing talking about it, but amazingly easy
to show,
as you have
P(m)/3^2 = (m^3 3^4 - 3m^2 3^2 + 3m) x^3 -
3(-1+m3^2 )x u^2 + 3u^3
where you can see that all the coefficients have a factor that
is 3.
The neat thing which I use by setting f=3, is that then it's
trivial
to show that the w's are constant valued.
So it's impossible for the w's to be functions of m, so that
when f is
coprime to 3, w_1 w_2 can vary from being coprime to f, to not
being
coprime, as you seem to *wish*, and it's trivially shown.
 So those posters who try to convince you that the w's are
actually
> dependent on m, like being functions of m, must now also
convince you
> that the w's make a decision, first looking to see if f=3
or have some
> non-unit factor in common with 3, and THEN they decide if
they're
> dependent on m.
> The w's are mathematical objects. They don't make decisions,
they have
> properties.
Yup. So if they were functions of m, then they'd maintain that
property, even when f=3.
The *mathematics* is straightforward.
Readers should see that it's also clear that posters for social
reasons are fighting the mathematical logic, which is
fascinating
behavior.
James Harris
===
Subject: Re: How I know, linchpin of my FLT proof
a rock and hollered:
> What I've seen are repeated attempts at distractions.
> That's what you think you see.
> But in reality your definition of object ring is still
circular, thus
your whole FLT proof is
> still invalid.
> So you cannot say nobody found an error in my FLT proof.
>Well I did find a problem with the definition of the object
ring that
>I'd given, and I've updated it.
>However, you still seem to not understand what a mathematical
proof is.
>It is a perfect argument that begins with a truth and
proceeds by
>logical steps to a conclusion which then must be true.
>So it's impossible to find an error in a proof.
Jimmy boy, I might know jack about math, but even I can tell
that you're
talking through your hat.
(followups set: sci.skeptic)
--
No collection of individuals is less vindictive than
an audience at amateur theatricals.
- P. G. Wodehouse, _The Intrusion of Jimmy_
===
Subject: Re: How I know, linchpin of my FLT proof
 I remember a post where someone was wondering how I could
keep
> claiming that I'm right with all these people posting
disagreement,
> and it seems to me that maybe explaining how might help.
> You are insane. You cannot counterargue fact, so you scream
some more
> as though reality never existed.
<deleted>
Nope, that's not why.
> You are loathsome and loud, <deleted>
No, that's not it either. Here's why.
Consider, in the ring of algebraic integers,
P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 -
3(-1+mf^2 )x u^2 + u^3 f).
Now using b_1, b_2, b_3, w_1, w_2, and w_3, I have the
factorization
P(m)/f^2 = (b_1 x + u w_1)(b_2 x + u w_2)(b_3 x + u w_3)
where w_1 w_2 w_3 = f, and
b_1 b_2 b_3 = (m^3 f^4 - 3m^2 f^2 + 3m),
and at m=0
P(0)/f^2 = 3xu^2 + u^3 f = u^2(3x + uf),
so two of the b's must equal 0, which means
P(0)/f^2 = w_1 w_2 u^2 (b_3 x + u w_3)
which is
P(0)/f^2 = u^2 (b_3 w_1 w_2 x + u f) = u^2(3x + uf)
proving that w_1 w_2 must equal 1, if f is coprime to 3, which
leaves
b_3 = 3.
Essentially objections to how f^2 divides off now come down to
claiming that the w's are functions of m, but consider that
w_1 w_2 =
1, when m=0, if f is coprime to 3.
But that was an arbitrary choice, so let f=3.
Now w_1 w_2 = 3^{2/3} WITHOUT REGARD TO m.
But before at m=0, they were coprime to f, now they are not
when f=3,
as they are constant. Clearly, they are constant in both cases
with
respect to m, without regard to the value of f. Which makes
sense as
f^2 is not a function of m, and it is what is being divided
off.
Therefore, the factorization is
P(m)/f^2 = (m^3 f^4 - 3m^2 f^2 + 3m) x^3 -
3(-1+mf^2 )x u^2 + u^3 f =
(b_1 x + u)(b_2 x + u)(b_3 x + uf)
where you'll notice that the b's are algebraic integers with
m=1,
f=sqrt(2), but that's a special case as generally they are
not, which
shows a problem with the ring of algebraic integers.
James Harris
===
Subject: Re: How I know, linchpin of my FLT proof
|Clearly, they are constant in both cases with
|respect to m, without regard to the value of f. Which makes
sense as
|f^2 is not a function of m, and it is what is being divided
off.
It seems to be a common tendancy for people writing proofs
(and not
always just amateurs) to use phrases like clearly, necessarily,
must
be true and so on at just the spot where they're failing quite
to make
an argument connect together.
We can tell you're essentially just bluffing and posturing
here. Logical
arguments don't rely upon the reader's sense of surprise to
close the
case. Now you write as if a function like (f-3)m were
something absurd.
After all, if I set f=3 this becomes a constant. But then,
miracle of
miracles, for f<>3 it suddenly depends on m. The argument
you're
trying to sell us doesn't make any more sense.
I always used to think (when very young) that one should be
able to
reach a point in a mathematical debate where you either agree,
or
the losing side is left just arbitrarily denying one of the
winning side's
points. It used to seem remarkable how long we could go with
you
moving very slowly toward the truth, until you finally
realized the latest
wonder proof was mistaken. Now I'm afraid you're seeming much
more like the person who simply asserts that the function he
wants
to be constant must be constant, with no need to examine
demonstrations to the contrary.
Keith Ramsay
===
Subject: Re: How I know, linchpin of my FLT proof
In sci.physics, James Harris
<jstevh@msn.com> Your definition of object ring is still circular,
thus your
whole FLT proof is invalid.
> Once again you choose not to address this issue.
> Simple math question - does your object ring contain
sqrt(2)?
> I bet you don't know.
> Tut, tut. Don't you JSH does not answer questions?
> Gib
> The answer is that sqrt(2) is an object.
> Very good.
> What I've seen are repeated attempts at distractions.
> That's what you think you see.
> But in reality your definition of object ring is still
circular, thus
your whole FLT proof is
> still invalid.
> So you cannot say nobody found an error in my FLT proof.
> Well I did find a problem with the definition of the object
ring that
> I'd given, and I've updated it.
> However, you still seem to not understand what a
mathematical proof
> is.
> It is a perfect argument that begins with a truth and
proceeds by
> logical steps to a conclusion which then must be true.
> So it's impossible to find an error in a proof.
Attempt at counterexample: I claim to prove that 1 = 2.
Let a = b = 1.
Then a^2 = ab.
a^2 - b^2 = ab - b^2.
(a+b)(a-b) = b(a-b)
Dividing by a-b we get
a+b = b.
1 = 2.
QED.
This is of course a claim of a proof only, and the error is
(hopefully) easily spotted. Many other claims have far more
obscure errors.
> However, a would-be discoverer *can* make errors in
describing a
> proof, or think they see a proof where none exists, and
potentially
> that can be found out by starting at the beginning of the
proof, and
> proceeding through it checking each step to make certain
that it is a
> logical one.
I submit you have a claim. Has it at least been peer-reviewed?
:-)
> James Harris
--
#191, ewill3@earthlink.net
It's still legal to go .sigless.
===
Subject: Re: How I know, linchpin of my FLT proof
> |Clearly, they are constant in both cases with
> |respect to m, without regard to the value of f. Which makes
sense as
> |f^2 is not a function of m, and it is what is being divided
off.
> It seems to be a common tendancy for people writing proofs
(and not
> always just amateurs) to use phrases like clearly,
necessarily,
must
> be true and so on at just the spot where they're failing
quite to make
> an argument connect together.
Clearly you deleted out the math that gives context.
> We can tell you're essentially just bluffing and posturing
here. Logical
> arguments don't rely upon the reader's sense of surprise to
close the
> case. Now you write as if a function like (f-3)m were
something absurd.
> After all, if I set f=3 this becomes a constant. But then,
miracle of
> miracles, for f<>3 it suddenly depends on m. The argument
you're
> trying to sell us doesn't make any more sense.
I have w's where w_1 w_2 = 1, when m=0, if f is coprime to 3,
but w_1
w_2 = 3^{2/3} if f has a factor that is 3.
Try that with something like
f(m) = (f-3)m + c
and the first case where w_1 w_2 equal 1 at m=0, would imply
that c be
a unit.
But the second case would imply that c=3^{1/3}.
Now I see a contradiction there Keith Ramsay.
It gets worse. It turns out that f doesn't have to equal 3,
but just
have 3 as a factor, so f=15, or f=3sqrt(17), or f=93, will
give the
same problem.
So clearly, you are wrong.
> I always used to think (when very young) that one should be
able to
> reach a point in a mathematical debate where you either
agree, or
> the losing side is left just arbitrarily denying one of the
winning
side's
> points. It used to seem remarkable how long we could go with
you
> moving very slowly toward the truth, until you finally
realized the
latest
> wonder proof was mistaken. Now I'm afraid you're seeming much
> more like the person who simply asserts that the function he
wants
> to be constant must be constant, with no need to examine
> demonstrations to the contrary.
> Keith Ramsay
There have been NO demonstrations to the contrary. If you have
one,
give it.
It is mathematics, so there's no need for debate, histrionics,
or
psychological commentary.
Consider, in the ring of algebraic integers,
P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 -
3(-1+mf^2 )x u^2 + u^3 f).
Now using b_1, b_2, b_3, w_1, w_2, and w_3, I have the
factorization
P(m)/f^2 = (b_1 x + u w_1)(b_2 x + u w_2)(b_3 x + u w_3)
where w_1 w_2 w_3 = f, and
b_1 b_2 b_3 = (m^3 f^4 - 3m^2 f^2 + 3m),
and at m=0
P(0)/f^2 = 3xu^2 + u^3 f = u^2(3x + uf),
so two of the b's must equal 0, which means
P(0)/f^2 = w_1 w_2 u^2 (b_3 x + u w_3)
which is
P(0)/f^2 = u^2 (b_3 w_1 w_2 x + u f) = u^2(3x + uf)
proving that w_1 w_2 must equal 1, if f is coprime to 3, which
leaves
b_3 = 3.
Essentially objections to how f^2 divides off now come down to
claiming that the w's are functions of m, but consider that
w_1 w_2 =
1, when m=0, if f is coprime to 3.
But that was an arbitrary choice, so let f have a factor that
is 3.
Now w_1 w_2 = 3^{2/3} WITHOUT REGARD TO m.
As I noted above, with f coprime to 3, w_1 w_2 = 1 at m=0, and
it's
not possible, if they were functions of m, that now they'd
equal
3^{1/3} for all m, including m=0. They are then not functions
of m.
Not surprisingly then as it is constant with respect to m, f^2
divides
off as a contant with respect to m.
Therefore, the factorization is
P(m)/f^2 = (m^3 f^4 - 3m^2 f^2 + 3m) x^3 -
3(-1+mf^2 )x u^2 + u^3 f =
(b_1 x + u)(b_2 x + u)(b_3 x + uf)
where you'll notice that the b's are algebraic integers with
m=1,
f=sqrt(2), but that's a special case as generally they are
not, which
shows a problem with the ring of algebraic integers.
James Harris
===
Subject: Re: How I know, linchpin of my FLT proof
> In sci.physics, James Harris
> <jstevh@msn.com>
<deleted However, you still seem to not understand what a
mathematical proof
> is.
 It is a perfect argument that begins with a truth and
proceeds by
> logical steps to a conclusion which then must be true.
 So it's impossible to find an error in a proof.
> Attempt at counterexample: I claim to prove that 1 = 2.
Which shows that like the poster I was answering before you
fail to
understand what a mathematical proof is. A mathematical proof
is a
*perfect* argument, so no counterexample exists. Possibly
you've been
programmed by social conventions where claims of proof are
called
proofs.
But it's like if I say I have proof that you are a dog.
My *saying* I have proof does not create a proof.
So if a person says they have proof you're a dog, does that
prove that
a proof can be in error?
No, it's just that they're in error, and do not have proof
you're a
dog.
If they did have proof you're a dog, then you'd be a dog.
> Let a = b = 1.
> Then a^2 = ab.
> a^2 - b^2 = ab - b^2.
> (a+b)(a-b) = b(a-b)
> Dividing by a-b we get
And given that a=b that's an attempt at dividing by zero in the
classic example.
This example only seems to work by *human* error as human
beings see
'a' and they see 'b' and think, different things, despite them
being
defined to be the same at the beginning.
Doing the substitution a=b, ignoring the 1 for the moment gives
a^2 = a^2
a^2 - a^2 = a^2 - a^2
(a+a)(a-a) = a(a-a)
Dividing by a-a would be an error, as a-a=0.
Now using the full substitution of a=b=1, you have
1^2 = 1^2
1^2 - 1^2 = 1^2 - 1^2
(1+1)(1-1) = 1(1-1)
and dividing by 1-1 would be an error as it equals 0.
> a+b = b.
> 1 = 2.
> QED.
> This is of course a claim of a proof only, and the error is
> (hopefully) easily spotted. Many other claims have far more
> obscure errors.
Given a claim of proof, you can test it by determining if the
argument
begins with a truth, and proceeds by logical steps to a
conclusion
which then must be true.
Unfortunately many people say proof when they mean claim of
proof,
so a lot of people believe that a math proof can be wrong, but
they
wouldn't believe that proof in any other context can be wrong,
as then
they realize it simply wasn't proof.
If you have proof that someone committed a crime, then you
have proof.
If it's not proof, then it's not proof.
That when math is stuck next to proof some people suddenly
think
something changes is problematic, and may be why some can
accept the
possibility of error in a math proof.
 However, a would-be discoverer *can* make errors in
describing a
> proof, or think they see a proof where none exists, and
potentially
> that can be found out by starting at the beginning of the
proof,
and
> proceeding through it checking each step to make certain
that it is a
> logical one.
> I submit you have a claim. Has it at least been
peer-reviewed? :-)
That's an interesting question and the answer is, I don't
know. I
have sent my work to math journals, and a paper is currently
at a math
journal, and I'm waiting to hear from them.
Have any of the other journals I've sent papers to actually
peer-reviewed?
I don't know.
What is important to remember though, is that math society is a
society, and I've already outlined the weird notion that a
proof can
be in error, where people actually believe that a *math proof*
can be
in error, when what they should realize is that a claim of
proof can
be in error.
Math proofs are perfect, just like any other proof that
actually is a
proof.
People, on the other hand, can say proof when in fact they
don't
have a proof.
Just like someone can say you are a dog, claim they have
proof, but it
be nonsense.
Hopefully I've cleared that issue up, and I've gone on about it
because it was being questioned!!!
Now here's a math proof. Those who doubt that fact can believe
it's a
claim of proof, but it's verified to be a proof by tracing the
argument out.
In this case, I begin with an expression. The expression
exists, so
that is the truth from which you start.
Consider, in the ring of algebraic integers,
P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 -
3(-1+mf^2 )x u^2 + u^3 f).
That is, I have the identity which defines P(m) in terms of
various
symbols, and it's all in the ring of algebraic integers, which
means
that the symbols can only represent numbers that are algebraic
integers.
Now using b_1, b_2, b_3, w_1, w_2, and w_3, I have the
factorization
P(m)/f^2 = (b_1 x + u w_1)(b_2 x + u w_2)(b_3 x + u w_3)
where w_1 w_2 w_3 = f, and
b_1 b_2 b_3 = (m^3 f^4 - 3m^2 f^2 + 3m),
and at m=0
P(0)/f^2 = 3xu^2 + u^3 f = u^2(3x + uf),
so two of the b's must equal 0, which means
P(0)/f^2 = w_1 w_2 u^2 (b_3 x + u w_3)
which is
P(0)/f^2 = u^2 (b_3 w_1 w_2 x + u f) = u^2(3x + uf)
proving that w_1 w_2 must equal 1, if f is coprime to 3, which
leaves
b_3 = 3.
Now that was a lot of steps, but each was a logical one.
First I introduced b_1, b_2, b_3, w_1, w_2, and w_3, which are
defined
by the factorization
P(m)/f^2 = (b_1 x + u w_1)(b_2 x + u w_2)(b_3 x + u w_3)
then I set m=0, and used the definition of P(m) to get P(0).
That told me that at m=0 two of the b's are 0, because then
P(0)/f^2 = 3xu^2 + u^3 f = u^2(3x + uf),
where the u^2 couldn't get there unless two of the b's are 0.
Then using that result I get from
P(m)/f^2 = (b_1 x + u w_1)(b_2 x + u w_2)(b_3 x + u w_3)
that
P(0)/f^2 = w_1 w_2 u^2 (b_3 x + u w_3)
and multiplying through by w_1 w_2 I have
P(0)/f^2 = u^2 (b_3 w_1 w_2 x + u f)
which with
P(0)/f^2 = 3xu^2 + u^3 f = u^2(3x + uf),
tells me that w_1 w_2 = 1, when m=0.
Essentially objections to how f^2 divides off now come down to
claiming that the w's are functions of m, but consider that
w_1 w_2 =
1, when m=0, if f is coprime to 3.
Now I'm focusing on what has been revealed to be an area of
confusion.
Apparently some people believe that when I divide off f^2 that
it can
divide off as a *function* of m, so that m=0 might be a
special case.
I'm now starting the argument to address that belief by noting
again
that w_1 w_2 = 1, when m=0, if f is coprime to 3. That is,
when f
doesn't have 3 as a factor.
But that was an arbitrary choice, so let f=3.
That is, I *said* f is coprime to 3 but in considering this
possibility it's worth it to relax that restriction and now
consider
what would happen if it equals 3.
Now w_1 w_2 = 3^{2/3} WITHOUT REGARD TO m.
Seeing that is as simple as looking at
P(m)/f^2 = (m^3 f^4 - 3m^2 f^2 + 3m) x^3 -
3(-1+mf^2 )x u^2 + u^3 f
with f=3 as then you have
P(m)/3^2 = (m^3 3^4 - 3m^2 3^2 + 3m) x^3 -
3(-1+m3^2 )x u^2 + 3u^3
so *every* coefficient has a factor that is 3, as you can tell
by
looking.
So with
P(m)/3^2 = (b_1 x + u w_1)(b_2 x + u w_2)(b_3 x + u w_3)
each of the b's and each of the w's has a factor that is
3^{1/3},
while the b's can have additional factors in common with 3,
the w's
cannot, as when 3 is separated out, notice you have
P(m)/3^2 = 3((m^3 3^3 - 3m^2 3 + m) x^3 -
(-1+m3^2 )x u^2 + u^3).
But before at m=0, they were coprime to f, now they are not
when f=3,
as they are constant. Clearly, they are constant in both cases
with
respect to m, without regard to the value of f. Which makes
sense as
f^2 is not a function of m, and it is what is being divided
off.
That is, if they were functions of m, so that w_1 w_2 = 1 at
m=0 as a
*function* of m, then it wouldn't matter if f had a factor of
3 or
not, you'd STILL get w_1 w_2 = 1 at m=0, without regard to the
value
of f.
But in fact, if f=3, you have w_1 w_2 = 3^{2/3} at m=0, which
only
works if the w's are independent of m, which they are.
It makes sense that they are anyway, as f^2 isn't a function
of m, but
I've seen that for some people the idea can take hold after
seeing m=0
highlighted.
But if the w's were functions of m, then w_1 w_2 would equal 1,
without regard to the value of f, but it does not.
Therefore, the factorization is
P(m)/f^2 = (m^3 f^4 - 3m^2 f^2 + 3m) x^3 -
3(-1+mf^2 )x u^2 + u^3 f =
(b_1 x + u)(b_2 x + u)(b_3 x + uf)
where you'll notice that the b's are algebraic integers with
m=1,
f=sqrt(2), but that's a special case as generally they are
not, which
shows a problem with the ring of algebraic integers.
And here I've packed in a lot of information as well.
First, with f coprime to 3, I now know that the factorization
is
P(m)/f^2 = (b_1 x + u)(b_2 x + u)(b_3 x + uf)
as the w's are constant with respect to m, so I can just check
at m=0,
which revealed that w_1 w_2 = 1. Now that doesn't necessarily
force
w_1 and w_2 to each equal 1, but even if they were factors of
1, i.e.
unit factors, that would only change b_1 and b_2.
So I have my factorization without regard to m in terms of
where the f
goes, and then I point out that you can actually check my work
using
m=1, f=sqrt(2), as then you get a polynomial which you can
factor
rather simply. So you can actually get the values for the b's
and
check them, and see that they are all algebraic integers, and
all are
coprime to 2.
However, usually, for f values that are coprime to 3, you
don't get
b's that are algebraic integers, which shows a problem with
the ring
of algebraic integers.
Now the nice thing about a mathematical proof is that if
someone
disagrees they have to find some misstep.
Unfortunately, people can *say* that proof is not a proof,
even when
it is, just like if you tried to say you were human, and not a
dog,
someone might dispute any proof you might give, claiming it
false.
James Harris
===
Subject: Re: How I know, linchpin of my FLT proof
> Have any of the other journals I've sent papers to actually
> peer-reviewed?
> I don't know.
It's unlikely that they could find one of your peers to
conduct the
review. Now, if you could find a journal that uses ignorant,
paranoid,
narcissistic drunks as reviewers, you might get a peer
review...
--
Wayne Brown | When your tail's in a crack, you improvise
fwbrown@bellsouth.net | if you're good enough. Otherwise you
give
| your pelt to the trapper.
e^(i*pi) = -1 -- Euler | -- John Myers Myers,
Silverlock
===
Subject: Re: How I know, linchpin of my FLT proof
> In sci.physics, James Harris
> <jstevh@msn.com <deleted However, you still seem to not understand what a
mathematical proof
> is.
 It is a perfect argument that begins with a truth and
proceeds by
> logical steps to a conclusion which then must be true.
 So it's impossible to find an error in a proof.
> Attempt at counterexample: I claim to prove that 1 = 2.
> Which shows that like the poster I was answering before you
fail to
> understand what a mathematical proof is. A mathematical
proof is a
> *perfect* argument, so no counterexample exists. Possibly
you've been
> programmed by social conventions where claims of proof are
called
> proofs.
> But it's like if I say I have proof that you are a dog.
> My *saying* I have proof does not create a proof.
> So if a person says they have proof you're a dog, does that
prove that
> a proof can be in error?
> No, it's just that they're in error, and do not have proof
you're a
> dog.
> If they did have proof you're a dog, then you'd be a dog.
Go away James. You're splitting hairs that: 1) don't need
to be split, and 2) don't illuminate any argument. Belaboring
pedantics such as this just tries peoples' patience.
-- Bob Day
===
Subject: Re: How I know, linchpin of my FLT proof
> In sci.physics, James Harris
> <jstevh@msn.com <deleted However, you still seem to not understand what a
mathematical proof
> is.
 It is a perfect argument that begins with a truth and
proceeds by
> logical steps to a conclusion which then must be true.
 So it's impossible to find an error in a proof.
 Attempt at counterexample: I claim to prove that 1 = 2.
> Which shows that like the poster I was answering before you
fail to
> understand what a mathematical proof is. A mathematical
proof is a
> *perfect* argument, so no counterexample exists.
Way too wide an opening there James. The obvious question is
why do counterexamples to your proofs abound? Could it
mean (gasp) your proof is less than perfect?
- Randy
===
Subject: Re: How I know, linchpin of my FLT proof
Visiting Assistant Professor at the University of Montana.
> [.snip.]
>Well I did find a problem with the definition of the object
ring that
>I'd given, and I've updated it.
>There have been at least two changes in recent memory, one
sort of
>announced, one done in silence. And then there was another
change in
>the past 36 hours, presumably what you are refering to here.
And yet another unannounced change has now occured.
Yesterday, the definition at
http://www.msnusers.com/AmateurMath/objectmathematic.msnw
was:
>The Object Ring is the set of all numbers where any member
that is a
>unit, i.e. factor of 1, and its multiplicative inverse are
units in
>all possible commutative rings in which either and all
integers are
>members, and where no non-unit member a is a factor of any two
>integers that are coprime.
The definition right now (4:12 MDT) is:
The Object Ring is the set of all numbers where 1 is the only
member
that is both a unit, i.e. factor of 1, and an integer, where no
non-unit member is a factor of any two integers that are
coprime.
You are still being sloppy in saying set of all numbers. I
suspect
that you mean to restrict yourself to complex numbers, if not
ALGEBRAIC numbers, and to give this set the inherited
structure. If
this is the case, then since -1 is both a unit and an integer
in any
subring of the complex numbers, it looks like you have
nothing, yet
again.
Oh, the second clause is still imprecise or empty. Coprimeness
is a
property that depends on the ring, although implications in one
direction may hold. You probably mean two integers that are
coprime
in the ring of integers, in which case the condition is
vacuously
true in any subring of C, as I noted yesterday.
Now, assuming you meant to say 1 and -1 are the only elements
which
are both units and integers, then you still must prove that
Object
ring under this definition specifies a unique such object.
Presumably, you want to say largest subring of the complex
numbers
such that..., because otherwise, the integers are The Object
Ring,
but so is any subring of the ring of all algebraic integers.
It would
be of paramount importance to make sure that it defines a
unique
thing, if you are going to call refer to it by using the
singular
I am also pretty certain that this definition includes way too
many
things that you do not want. But it is obvious that once again
all you
are doing is trying to fix, by fiat, the problems that plagued
your
original proof of two years ago.
I must, however, confess that I am flabbergasted at your
brilliance:
here we have what, by your own account, is the key, central,
germain,
touchstone, concept of your approach. And even though you have
been
able to change the definition in significant ways over the
past 8
months, yet your proof is so solid that changing this key
definition
does not require you to change even a single word of the rest
of your
developement to take into account these changes. Truly, a work
of
genius.
Why do you take so much trouble to expose such a reasoner as
Mr. Smith? I answer as a deceased friend of mine used to answer
on like occasions - A man's capacity is no measure of his power
to do mischief. Mr. Smith has untiring energy, which does
something; self-evident honesty of conviction, which does more;
and a long purse, which does most of all. He has made at least
ten publications, full of figures few readers can critize. A
great
many people are staggered to this extend, that they imagine
there
must be the indefinite something in the mysterious all this.
They are brought to the point of suspicion that the
mathematicians
ought not to treat all this with such undisguised contempt,
at least.
-- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan
Arturo Magidin
magidin@math.berkeley.edu
===
Subject: Re: How I know, linchpin of my FLT proof
I'm going to attempt a cold-reading of the subject, because
I often pretend to be my own college career counselor --
just kidding.
on second thought, this interface is new to me, so that
I'll just jump to the end by deleting this passage
from the 10year mission (henceforth, TTYM, which is actually
short,
for the 10y. Mission to Pummel Fermat et Son Fil
into an elementary lesson on differential equations, Yeeha),
and putting m thoughts on the matter, herein.
oops.
> The *mathematics* is straightforward.
> Readers should see that it's also clear that posters for
social
> reasons are fighting the mathematical logic, which is
fascinating
> behavior.
--les ducs d'Enron!
http://members.tripod.com/~american_almanac
===
Subject: Re: How I know, linchpin of my FLT proof
> [.snip.]
>Well I did find a problem with the definition of the
object ring that
>I'd given, and I've updated it.
>There have been at least two changes in recent memory, one
sort of
>announced, one done in silence. And then there was another
change in
>the past 36 hours, presumably what you are refering to here.
> And yet another unannounced change has now occured.
> Yesterday, the definition at
> http://www.msnusers.com/AmateurMath/objectmathematic.msnw
> was:
>The Object Ring is the set of all numbers where any member
that is a
>unit, i.e. factor of 1, and its multiplicative inverse are
units in
>all possible commutative rings in which either and all
integers are
>members, and where no non-unit member a is a factor of any
two
>integers that are coprime.
> The definition right now (4:12 MDT) is:
> The Object Ring is the set of all numbers where 1 is the
only member
> that is both a unit, i.e. factor of 1, and an integer, where
no
> non-unit member is a factor of any two integers that are
coprime.
> You are still being sloppy in saying set of all numbers. I
suspect
> that you mean to restrict yourself to complex numbers, if not
> ALGEBRAIC numbers, and to give this set the inherited
structure. If
> this is the case, then since -1 is both a unit and an
integer in any
> subring of the complex numbers, it looks like you have
nothing, yet
> again.
Hey, you're right. Good catch. I'll update the page.
James Harris
===
Subject: How to create an equation?
Hey,
I have a small problem creating a equation which would return
following values:
3/8 = 1/4
--
5/8 -
6/8 | = 2/4
7/8 -
--
9/8 -
. | = 4/4
15/8 -
--
17/8 -
. | = 8/4
31/8 -
--
33/8 -
. | = 16/4
63/8 -
........
I hope it is clear what I mean - basicly everytime I increment
the x
of the x/8 and if x/8 a non float there is no result.
I have no idea how to set up this equation...
It would be great if someone knows how to approach this
problem.
Nikolai Onken
===
Subject: Re: How to create an equation?
> Hey,
> I have a small problem creating a equation which would return
> following values:
> 3/8 = 1/4
> --
> 5/8 -
> 6/8 | = 2/4
> 7/8 -
> --
> 9/8 -
> . | = 4/4
> 15/8 -
> --
> 17/8 -
> . | = 8/4
> 31/8 -
> --
> 33/8 -
> . | = 16/4
> 63/8 -
> ........
> I hope it is clear what I mean - basicly everytime I
increment the x
> of the x/8 and if x/8 a non float there is no result.
> I have no idea how to set up this equation...
> It would be great if someone knows how to approach this
problem.
> Nikolai Onken
I'm not sure that I understand. It seems to me that what you
want is a
_function_ not an equation, and the domain of that function is
{3,5,6,7,9,10,11,12,13,14,15,17,18,19,20,21,22,23,**24**,
25,26,27,28,29,30,3
1,33,...,63}.
Now, _do_ you want 24 in there since 24/8 is an integer? Also
what
about the x in [33,63] such that x/8 is an integer, such as
40, 48, 56?
Or are you trying to exclude the powers x = 2^n for n =
2,3,4,5? Apart
from these unclearnesses, you have almost defined the function
yourself
by cases.
Since you use the word float me thinks that this may be a
programming
problem. Is it?
--
G.C.
===
Subject: Re: How to create an equation?
> I'm not sure that I understand. It seems to me that what you
want is a
> _function_ not an equation, and the domain of that function
is
>
{3,5,6,7,9,10,11,12,13,14,15,17,18,19,20,21,22,23,**24**,
25,26,27,28,29,30,31
,
> 33,...,63}.
> Now, _do_ you want 24 in there since 24/8 is an integer?
Also what
> about the x in [33,63] such that x/8 is an integer, such as
40, 48, 56?
> Or are you trying to exclude the powers x = 2^n for n =
2,3,4,5? Apart
> from these unclearnesses, you have almost defined the
function yourself
> by cases.
It sounds to me like he's trying to find a general expression
for the
pattern expressed by his cases. It could be written as
{ 2^floor(log2(a)) if log2(a) not an integer
f(a) = {
{ undefined otherwise
If you don't like writing down the special case, it can be
built in
by writing it as (2^F)*(L-F)/(L-F), where L=log2(a) and
F=floor(L).
===
Subject: Re: How to create an equation?
> Hey,
> I have a small problem creating a equation which would return
> following values:
> 3/8 = 1/4
> --
> 5/8 -
> 6/8 | = 2/4
> 7/8 -
> --
> 9/8 -
> . | = 4/4
> 15/8 -
> --
> 17/8 -
> . | = 8/4
> 31/8 -
> --
> 33/8 -
> . | = 16/4
> 63/8 -
> ........
> I hope it is clear what I mean - basicly everytime I
increment the x
> of the x/8 and if x/8 a non float there is no result.
> I have no idea how to set up this equation...
> It would be great if someone knows how to approach this
problem.
> Nikolai Onken
equality symbol it must be true that whatever value is on the
left of the
equal sign must be the same value on the right hand side.
Always. Since 3/8
> 1/4 it follows that (3/8) /= (1/4). Be careful, one must
never abuse an
equal sign.
===
Subject: Re: How to create an equation?
> I have a small problem creating a equation which would return
> following values:
> 3/8 = 1/4
> --
> 5/8 -
> 6/8 | = 2/4
> 7/8 -
> --
> 9/8 -
> . | = 4/4
> 15/8 -
> --
> 17/8 -
> . | = 8/4
> 31/8 -
> --
> 33/8 -
> . | = 16/4
> 63/8 -
> ........
Let x be the value on the left, then you want f(x) which
gives the value on the right. How about
f(x) = 2 ^ ([log(x * 8) / log (2)] - 1) / 4
where ^ is exponentiation and [] is greatest integer function.
There may be other functions, but I think this one does
what you want.
If you are writing a C program, let x be the numerator of the
value on the right as an integer value. Then do
int k, z, i ;
k = (log((float) x) + 0.0000001) / log(2.0) - 1;
z = 1 ;
for (i = 0; i < k; i++)
z = 2 * z ;
Then z should be the numerator of the expression on the right.
> I hope it is clear what I mean - basicly everytime I
increment the x
> of the x/8 and if x/8 a non float there is no result.
> I have no idea how to set up this equation...
> It would be great if someone knows how to approach this
problem.
> Nikolai Onken
===
Subject: Re: How to read vector in MOG format?
> I am reading Conway & SLOANE's Sphere Packings, Lattices and
Groups
> in this book, he expresses a 24 dimensional vector in MOG
format
> Where can i find some information about reading this format?
> In the book they express the vector like
> u = |-------------------|
> |-2 2 | 4 0 | 0 0 |
> | 2 2 | 0 0 | 0 0 |
> | 2 2 | 0 0 | 0 0 |
> | 2 2 | 0 0 | 0 0 |
> |-------------------|
> What's u?
As I recall u can be interpreted as a code word, (Hamming,
length 24?)
a vector in the Leech lattice, and an element
of the Mathieu group M24.
MOG is the Miracle Octad Generator. They give an
exposition and a tutorial, but since it is not in front
of me now I cannot give you chapter and verse.
--
Michael Press
===
Subject: Re: Ideal Class Group
===
Subject: Re: Ideal Class Group
> Let R be an integral domain or a ring of integers,
> I = (a) a principal ideal of R,
> b,d two nonzero elements of R and J an ideal with bI = dJ.
> Thus (ab) = dJ. I'm assured J is a principal ideal.
> May I have a hint or two how to show this?
>ba is in bI and so in dJ.
>In other words ba = dj for some j in J.
>It is easy enough to see that J = (j).
>In effect, if x is in J then dx is in bI = (ab),
>ie dx = aby = djy.
>Henc x = jy since R is an integral domain.
ideals. The last part needed to establish ideal class groups
is given an ideal I, find an ideal J such that
bIJ = (a)
for some nonzero a,b where
IJ = { sum(i=1..n) a_i b_i | n in N, a_i in I, b_i in J }
the ideal generated by { ab | a in I, b in J }
For this I request some help. I think is the crutial part
requiring for
the first time, more structure of R than just being an
integral domain.
Doesn't R need to be a ring of integers or a ring of integers
of a number
field?
----
===
Subject: Re: Ideal Class Group
> ideals. The last part needed to establish ideal class groups
> is given an ideal I, find an ideal J such that
> bIJ = (a)
> for some nonzero a,b where
> IJ = { sum(i=1..n) a_i b_i | n in N, a_i in I, b_i in J }
> the ideal generated by { ab | a in I, b in J }
> For this I request some help. I think is the crutial part
requiring for
> the first time, more structure of R than just being an
integral domain.
> Doesn't R need to be a ring of integers or a ring of
integers of a number
> field?
The result certainly does not hold for all integral domains.
On the other hand, R does not need to be a number ring,
eg the result would hold (trivially) for all principal ideal
domains,
eg the ring k[x] of polynomials over a field k.
An interesting class of integral domains are the rings F[x]
of polynomials over a finite field F,
and the finite algebraic extensions of these.
These rings behave very much like number rings;
almost every result for number rings holds also for these
function-rings.
In particular, they have finite ideal class groups.
But to return to your question;
two ideals I,J in R are said to be in the same ideal class
if aI = bJ for some non-zero a,b in R.
It is easy to show that the ideal classes form an abelian
semigroup.
The condition you give is the condition that the ideal classes
should form a group.
I believe that a necessary and sufficient condition for this
is that R should be a Dedekind domain,
ie ever ideal in R is uniquely expressible as a product of
prime ideals.
(Robin Chapman will correct me if I'm wrong!)
An alternative way (neater in my opinion) to view this
is to consider fractional ideals, ie subsets of the
quotient-field k of R
of the form xI, where I is an ideal in R and x is in k
(ie x = a/b with a,b in R).
Then we set I^{-1} = {x in k: xI < R};
and the condition you give is equivalent to I I^{-1} = R.
--
Timothy Murphy
e-mail: tim@birdsnest.maths.tcd.ie
tel: +353-86-233 6090
s-mail: School of Mathematics, Trinity College, Dublin 2,
Ireland
===
Subject: Re: inequality
> Could someone explain this to me...
 1) if a<b then a<=b [true]
> 2) if a<=b then a<b [False]
 My text says this is right, but has no explaination for
it. A proof
> or something might be nice...
 Steve, I can sort of grasp the either/or of the matter, but
if in
> case 2)
> the possibility of a=b is not met, as in 2<=3, why is the
statement if
> 2<=3 then 2<3 false? Have I muddled something up?
> Josh
Josh,
In this context, to say 2) is true means that it needs to be
true for
*any possible* values of a and b. Sure, 2<=3 and 2<3 are both
true,
but 2) is not true *in general*, which is intended here. Get
it?
In this, you might stumble over if 3<2 then 3<=2, which is
regarded
as a true conditional since the premise is false. I can only
suggest
that such conditionals are best thought of as *vacuously
true*, true
for purposes like stamping 1) true for all possible values of
a and
b. But all we really care about are the values of a and b
where the
premise a<=b is really true.
Bob
===
Subject: Re: is Principal Of Induction theorem a theorem or an
axiom??
> The Principal Of Induction which is:
> a statement P is true for P=1,
> and true for P=k+1 if k was true,
> then P is true
> Is it the axiom or theorem??
> In my text it says it lies at a different level than the
axioms,
> together with this rule.
> If F and F->G then G
> the rule of induction is like the rule of inference, they
are added to the
axioms.
> But I don't see any formal reason for its special case, at
this point it
looks
> likely as a step to complete formalism, but we are short on
categorising
many
> other proof mechanisms.
> Herc
Get a better text. Some rules of inference have one
antecedent; some
have two; some have three or even more (see Louis Carroll on
modus
ponens); some have zero antecedents. The last we call axioms.
As for the original question, you can indeed either take it as
an
axiom or assume even stronger axioms and derive it as a
theorem. Me,
I prefer to think of it as an axiom of the natural numbers.
===
Subject: Re: is Principal Of Induction theorem a theorem or an
axiom??
> The Principal Of Induction which is:
 a statement P is true for P=1,
> and true for P=k+1 if k was true,
> then P is true
 Is it the axiom or theorem??
 In my text it says it lies at a different level than the
axioms,
> together with this rule.
> If F and F->G then G
> the rule of induction is like the rule of inference, they
are added to
the axioms.
> But I don't see any formal reason for its special case, at
this point it
looks
> likely as a step to complete formalism, but we are short
on categorising
many
> other proof mechanisms.
> Herc
> Get a better text. Some rules of inference have one
antecedent; some
> have two; some have three or even more (see Louis Carroll on
modus
> ponens); some have zero antecedents. The last we call axioms.
> As for the original question, you can indeed either take it
as an
> axiom or assume even stronger axioms and derive it as a
theorem. Me,
> I prefer to think of it as an axiom of the natural numbers.
modus ponens and induction formula together to output an
inducted proof?
NO, because the induction looks like an axiom but
instantiating P(x)->P(sx)
is an intellectual feat. Where's an axiom for diagonalization
and the 100
other succint techniques used in this forum as proofs every
day?
Herc
sorry but noone knocks A.K. Dewdney (especially to a
numerologist)
===
Subject: Re: Is there a bijection from R^n to R?
> N^n injects into N? How is this possible? Does this mean
that Z^n also
> injects into Z?
> This seems ridiculous to me, but I will have to think about
it more after
I
> read all of the
> responses.
Yes there is an injection Z^n -> Z. Even Z^n -> N. And even
(as I said earlier) Q^n -> N. All assuming n is a natural
number, i.e. finite.
A complete enumeration of Z^2 is easy to visualize. Start at
(0,0). Around that point there is a square of eight points;
traverse them counterclockwise starting at the x-axis:
(1,0), (1,1), (0,1), (-1,1), (-1,0), (-1,-1), (0,-1), (1,-1).
Around this square there is a larger square of 16 points which
we traverse by the same rule:
(2,0), (2,1), (2,2), (1,2), (0,2), (-1,2), (-2,2), (-2, 1),
(-2,0), (-2,-1), (-2,-2), (-1,-2), (0,-2), (1,-2), (2,-2),
(2,-1)
thus you see we have a bijection between all points in this
subset of Z^2 and the first 25 elements of N -- and of course
we can continue in the same vein to ever larger squares, and
thereby cover all of Z^2 in sequence. Adding dimensions only
makes our rule more complicated (and our progress outward
slower) but the basic idea still works.
My example is clearly a bijection, but with Cantor-Bernstein
all we really need to show is an injection; hence the other
example you were given -- f(a,b) = 2^a * 3^b -- is also a
satisfactory answer for the case N^2 -> N.
===
Subject: Re: Is there a bijection from R^n to R?
>But this doesn't seem to be possible, R^n has dimension n and
R has
>dimension 1.
>You can't create an injection here because of different
dimensions,
however
>surjection is easy.
>Am I missing something?
Could it be that you forgot to add the word continuous
somewhere? The
bijections between R^n and R don't have anything to do with the
algebraic or metric properties, and will be wildly
discontinuous.
> I can't seem to construct one. Is there such a mapping?
(R is the
real
> numbers)
> The Cantor-Bernstein theorem is your friend. All you need
is an
> injection f: R^n -> R and another going the opposite
direction, but the
> latter is trivial. To construct the injection f, try an
> interleave-the-digits trick. You can choose your decimal
representations
> to avoid any that terminate in all 9's.
> --
> Dave Seaman
> Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling.
>
<http://www.commoncouragepress.com/index.cfm?action=book&
bookid=228>
Larry
(this space unintentially left blank .....
make obvious deletion for email
===
Subject: Re: Is there a bijection from R^n to R?
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at 06:34 PM, Tim <hobiewon@hotmail.com> said:
>I can't seem to construct one. Is there such a mapping?
Of course, as long as you don't require continuity or
linearity.
--
Shmuel (Seymour J.) Metz, SysProg and JOAT
Any unsolicited bulk E-mail will be subject to legal action. I
reserve the
right to publicly post or ridicule any abusive E-mail.
Reply to domain Patriot dot net user shmuel+news to contact
me. Do not
reply
to spamtrap@library.lspace.org
===
Subject: Re: Is there a bijection from R^n to R?
===
Subject: Re: Is there a bijection from R^n to R?
> If f is a _continuous_ map from (0,1) _onto_ (0,1)^2,
> then f cannot be injective.
>Proof: Suppose f is injective. Then f is a homeomorphism on
each
>compact subinterval [a,b] of (0,1). By Baire, some f([a,b])
will
>contain an open disc. f is then a homeomorphism from some
interval
>onto an open disc, a contradiction because the removal of one
point
>can disconnect the interval but not the disc.
To amend your oversight, restrict a,b to rationals. Thus
(0,1)^2 is a
countable union of compact closed sets f([a,b]). As R^2 is a
Baire space,
at least one f([a,b]) can't have empty interior, hence
contains an open
ball.
--
> there are no *continuous* injections from R^n -> R.
>The map would have to be a homeomorphism from the closed unit
ball
>onto an interval. Removing an interior point from the
interval would
>disconnect it, but that is false for the closed unit ball.
This is slightly simpler.
The image of a continuous injection f:R^n -> R is a interval.
Remove from f an interior point of the interval and it's
preimage.
The resulting continuous map from a connected space onto a
disconnected interval is a contradiction.
----
===
Subject: Re: Is there a bijection from R^n to R?
> If f is a _continuous_ map from (0,1) _onto_ (0,1)^2,
> then f cannot be injective.
>Proof: Suppose f is injective. Then f is a homeomorphism on
each
>compact subinterval [a,b] of (0,1). By Baire, some f([a,b])
will
>contain an open disc. f is then a homeomorphism from some
interval
>onto an open disc, a contradiction because the removal of
one point
>can disconnect the interval but not the disc.
> To amend your oversight, restrict a,b to rationals. Thus
(0,1)^2 is a
> countable union of compact closed sets f([a,b]). As R^2 is a
Baire
space,
> at least one f([a,b]) can't have empty interior, hence
contains an open
> ball.
There was no oversight.
===
Subject: Re: JSH: Challenge of the Century
>You're being remarkably dense Nora Baron.
> [etc, etc...]
> Did JSH really write this? I'm not so sure. New e-mail
address;
unusual
> posting time; language seems subtly different.
> JSH has posted in alt.fiction.original that it isn't him.
In that post, JSF not only says it wasn't him, he says the
poster was
*convincing* and gives one the impression he himself usually
isn't:
However as someone has done so and cross-posted to here in a
way
that I guess is convincing, I thought it worth it to tell you
all
that it's not me.
-jiw
===
Subject: Re: learning math outside of high school
<3f301136$8$fuzhry+tra$mr2ice@news.patriot.net>
<oii0jv0o77olo5blua03lbp0c64v23lge0@4ax.com>
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X-Treme: C&C,DWS
<lawrence.delete.mcknight@sbcglobal.net> said:
>another book.
There seem to have been a number of books published under that
name,
of widely differing character and size. The one that I used
*did*
define limit before derivative, and was considerably larger
than the
recent ones I've seen under that name. I don't, however,
remember
whether it had a proper set of axioms for the Real Line; it's
been too
long since the 1950s.
--
Shmuel (Seymour J.) Metz, SysProg and JOAT
Any unsolicited bulk E-mail will be subject to legal action. I
reserve the
right to publicly post or ridicule any abusive E-mail.
Reply to domain Patriot dot net user shmuel+news to contact
me. Do not
reply
to spamtrap@library.lspace.org
===
Subject: Re: Linear Algebra and Summation Notation
> Yesterday, Nemo said:
===
> Subject: Re: Linear Algebra and Summation Notation
> snip
> What I want to know is: has this relationship and the
interconversion
> of these families of equations been formalized in any
significant
> way? Is there any symbolic formalism for taking an
equation in
> summation notation and rewriting it as standard
matrix/vector
> operations (or vice versa, although this way is easier
to figure out
> intuitively at least for me)?
> Looking at some basic matrix operations, it's easy (though
sometimes
> painstaking) to see what individual scalar arithmetic is
going on, and
> convert the operation to series of summations or other long
winded
> operations. What interests me is the possibility of looking
at some of
> the more longwinded methods for dealing with large ordered
set
operations,
> like summation, and using generalized rules for replacing
those
operations
> with standard matrix operations.
> I'll check out the languages you mentioned though; sound
interesting. In
> my field MatLab and Fortran are kind of the standards for
dealing with
> multidimensional data.
I don't actually see what is bothering you. Summation
convension,
explicitly written down sums (using the sigma notation), matrix
Just different people prefer to say it in different ways (then
you'll
hear about dual spaces, inner products, and so on.... but the
basic
stuff, in the sense you asking the question, is identical.
I should elaborate a little more. Take
sigma(i = 1..n) A_i B_i
for instance. The summation convention recognizes the fact
that i is
as
A_i B_i, assuming the reader knows what values are assigned to
i.
Dot product *is* the same stuff, that makes you think of A_i's
a whole
entity, namely a vector. The difference is entirely
conceptual. And
it's not worth saying it's a conceptual revolution or
something.
Matrix multiplication is *defined* in terms of dot product, so
to
speak. It fascilitates this kindda stuff. That's all there is
to it.
This thing is so basic that different people working in
different
fields like to think in different conceptual framework, that
ultimately boils down to nothing worth it.
Imam
===
Subject: math solution to puzzle? (simple)
If I want to solve a puzzle such as the one below, is there a
mathematical
way of doing it? I'm guessing it could be done by algebra or
something
similar? As you may have guessed I'm not a maths expert by a
long shot...
8)
if andrew, bob, charles, dave have 4963 between then, and
andrew
has 598
more than bob, and bob has 415 more than charles, and charles
has
twice as
much as dave, how much does dave have?
A+B+C+D = 4963
A=B+598
B=C+415
D=C/2
What number is D?
Now it took me about 60 seconds to enter the formular in Excel
and type
numbers into it until I got the answer of 505, but that is
kind of
cheating - what is the proper way to solve these puzzles?
===
Subject: Re: math solution to puzzle? (simple)
> A+B+C+D = 4963
> A=B+598
> B=C+415
> D=C/2
> What number is D?
> Now it took me about 60 seconds to enter the formular in
Excel and type
> numbers into it until I got the answer of 505, but that is
kind of
> cheating - what is the proper way to solve these puzzles?
This is a system of 4 linear equations in 4 unknows. There are
a lot of
ways
to solve these. A particularly simple one in this case is the
method of
substitution, in which you solve 1 of the equations and then
substitute
this
expression, which in general still has other unknows in it, in
all
remaining
equations.
I solve (4) for D, that's just D=C/2. Then I substitute this
in the (1) (2)
and (3)
A+B+C+C/2 = 4963
A=B+598
B=C+415
3 linear equations in 3 unknows. Repeat the same trick.
A+(C+415)+C+C/2=4963 -> A + 5C/2 = 4548
A=(C+415)+598 -> A = C+1013
2 linear equations in 2 unknows. Repeat.
(C+1013) + 5C/2 = 4548 -> 7C/2 = 3535 -> C=1010
Now you have one of the unknows. Work back to get the other
unknows
C=1010 -> A = 2023 - > B=1425 -> D=505
===
Subject: Re: math solution to puzzle? (simple)
>If I want to solve a puzzle such as the one below, is there a
mathematical
>way of doing it? I'm guessing it could be done by algebra or
something
>similar? As you may have guessed I'm not a maths expert by a
long shot...
>if andrew, bob, charles, dave have 4963 between then, and
andrew
has 598
>more than bob, and bob has 415 more than charles, and charles
has twice as
>much as dave, how much does dave have?
>A+B+C+D = 4963
>A=B+598
>B=C+415
>D=C/2
>What number is D?
>Now it took me about 60 seconds to enter the formular in
Excel and type
>numbers into it until I got the answer of 505, but that is
kind of
>cheating - what is the proper way to solve these puzzles?
Use Solver.
4 equations in 4 unknowns, you basically have two choices.
1) Substitution. Solve for one in terms of the others,
substitute,
repeat.
2) For linear problems, you can set it up like:
A + B + C + D = 4963 (eq 1)
A - B = 598 (eq 2)
B - C = 415 (eq 3)
.5C - D = 0 (eq 4)
and so forth.
You can see that (adding the first equation to the second one)
2A + C + D = 5561 (eq 1a)
(subtracting the 3rd from the 1st)
A + 2C + D = 4548 (eq 2a)
Now you've got 3 equations (your two new plus the old 4th)
equations in
3 unknowns.
.5C - D = 0 (eq 3a)
If you subtract 2*(eq 3a) from eq 1a and 4*(eq 3a) from eq 2a,
you'll
get two equations (1b and 2b, of course) with only two
unknowns. Keep
going until you're done.
There are more efficient ways to do this (why should you do
all the
work? what are machines for? [or, in the old days, banks of
calculators, who were people who calculated, not machines that
calculated]). But for puzzles, the most efficient way might be
to
just look at it and see what needs to be eliminated. Of
course, if you
are modelling the aquifer in Kansas for water projections
(10,000 eqns
in 10,000 unknowns) or a nuclear reactor core (100,000 eqns in
100,000
unknowns), you need to have a machine to do the calculations.
As to your comment about cheating, I don't think I agree.
Guess and
check or check every possible solution [try it for Rubik's
cube] are
ugly, but they work. The object of puzzles is, of course, to
find an
elegant solution (otherwise, they aren't recreation, they're
work). But
a puzzle that requires more than purely minimal computation
(in the eyes
of the beholder, of course) isn't very interesting. (Play is
what you
get to do, work is what you have to do.)
Jon Miller
===
Subject: Re: math solution to puzzle? (simple)
> A+B+C+C/2 = 4963
> A=B+598
> B=C+415
> 3 linear equations in 3 unknows. Repeat the same trick.
> A+(C+415)+C+C/2=4963 -> A + 5C/2 = 4548
> A=(C+415)+598 -> A = C+1013
> 2 linear equations in 2 unknows. Repeat.
> (C+1013) + 5C/2 = 4548 -> 7C/2 = 3535 -> C=1010
> Now you have one of the unknows. Work back to get the other
unknows
> C=1010 -> A = 2023 - > B=1425 -> D=505
===
Subject: Re: math solution to puzzle? (simple)
> As to your comment about cheating, I don't think I agree.
Guess and
> check or check every possible solution [try it for Rubik's
cube]
are
> ugly, but they work. The object of puzzles is, of course, to
find an
> elegant solution (otherwise, they aren't recreation, they're
work). But
> a puzzle that requires more than purely minimal computation
(in the eyes
> of the beholder, of course) isn't very interesting. (Play is
what you
> get to do, work is what you have to do.)
I saw the question in a newspaper giving a sample exam
question, which was
to be worked out without calculators let alone computers.
Therefore I
wanted to know how to do it on paper. Also, I am trying to
improve my
algebra skills, which are currently very basic...8)
Unfortunately when I was at school I didn't really pay much
attention
(but
miss, why do we have to learn algebra, we'll never use it in
real life),
ironically I use it (basic algebra) all the time now...
===
Subject: Re: math solution to puzzle? (simple)
That branch of math is called linear algebra.
===
Subject: More on the the primality of random integers
[My apologies is this is a repeat. My first posting shows up
on none of
my sources for newsgroups.]
What has been done on the measure of the set primes from a
given
probability measure on the natural numbers? For example, if
random
variable X has Poisson(1) distribution, is there a known
constant for
P{X is prime}?
For a probability measure P on N, let r(P) = P{x : x is prime}.
More specific questions:
- Is anything known about r for general Poisson distributions?
Geometric? Zeta (P{n} proportional to n^-a)? For example, what
is the
asymptotic behavior of r(Geom(p)) as p approaches 0? Of
r(Zeta(a)) as a
approaches 1?
- Is r(Poisson()) of bounded variation on intervals?
This much we do know:
For U_n the uniform measure on {1, 2,..., n}, r(U_n) is
asymptotically
1/log n.
It is somewhat easy to get bounds. r(P) >= PA for any set A of
prime
numbers, easily computable when A is finite. For an upper
bound, use
P(nN) = sum{k=0..n-1, c(2 pi k / n)) / n,
where c is the Fourier transform (characteristic function) of
P. (I have
cited this result in several previous postings to this
newsgroup.) Then
apply inclusion/exclusion to the right-hand side of
r(P) <= 1 - PU{nN: n in A} + PA
where A is a finite set of primes.
Clearly (and sillily), r(P) = 1 if P is supported by the
primes.
--
Stephen J. Herschkorn herschko@rutcor.rutgers.edu
--
Stephen J. Herschkorn herschko@rutcor.rutgers.edu
===
Subject: Re: More on the the primality of random integers
>[My apologies is this is a repeat. My first posting shows up
on none of
>my sources for newsgroups.]
>What has been done on the measure of the set primes from a
given
>probability measure on the natural numbers? For example, if
random
>variable X has Poisson(1) distribution, is there a known
constant for
>P{X is prime}?
>For a probability measure P on N, let r(P) = P{x : x is
prime}.
>More specific questions:
>- Is anything known about r for general Poisson distributions?
>Geometric? Zeta (P{n} proportional to n^-a)? For example,
what is the
>asymptotic behavior of r(Geom(p)) as p approaches 0? Of
r(Zeta(a)) as a
>approaches 1?
It seems likely that all these things can be determined easily
from
the Prime Number Theorem. I just say likely because I don't
recall the definitions of any of those distributions. But you
tell
us what Zeta(a) is, so I should be able to figure that one out:
Define Zeta(a)({n}) = m(a) n^(-a) (note that if I did the
arithmetic
right m(a) ~ a - 1 as a -> 1.)
Fix a number L > 1, and let I_j be the positive integers
between
L^(j-1) and L^j. The point is that I_j is long enough that the
prime number theorem says something about the number
of primes in I_j (at least for large j, where the meaning of
large depends on L), but I_j is short enough that n^(-a)
is essentially constant on I_j (we're going to get upper and
lower estimates that differ by a factor of a power of L; then
at the very end we let L -> 1.)
Now if j is large enough the PNT shows that pi(L^j) is
larger than pi(L^(j-1)), so we can subtract the two estimates;
looks to me like you get that the number of primes in I_j
is asymptotic to p_j = L^(j-1)(L-1)/(j log(L)) as j ->
infinity.
Hence if eps > 0 and mu_j is Zeta(a)({primes in I_j}) then
for j > J(L, eps) you have
(1-eps)m(a) p_j L^(-ja) <= mu_j <= (1+eps)m(a) p_j L^(-(j-1)a).
Those are geometric series that you can add. Note that
J(L, eps) does not depend on a; let a -> 1 at this point,
note that the sum of mu_j for j < J certainly tends to 0
as a -> infinity, and you get bounds on the behavior
of Zeta(a)({primes}) as a -> 1. Now let L -> 1 and eps -> 0
and you should get something.
That may need some revising, but I'd be very surprised
if one couldn't work it out this way.
>- Is r(Poisson([Eth])) of bounded variation on intervals?
>This much we do know:
>For U_n the uniform measure on {1, 2,..., n}, r(U_n) is
asymptotically
>1/log n.
>It is somewhat easy to get bounds. r(P) >= PA for any set A
of prime
>numbers, easily computable when A is finite. For an upper
bound, use
>P(nN) = sum{k=0..n-1, c(2 pi k / n)) / n,
>where c is the Fourier transform (characteristic function) of
P. (I have
>cited this result in several previous postings to this
newsgroup.) Then
>apply inclusion/exclusion to the right-hand side of
>r(P) <= 1 - PU{nN: n in A} + PA
>where A is a finite set of primes.
>Clearly (and sillily), r(P) = 1 if P is supported by the
primes.
>--
>Stephen J. Herschkorn herschko@rutcor.rutgers.edu
************************
David C. Ullrich
===
Subject: Re: More on the the primality of random integers
>[My apologies is this is a repeat. My first posting shows up
on none of
>my sources for newsgroups.]
>What has been done on the measure of the set primes from a
given
>probability measure on the natural numbers? For example, if
random
>variable X has Poisson(1) distribution, is there a known
constant for
>P{X is prime}?
In other words, exp(-1) sum_{p prime} 1/p!. Well, according to
Maple
that's approximately .248391624018571539518614035610. The
Inverse
Symbolic Calculator
<http://www.cecm.sfu.ca/projects/ISC/ISCmain.html>
doesn't find anything interesting for it, although it does
identify
sum_{p prime} 1/p! (approximately
.675198437911114341900561607591):
Sum(Annnnnn(n)/(n-1)!,n=1..inf), Annnnnn from the Enc. Integer
Seqs.
6751984379111143 = sum(A010051(n)/(n-1)!) from E.I.S.
Guess what sequence A010051 from E.I.S. is?
>For a probability measure P on N, let r(P) = P{x : x is
prime}.
>- Is r(Poisson([Eth])) of bounded variation on intervals?
Of course: all analytic functions are of bounded variation on
intervals.
Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia
Vancouver, BC, Canada V6T 1Z2
===
Subject: Re: MY GOD. I love this story. It is the most
interesting thing I
have read in so long.
>Probably becuse its a recipe for confusion among British
readers.
Why would British readers be confused by it? Why would
_anyone_ be
confused by it?
Granted, to someone to whom it was new it might take a moment's
thought to interpret: Ah, the year is first so this must be
numbered from larger to smaller time units; therefore 8 is the
month
and 6 is the day. But there would be no _ambiguity_, as there
was
with the other format.
--
Stan Brown, Oak Road Systems, Cortland County, New York, USA
http://OakRoadSystems.com
Fortunately, I live in the United States of America, where we
are
gradually coming to understand that nothing we do is ever our
fault, especially if it is really stupid. --Dave Barry
===
Subject: Re: MY GOD. I love this story. It is the most
interesting thing I
have read in so long.
>Probably becuse its a recipe for confusion among British
readers.
>Why would British readers be confused by it? Why would
_anyone_ be
>confused by it?
Yes, I was wondering why he thought it would confuse British
readers
specifically. Maybe he believes that Brits are more than
averagely dense?
Derek Holt.
>Granted, to someone to whom it was new it might take a
moment's
>thought to interpret: Ah, the year is first so this must be
>numbered from larger to smaller time units; therefore 8 is
the month
>and 6 is the day. But there would be no _ambiguity_, as there
was
>with the other format.
>--
>Stan Brown, Oak Road Systems, Cortland County, New York, USA
> http://OakRoadSystems.com
>Fortunately, I live in the United States of America, where we
are
>gradually coming to understand that nothing we do is ever our
>fault, especially if it is really stupid. --Dave Barry
===
Subject: Re: MY GOD. I love this story. It is the most
interesting thing I
have
read in so long.
>Probably becuse its a recipe for confusion among British
readers.
>Why would British readers be confused by it? Why would
_anyone_ be
>confused by it?
> Yes, I was wondering why he thought it would confuse British
readers
> specifically. Maybe he believes that Brits are more than
averagely dense?
...
Well, now you've managed to confuse me, a US reader. Can
you explain the exact meaning of more than averagely or
more than averagely dense ?
-jiw
>Fortunately, I live in the United States of America, where
we are
>gradually coming to understand that nothing we do is ever
our
>fault, especially if it is really stupid. --Dave Barry
===
Subject: my question is no news from genius
if we have solved the geodesic curvature,
for the unit velocity curve a
I have the use of the a``= kn U +kg
V (T V =
U)
or kg = a`` V
or kg = a`
a``U
--------------------------------------
but i know that ( kg = k B
U ) for geodesic curvature formula.
now, my question is how to derive to ( kg = k
B U )
k:curvature
B : unit binormal vector
U : unit normal vector
: inner product
please, sir.
===
Subject: Re: my question is no news from genius
> if we have solved the geodesic curvature,
> for the unit velocity curve a
This sounds like a question on general relativity. Is
it? If so, it belongs in sci.physics.
> I have the use of the a``= kn U +kg
V (T V =
U)
> or kg = a`` V
> or kg = a`
a``U
The rest of this is incomprehensible because it used
special characters that got translated as garbage characters.
Please try again using standard ASCII.
- Randy
===
Subject: Re: Nora Baron is a palindrome, Not a person
> In asking the question of why would mathematicians lie, we
have two
> possibilities.
> One possibility is that they are not lying, but really can't
> understand my work.
Your work is incoherent. Your so-called proofs are parodies of
proofs.
> This September will mark eight and half years since I made
the fateful
> decision to pursue a proof of Fermat's Last Theorem.
Are you familiar with Zeno's paradox of Achilles and the
Totoise? When
you started persuing an elementary proof of FLT and you have
not caught
it yet.
I will tell you James, reading your stuff macht mir schwartz
in die augen.
Bob Kolker
===
Subject: Re: Nora Baron is a palindrome, Not a person
> In asking the question of why would mathematicians lie, we
have two
> possibilities.
> One possibility is that they are not lying, but really can't
> understand my work.
Actually, this is a VERY likely possibility, since your work
is simply
incomprehensible.
> There was always the global truth of whether or not I
> had found the short FLT Proof
Right, the answer is (quite obviously): no, you haven't found
the
short FLT Proof. (Actually it's not clear if there is such a
proof.)
> and the local truths of the actual mathematical arguments.
Ehrrr? Shouldn't your proof consist of mathematical arguments?
(As well
as any other math _proof_?)
F.
===
Subject: Re: Nora Baron is a palindrome, Not a person
A8<L{-iZAH|#`e@d1Dd.&:lu`[O}2^rhg@6(x'2l2qK1>EwTYfhf*u~,Eu,tf6
$HN*MY&)u0G
=N'
x<%)/s=GZ_BD2Qz9m=S%4v^I+>T|'1{w70ZY=ih,=)kMY_}?{%)x0)];K~@
J6m5.EN?>Zh
Xh;Y
V|',x(js'Jfq02joVpj|#x
> In asking the question of why would mathematicians lie, we
have two
> possibilities.
[...]
These forgeries are getting more offensive as they get more
difficult
to detect. The forger is no longer simply re-posting James's
words,
but writing in his voice (convincingly enough, until he
mentions the
robot program).
Whether or not one likes JSH, surely this behavior is
repugnant.
--
And yes, for those who think that just maybe I did find a
short proof
of Fermat's Last Theorem, and THE prime counting function, if I
succeed at what I'm working on now world economy as you know
it will
be gone. -- James Harris branches out.
===
Subject: Re: normal operators
>Killing flies with sledge-hammers again, are we?
>People talk like killing a fly with a sledge hammer is
easy.
>Try it sometime.
> Hey, if I ever see a fly with a sledge hammer, I'm not
going to
> stick around and try to kill it.
> I've never seen a fly with a sledge hammer.
Neither have I. I've seen one with a magnifying glass, though.
> I never hope to see one.
> I cannot even tell you if
> I'd rather see than be one.
Aren't you channeling the wrong Nash for this newsgroup?
--
Wayne Brown | When your tail's in a crack, you improvise
fwbrown@bellsouth.net | if you're good enough. Otherwise you
give
| your pelt to the trapper.
e^(i*pi) = -1 -- Euler | -- John Myers Myers,
Silverlock
===
Subject: Re: normal operators
>Killing flies with sledge-hammers again, are we?
>People talk like killing a fly with a sledge hammer is
easy.
>Try it sometime.
> Hey, if I ever see a fly with a sledge hammer, I'm not
going to
> stick around and try to kill it.
> I've never seen a fly with a sledge hammer.
> Neither have I. I've seen one with a magnifying glass,
though.
> I never hope to see one.
> I cannot even tell you if
> I'd rather see than be one.
> Aren't you channeling the wrong Nash for this newsgroup?
Although commonly ascribed to Ogden Nash,
Purple Cow was actually written by Gillette Burgess.
which goes very approximately like this:
But I can tell you, anyhow,
I'll sue you if you quote it.
--
Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email)
===
Subject: number of seatings
In how many ways can n people be seated in m chairs if there
is 1 person
per
chair and its not necessary to seat all/any of the people? I
have a
solution
but it's not near as pretty as I'd like. I'm looking for a
formula that
does
not require summation.
Using recursion results in an equation with two variables (n
people and m
chairs):
a(n,m) = a(n,m-1) + n*a(n-1,m-1) = a(n-1,m) + m*a(n-1,m-1)
which also
yields
(m-n)*a(n,m) = =m*a(n,m-1) - n*a(n-1,m)
For example, such a seating with 3 people and 2 chairs would
result in
a(3,2) = a(3,1) + 3*a(2,1) = 4 + 3*3 = 13.
Using the fact that a(i,j)=a(j,i)and a(n,1) = n+1, I figured I
could make
the assumption that n>m and reduce the recursion algebraically
for m-1
steps
to get an answer but it gets ugly very quickly. It's been a
long time since
I've worked with characteristic equations with more than one
index and I'm
a
bit rusty. A solution using this would be greatly appreciated.
Heck, even a
URL describing how to do multivariable generating functions
would be
appreciated.
I can solve it using counting techniques as:
a(n,m) = Sumfrom_0_to_min{m,n}[C(n,i)*P(m,i)] so that
a(3,2) = C(3,0)P(2,0) + C(3,1)P(2,1) + C(3,2)P(2,2) = 1+6+6 =
13 as above.
This requires 2*(min{m,n}+1) calulations. By the way, for you
graph-theorists, this is the same as the number of subgraphs
of K(m,n)
(distinct vertices) with no vertex having degree more than 1.
Anyhow, my
gut
tells me:
a)there is a simpler way of doing it that results in a simple
calculation
or
two and,
b)I'll probably be embarassed when I see it.
Here is some sample data (I'll bet the columns don't line up,
sorry):
C H A I R S
0 1 2 3 4 5 6
7 8
0 1 1 1 1 1 1 1
1 1
1 1 2 3 4 5 6 7
8 9
P 2 1 3 7 13 21 31 43
57 73
E 3 1 4 13 34 73 136 229
358 529
O 4 1 5 21 73 209 501 1045 1961
3393
P 5 1 6 31 136 501 1546 4051 9276
19081
L 6 1 7 43 229 1045 4051 13327 37633
93289
E 7 1 8 57 358 1961 9276 37633 130922
394353
===
Subject: OT: redirect
Distribution: world
> It's a redirect. The first URL should always work; the
second might
> not.
Did you tried it? The redirect doesn't work for me. I've
tested it with
Internet Explorer and Mozilla.
--
Frank Bu, fb@frank-buss.de
http://www.frank-buss.de, http://www.it4-systems.de
===
Subject: Re: Pareto-optimal world
> Lust and love are not opposites. You aren't very bright, are
you?
no reason to be rude when teaching a lesson, but I agree with
you..even
though it is probably unprovable in either direction.....
===
Subject: Re: Pareto-optimal world
>Lust and love are not opposites. You aren't very bright, are
you?
> no reason to be rude when teaching a lesson, but I agree
with you..even
> though it is probably unprovable in either direction.....
Does love have an opposite? What would it be?
Certainly not hate.
Cold indifference?
Dispassionate murderous intent?
No, I'm not at all clear that love has an opposite.
OK, here's another one. What is the opposite of the taste of
mustard?
===
Subject: Re: Pareto-optimal world
>Lust and love are not opposites. You aren't very bright,
are you?
> no reason to be rude when teaching a lesson,
You don't know bob or his bobisms too well, huh?
> but I agree with you..even
> though it is probably unprovable in either direction.....
I see what you guys are saying, but try to open your minds to
see what I am
saying.
Hard for some reading...I'm sure.
just love or the ultimate combo of the two.
> Does love have an opposite? What would it be?
> Certainly not hate.
Nope.
> Cold indifference?
If it is just indifference it comes as close to opposite of
love I ever
knew. Put the cold in there and well, it is an emotion close
to hate.
> Dispassionate murderous intent?
> No, I'm not at all clear that love has an opposite.
> OK, here's another one. What is the opposite of the taste of
mustard?
Sweet honey B-B-Q sauce?
===
Subject: Re: Pareto-optimal world
> Does love have an opposite? What would it be?
Indifference.
Bob Kolker
===
Subject: Re: Pareto-optimal world
> Your universe or mine?
> Now you go.
It is MY universe and I can make of it anything I want to.
Now your turn.
M.
===
Subject: Re: Pareto-optimal world
j2thomas@cavtel.net postulates...
> OK, here's another one. What is the opposite of the taste of
mustard?
for me, it evocates pickled cucumber
===
Subject: Re: Pareto-optimal world
> Schiller has defined beauty this way: the accord between
reason and
> sensuousness.
>What about a person who is not sensuous, according to his
theory, a
person
>who lacks this would not be able to recognize or appreciate
beauty and I
>think that is untrue.
It doesn't mean they wouldn't be able to appreciate it, it
just means
they wouldn't have it.
> You mean someone like you?
Is that why you wanted her to move in with you?
This.. befriending people, then berating them, smacks of
psychosis.
You're like a radioactive balloon that nobody knows when will
explode.
> the more the opposite forces work with one another
intelligently
> the more beautiful is the result.
>Opposite would be lust verses love and nothing beautiful in
that.
> Lust and love are not opposites. You aren't very bright, are
you?
They are not opposites, they are complements. She would not
see them
as opposites either.
===
Subject: Re: Pareto-optimal world
> Schiller has defined beauty this way: the accord between
reason and
> sensuousness.
>What about a person who is not sensuous, according to his
theory, a
person
>who lacks this would not be able to recognize or
appreciate beauty and
I
>think that is untrue.
>It doesn't mean they wouldn't be able to appreciate it, it
just means
>they wouldn't have it.
> You mean someone like you?
>Is that why you wanted her to move in with you?
I never wanted her to move in with me. We joked about her
living in a
tent in my backyard or sharing a room with my daughter but
those were
jokes. Perhaps I don't tell jokes any better than Jonah. I've
never
talked to her on the phone let alone met her face to face. We
were
silly for a while in threads and it lead to some emailing. She
started
trying to force advice down my throat. I told her to stop. She
wasn't
able to stop. I told her to  off. Now we don't like each
other.
She makes it a point to taunt me. Each time she does I make it
a point
to lambast her. She must like it or she'd stop taunting.
>This.. befriending people, then berating them, smacks of
psychosis.
Ilya talking to me about psychosis. At least I got to start
the day
with a laugh.
>You're like a radioactive balloon that nobody knows when will
explode.
Maybe I need an enema.
> the more the opposite forces work with one another
intelligently
> the more beautiful is the result.
>Opposite would be lust verses love and nothing beautiful
in that.
> Lust and love are not opposites. You aren't very bright,
are you?
>They are not opposites, they are complements. She would not
see them
>as opposites either.
I was wondering when you'd throw in your two cents.I was
thinking
about you this morning. In fact, it's the reason I couldn't
sleep.
My life is a waking nightmare.
===
Subject: Re: Pareto-optimal world
ishambat@hotmail.com postulates...
> Schiller has defined beauty this way: the accord
between reason and
> sensuousness.
What about a person who is not sensuous, according to his
theory, a
person
>who lacks this would not be able to recognize or
appreciate beauty and
I
>think that is untrue.
> It doesn't mean they wouldn't be able to appreciate it, it
just means
> they wouldn't have it.
> You mean someone like you?
> Is that why you wanted her to move in with you?
> This.. befriending people, then berating them, smacks of
psychosis.
there are whole communities that do nothing but this....that
feel
fulfilled by doing this...except when they're on the receiving
side...then they ask you for 'proof'...
you may see an example of this after i x-post to
alt.religion.kibology
schwann
www.webtrance.co.za
===
Subject: Re: Pareto-optimal world
> I was wondering when you'd throw in your two cents.I was
thinking
> about you this morning. In fact, it's the reason I couldn't
sleep.
> My life is a waking nightmare.
i'm sorry to hear that you can't sleep, and that things are
tough for
you...sleep deprivation leads to bad decisions...good luck!!
schwann
===
Subject: Re: Pareto-optimal world
>ishambat@hotmail.com postulates...
> Schiller has defined beauty this way: the accord
between reason and
> sensuousness.
>What about a person who is not sensuous, according to
his theory, a
person
>who lacks this would not be able to recognize or
appreciate beauty and
I
>think that is untrue.
> It doesn't mean they wouldn't be able to appreciate it, it
just means
> they wouldn't have it.
> You mean someone like you?
> Is that why you wanted her to move in with you?
> This.. befriending people, then berating them, smacks of
psychosis.
>there are whole communities that do nothing but this....that
feel
>fulfilled by doing this...except when they're on the receiving
>side...then they ask you for 'proof'...
>you may see an example of this after i x-post to
alt.religion.kibology
Welcome back to alt.religion.kibology, Schwann. Long time no
see. I
hope all is well with you and the wife and kids.
Best wishes,
--
Kevin S. Wilson
Tech Writer at a University Somewhere in Idaho
Who put these fingerprints on my imagination?
===
Subject: Re: Pareto-optimal world
O`wfV?bcyO,O{ORK){SV?:=>TfEdk=@S$k_zy&n_LMIM<;.CZ^$E^!&pc!%
duFJBd_
r8D5$ZIs1i3_&2Hw7wldan8?uo/wZH'a-SAlUy?Vyj&N)]mPBPk$tn#n1fE+99S
>there are whole communities that do nothing but
this....that feel
>fulfilled by doing this...except when they're on the
receiving
>side...then they ask you for 'proof'...
>you may see an example of this after i x-post to
alt.religion.kibology
> Welcome back to alt.religion.kibology, Schwann. Long time no
see. I
> hope all is well with you and the wife and kids.
> Best wishes,
Oooh, kismet! I was just thinking t'other day how I missed
seeing
Schwinn's posts here! Your Welcome Back! is heartily seconded!
Wmst.
N. Gergen
--
Albert the Pintpot explains relativity with trainsets and
planets
and atoms and ether, and how in the end everything's politics.
He says, 'Have you met my relatives? You wouldn't like them
either'
===
Subject: Re: Pareto-optimal world
Q;aBGy<I)&RX5`15K2Tvl,Brp^&.n{NpTUGS8/AK;2h~,6qcx^T$(k|DBcki{
WMvI;C
3U;{jJ1Dh2pT6,ZB~tB7KiGRM[mvrHiO`.0Z]9mnm!S%E
> ishambat@hotmail.com postulates...
> Is that why you wanted her to move in with you?
> This.. befriending people, then berating them, smacks of
psychosis.
> there are whole communities that do nothing but this....that
feel
> fulfilled by doing this...except when they're on the
receiving
> side...then they ask you for 'proof'...
__
POST PROOF OR RETRACT! o--) o(
/:|::|
/::|::|
> you may see an example of this after i x-post to
alt.religion.kibology
__
)o (--o A TOUCH, A TOUCH, I DO CONFESS IT!
|::|:
|::|::
-jwgh
--
Debate is when people can no longer throw chairs.
===
Subject: peano's postulates for the reals
If you continuously use the successor function on a natural
number,
the result will be infinite, or you cover the span of the
naturals. This
is
the essense of Peano's encapsulation of naturals.
I believe there is an equivalent version for reals if you
continuously add
reals
to a real number.
n = 0
repeat
n = succ(n)
until n = inf shorthand for AN, n>N
n = 0
repeat
s = 0
repeat
t = random (Reals)
s = s + t
until s does not converge
n = n + s
until n = inf
The only constraint would be that s!=0, which is similar for
a Peano axiom anyhow, succ(0) != 0.
The inner loop being independant guarantees n does not
converge,
The inner loop should be computable and perhaps further
specifiable
by detecting any increase in absolute change, t would be
constrained to >0.
Pardon the monte carlo syntax, random justs means for any real,
you'll see it makes a 'grain' for the reals,
Herc
===
Subject: Re: peano's postulates for the reals
> If you continuously use the successor function on a natural
number,
> the result will be infinite, or you cover the span of the
naturals. This
is
> the essense of Peano's encapsulation of naturals.
> I believe there is an equivalent version for reals if you
continuously add
reals
> to a real number.
> n = 0
> repeat
> n = succ(n)
> until n = inf shorthand for AN, n>N
> n = 0
> repeat
> s = 0
> repeat
> t = random (Reals)
> s = s + t
> until s does not converge
> n = n + s
> until n = inf
> The only constraint would be that s!=0, which is similar for
> a Peano axiom anyhow, succ(0) != 0.
> The inner loop being independant guarantees n does not
converge,
> The inner loop should be computable and perhaps further
specifiable
> by detecting any increase in absolute change, t would be
constrained to
>0.
> Pardon the monte carlo syntax, random justs means for any
real,
> you'll see it makes a 'grain' for the reals,
> Herc
So what is succ(0)? How about succ(pi)?
===
Subject: Re: peano's postulates for the reals
> If you continuously use the successor function on a
natural number,
> the result will be infinite, or you cover the span of the
naturals.
This is
> the essense of Peano's encapsulation of naturals.
> I believe there is an equivalent version for reals if you
continuously
add reals
> to a real number.
> n = 0
> repeat
> n = succ(n)
> until n = inf shorthand for AN, n>N
> n = 0
> repeat
> s = 0
> repeat
> t = random (Reals)
> s = s + t
> until s does not converge
> n = n + s
> until n = inf
> The only constraint would be that s!=0, which is similar
for
> a Peano axiom anyhow, succ(0) != 0.
> The inner loop being independant guarantees n does not
converge,
> The inner loop should be computable and perhaps further
specifiable
> by detecting any increase in absolute change, t would be
constrained to
>0.
> Pardon the monte carlo syntax, random justs means for any
real,
> you'll see it makes a 'grain' for the reals,
> Herc
> So what is succ(0)? How about succ(pi)?
you're thinking in terms of integers.
the rsucc is *any* number added.
Theorem : for any real, a point below and above using rsucc
can be
forumulated.
__________________________________
small enhancement to the algorithm :
n = 0
repeat
s = 0
repeat
t = abs(Random(reals)) **
s = s + t
until s does not converge
n = n + s
until n = infinity
unfortunately this fails to capture my meaning as its
equivalent to the
simpler:
n = 0
repeat
s = abs(Random(reals))
n = n + s
until n = infinity
using the same Random sequence they both give identical growth
for n.
so working backwords, the addition at s = s + t
has to be fundamentally different (or a cheaper operation)
than n = n + s (which is ordinary real addition).
I see it as a search for a number to an arbitrary depth
that eventually gives away its magnitude once it has converged.
_____________________________
If we take the inner loop as a discrete step then the number
line can be
arbitrarily plotted using the notion of convergence,
more specifically: Repetitive Accumulative Non Convergence.
(Rank!)
The unit number has no part in the mapping, as long as
s is not 0, which it can't be because s would have to
converge. Whether the unspecified Random distribution
yields billionths or billions, both counting processes
will terminate with n greater than any given number.
Herc
===
Subject: Re: Peter Lynds
I can't tell you if Peter Lynds is correct, but I believe he
is on to
something here. A problem I've had for a while now is that
with our
current understanding of time and physical matter, something
must have
come from nothing. Think about it, current theory is that it
all
started with a big bang. Where did the matter come from? Some
say it
was a contraction of a previous universe and maybe that is so,
but if
you follow it back to the beginning, you have to conclude that
something came from nothing or that the matter always existed.
Neither concept can be understood within the current framework
of
physics or philosophy. Now, if our concept of time is
incorrect then
maybe we can start to understand more about our origins.
As human beings, we will have a very hard time grasping these
concepts
of time and motion. We are hard wired to see things in a
particular
way and everything we see reinforces these beliefs.
Good luck Peter.
===
Subject: Re: Poincar.8e and algebraic varieties?
> Never mind, I have found the answer.
> -mb
So what is the answer?
===
Subject: Re: Poincar.8e and algebraic varieties?
> So what is the answer?
Apparently Poincar.8e applied the rudimentary concepts of what
is now
algebraic topology to cycles on certain complex surfaces in
his Analysis
Situs (and following writings). I have been told, that he was
not all that
careful about the distiction between manifolds and, what we
now call,
algebraic varieties. At the time, perhaps most of the spaces
you could
study arose as solutions to equations - among them polynomial
equations.
Martin
--
Tout ce qu'il y a de b.8eb.90te. -- Grothendieck
===
Subject: Re: Problem with Ring of Algebraic Integers, plus fix
Originator: a@shell3.shore.net (a)
>I found need to make minor changes. ___JSH
>Consider, in the ring of algebraic integers,
> P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 -
> 3(-1+mf^2 )x u^2 + u^3 f).
>Now using b_1, b_2, b_3, w_1, w_2, and w_3, I have the
factorization
> P(m)/f^2 = (b_1 x + u w_1)(b_2 x + u w_2)(b_3 x + u w_3)
>where w_1 w_2 w_3 = f, and
> b_1 b_2 b_3 = (m^3 f^4 - 3m^2 f^2 + 3m),
>and at m=0
> P(0)/f^2 = 3xu^2 + u^3 f = u^2(3x + uf),
>so two of the b's must equal 0, which means
> P(0)/f^2 = w_1 w_2 u^2 (b_3 x + u w_3)
>which is
> P(0)/f^2 = u^2 (b_3 w_1 w_2 x + u f) = u^2(3x + uf)
>proving that w_1 w_2 must equal 1, if f is coprime to 3,
which leaves
>b_3 = 3.
>Essentially objections to how f^2 divides off now come down to
>claiming that the w's are functions of m, but consider that
w_1 w_2 =
>1, when m=0, if f is coprime to 3.
>But that was an arbitrary choice, so let f=3.
>Now w_1 w_2 = 3^{2/3} WITHOUT REGARD TO m.
>That is, the w's are now all constant with regard to m and
have the
>same value no matter what the value of m is, so they are also
constant
>with f coprime to 3.
>Therefore, the factorization is
> P(m)/f^2 = (m^3 f^4 - 3m^2 f^2 + 3m) x^3 -
> 3(-1+mf^2 )x u^2 + u^3 f =
> (b_1 x + u)(b_2 x + u)(b_3 x + uf)
>where you'll notice that the b's are algebraic integers with
m=1,
>f=sqrt(2), but that's a special case as generally they are
not, which
>shows a problem with the ring of algebraic integers.
>I've found the Ring of Objects which includes the ring of
algebraic
>integers, and does not have this problem, as the b's are all
included
>in it.
>The Ring of Objects is the set of all numbers where 1 is the
only
>member that is both a unit, i.e. factor of 1, and an integer,
where no
>non-unit member is a factor of any two integers that are
coprime.
>That definition and more is linked to from my primary website
> http://groups.msn.com/AmateurMath
>where you can also find information on my other math research.
Why do you assume that what's true when m equals zero is also
true when m does not equal zero?
--
The simple result of that was I freaked out, did a good deal of
drinking, and sang pieces of popular songs quietly, and at
times
loudly to myself, others, and the walls.
===
Subject: Re: Problem with Ring of Algebraic Integers, plus fix
<deletedEssentially objections to how f^2 divides off now come down
to
>claiming that the w's are functions of m, but consider that
w_1 w_2 =
>1, when m=0, if f is coprime to 3.
>But that was an arbitrary choice, so let f=3.
>Now w_1 w_2 = 3^{2/3} WITHOUT REGARD TO m.
>That is, the w's are now all constant with regard to m and
have the
>same value no matter what the value of m is, so they are
also constant
>with f coprime to 3.
<deleted Why do you assume that what's true when m equals zero is also
> true when m does not equal zero?
I don't. I've deleted out to focus on the relevant section
which you
possibly skimmed past or you wouldn't have asked your question,
assuming you understand basic mathematical reasoning, of
course.
I'm also including it here, so that you can see it again.
>Essentially objections to how f^2 divides off now come down
to
>claiming that the w's are functions of m, but consider that
w_1 w_2 =
>1, when m=0, if f is coprime to 3.
>But that was an arbitrary choice, so let f=3.
>Now w_1 w_2 = 3^{2/3} WITHOUT REGARD TO m.
>That is, the w's are now all constant with regard to m and
have the
>same value no matter what the value of m is, so they are
also constant
>with f coprime to 3.
Now the math is rather basic, but if you can't follow
mathematical
reasoning, then of course, that won't matter.
James Harris
===
Subject: *proportional* Venn diagrams
I made a web page for creating proportional Venn diagrams:
http://www.venndiagram.tk.
The proportional thing (every circle and every overlap is as
large as
the number it represents) is missing in every program I know
that is
able to create Venn diagrams, so it's quite new!
But I have a problem with the three-circle overlap: could
anybody tell
me if it is always possible to draw this one proportionally
too, given
that the circles must be real circles and not ellipses? At the
moment
all of the three two-circle overlaps are ok, but this doesn't
make the
three-circle overlap proportional...
Thnx,
Tim Hulsen.
===
Subject: Re: *proportional* Venn diagrams
> I made a web page for creating proportional Venn diagrams:
> http://www.venndiagram.tk.
> The proportional thing (every circle and every overlap is as
large as
> the number it represents) is missing in every program I know
that is
> able to create Venn diagrams, so it's quite new!
> But I have a problem with the three-circle overlap: could
anybody tell
> me if it is always possible to draw this one proportionally
too, given
> that the circles must be real circles and not ellipses?
No. Trivial counterexample: Suppose A ^ B ^ C = B ^ C, but
A ^ B ^ C is smaller than from A ^ B and A ^ C. That is,
you have to make the three-circle overlap occupy all of
one of the two-circle overlaps. (I'm using the caret for
intersect).
Another way of showing your problem is overdetermined:
Areas A, B and C are fixed, thus so are their radii.
Your only options are to place the centers.
You have the following degrees of freedom:
1. Distance of center of B from center of A.
2. Distance of center of C from center of A.
3. Location (angle) of center of C relative to
the line of A-B (centers).
That's it. Three degrees of freedom. Now you have four
constraints: area of A^B, area of A^C, area of B^C and
area of A^B^C. Four equations, three unknowns. As you've
found, you can match the conditions three of them.
The classic approach to overdetermined systems is to
look for an approximate solution which is a best approximation
in some sense, usually least square error. Given that you're
going to have to back off from matching all the proportions,
you might still be able to approximate them well by
fudging a little.
The least-squares problem could be solved numerically
by setting it up as an unconstrained nonlinear optimization
problem in those three variables.
- Randy
===
Subject: Question about Lambert W function
Hi all,
I've recently discovered how useful the Lambert W function is
in
solving transcental equations involving exponentials. However,
the following equation has stumped me so far:
(x^2 + K)e^x = y
where K is any real number. So far, I have only been able to
solve for x (in terms of W and y) when K=0.
Is it actually possible to solve this equation in terms of W
for
non-zero K? If not, I hope to understand why not, esp. since
(x^n)e^x = y *does* have closed-form solutions in terms of W
for
all n.
===
Subject: Re: Question about Lambert W function
>I've recently discovered how useful the Lambert W function is
in
>solving transcental equations involving exponentials. However,
>the following equation has stumped me so far:
>(x^2 + K)e^x = y
>where K is any real number. So far, I have only been able to
>solve for x (in terms of W and y) when K=0.
>Is it actually possible to solve this equation in terms of W
for
>non-zero K? If not, I hope to understand why not, esp. since
>(x^n)e^x = y *does* have closed-form solutions in terms of W
for
>all n.
As far as I know (and as far as Maple knows), your equation
has no
closed-form solution for general K. Of course, most
transcendental equations don't, so you shouldn't be surprised.
x^n e^x = y is rather special, since taking logarithms you have
a linear equation involving x and log(x).
If u = 2 LambertW((+/-)sqrt(y)/2), a solution of u^2 e^u = y,
then
there's a series expansion in powers of K for a solution of
your equation:
2 3
1 -2 + u 2 (u - 1) (u - 6 u - 6) 3
x = u - --------- K + ------------- K - ----------------------
K
u (2 + u) 3 3 5 5
2 u (2 + u) 3 u (2 + u)
5 4 3 6 2
-60 - 8 u - 33 u + 20 u + 3 u + 90 u 4 5
+ ----------------------------------------- K - (356 u
7 7
12 (2 + u) u
4 3 2 8 7 6
+ 990 u - 420 u - 1680 u + 840 + 12 u - 58 u - 181 u )
/ 9 9 5 6
/ (60 u (2 + u) ) K + O(K )
/
Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia
Vancouver, BC, Canada V6T 1Z2
===
Subject: Re: Question about Lambert W function
>I've recently discovered how useful the Lambert W function
is in
>solving transcental equations involving exponentials.
However,
>the following equation has stumped me so far:
>(x^2 + K)e^x = y
> As far as I know (and as far as Maple knows), your equation
has no
> closed-form solution for general K. Of course, most
> transcendental equations don't, so you shouldn't be
surprised.
> x^n e^x = y is rather special, since taking logarithms you
have
> a linear equation involving x and log(x).
In which case, I'm curious whether there is any known general
form
of transcendental equations that can be solved in terms of W.
Obviously, it must be possible to arrange the equation into
the form
f(x,y)e^f(x,y)=g(y), but I wonder if there is a more precise
characterization of W-solvable equations.
E.g., if the equation is of the form p(x)e^x = y, where p(x)
is a
polynomial, what are the conditions on p(x) so that the
equation is
W-solvable? Or, to approach it from another way, if p(x) is an
unconstrained polynomial of degree n, what is the minimal
number of
transcendental inverses (in the spirit of W, e.g.,
p(V(x))e^x=y)
needed such that p(x)e^x = y can be solved in terms of
combinations
of these inverses?
For n=1, it seems that W is the only necessary function, since
(ax + b)e^x = y
is equivalent to
(x + b/a)e^(x + b/a) = (y/a)e^(b/a)
giving
x = (a/b)W((y/a)e^(b/a)).
But for n not equal to 1, it seems that only very constrained
forms
of p(x) admit closed-form solutions in terms of W. It would be
interesting to find out how many other transcendental inverse
functions would be required so that all equations where, say,
n=2,
have closed-form solutions. It would be especially interesting
if
only a finite number of inverses are required for each
specific n,
which seems to be a rather compelling proposition.
===
Subject: Re: Question on Godel Incompleteness
>The predicate would be simple to implement in code. When
the
>search-process completes and does (or does not) reach the
statement
>in question, it might set an indicator (ProvedFlag==true or
>ProvedFlag==false). This indicator can be used by R.
> The problem is that the search process (searching for a
proof of some
> statement) does _not_ complete if the statement is not
provable.
>Good point. Although for simpler systems (ones with limited
number of
>simpler axioms) the combinations in the search is ceartainly
finite.
>There are also ways to logically tell if going down this path
of
>combinations will not have any chance of proving A.
>So the question then becomes, are these simpler systems
powerful
>enough to support the Godel numbering. If they are, then
*maybe* they
>can support R and hence to NOT succomb to Godel's
incompleteness ?
Godel's Theorem is a theorem. For consistent systems to which
it applies,
provability is not a recursive function. There is no algorithm
to tell
in a finite number of steps whether a statement is provable in
such
systems. There are some systems for which provability is
decidable
(e.g. Pressburger arithmetic, which is basically the theory of
arithmetic
with addition but not multiplication).
Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia
Vancouver, BC, Canada V6T 1Z2
===
Subject: Re: Question on Godel Incompleteness
>The predicate would be simple to implement in code. When
the
>search-process completes and does (or does not) reach the
statement
>in question, it might set an indicator (ProvedFlag==true
or
>ProvedFlag==false). This indicator can be used by R.
> The problem is that the search process (searching for a
proof of some
> statement) does _not_ complete if the statement is not
provable.
> Godel's Theorem is a theorem. For consistent systems to
which it
applies,
> provability is not a recursive function. There is no
algorithm to tell
> in a finite number of steps whether a statement is provable
in such
> systems. There are some systems for which provability is
decidable
> (e.g. Pressburger arithmetic, which is basically the theory
of arithmetic
> with addition but not multiplication).
I have 2 ideas here.
The first is that my formal system is equiped with a predicate
that
implements the Cantor Diagonal for the axioms etc. to
determine if S
is provable.
The second is I wonder if one could make a SPECIFIC
hand-crafted
formal system with simpler axioms that supports the Godel
Numbering...and furthermore the provability search-space is
finite.
For such a hand-crafted formal system, R could be used...and
the
formal system would then be complete. Granted, it would not be
a very
powerful formal-system, but who cares.
I get the feeling that my R idea is almost dead !!
Oh well...its been fun.
===
Subject: Re: Question on Godel Incompleteness
> The Fourth Puzzle.
 You encounter an inhabitant of the island who tells
> you, I am not an established knight. What, if
> anything, can be determined about this individual?
> SOLUTION.
 The surprising solution to the Fourth Puzzle is that the
native
> must be an unestablished knight! Replacing knight with
true
> statement and established knight with provable statement
> (and established knave with refutable statement) we see
that
> this is really a Godelian puzzle in disguise.
> Yeah, I see some of the similarity, maybe not all of it.
At the risk
> of being wrong, I will guess that in the 4th puzzle, the
talker must
> be a knight, not a knave because knaves must lie and that
statement
> would not be lie relative to a knave. Thus it must be a
knight or
> established-knight, both of whom tell the truth, thus we
take that
> statement as true: he says he is not an
established-knight. Only
> remaining possibility is that he is a knight. Finally, we
can say
> that the statment is TRUE.
> Hmm...I think this is different than the Godel case. I
think a formal
> system could reach the true conclusion. But before we
discuss this
> further, Arthur, is my conclusion above correct ? Is he a
knight and
> is it TRUE ?
> Yes.
> You are allowed to read the solutions, you know. I marked
them
> with the word SOLUTION in all caps.
Uh duh, yeah I looked closer and the answer was right there.
Oh well,
at least you could see that I worked through the axioms to
arrive at
the correct solution. Arthur, do you feel that a formal-system
could
NOT work from the axioms as I did to arrive at the answer ??
===
Subject: Question on Number Fields
This question is rather open. Suppose we have a number field
K. If K
is a quadratic field (i.e. [K:Q] = 2), then it is uniquely
determined
by its discriminant disc(K/Q). In fact, the discriminant can be
used to determine the algebraic order O < K.
What about for cubic fields or fields of higher degree? Any
similar
===
Subject: Questions on Profinite Completion
Apologies for posting 2 questions in a row.
If G is any group, we let G^ be its profinite completion
(inverse limit
of { G/N } for all finite G/N).
(1) Suppose G = F_n is a free group with n generators. What
kind of
useful universal property does G^ satisfy?
(2) Suppose G = F_n / <R>, where R is a set of relations on n
generators. Then when is G^ = F_n^ / <R>? For instance if G is
the
braid group on n+1 braids, and R are the usual braid relations
(see http://mathworld.wolfram.com/BraidGroup.html for example),
then G^ is exactly F_n^ / <R>.
===
Subject: Re: Quick terminology question
> |CUSPIDAL: (1) Belonging to the apex (of a cone).
> | (2) Having, relating to, or of the nature of, a cusp.
> In the study of modular forms, cuspidal has a technical
meaning.
> Maybe it can be covered by (2), but it's kind of a stretch!
I may be showing my ignorance here, but I feel compelled to
ask:
Isn't a cuspidal modular form (or sometime I hear cusp forms,
which I
think is synonymous) related to a Riemann surface with cusp
points? And
isn't much information about the form given by examining these
cusps?
If so, I think (2) above is perfectly appropriate and not that
much of a
stretch. Feel free to correct me!
===
Subject: Re: Quick terminology question
> |CUSPIDAL: (1) Belonging to the apex (of a cone).
> | (2) Having, relating to, or of the nature of, a cusp.
> In the study of modular forms, cuspidal has a technical
meaning.
> Maybe it can be covered by (2), but it's kind of a stretch!
>I may be showing my ignorance here, but I feel compelled to
ask:
>Isn't a cuspidal modular form (or sometime I hear cusp forms,
which I
>think is synonymous) related to a Riemann surface with cusp
points? And
>isn't much information about the form given by examining
these cusps?
>If so, I think (2) above is perfectly appropriate and not
that much of a
>stretch. Feel free to correct me!
You're right as far as you go, which isn't as far as the
theory goes.
If you're studying modular forms related to SL(2,Z), which is
of course
the classical and best understood case, then cusp forms are
indeed
intimately related to cusps (orbifold points with non-trivial
group) on the natural compactification of the non-compact
Riemann
surface you get by taking the quotient of the upper half plane
by (your favorite arithmetic subgroup of) SL(2,Z). When you
want
to generalize to groups of...is the term higher rank?
whatever...
like SL(n,Z), you're not dealing with the upper half plane any
more,
and there's much more hassle (and considerably less uniqueness
and
naturality) in finding a (manifold or orbifold)
compactification
of the quotient of Whatever It Is (which might be the cone of
positive definite matrices modulo homothety, if you're doing
SL(n,Z); or the product of two upper half planes if you're
doing
Siegel modular forms, whatever they are; and so on) by your
discrete group. The stuff you throw in to compactify is still
called cusps for old times' sake, and forms which have
appropriate
asymptotic behavior with respect to the cusps are still called
cusp forms, but I think Keith's phrase a bit of a stretch has
considerable merit.
Lee Rudolph
===
Subject: Radial Positive Definite Function
Does anyone know any references or ideas about the conditions
that
Hsuan-Tien
===
Subject: Random Walk absolute Deviation
I am looking for the solution for the following problem:
Let say, I have 1-dim symmetric random walk (each step +1 or
-1 with
equal probability). I would like to find E(abs(x)). None of my
book in
probability theory has a solution. I've tried some
combinatorics
identities but apparently got the wrong result.
Can anybody help me and point to the right direction,please?
Any help would be appreciated,
Sergey
===
Subject: Re: Random Walk absolute Deviation
>Let say, I have 1-dim symmetric random walk (each step +1 or
-1 with
>equal probability). I would like to find E(abs(x)). None of
my book in
>probability theory has a solution. I've tried some
combinatorics
>identities but apparently got the wrong result.
>Can anybody help me and point to the right direction,please?
Let X_n be your random walk after n steps with X_0 = 0. Note
that E[abs(X_{n+1})|X_n] = X_n if X_n <> 0, X_n + 1 if X_n =
0. So
E[abs(X_n)] is the expected number of times the random walk is
at 0
from time 0 to time n-1. For odd j, X_j is never 0; for even j,
Pr{X_j = 0} = (j choose j/2) 2^(-j). So
E[abs(X_{2m})] = E[abs(X_{2m-1})]
and E[abs(X_{2m+1})] = sum_{k=0}^m (2k choose k) 2^(-2k)
which Maple says is 2 (m+1) (2m+2 choose m+1) 2^(-2m-2)
Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia
Vancouver, BC, Canada V6T 1Z2
===
Subject: References requested: subrings of the Hamiltonian
quaternions
Hi all,
I am looking for non-commutative subrings R of the usual
quaternions that
also form a finitely generated free abelian group. The
examples I can think
of are the following. We can think of the quaternions H as a
2-dimensional
vector space over C with basis B={1,j}. If K is a number field
(finite
dimensional over Q) that is stable under the usual complex
conjugation,
and if O is its ring of integers, then the set of quaternions
H_O={u + v j | u,v in O}
forms a ring of the desired type (stability under the usual
conjugation
makes this work, as v*j=j*(bar v) for all complex numbers v).
That's nice and I can use these, but are there others that
would be
essentially different? I mean that obviously one can apply any
of the
automorphisms of H to get a continuum of rings isomorphic to
H_0, but
are there others that cannot be gotten in this way?
If anybody knows of relevant references to litterature, I
would be most
grateful. Is the theory of ideals of such rings well
developed/ developed
at all?
Jyrki Lahtonen, University of Turku, Finland
===
Subject: Re: Relative change calculation
> However I think with most (serious) currencies the
stochastic element
> outweighs the exponential so they tend to just graph
values.
> Now, this sentence seems to be a description to the news
> element of currencies ('The yen is down today') rather
> than long-term, mathematically convoluted commentary.
Yep, I think most of the long-term math stuff is done with the
forward
interest rates, which of course can be use to imply the
forward FX spot
rates.
===
Subject: Rosser and logic
I've read on the Web that B. Rosser published a paper in the
1930's
that improved or extended Godel's theorem.
What was Rosser's contribution to the theory of undecidability?
Also, there is the Church-Rosser Theorem, but I don't know
why it's important.
David Bernier
===
Subject: Re: Rosser and logic
|I've read on the Web that B. Rosser published a paper in the
1930's
|that improved or extended Godel's theorem.
|
|What was Rosser's contribution to the theory of
undecidability?
He produced a statement which is independent of the system in
question
under weaker conditions on the system-- the system doesn't
have to be
assumed to have a property known as omega-consistency, which
I'll explain.
For systems satisfying Goedel's basic assumptions (roughly
that the system
S includes arithmetic), his proof constructs a statement G(S)
such that
G(S) is true if and only if G(S) cannot be proven in S. Hence
if G(S) can
be proven in S, then G(S) is false (and this turns out also to
be provable
in S). So if G(S) can be proven in S, then S is inconsistent.
If G(S) can't
be proven in S, then G(S) is true but unprovable in S.
That leaves open the possibility that G(S) is true but that
there exists a
(fallacious) disproof of G(S) in S. There are consistent
systems S where
this occurs. But they have something wrong with them. The
falsity of G(S)
says that there exists a natural number n with a certain
property (namely
encoding a proof in S of G(S)), whereas for each individual n
there is a
proof that *that* n does not have this property. Goedel termed
a system
which has a theorem saying there exists an n with a certain
property, but
a disproof of that property for each n, an omega-inconsistent
system
(unless maybe someone else had coined the term already).
Taking Goedel's second incompleteness theorem into
consideration, the
formal systems, satisfying the usual initial requirements,
which disprove
their Goedel sentences, basically are the systems implying
their own
inconsistency (even though some of them actually are
consistent). Starting
with a reasonable system S, S* = S + S is inconsistent is this
type
of
system. The falsity of G(S*) is a theorem of S* because it's
implied by
the inconsistency of S*: in an inconsistent system, G has a
proof, but G
is equivalent to its own unprovability.
Rosser adjusted the proof by making a statement R such that R
is true
when if R has a proof in S, then R has a shorter disproof in
S. If R is
provable, then there are a finite number of shorter proofs to
check. So
either R has a shorter disproof (which makes the system
inconsistent) or R
does not have a shorter disproof. If there's no shorter
disproof, then a
tedious verification of that fact shows that R is in fact
false. Because S
is assumed to include a certain base of arithmetic in it, this
tedious
verification can be formalized in S, so in fact R can be
disproven in S as
well. So to sum up, if R has a proof in S, then either it has
a shorter
disproof in S, or it has a lengthy disproof in S. So either
way, if R can
be proven in S then S is inconsistent.
On the other hand, suppose R has a disproof in S. Then we can
again check
the proofs which are no longer than that disproof. If R also
has a proof
within that length bound, the system is inconsistent again. If
R does not
have a proof within that length bound, then R is demonstrably
true, because
whether or not S can prove R the disproof of R is shorter. But
that's again
the kind of proof S can do, by assumption, showing that S has
a tedious
proof of R.
To sum up, then, if the system contains either a proof of R or
a disproof
of R, then the system is inconsistent. That way Rosser was
able to exhibit
a statement whose independence from S requires only the
assumption that S
is consistent. (To say X is independent of S means that there
isn't a proof
of X in S and there isn't a disproof of X in S either.) That's
a bit more
than what Goedel got.
The Church-Rosser theorem says that in a certain
term-rewriting system,
if a given term t can be rewritten in a sequence of steps to
each of two
terms u1 and u2, then u1 and u2 can both be rewritten in a
sequence of
steps to some term v. The particular term-rewriting system is
interesting
because of it's relationship with basic models of computation.
The concept
of having the Church-Rosser property is interesting as far as
I can
tell
because it's a sort of half-way step toward the system having
canonical
forms for terms which can be normalized (by being rewritten
until there
are no more reductions to make). A logician would know more.
Keith Ramsay
===
Subject: Re: Rosser and logic
> I've read on the Web that B. Rosser published a paper in the
1930's
> that improved or extended Godel's theorem.
> What was Rosser's contribution to the theory of
undecidability?
> Also, there is the Church-Rosser Theorem, but I don't know
> why it's important.
> David Bernier
Keith Ramsay's explanation is very clear, but you might also
like to
look at Martin Davis's book The Undecidable: Basic Papers on
Undecidable Propositions, Unsolvable Problems and Computable
Functions. It contains Rosser's paper Extensions of Some
Theorems of
Godel and Church and also an expository paper of his entitled
An
Informal Exposition of Proofs of Godel's Theorem and Church's
Theorem
in which he gives an explanation of his own extension of
Godel's result.
PS My spell checker tells me that Unsolvable Problems should
read
Unlovable Problems.
--
G.C.
===
Subject: Re: Rule 90 and left-to-right reversal
Distribution: world
> When playing with 1D cellular automata, I have found the
following
> property of the Rule 90 automaton. On states of certain
finite lengths
> (powers of two, it appears), surrounded at the sides by
zeroes (which
> themselves don't undergo change), the whole state is reversed
> left-to-right after a number of steps equal to the length of
the
> state.
> Example, for a 16-cell state:
>|## ##### # # |
>|## # ## # #|
>|## # ##### # # |
>|#### # # #|
>|# ### # # # # |
>| ### # ## #|
>|## # # ### |
>|## ## # ## ## |
>|###### #### ###|
>|# #### # # #|
>| # ## ### |
>|# ####### ## |
>| # # ### |
>| # # # # ## |
>|# # # ## ### |
>| # # ##### ##|
> The first state (random):
>|## ##### # # |
> The final state:
>| # # ##### ##|
> I couldn't find anything regarding this property on the
'net, or in
> Wolfram's NKS. Is this a known property?
I don't know, but I've checked it for other lengths and it
looks like it
is possible for other even lengths, too:
length, steps needed for reverse:
2,2
4,4
6,8
8,8
10,32
12,64
14,16
16,16
18,512
20,64
22,2048
24,1024
26,512
28,16384
30,32
32,32
34,4096
36,87382
38,4096
40,1024
42,128
44,4096
46 looks like it has many steps or it is not possible, because
my Haskell
program (see at the end of this posting), with which I've
tested it, is
still calculating. For the other lengths I'm not sure, because
I've
tested it with one pattern, only.
Perhaps someone in sci.math can find a formula for the number
of steps or
can prove, that for length 2^n the number of steps is n, which
should be
easy.
Rule 90 description: A one dimension cellular automaton. The
next state
of a cell is 1, if the left or the right neighbour is 1. If
both
neighbours are 1 or both neighbours are 0, the next state is 0.
The Haskell test program:
charToBool x = map (=='#') x
boolToChar xs = [if x then '#' else ' ' | x<-xs]
rule90 True _ False = True
rule90 False _ True = True
rule90 _ _ _ = False
applyRuleBool _ _ [] _ = []
applyRuleBool f left (x:[]) right = [f left x right]
applyRuleBool f left (x1:xs'@(x2:xs)) right =
f left x1 x2 : applyRuleBool f x1 xs' right
applyRule f xs = boolToChar $ applyRuleBool f False
(charToBool xs) False
iterateRule90 x = iterate (applyRule rule90) x
checkRule90 x = length (takeWhile (/=(reverse x))
(iterateRule90 x)) + 1
test = [(length x, checkRule90 x)|x<-(iterate (++ ) #)]
--
Frank Bu, fb@frank-buss.de
http://www.frank-buss.de, http://www.it4-systems.de
===
Subject: Re: Rule 90 and left-to-right reversal
> length, steps needed for reverse:
> 2,2
> 4,4
> 6,8
> 8,8
> 10,32
> 12,64
> 14,16
> 16,16
> 18,512
> 20,64
> 22,2048
> 24,1024
> 26,512
> 28,16384
> 30,32
> 32,32
> 34,4096
> 36,87382
> 38,4096
> 40,1024
> 42,128
> 44,4096
> 46 looks like it has many steps or it is not possible,
because my Haskell
> program (see at the end of this posting), with which I've
tested it, is
> still calculating. For the other lengths I'm not sure,
because I've
> tested it with one pattern, only.
The above numbers are the generation number at which the
reversal
appears, counting from generation 1, so are one greater than
the
actual reversal period.
Definitely related to what's on
<http://www.richholmes.net/games/cellaut/r90.html>, then. I
tabulate
the replication (not reversal) period for a pattern consisting
of a
single live cell at the end of the line, and (except for
length 2) I
get twice the above numbers minus two (i.e. twice the actual
reversal
period). In which case the number you should find for 46 would
be
8388608. I didn't come up with a formula but I did find an
easier way
to calculate a recurrence length which is either the period or
a
multiple of the period; for all lengths up to 254 the
calculated
recurrence length *is* the period -- *except* for 36, where I
calculate 524286 while the period is 174762, a factor of 3
smaller.
I didn't generalize this to arbitrary patterns, but evidently
it works
out the same.
===
Subject: Re: Rule 90 and left-to-right reversal
|
|> When playing with 1D cellular automata, I have found the
following
|> property of the Rule 90 automaton. On states of certain
finite lengths
|> (powers of two, it appears), surrounded at the sides by
zeroes (which
|> themselves don't undergo change), the whole state is
reversed
|> left-to-right after a number of steps equal to the length
of the
|> state.
|> Example, for a 16-cell state:
|>|## ##### # # |
|>|## # ## # #|
|>|## # ##### # # |
|>|#### # # #|
|>|# ### # # # # |
|>| ### # ## #|
|>|## # # ### |
|>|## ## # ## ## |
|>|###### #### ###|
|>|# #### # # #|
|>| # ## ### |
|>|# ####### ## |
|>| # # ### |
|>| # # # # ## |
|>|# # # ## ### |
|>| # # ##### ##|
|> The first state (random):
|>|## ##### # # |
|> The final state:
|>| # # ##### ##|
|> I couldn't find anything regarding this property on the
'net, or in
|> Wolfram's NKS. Is this a known property?
|
|I don't know, but I've checked it for other lengths and it
looks like it
|is possible for other even lengths, too:
|
|length, steps needed for reverse:
|
|2,2
|4,4
|6,8
|8,8
|10,32
|12,64
|14,16
|16,16
|18,512
|20,64
|22,2048
|24,1024
|26,512
|28,16384
|30,32
|32,32
|34,4096
|36,87382
|38,4096
|40,1024
|42,128
|44,4096
|
|
|46 looks like it has many steps or it is not possible,
because my Haskell
|program (see at the end of this posting), with which I've
tested it, is
|still calculating. For the other lengths I'm not sure,
because I've
|tested it with one pattern, only.
|
|Perhaps someone in sci.math can find a formula for the number
of steps or
|can prove, that for length 2^n the number of steps is n,
which should be
|easy.
i took the first 11 period lengths you listed:
2,4,8,8,32,64,16,16,512,64,2048
and then took their logs base 2:
1,2,3,3,5,6,4,4,9,6,11
and then fed them into sloane's on-line encycploedia of integer
sequences:
<http://www.research.att.com/~njas/sequences/index.html>.
this yielded:
ID Number: A003558
URL: http://www.research.att.com/projects/OEIS?Anum=A003558
Sequence:
1,2,3,3,5,6,4,4,9,6,11,10,9,14,5,5,12,18,12,10,7,12,23,21,8,
26,20,9,29,30,6,6,33,22,35,9,20,30,39,27,41,8,28,11,12,10,
36,24,15,50,51,12,53,18,36,14,44,12,24,55
Name: Least number m such that 2^m = +- 1 mod 2n + 1.
See also: Sequence in context: A023160 A085312 A046530
this_sequence
A072451
A023156 A051599
Adjacent sequences: A003555 A003556 A003557 this_sequence
A003559
A003560 A003561
Keywords: nonn
Offset: 0
Author(s): njas
so evidently something's going on here. you sure about that
87382,
though? the encyclopedia predicts 2^18.
--
[e-mail address jdolan@math.ucr.edu]
===
Subject: Re: Rule 90 and left-to-right reversal
Distribution: world
> The above numbers are the generation number at which the
reversal
> appears, counting from generation 1, so are one greater than
the
> actual reversal period.
> Definitely related to what's on
> <http://www.richholmes.net/games/cellaut/r90.html>, then.
Your site has moved:
http://web.syr.edu/~rsholmes/games/cellaut/r90.html
> I didn't generalize this to arbitrary patterns, but
evidently it works
> out the same.
I've enhanced the Haskell program and verified it for all
patterns for
the first 8 lengths (2, 4, 6, 8, ... 16):
numberOfSteps = [(length x, checkRule90 x)|x<-(iterate (++ )
#)]
intToPattern 0 = []
intToPattern i =
(if i `mod` 2 == 1 then '#' else ' ') : (intToPattern (i `div`
2))
createPattern i len = pattern ++ replicate (len - (length
pattern)) ' '
where pattern = intToPattern i
createPatterns bits = [createPattern x bits | x<-[0..2^bits-1]]
checkAllPatterns len steps =
foldl1 (&&) (map (x->iterateRule90 x !! (steps-1) == reverse x)
(createPatterns len))
checkAll =
[(len, steps, checkAllPatterns len steps) |
(len, steps) <- numberOfSteps]
--
Frank Bu, fb@frank-buss.de
http://www.frank-buss.de, http://www.it4-systems.de
===
Subject: Re: Rule 90 and left-to-right reversal
Distribution: world
> so evidently something's going on here. you sure about that
87382,
> though? the encyclopedia predicts 2^18.
Yes. Testing it for all values would take some 200 years, if I
could test
1 million steps per second, but I've verified it for 100000
random values
with this Java program:
import java.util.*;
public class Test {
public static void main(String args[]) {
Random r = new Random();
int len = 36;
long count = 1;
for (int i = 0; i < len; i++)
count *= 2;
int max = 0;
long checks = 0;
while (true) {
// generate random test value
long value = Math.abs(r.nextLong()) % count;
// reverse value
long reverse = 0;
for (long b1 = 1, b2 = count / 2; b2 > 0; b1 *= 2, b2 /= 2) {
if ((value & b1) != 0)
reverse |= b2;
}
// count number of steps for reversing
int steps = 0;
long current = value;
while (current != reverse) {
steps++;
current = (current > 1) ^ (current << 1) & (count - 1);
}
// check, if more than the current maximum
if (steps > max)
max = steps;
// print every 1000 steps
if ((checks++ % 1000) == 0)
System.out.println(
length:
+ len
+ , steps:
+ max
+ , checks:
+ checks);
}
}
}
--
Frank Bu, fb@frank-buss.de
http://www.frank-buss.de, http://www.it4-systems.de
===
Subject: Re: Rule 90 and left-to-right reversal
> The above numbers are the generation number at which the
reversal
> appears, counting from generation 1, so are one greater
than the
> actual reversal period.
 Definitely related to what's on
> <http://www.richholmes.net/games/cellaut/r90.html>, then.
> Your site has moved: http:...
It's a redirect. The first URL should always work; the second
might
not.
===
Subject: Re: Rule 90 and left-to-right reversal
> ID Number: A003558
> URL: http://www.research.att.com/projects/OEIS?Anum=A003558
> Sequence:
1,2,3,3,5,6,4,4,9,6,11,10,9,14,5,5,12,18,12,10,7,12,23,21,8,
> 26,20,9,29,30,6,6,33,22,35,9,20,30,39,27,41,8,28,11,12,10,
> 36,24,15,50,51,12,53,18,36,14,44,12,24,55
> Name: Least number m such that 2^m = +- 1 mod 2n + 1.
> See also: Sequence in context: A023160 A085312 A046530
this_sequence
A072451
> A023156 A051599
> Adjacent sequences: A003555 A003556 A003557 this_sequence
A003559
> A003560 A003561
> Keywords: nonn
> Offset: 0
> Author(s): njas
EIS, but evidently not this.
> so evidently something's going on here. you sure about that
87382,
> though? the encyclopedia predicts 2^18.
That's the length = 36 anomaly I mentioned in another post. The
procedure I came up with proves the pattern recurs in 524286
(2^19-2)
steps, and in fact it does, but for the third time -- the
recurrence
period is 174762 ((2^19-2)/3).
===
Subject: Re: Rule 90 and left-to-right reversal
3QLpj-NoP*NzsIC,boYU]bQ]H'<P%JD/,.J>y<#4ga3$21:
> Definitely related to what's on
> <http://www.richholmes.net/games/cellaut/r90.html>, then.
 Your site has moved: http:...
> It's a redirect. The first URL should always work; the
second might
> not.
I get a 404 from the URL above.
--
David Eppstein http://www.ics.uci.edu/~eppstein/
Univ. of California, Irvine, School of Information & Computer
Science
===
Subject: Re: Rule 90 and left-to-right reversal
<snip I didn't come up with a formula but I did find an easier way
> to calculate a recurrence length which is either the period
or a
> multiple of the period; for all lengths up to 254 the
calculated
> recurrence length *is* the period -- *except* for 36, where I
> calculate 524286 while the period is 174762, a factor of 3
smaller.
524286 = 2*(2^18-1), while 18 = 36/2.
Just skimmed this thread - not doing research on this.
HTH.
===
Subject: Re: Rule 90 and left-to-right reversal
Uh oh... yes, something's misconfigured.
So indeed, use
<http://web.syr.edu/~rsholmes/games/cellaut/r90.html>.
===
Subject: Short FLT Proof, Analysis of recent objections
I say I have a proof. The math should be trivial for
mathematicians.
The work is available online 24 hours a day around the world.
Why is there still a debate?
Because the truth is that I'm right. Mathematics is being
taught that
is false, and it has been taught for quite some time.
Mathematicians
claim that they don't have any errors in core mathematics, but
here
is one. Also if they admit the error then they have to
acknowledge
me, then my proof of Fermat's Last Theorem and my prime
counting work
should come out as well.
And yes, the Hammer has arrived and is in full swing. I have
the
momentum I've been looking for, so it's time to change the
establishment, for the betterment of all.
And someone brought up your current crop of great
mathematicians
which included Ribet, Wiles, Taylor, Frey, and some other guy,
and I'm
now speaking directly to them--You should be ashamed of
yourselves,
and you should have known the day of reckoning was coming soon.
I've been looking for a simple solution using elementary
methods, as a
hobby, for almost seven years. Despite having started from
scratch, I
think I made a little progress and I'm talking about it.
Over the span of time I've been pursuing my little hobby, I've
created
a lot of enemies on this newsgroup by jumping to my desired
conclusion
and talking about it, only to find out later I was wrong. At
times,
I've also questioned the morals or competency of those enemies
(especially when they were calling me names, questioning my
sanity, or
otherwise being obnoxious).
So in the meantime the debate continues. Some of you now know
that
mathematicians are worse than not being quite what you might
have
thought they were. But the disillusionment may soon get worse.
They are people who in not admitting they are wrong are
apparently
willing to continue to teach false mathematics to students who
trust
them because that's an inevitable consequence of ignoring my
work.
They ignore that paper; then they'll be teaching false
mathematics.
I'm waiting for them to do it, so hopefully the federal
authorities
can pounce on them for fraud. But I'm warning like this post
because
I don't think mathematicians believe that they are subject to
the
rules of society.
I think they'll read this post and think they can get away
with it.
It turns out that destructive ideas, what I call hostile
memes, can
take over the human mind. They are like viruses and can remove
the
ability to think rationally.
People under the influence of hostile memes can behave as if
possessed.
They do odd things like attack countries that are from all
appearances
actually trying to comply with the international mood.
They also do interesting things like proclaim that they are
experts
about diseases which are also called mysterious.
More interestingly to me people under the influence of hostile
memes
can start a war claiming they are trying to help and free
people they
are attacking!!!
These hostile memes can be the tip of the iceberg for groups
of ideas
that in their totality are more sentient than homo sapiens
sapiens.
They like you though, and have endless fun playing with you,
and some
of you call them demons or devils.
You all depend on me shutting up, so that people won't know
the truth.
Insults, including talk of racial slurs, and continual
references
back to the rest of sci.math with the claim that no one
believes me
are apparently efforts to get me to quiet down by using
intimidation
before the world finds out that there are mathematicians who
will not
only will lie about important mathematics, but who seem to
live in
their own little world where they make up their own rules.
They are immortal. And they have been around for longer than
you
have, and will be here after you're gone. However, they play by
rules, unlike many people.
So I put it out there so that when they're facing the public,
you know
the truth. If they whine about their importance to society, as
if
that means they should be able to get away with betraying it,
think of
the young people they were willing to teach false mathematics
to, and
consider their contempt for those young minds, and the future
they
represent.
I'm curious about how some of you would react if you found out
that
indeed I was right, and that for all these months there's been
a short
proof of Fermat's Last Theorem known, but resisted by
mathematicians.
Would you care?
Would it matter to you if they were confused or deliberately
hiding
the truth?
Do you think it'd matter to you if it turned out it was just a
few
people who've been posting here or if a bigger number of
mathematicians than you supposed knew the truth but kept quiet?
If you're a mathematician, do you think it'd have any impact
on you
personally?
Professionally?
If you're not a mathematician, do you think it'd have any
impact on
your trust of things mathematicians say or have said?
Some of you may know that I also recently found what I've
called the
functional definition of the prime counting function.
Do you see any significance in my using the term functional?
If mathematicians have been avoiding an important bit of work
in prime
number theory do you think they would be doing so because they
*believe* it's unimportant, or shockingly important?
If you find out that it is important work, but a large number
of
mathematicians deliberately ignored it even though it was
brought to
their attention in private communications, would you be more
or less
likely to trust mathematicians specifically about prime
numbers?
What if you found out that I had information that proved my
case
conclusively but was instead waiting to see if mathematicians
would
act in a way that showed they would lie for their own
interests.
Do you think I would have justification for witholding this
information to see if they'd tell the truth?
Would you feel better if I held this information until they
told the
truth, waited a while and then produced it whether they told
the truth
or not, or would you just as soon I shut-up whether I'm right
or not
because you're just sick of me, and you couldn't care less how
important the math I've discovered is?
Do you believe that if I did have important mathematical work
that I
could just send it to a math journal as you feel confident
that a
journal would consider it and report the information to the
world if
it were correct?
If you find out that even journals failed in this case, would
you find
yourself more or less likely to trust pronouncements made in
journals
in the future?
How about science journals versus math journals?
Would you consider a very large failure to tell the truth in
the math
field when looking at result in other fields?
If I tell you now to buy futures in the natural gas market,
and that's it,
am I not making an assertion about my expertise?
In the regular world, you'd probably have context to help you
evaluate my
true expertise but this is a newsgroup on the INTERNET, and
it's a far more
difficult proposition.
So, in the past I've told you NOT to just trust me but to
check the math,
and I've often provided math for you to check.
where I was using my name, and various words like prime, prime
counting, and prime counting function, when I noticed
something odd
using just prime counting, which was that links to some of my
posts
were coming up as high as number 4 in a list of over 100,000
search
results.
It turned out that only MathWorld was beating me out when it
came to
the subject of counting primes.
I found that fascinating, and contemplated it.
I can understand that you'd be perturbed at the idea that you
should
question Galois Theory (or better yet your own work which you
claim
depends on it) as that is probably an idea that gets a very
emotional
reaction from you.
However the choice is clear, given that polynomials *are*
reducible,
and the simplicity of my argument where ultimately reliance is
on the
distributive principle.
I've mentioned that I've emailed leading mathematicians and
discussed
my prime counting work. I mentioned this interesting oddity to
one of
them, and it stopped. Well, I should say the behavior
*changed*, as
will find it more difficult to see what I've actually said,
while a
link to a flame page against me now gets top billing (all still
amazing high in the search list).
(I'd appreciate verification from someone else, as I'm not
certain
name is being used in searches when I do it. Um, you might
want to
hurry though, as I'm still wondering about the speed of the
last
change, which may have been a coincidence, but after this
post, things
may change again.)
Oh yeah, another leading mathematician told me that one out of
five
graduate students who do work in the area find something like
my prime
counting function. His opinion apparently being that their
work had
not been worth publishing either.
So here you have my claim that I've found a *short* proof of
Fermat's
Last Theorem, where the methods used involve factoring
polynomials
into non-polynomial factors, which you can't find in all of
mathematics outside of my work, meanwhile I face a lot of
hostility
over my work, from people who can't show an error within the
work
itself.
So the gist of it is that the short FLT Proof, which can be
found at
my website for my math discoveries, which is
http://www.msnusers.com/AmateurMath is currently unwanted and
unloved,
except by me of course, as I think it's really neat!!!
Along with it is a prime counting function, which you can't
find in
any established math reference or even online if it's not
connected to
me (last time I checked), and it's unwanted as well.
So you have all these math bigwigs ganging up on my *short*
math
results, and strutting around the newsgroup, and I do admit
that I'd
like them cut down to size.
But it looks like it'll take me some time.
However, now that I have the Hammer and am getting a feel for
its heft
and weight, it might finally be giant thumping time!!!
I talk of the FLT Proof as Thor's Hammer because, yes I'm a
mythology
buff.
And it is massively incredible to have such a thing as your own
discovery of a short FLT Proof, which is indestructible, and
quite
powerful, but also light and sublime.
I rely heavily on its power to get me through these dark days.
Still, I guess it's really not mine, but it does make me feel
quite
powerful, like Thor, while my pitiful mortal frame wields it
for a
time.
James Harris
===
Subject: Re: Short FLT Proof, Analysis of recent objections
> I say I have a proof. The math should be trivial for
mathematicians.
> The work is available online 24 hours a day around the world.
> Why is there still a debate?
> Because the truth is that I'm right. Mathematics is being
taught that
> is false, and it has been taught for quite some time.
Mathematicians
> claim that they don't have any errors in core mathematics,
but here
> is one. Also if they admit the error then they have to
acknowledge
> me, then my proof of Fermat's Last Theorem and my prime
counting work
> should come out as well.
> And yes, the Hammer has arrived and is in full swing. I have
the
> momentum I've been looking for, so it's time to change the
> establishment, for the betterment of all.
> And someone brought up your current crop of great
mathematicians
> which included Ribet, Wiles, Taylor, Frey, and some other
guy, and I'm
> now speaking directly to them--You should be ashamed of
yourselves,
> and you should have known the day of reckoning was coming
soon.
> I've been looking for a simple solution using elementary
methods, as a
> hobby, for almost seven years. Despite having started from
scratch, I
> think I made a little progress and I'm talking about it.
> Over the span of time I've been pursuing my little hobby,
I've created
> a lot of enemies on this newsgroup by jumping to my desired
conclusion
> and talking about it, only to find out later I was wrong. At
times,
> I've also questioned the morals or competency of those
enemies
> (especially when they were calling me names, questioning my
sanity, or
> otherwise being obnoxious).
> So in the meantime the debate continues. Some of you now
know that
> mathematicians are worse than not being quite what you might
have
> thought they were. But the disillusionment may soon get
worse.
> They are people who in not admitting they are wrong are
apparently
> willing to continue to teach false mathematics to students
who trust
> them because that's an inevitable consequence of ignoring my
work.
> They ignore that paper; then they'll be teaching false
mathematics.
> I'm waiting for them to do it, so hopefully the federal
authorities
> can pounce on them for fraud. But I'm warning like this post
because
> I don't think mathematicians believe that they are subject
to the
> rules of society.
> I think they'll read this post and think they can get away
with it.
> It turns out that destructive ideas, what I call hostile
memes, can
> take over the human mind. They are like viruses and can
remove the
> ability to think rationally.
> People under the influence of hostile memes can behave as if
> possessed.
> They do odd things like attack countries that are from all
appearances
> actually trying to comply with the international mood.
> They also do interesting things like proclaim that they are
experts
> about diseases which are also called mysterious.
> More interestingly to me people under the influence of
hostile memes
> can start a war claiming they are trying to help and free
people they
> are attacking!!!
> These hostile memes can be the tip of the iceberg for groups
of ideas
> that in their totality are more sentient than homo sapiens
sapiens.
> They like you though, and have endless fun playing with you,
and some
> of you call them demons or devils.
> You all depend on me shutting up, so that people won't know
the truth.
> Insults, including talk of racial slurs, and continual
references
> back to the rest of sci.math with the claim that no one
believes me
> are apparently efforts to get me to quiet down by using
intimidation
> before the world finds out that there are mathematicians who
will not
> only will lie about important mathematics, but who seem to
live in
> their own little world where they make up their own rules.
> They are immortal. And they have been around for longer than
you
> have, and will be here after you're gone. However, they play
by
> rules, unlike many people.
> So I put it out there so that when they're facing the
public, you know
> the truth. If they whine about their importance to society,
as if
> that means they should be able to get away with betraying
it, think of
> the young people they were willing to teach false
mathematics to, and
> consider their contempt for those young minds, and the
future they
> represent.
> I'm curious about how some of you would react if you found
out that
> indeed I was right, and that for all these months there's
been a short
> proof of Fermat's Last Theorem known, but resisted by
mathematicians.
> Would you care?
> Would it matter to you if they were confused or deliberately
hiding
> the truth?
> Do you think it'd matter to you if it turned out it was just
a few
> people who've been posting here or if a bigger number of
> mathematicians than you supposed knew the truth but kept
quiet?
> If you're a mathematician, do you think it'd have any impact
on you
> personally?
> Professionally?
> If you're not a mathematician, do you think it'd have any
impact on
> your trust of things mathematicians say or have said?
> Some of you may know that I also recently found what I've
called the
> functional definition of the prime counting function.
> Do you see any significance in my using the term functional?
> If mathematicians have been avoiding an important bit of
work in prime
> number theory do you think they would be doing so because
they
> *believe* it's unimportant, or shockingly important?
> If you find out that it is important work, but a large
number of
> mathematicians deliberately ignored it even though it was
brought to
> their attention in private communications, would you be more
or less
> likely to trust mathematicians specifically about prime
numbers?
> What if you found out that I had information that proved my
case
> conclusively but was instead waiting to see if
mathematicians would
> act in a way that showed they would lie for their own
interests.
> Do you think I would have justification for witholding this
> information to see if they'd tell the truth?
> Would you feel better if I held this information until they
told the
> truth, waited a while and then produced it whether they told
the truth
> or not, or would you just as soon I shut-up whether I'm
right or not
> because you're just sick of me, and you couldn't care less
how
> important the math I've discovered is?
> Do you believe that if I did have important mathematical
work that I
> could just send it to a math journal as you feel confident
that a
> journal would consider it and report the information to the
world if
> it were correct?
> If you find out that even journals failed in this case,
would you find
> yourself more or less likely to trust pronouncements made in
journals
> in the future?
> How about science journals versus math journals?
> Would you consider a very large failure to tell the truth in
the math
> field when looking at result in other fields?
> If I tell you now to buy futures in the natural gas market,
and that's
it,
> am I not making an assertion about my expertise?
> In the regular world, you'd probably have context to help
you evaluate my
> true expertise but this is a newsgroup on the INTERNET, and
it's a far
more
> difficult proposition.
> So, in the past I've told you NOT to just trust me but to
check the math,
> and I've often provided math for you to check.
> where I was using my name, and various words like prime,
prime
> counting, and prime counting function, when I noticed
something odd
> using just prime counting, which was that links to some of
my posts
> were coming up as high as number 4 in a list of over 100,000
search
> results.
> It turned out that only MathWorld was beating me out when it
came to
> the subject of counting primes.
> I found that fascinating, and contemplated it.
> I can understand that you'd be perturbed at the idea that
you should
> question Galois Theory (or better yet your own work which
you claim
> depends on it) as that is probably an idea that gets a very
emotional
> reaction from you.
> However the choice is clear, given that polynomials *are*
reducible,
> and the simplicity of my argument where ultimately reliance
is on the
> distributive principle.
> I've mentioned that I've emailed leading mathematicians and
discussed
> my prime counting work. I mentioned this interesting oddity
to one of
> them, and it stopped. Well, I should say the behavior
*changed*, as
> will find it more difficult to see what I've actually said,
while a
> link to a flame page against me now gets top billing (all
still
> amazing high in the search list).
> (I'd appreciate verification from someone else, as I'm not
certain
> name is being used in searches when I do it. Um, you might
want to
> hurry though, as I'm still wondering about the speed of the
last
> change, which may have been a coincidence, but after this
post, things
> may change again.)
> Oh yeah, another leading mathematician told me that one out
of five
> graduate students who do work in the area find something
like my prime
> counting function. His opinion apparently being that their
work had
> not been worth publishing either.
> So here you have my claim that I've found a *short* proof of
Fermat's
> Last Theorem, where the methods used involve factoring
polynomials
> into non-polynomial factors, which you can't find in all of
> mathematics outside of my work, meanwhile I face a lot of
hostility
> over my work, from people who can't show an error within the
work
> itself.
> So the gist of it is that the short FLT Proof, which can be
found at
> my website for my math discoveries, which is
> http://www.msnusers.com/AmateurMath is currently unwanted
and unloved,
> except by me of course, as I think it's really neat!!!
> Along with it is a prime counting function, which you can't
find in
> any established math reference or even online if it's not
connected to
> me (last time I checked), and it's unwanted as well.
> So you have all these math bigwigs ganging up on my *short*
math
> results, and strutting around the newsgroup, and I do admit
that I'd
> like them cut down to size.
> But it looks like it'll take me some time.
> However, now that I have the Hammer and am getting a feel
for its heft
> and weight, it might finally be giant thumping time!!!
> I talk of the FLT Proof as Thor's Hammer because, yes I'm a
mythology
> buff.
> And it is massively incredible to have such a thing as your
own
> discovery of a short FLT Proof, which is indestructible, and
quite
> powerful, but also light and sublime.
> I rely heavily on its power to get me through these dark
days.
> Still, I guess it's really not mine, but it does make me
feel quite
> powerful, like Thor, while my pitiful mortal frame wields it
for a
> time.
> James Harris
Speaking of you with a hammer trying to nail down some
mathematics...
http://www.edu-observatory.org/eo/hammercons.gif
===
Subject: Re: Short FLT Proof, Analysis of recent objections
> However, now that I have the Hammer and am getting a feel
for its heft
> and weight, it might finally be giant thumping time!!!
May the force be with you!
F.
===
Subject: Re: Simple Series Question
>Another series question. Does this series converge and if so
how do I show
>it.
>sum_{n=1}^{infty} (1 - 1/n)^{n^2}
Somewhere in the book it should say something about the
limit of (1 - 1/n)^n; if you know what that limit is you can
use it here...
************************
David C. Ullrich
===
Subject: Re: Simple Series Question
> Another series question. Does this series converge and if so
how do I
show
> it.
> sum_{n=1}^{infty} (1 - 1/n)^{n^2}
Take advantage of the circumstances. (The summation actually
takes place from n=2 to infinity, but that's beside the point.)
Observe that (1-1/n)^n converges to 1/e (it looks like you
covered this limit), so sooner or later the (positive) terms of
the series become less than (2/e)^n. Will it help?
(A closer look will reveal that all the terms are already less
than (1/e)^n. )
===
Subject: Solution of linear equations with constraints
I need to solve a class of simple equations, such as
a = b + 2c
where the variables involved are integers which are
constrained in
their allowable values. For example, in this case, 'a' must be
in the
range [1..9], and both b and c must be greater than zero, so
there are
a fixed number (16, in this case) of discrete solutions.
Can anyone tell me if there's an algebraic method for solving
this
sort of problem? If so, can you point me to any further
reading?
Evan
__________________________________________
To get a valid mail address: s/spam/spam1/
===
Subject: Re: Solution of linear equations with constraints
>I need to solve a class of simple equations, such as
>a = b + 2c
>where the variables involved are integers which are
constrained in
>their allowable values. For example, in this case, 'a' must
be in the
>range [1..9], and both b and c must be greater than zero, so
there are
>a fixed number (16, in this case) of discrete solutions.
>Can anyone tell me if there's an algebraic method for solving
this
>sort of problem? If so, can you point me to any further
reading?
Integer linear programming, in general. In particular cases
there
may be shortcuts which make it easier. E.g. in this case:
Start with
{a = b + 2 c, 1 <= a <= 9, b >= 1, c >= 1}
Replace a with b + 2 c, and you have
{ 1 <= b + 2 c, b + 2 c <= 9, b >= 1, c >= 1}
The inequalities involving b can be written as
1 - 2 c <= b
1 <= b
b <= 9 - 2 c
Each lower bound for b is <= each upper bound for b, so
1 - 2 c <= 9 - 2 c (true)
1 <= 9 - 2 c, i.e. c <= 4
Now all you have left is
{ c <= 4, c >= 1}, so c could be 1,2,3, or 4. Then work
backwards,
e.g. for c=2 the bounds for b say 1 <= b <= 5, and for each
integer b in
this interval you get an a.
Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia
Vancouver, BC, Canada V6T 1Z2
===
Subject: Re: Solving diophantine equation of special form
> By 'very large' I mean that cofficients
> and variables may have several
> hundred digits.
> But if factoring is required... I guess
> I'm out of luck, right?
Yep.
I haven't thought it through, but in converting an equation
bxy + ex + fy =
n
to (Ax + B)(Cy + D) = N, there might be ways that result in
coefficients
smaller than those per the method on my website. If nothing
else, you can
apply the Euclidean algorithm to A and B, and divide out by
the resulting
gcd.
Ditto C and D. Maybe that's the best you can do. At some
point, you will
have
to factor an N.
John Robertson
recently fixed a few errors in the page on Matthews' method
for solving
binary
quadratic form equations?
===
Subject: Re: Test for quasi (or pseudo) convexity
>Is anyone aware of a test for quasiconvexity (or
pseudoconvexity) of
a
>function over a (compact, possibly simply connected, and
even convex
>if we restric ourselves) domain in R^2? We are aware of
a test based
>on the Bordered Hessian, although I don't think we can
evaluate the
>derivatives at every point in our domain. What would be
ideal is
some
>sort of test we could perform on the boundary of our
domain.
> I don't see how a test on the boundary of the domain
could tell you
> anything about the behaviour of the function in the
interior of the
> domain (except for special cases such as harmonic
functions where the
> function is determined by its boundary values).
>Yes - of course...I forgot to mention our function is
subharmonic
> First, I have no idea what a pseudoconvex function would be;
the
> pseudoconvexity I know is a property of domains in C^n, not
of
> functions. I've seen f called quasi-convex if it's defined
on a convex
> set and
> f(tx + (1-t)y) <= max{f(x), f(y)} (0 < t < 1);
> is that what you mean by quasi-convexity?
This is the quasi-convexity we're looking for. I've seen
Pseudo-convexity defined as:
<grad f(x1), x2-x1>= 0 implies f(x2) >= f(x1) (increasing
locally
in a direction implies increasing globally in that direction)
On
second thought, I can't differentiate this from the definition
of
quasi-convexity...(or at least come up with an example of one
without
the other) although I've seen it stated that quasiconvex
implies
pseudoconvex but not vice-versa.
> If so, then the boundary values of a subharmonic function
> do not suffice to determine whether the function is
> quasi-convex. For example, let u = 0 in the unit disk; this
> is quasi-convex. But there exist subharmonic v in the disk
> with the same boundary values which equal -infinity at
> a few points, hence are certainly not quasi-convex
> (let x, y in the definition equal points where x =
-infinity.)
Absolutley true...that's why we need a test, since it's not
true in
general. The truth is we have only one such singularity, but
you
could certainly imagine functions with multiple local minima
which
violate quasiconvexity. In the end, what we want is to test for
unique local minimum where quasiconvexity is a sufficient
condition.
>(actually Laplacian(U)=2 Pi)
> It's _possible_ that this makes a difference, although I
> still tend to doubt that you're going to find the sort of
> answer you want. The reason it's not clearly impossible
> is that if you know that Laplacian(U) = 2 Pi then U _is_
> determined by its boundary values (because then
> U(z) - (Pi/2) |z|^2 is harmonic.)
This was my intuition, but I hadn't laid it out
explicitly...I'll have
to see what U(z)-(Pi/2) |z|^2 means physically in out case. In
case
anyone is interested, there is a physical situation here: we
have a
2-D air flow field generated by a single flow sink (suction
point).
The velocity (under certain assumptions) is the -gradient of
u=log(r)
where r is the distance from the suction point. The air flows
over
the surface of an object (our region) generating a shear force
per
unit area proportional to the velocity. The net effect is that
(without rotation) the object experiences a net force:
F=int int -grad(u) dA
and hence a lifted potential field
U=int int u dA
where
F=-grad(U)
In this case, U is a function of the position of some
reference point
fixed on the object as the object moves. It turns out that
while the
singularity is in the interior of the object,
Laplacian(U)=2pi, and
Laplacian(U)=0 (obviously) otherwise.
We want to either show something like all convex objects
experience a
single equilibrium position (at fixed orientation) on this
flow field,
or come up with a test for any object to see if it has a unique
equilibrium. (We do know that at least one equilibrium does
exist,
and that it is (of course) when the sinularity lies in the
interior of
the object.
>-Jonathan Luntz
> ************************
> David C. Ullrich
-Jonathan Luntz
===
Subject: Re: Test for quasi (or pseudo) convexity
> First, I have no idea what a pseudoconvex function would
be; the
> pseudoconvexity I know is a property of domains in C^n, not
of
> functions. I've seen f called quasi-convex if it's defined
on a convex
> set and
> f(tx + (1-t)y) <= max{f(x), f(y)} (0 < t < 1);
> is that what you mean by quasi-convexity?
>This is the quasi-convexity we're looking for. I've seen
>Pseudo-convexity defined as:
> <grad f(x1), x2-x1>= 0 implies f(x2) >= f(x1) (increasing
locally
>in a direction implies increasing globally in that direction)
On
>second thought, I can't differentiate this from the
definition of
>quasi-convexity...(or at least come up with an example of one
without
>the other) although I've seen it stated that quasiconvex
implies
>pseudoconvex but not vice-versa.
It's the other way around.
An example of a quasiconvex function that isn't pseudoconvex is
f(x) = x^3 on the real line (take x1 = 0).
On the other hand, suppose f is not quasiconvex, say
f(tx + (1-t) y) > max(f(x), f(y)) with 0 < t < 1.
For convenience let g(s) = f(sx + (1-s) y). Quasiconvexity
would
say g'(s1) >= 0 implies g(s2) >= g(s1) for s2 >= s1. But
since g(t) > max(g(0), g(1)), g has a maximum at some point s1
in
(0,1), and we get a contradiction with s2 = 1.
Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia
Vancouver, BC, Canada V6T 1Z2
===
Subject: Textbook Rant
I just received a calculus textbook I ordered for self study.
Unfortunately, I didn't get the CD containing the worked out
solutions
to the old numbered exercises in the textbook which I had
thought was
included in the CD attached to the textbook.
This incident brings me to the observation, which I suspect is
common
in college math textbooks, of requiring students to pay for the
additional solutions manual instead of either (i) including
the worked
out solutions in the back of the textbook or (ii) including the
solutions in the CDROM that already comes with each textbook.
In the
case of my textbook, the solutions are already in the back of
the
textbook. It would only require maybe 20 additional pages to
include
all the worked out examples, instead of just the solutions
since the
font type used is already small. In the case of the CDROM
which also
comes with my textbook, there is more than sufficient space
remaining
on the CDROM to include the solutions manual in PDF format
(the CDROM
for the textbook had 3 MB used).
The point that I am trying to make is that for very little
additional
cost to the publisher, the worked out solutions could be
included as a
few additional pages in the text or for no additional cost on
the
accompanying CDROM. Instead, the publisher charges a
rediculous price
for the solutions manual (in my case, over $50) most of which
is
additional profit.
It makes me angry that after spending over $120 for a
textbook, I need
to spend over another $50 to pay for the solutions manual when
the
publisher could have added the solutions manual to the
textbook for
virtually no additonal cost (i.e. included in the attached
CDROM). I
think that the publishers are abusing their situation and
therefore
students should behave accordingly towards publishers when the
opportunity arises.
===
Subject: Re: Textbook Rant
> I just received a calculus textbook I ordered for self study.
> Unfortunately, I didn't get the CD containing the worked out
solutions
> to the old numbered exercises in the textbook which I had
thought was
> included in the CD attached to the textbook.
> This incident brings me to the observation, which I suspect
is common
> in college math textbooks, of requiring students to pay for
the
> additional solutions manual instead of either (i) including
the worked
> out solutions in the back of the textbook or (ii) including
the
> solutions in the CDROM that already comes with each
textbook. In the
> case of my textbook, the solutions are already in the back
of the
> textbook. It would only require maybe 20 additional pages to
include
> all the worked out examples, instead of just the solutions
since the
> font type used is already small. In the case of the CDROM
which also
> comes with my textbook, there is more than sufficient space
remaining
> on the CDROM to include the solutions manual in PDF format
(the CDROM
> for the textbook had 3 MB used).
> The point that I am trying to make is that for very little
additional
> cost to the publisher, the worked out solutions could be
included as a
> few additional pages in the text or for no additional cost
on the
> accompanying CDROM. Instead, the publisher charges a
rediculous price
> for the solutions manual (in my case, over $50) most of
which is
> additional profit.
> It makes me angry that after spending over $120 for a
textbook, I need
> to spend over another $50 to pay for the solutions manual
when the
> publisher could have added the solutions manual to the
textbook for
> virtually no additonal cost (i.e. included in the attached
CDROM). I
> think that the publishers are abusing their situation and
therefore
> students should behave accordingly towards publishers when
the
> opportunity arises.
I feel for you. Textbooks were actually affordable when I went
to school.
Talk to your professor. I assume (s)he selected the textbook.
You would think the professor would have the interests of
his/her
students in mind.
Nemo
===
Subject: Re: Textbook Rant
To learn the material, work out the solutions to the exercises
yourself. Including solutions with the book means that lazy
students
will simply copy them, learning very little...
Someone once said: College is the place where you pay for
something,
and then you try as hard as you can NOT to get your money's
worth.
===
Subject: Re: Textbook Rant
> It makes me angry that after spending over $120 for a
textbook, I need
> to spend over another $50 to pay for the solutions manual
when the
> publisher could have added the solutions manual to the
textbook for
> virtually no additonal cost (i.e. included in the attached
CDROM). I
> think that the publishers are abusing their situation and
therefore
> students should behave accordingly towards publishers when
the
> opportunity arises.
The problem with the economics of textbook publishing is
volume. There
simply aren't enough copies sold to pay for the fixed cost of
publishing. So textbooks are ridiculously expensive compared
to best
sellers.
That said, if you're doing self-study, Schaum's outlines are
very good.
Also Dover publishes a lot of cheap math and science books
that are very
good.
===
Subject: Re: Textbook Rant
> To learn the material, work out the solutions to the
exercises
> yourself. Including solutions with the book means that lazy
students
> will simply copy them, learning very little...
Yeah, then the lazy ones get their asses kicked on the tests.
===
Subject: Re: Textbook Rant
>Talk to your professor. I assume (s)he selected the textbook.
>You would think the professor would have the interests of
his/her
>students in mind.
Calc books tend to be chosen by committees at larger schools,
and they all
cost
about the same. On the other hand, they're good for 3
semesters (more if
the
student fails a couple of times).
--irascible since 1957
===
Subject: Re: Textbook Rant
> I just received a calculus textbook I ordered for self study.
> Unfortunately, I didn't get the CD containing the worked out
solutions
> to the old numbered exercises in the textbook which I had
thought was
> included in the CD attached to the textbook.
> This incident brings me to the observation, which I suspect
is common
> in college math textbooks, of requiring students to pay for
the
> additional solutions manual instead of either (i) including
the worked
> out solutions in the back of the textbook or (ii) including
the
> solutions in the CDROM that already comes with each
textbook. In the
> case of my textbook, the solutions are already in the back
of the
> textbook. It would only require maybe 20 additional pages to
include
> all the worked out examples, instead of just the solutions
since the
> font type used is already small. In the case of the CDROM
which also
> comes with my textbook, there is more than sufficient space
remaining
> on the CDROM to include the solutions manual in PDF format
(the CDROM
> for the textbook had 3 MB used).
> The point that I am trying to make is that for very little
additional
> cost to the publisher, the worked out solutions could be
included as a
> few additional pages in the text or for no additional cost
on the
> accompanying CDROM. Instead, the publisher charges a
rediculous price
> for the solutions manual (in my case, over $50) most of
which is
> additional profit.
> It makes me angry that after spending over $120 for a
textbook, I need
> to spend over another $50 to pay for the solutions manual
when the
> publisher could have added the solutions manual to the
textbook for
> virtually no additonal cost (i.e. included in the attached
CDROM). I
> think that the publishers are abusing their situation and
therefore
> students should behave accordingly towards publishers when
the
> opportunity arises.
Welcome to the world of capitalism.
Steve
Socialist
===
Subject: Re: Textbook Rant
> To learn the material, work out the solutions to the
exercises
> yourself. Including solutions with the book means that lazy
students
> will simply copy them, learning very little...
> Someone once said: College is the place where you pay for
something,
> and then you try as hard as you can NOT to get your money's
worth.
If the student short circuits the learning process by copying
the
solutions to the exercises to complete homeowork assignments,
s/he is
just wasting money and time. I use the solutions as feedback to
determine errors in comprehension, especially since I am
studying on
my own.
The point of my original post was that textbook publishers are
charging significant fees for solution manuals when they could
have
included them with the text for very little or no additional
cost. I
think the publishers are taking advantage of the situation and
I think
this is wrong.
===
Subject: Re: Textbook Rant
> To learn the material, work out the solutions to the
exercises
> yourself. Including solutions with the book means that
lazy students
> will simply copy them, learning very little...
> Someone once said: College is the place where you pay for
something,
> and then you try as hard as you can NOT to get your
money's worth.
> If the student short circuits the learning process by
copying the
> solutions to the exercises to complete homeowork
assignments, s/he is
> just wasting money and time. I use the solutions as feedback
to
> determine errors in comprehension, especially since I am
studying on
> my own.
> The point of my original post was that textbook publishers
are
> charging significant fees for solution manuals when they
could have
> included them with the text for very little or no additional
cost. I
> think the publishers are taking advantage of the situation
and I think
> this is wrong.
They could answer the odd numbers, but then the learning style
is
all by example, and there would still be a seperate book for
the even
answers
with even lower publishing numbers.
Answer books are so small they could go online for people to
printout.
I was lucky to get 2 textbooks a semester and look how I
turned out!
Herc
===
Subject: Re: Textbook Rant
>To learn the material, work out the solutions to the exercises
>yourself. Including solutions with the book means that lazy
students
>will simply copy them, learning very little...
That is not the way to use solution keys or solution manuals.
The lazy
student
who uses them as you describe will not achieve much benefit. A
student
must
work with the problem, recheck the concepts discussed in the
book,
re-examine
his own efforts, and then when he wants to compare his work
and results
with
the key in the book, THEN he can check it and compare.
The lazy student who simply copies from the solution manual
will not learn
anything, except how to cheat and how to fail.
G C
===
Subject: Re: Textbook Rant
> It makes me angry that after spending over $120 for a
textbook, I need
> to spend over another $50 to pay for the solutions manual
when the
> publisher could have added the solutions manual to the
textbook for
> virtually no additonal cost (i.e. included in the attached
CDROM). I
> think that the publishers are abusing their situation and
therefore
> students should behave accordingly towards publishers when
the
> opportunity arises.
> The problem with the economics of textbook publishing is
volume. There
> simply aren't enough copies sold to pay for the fixed cost of
> publishing. So textbooks are ridiculously expensive compared
to best
> sellers.
I'll buy that explanation for advanced books. But what about
books on
beginning algebra, trig, calculus -- courses taken by a huge
number of
students?
<rant> I've seen books on basic algebra cost upwards of $100.
That is
absolutely outrageous. Nothing new has been said in a beginning
algebra book in, what, 60 years? 160 years? This stuff has
been done
over and over again in all possible permutations. Why in the
hell is
there a need to come out with yet another rehash of the same
old stuff
with a $100 price tag -- especially when you can get a Schaum's
outline for under $20? There has to be something else than
copies sold
vs. fixed costs. And, to add insult to injury, most of the
math books
I've seen at the introductory level are garbage. Here is the
recipe
for a modern beginning algebra book:
Remove half of the pages of an old algebra book, and put the
remainder
in a blender.
Add 157 color photographs and 5 interviews with people who use
math at
work.
Hit frappe for 3 minutes.
Apply $100 price tag and serve. Feeds 1 (because a new edition
will
come out next semester)
I could go on for DAYS about this crap. I remember once taking
a
graduate level course in topology, back in '96 I think. The
text was
George Cain's An Introduction to General Topology at $80. Now,
IMHO
this book rates undergraduate rather than graduate level.
Opinion
aside, I already had a Dover paperback edition of Michael
Gemignani's
Elementary Topology which had only cost $7.95. Gemignani's book
covered more material in greater depth than Cain's book -- at
10% of
the cost! I returned Cain for a refund, and borrowed from
another
student to get the homework problems.
Another time, as an undergraduate, I bought a General
Psychology text,
4th ed. hot off the press, completed the class, and went to
sell it
back (at half price). But Oh! They wouldn't take it! The 5th
edition
was coming out next semester! Man! The field of psychology
must really
be hot hot hot! They needed to update their freshman
introductory text
every semester! Come on!! How many ways can you tell the tale
of
Pavlov's dog?
</rant That said, if you're doing self-study, Schaum's outlines are
very good.
> Also Dover publishes a lot of cheap math and science books
that are very
> good.
Exactly. And if Dover and Schaum's can do it, so can everybody
else.
But they don't. Why? The textbook industry is a RACKET --
plain and
simple.
===
Subject: Re: Textbook Rant
> I just received a calculus textbook I ordered for self study.
Let me get this right, that book had _not_ been forced upon
you as a
selected text for a course? You bought it voluntarily?
> Unfortunately, I didn't get the CD containing the worked out
solutions
> to the old numbered exercises in the textbook which I had
thought was
> included in the CD attached to the textbook.
>[... more details skipped ...]
> The point that I am trying to make is that for very little
additional
> cost to the publisher, the worked out solutions could be
included as a
> few additional pages in the text or for no additional cost
on the
> accompanying CDROM. Instead, the publisher charges a
rediculous price
> for the solutions manual (in my case, over $50) most of
which is
> additional profit.
> It makes me angry that after spending over $120 for a
textbook, I need
> to spend over another $50 to pay for the solutions manual
when the
> publisher could have added the solutions manual to the
textbook for
> virtually no additonal cost (i.e. included in the attached
CDROM). I
> think that the publishers are abusing their situation and
therefore
> students should behave accordingly towards publishers when
the
> opportunity arises.
Well, obviously the publisher can safely assume that whoever
spends $120
on their stupid calculus book, does not mind spending $50 on
additional
material.
In fact, these books are targeted at groups or individuals
inside the
faculty who decide on the adoption of a particular text for a
course.
In this case the students are captive consumers and the
relation between
price and quality is not an issue.
Moreover students rarely need the solutions manual because
there are
T.A.s and professors available.
Why anyone would buy such books without being forced is beyond
me.
You should probably try these steps:
(1) sell the book to someone
(2) use the library until you find a book you can work well
with
(3) if it is convenient, buy the book you found in step (2)
Marc
===
Subject: Re: Textbook Rant
[...]
> Why anyone would buy such books without being forced is
beyond me.
> You should probably try these steps:
> (1) sell the book to someone
> (2) use the library until you find a book you can work well
with
> (3) if it is convenient, buy the book you found in step (2)
> Marc
I just did a search for calculus at eBay; there are several
books
priced under $10.00 ...
David Bernier
===
Subject: Re: Textbook Rant
>To learn the material, work out the solutions to the exercises
>yourself. Including solutions with the book means that lazy
students
>will simply copy them, learning very little...
I don't see that at all. Firstly, you are assuming that the
exercises
are going to be set as homework - in which case the lazy one is
actually the course teacher, who is copying exercises out of a
book.
Secondly, even if exercises from a book are set for homework -
surely
it is easy to spot the difference between a copied answer and
a worked
out one - most textbooks that give solutions just give a final
numerical answer, not all of the reasoning behind the answer.
I'm currently studying for a maths degree by self study. The
University assigns textbooks for each course, and provides a
course
outline - but there is no tuition. I was pretty annoyed to
find that
the primary text for one course didn't have any solutions.
Gareth
===
Subject: Re: Textbook Rant
> To learn the material, work out the solutions to the
exercises
> yourself. Including solutions with the book means that lazy
students
> will simply copy them, learning very little...
>Yeah, then the lazy ones get their asses kicked on the tests.
But who cares about the lazy ones anyway? Why is something I'd
find
useful omitted from a book just because someone else doesn't
actually
want to learn maths?
Gareth
===
Subject: Re: Textbook Rant
> I just received a calculus textbook I ordered for self
study.
>Let me get this right, that book had _not_ been forced upon
you as a
>selected text for a course? You bought it voluntarily?
He didn't say that....there are self-study routes to getting a
degree
which still have required reading lists. If course, in that
case, the
primary fault is with whoever set the book...
Gareth
===
Subject: Re: Textbook Rant
> Why anyone would buy such books without being forced is
beyond me.
> You should probably try these steps:
> (1) sell the book to someone
Maybe return it to the seller? Then you're only out the
shipping fees.
Failing that, there are online services. The prices are
artificially
(naturally?) depressed because the purchaser can't examine the
book
before buying.
> (2) use the library until you find a book you can work well
with
> (3) if it is convenient, buy the book you found in step (2)
> I just did a search for calculus at eBay; there are several
books
> priced under $10.00 ...
But how good are they?
Although, for under $10, you can make 11 errors for each good
catch, and
still come out even.
Jon Miller
===
Subject: Re: Textbook Rant
NNTP-Posting-User: [mxpPujnWoCsyWiSEL8efHmlb+R3alXtU]
>The point of my original post was that textbook publishers are
>charging significant fees for solution manuals when they
could have
>included them with the text for very little or no additional
cost. I
>think the publishers are taking advantage of the situation
and I think
>this is wrong.
Speaking of textbook publishers charging outrageous amounts of
money, has
anyone noticed that there is now a 5th Edition of Stewart's
_Calculus_?
Looking through both copies, I have so far found nothing more
than minor
changes (mostly in the problems). With so many schools
requiring this text
for calc classes, James Stewart seems to be a one-man
industry. I
half-wonder if they just release a new edition every time
sales slow down
from a glut of used copies.
-Davis
===
Subject: Re: Textbook Rant
> I just received a calculus textbook I ordered for self study.
> Unfortunately, I didn't get the CD containing the worked out
solutions
> to the old numbered exercises in the textbook which I had
thought was
> included in the CD attached to the textbook.
> ...
Perhaps the book is more appealing to lecturers if the
solutions aren't
there; and if a lecturer adopts it for a course that
guarantees sales
and the publisher does not have to bother about the likes of
you (with
whom I sympathize).
> think that the publishers are abusing their situation and
therefore
> students should behave accordingly towards publishers when
the
> opportunity arises.
So you must tell us who the publisher is.
--
G.C.
===
Subject: Re: Textbook Rant
 I just received a calculus textbook I ordered for self
study.
> Unfortunately, I didn't get the CD containing the worked
out solutions
> to the old numbered exercises in the textbook which I had
thought was
> included in the CD attached to the textbook.
Would that be the ODD numbered exercises?
===
Well, I just would like to write down what I think. In fact, I
have
certainly not the ability to be a mathematician. I am very
respect to all
mathematicians.
Since I am interested in mathematics, that is why I posted
this concise
proof, I would like to make a little contribution in
mathematics.
Well, I don't mind that my concise proof seems a nothing.
Michael Leung
===
Subject: Re: The basic idea behind my great forthcoming proof
>What difference does the existence of P(N) make?
Mainly the consequences one can deduce from it, which
depend on what else you are assuming.
Keith Ramsay
===
Subject: Re: The basic idea behind my great forthcoming proof
|It occurred to me recently that I have almost no instinct for
what can
|be proved without using AC.
I'm guessing that it's fairly common among research
mathematicians
not to have a good sense for when they're using AC, which is
what a
thorough familiarity with ZFC should include, I think. How
much of it
can be proven otherwise is a tougher assignment, and being
able to
do independence proofs is more strictly a mathematical logic
skill.
Here, at least you recognize that the usual disproof is using
a result
that isn't a theorem of ZF:
|For example, can you prove that there is no group G with
|Aut(G)| = 3
|(a problem that arose recently on sci.math) without using AC ?
|
|You reduce easily to an infinite abelian group G in which all
elements
have
|order 2 or, equivalently, an infinite vector space over the
field of order
|2. Without AC, can you say anthing at all about Aut(G) ?
I'm guessing you can't say much. It might be worth learning
this
permutation model method to prove it.
Keith Ramsay
===
Subject: Re: The basic idea behind my great forthcoming proof
|But, suppose here we have someone who thinks that infinite
sets do
|not exist at all, let alone P(N). Would such a person be
likely to
|graduate, or even to start studying math and persevere?
|Still, such people may still be good mathematicians.
I don't know about likely, but whether they succeeded might
depend
on whether they spent their time trying to refute these
things, or on
topics (certain kinds of combinatorics, for example) which
would suit
them.
Keith Ramsay
===
Subject: Re: The basic idea behind my great forthcoming proof
>It makes a large difference, however, whether you think of N
and P(N)
>as a fixed, completed totality, or that you try to do justice
to its
>constructive character.
Those two things are somewhat independent of each other. The
only
way I can see to define completed totality so that treating a
set as
a completed totality implies that one is treating it
nonconstructively
is by defining completed totality to include the validity of
using
nonconstructive reasoning on it.
In a certain sense Brouwer's free choice sequences are a
typical
notion of a non-completed totality. But some other
constructions
of sets, sequences or functions treat them essentially as being
rules, which once given are given completely.
Keith Ramsay
===
Subject: Re: The basic idea behind my great forthcoming proof
> |And, be fair: can anyone become a mathematician, nowadays,
> |without being turned into a ZFC-talker?
> As often as people on usenet refer to ZFC, I think the
typical
> mathematician knows very little about it, nor does it make
> much difference to them.
> Keith Ramsay
> You're right.
> But, suppose here we have someone who thinks that infinite
sets do
> not exist at all, let alone P(N). Would such a person be
likely to
> graduate, or even to start studying math and persevere?
> Still, such people may still be good mathematicians.
Well, we do have someone who doesn't believe infinite sets
exist, and goes
so far as to say sets don't exist at all! That person is me.
I don't believe in the Platonic realm or whatever, but I do
assign greater
degrees of realness to mathematical objects. Sets have my
lowest
realness rating.
My experience is that most do not care about foundations or
even have
enough of an opinion on the subject to talk about it, as Keith
has
indicated. But I've been surprised enough by the number of
times that
people agree with my views. So I would sum up my listeners'
reactions as
being either of indifference or (at least mild) agreement,
which is a far
cry from the picture painted by some people in this discussion.
Disclaimer: I study a field (3-manifolds) in which most
practitioners have
probably never seen any complete set of axioms for set theory,
like ZFC,
and in which people actually expend effort to avoid talking
like set
theorists. So it may be that the kind of view I harbor may be
detrimental
in other areas of mathematics.
===
Subject: Re: The basic idea behind my great forthcoming proof
>|It occurred to me recently that I have almost no instinct
for what can
>|be proved without using AC.
>I'm guessing that it's fairly common among research
mathematicians
>not to have a good sense for when they're using AC,
This is not so clear to me. I had no idea for a _long_ time
that when
I said that a countable union of countable sets was countable
I was
using AC - I see this given as a theorem in the introductory
material
in one of my favorite books on real analysis, without mention
of AC,
although the author _seems_ to intend to say so when he's using
AC because he does so elsewhere; I've had many colleagues
react with disbelief when I told them this needed AC, etc.
(Yes, the typical application doesn't really need AC because
it's
a countable union of counted sets. But regardless...)
>which is what a
>thorough familiarity with ZFC should include, I think. How
much of it
>can be proven otherwise is a tougher assignment, and being
able to
>do independence proofs is more strictly a mathematical logic
skill.
>Here, at least you recognize that the usual disproof is using
a result
>that isn't a theorem of ZF:
>|For example, can you prove that there is no group G with
|Aut(G)| = 3
>|(a problem that arose recently on sci.math) without using AC
?
>|You reduce easily to an infinite abelian group G in which
all elements
have
>|order 2 or, equivalently, an infinite vector space over the
field of
order
>|2. Without AC, can you say anthing at all about Aut(G) ?
>I'm guessing you can't say much. It might be worth learning
this
>permutation model method to prove it.
>Keith Ramsay
************************
David C. Ullrich
===
Subject: Re: The basic idea behind my great forthcoming proof
Jurjus
> |And, be fair: can anyone become a mathematician,
nowadays,
> |without being turned into a ZFC-talker?
> As often as people on usenet refer to ZFC, I think the
typical
> mathematician knows very little about it, nor does it make
> much difference to them.
> Keith Ramsay
> You're right.
> But, suppose here we have someone who thinks that infinite
sets do
> not exist at all, let alone P(N). Would such a person be
likely to
> graduate, or even to start studying math and persevere?
> Still, such people may still be good mathematicians.
> Well, we do have someone who doesn't believe infinite sets
exist, and
goes
> so far as to say sets don't exist at all! That person is me.
> I don't believe in the Platonic realm or whatever, but I do
assign
greater
> degrees of realness to mathematical objects. Sets have my
lowest
> realness rating.
> My experience is that most do not care about foundations or
even have
> enough of an opinion on the subject to talk about it, as
Keith has
> indicated. But I've been surprised enough by the number of
times that
> people agree with my views. So I would sum up my listeners'
reactions as
> being either of indifference or (at least mild) agreement,
which is a far
> cry from the picture painted by some people in this
discussion.
> Disclaimer: I study a field (3-manifolds) in which most
practitioners
have
> probably never seen any complete set of axioms for set
theory, like ZFC,
> and in which people actually expend effort to avoid talking
like set
> theorists. So it may be that the kind of view I harbor may be
detrimental
> in other areas of mathematics.
A little story.
A few years ago, i was attending a lecture. the audience
consisted of
about 30 or 40 professional mathematicians of the kind you
describe:
from all kinds of 'normal' mathematical subjects, not
especially interested
in set theory or foundations.
During the lecture, there was some talk of AD, and how strange
it was
that AD is 'not true' (most people accept ZFC as true, and ZFC
+ AD
is inconsistent, so a consequence is, they think AD is not
true).
There was some noise yeah, yeah, it's strange, but the
shrugging of
shoulders
continued, nonetheless.
For those who don't know AD, the following variant of AD is
already
contradicting ZFC:
Given a set A subset of N^N, and two players 1, and 2 playing
a game.
1 chooses a natural number, then 2 chooses one, then 1 chooses
one, etc.
After infinitely many steps, the result is an element of N^N.
If this element is in A, 1 wins, otherwise, 2 wins.
The axiom says: either 1 has a winning strategy or 2 has.
(Either 1 can manipulate the end-result into A, or 1 cannot,
that is:
2 can prevent the en-result to be in A.)
Now, if you think a little bit about this principle, it is
just as
plausible
a principle
as AC, if not much more plausible. Finite games are decided,
why aren't
infinitary
games decided? It is really strange. But nobody seemed to care.
Herman Jurjus
===
Subject: Re: The basic idea behind my great forthcoming proof
> Well, we do have someone who doesn't believe infinite sets
exist, and
goes
> so far as to say sets don't exist at all! That person is me.
> I don't believe in the Platonic realm or whatever, but I do
assign
greater
> degrees of realness to mathematical objects. Sets have my
lowest
> realness rating.
But you do talk and think about them, right? Perhaps only in
disguised
form.
And when you do, you probably think about them *as if* they
are fixed,
completed totalities? In that case, you are assuming lots of
properties
about them....
And, indeed, most mathematicians are not aware that they do
exactly that:
assuming lots of things silently. They don't even care that
they do.
But what can be done about it? Sigh.
Herman Jurjus
===
Subject: Re: The basic idea behind my great forthcoming proof
Dave Seaman says...
> Well, I think you're missing the point.
> Let's roll back to my use of the phrase mathematical
objects, which
you
> challenged, which I borrowed from a recent Robin Chapman
post
> in this thread.
>You were asking whether all mathematical objects exist. My
point is
>that you haven't gained anything by introducing that
terminology.
You are certainly missing the point. fpluser was responding to
Herman Jurjus, who seemed to reject ZFC because he didn't
accept
that infinite objects (such as P(N)) exist. fpluser was trying
to
understand what Herman meant by that, not what *you* mean.
Herman
in doubting that P(n) exists is not saying that he doubts that
it
is provable in ZFC that P(n) exists. He knows that ZFC proves
that,
and he thinks that, therefore, ZFC is wrong---it asserts that
things
exist that really don't exist. It doesn't do any good for you
to ask
fpluser to clarify what he means by such things, because he
doesn't
know---he's trying to clarify exactly that.
fpluser didn't introduce the terminology---other people did,
and
fpluser is trying to understand what they mean.
--
Daryl McCullough
Ithaca, NY
===
Subject: Re: The basic idea behind my great forthcoming proof
> Dave Seaman says...
> Well, I think you're missing the point.
> Let's roll back to my use of the phrase mathematical
objects, which
you
> challenged, which I borrowed from a recent Robin Chapman
post
> in this thread.
>You were asking whether all mathematical objects exist. My
point is
>that you haven't gained anything by introducing that
terminology.
> You are certainly missing the point. fpluser was responding
to
> Herman Jurjus, who seemed to reject ZFC because he didn't
accept
> that infinite objects (such as P(N)) exist. fpluser was
trying to
> understand what Herman meant by that, not what *you* mean.
Herman
> in doubting that P(n) exists is not saying that he doubts
that it
> is provable in ZFC that P(n) exists.
I didn't think he was. I assumed he was questioning whether we
should
accept ZFC.
>He knows that ZFC proves that,
> and he thinks that, therefore, ZFC is wrong---it asserts
that things
> exist that really don't exist. It doesn't do any good for
you to ask
> fpluser to clarify what he means by such things, because he
doesn't
> know---he's trying to clarify exactly that.
> fpluser didn't introduce the terminology---other people did,
and
> fpluser is trying to understand what they mean.
I haven't been debating Herman, but I am not aware of anything
he has
said that is inconsistent with what I said about existence.
--
Dave Seaman
Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling.
<http://www.commoncouragepress.com/index.cfm?action=book&
bookid=228>
===
Subject: Re: The basic idea behind my great forthcoming proof
Dave Seaman says...
>I haven't been debating Herman, but I am not aware of
anything he has
>said that is inconsistent with what I said about existence.
I think that he rejects ZFC *because* he doesn't believe in the
existence of infinite sets. That contradicts the idea that
existence means provable existence inside some theory.
People choose their axioms because they believe they capture
some intuitions about mathematical objects. So there is some
notion of mathematical object that is prior to formalization,
and helps guide the formalization.
Of course, there is feedback that goes on, as well. You come
up with axioms to capture your intuitions, and then working
with the axioms (or working with models of those axioms)
can cause you to change your intuitions. (which might suggest
different axioms).
--
Daryl McCullough
Ithaca, NY
===
Subject: Re: The basic idea behind my great forthcoming proof
>This has been bothering me lately.
>What difference does the existence of P(N) make?
>For that matter what difference does the existence
>of 7 make?
>I've not yet seen a careful definition that
>establishes exactly what existence means in
>this context (yet some people talk about it
>often), and more importantly why it should
>matter to me. In the most general sense, all
>mathematical objects exist because they
>can be imagined. Right?
>Perhaps so, but then the question becomes, What is a
mathematical
>object?
>I don't think the phrase mathematical object in this
context
>warrants much investigation, whatever type of objects
Herman
>meant to be discussing are the kind I meant. I thought
>mathematical object was generic enough, perhaps not.
>My point was, the word exist was being used to
differentiate
between
>two classes of things that can be expressed in
mathematics; I wasn't
>the one making the distinction.
>I'm afraid I don't follow. What do you mean by something
that can be
>expressed in mathematics?
>Well, I think you're missing the point.
>Let's roll back to my use of the phrase mathematical
objects, which
you
>challenged, which I borrowed from a recent Robin Chapman post
>in this thread.
> You were asking whether all mathematical objects exist. My
point is
> that you haven't gained anything by introducing that
terminology. If it
> is actually a mathematical object, then of course it exists,
but the
> question still remains, exactly what is a mathematical
object? You
> ignored my answer.
>He said:
> ``Mathematics is not a theory of the universe. It is the
study
of
> mathematical objects.''
>Now why didn't you question the notion of mathematical
objects in
>that post? Surely if you didn't know what mathematical
objects
>were, you couldn't comfortably agree that Mathematics is
>the study of them? No ?
> Something doesn't add up here.
Yes, I don't think you're putting much effort into
comprehending what I've written, and I don't mean
to be confrontational but I've no idea how to put
it another way.
> You are the one who keeps claiming that
> you don't understand what it means for something to exist in
> mathematics, and when someone endeavors to explain to you
what the word
> means, you accuse him of not knowing what a mathematical
object is.
> Where did I say that?
If I claimed anything it was that I didn't understand what
Herman's
definition of exist was.
For example, lots of people in recent threads have asked JSH
what he
means by factor... obviously, any one of them can go look up
exactly what factor means. I'm certain most if not all of the
people who asked already know what factor means, but they are
asking because they want to know exactly what JSH means by it.
> Let's go back and compare answers:
> Your answer. An object exists if it is a mathematical object.
> End of story.
I never said this.
Can you try to make a better effort understanding what
I have written, please.
I've not put much effort into describing what exists means,
I've tried just the opposite. I really don't want to define
what exists means, but if someone says N or P(N) doesn't
exist, and 1, 2, 3 does, I'd like to know the distinction.
> My answer. An object exists if its existence can be deduced
from the
> axioms in whatever system we are using. I might add,
sometimes
things
> exist only hypothetically, as in if a measurable cardinal
exists,
> then .... A mathematical object is something that exists in
some
> system.
> I submit that my answer explains what existence means, while
yours does
> not.
You've submitted a definition of exist that seems consistent,
though it doesn't answer my question, because I don't think
this
is what Herman, or his hypothetical student had in mind (of
course I could be mistaken).
>If you've been reading this thread you're aware that my
>questions are directed at Herman, who has raised issue,
>indirectly that perhaps N and P(N) don't exist.
> In some particular system.
No - regardless of the system.
The actual quote (that I responded to) from Herman was:
But, suppose here we have someone who thinks that
infinite sets do not exist at all, let alone P(N)
Herman's student, refuses to accept that N exists.
He didn't say the student doesn't accept P(N) in ZF
only, and accepts P(N) in some other system...
He didn't say that his student rejects the notion
of P(N) in all known theories but reserves the
possibility that he might be wrong, and that
P(N) may exist after all...
His student, as a matter of fact asserts that N,
P(N) (any infinite set) doesn't exist (and no
axioms shall sway him) which implies his usage
of exist is more along the lines of a natural
language statement, which also seems (to me) to
fit the constructive mindset that has been
expressed previously in this thread.
>So, please tell me. When you respond to Herman, and
>talk of things like N and P(N) or any X in the context of
>mathematics that may or may not exist, generically, what
phrase
>would you use to characterize those things?
> I don't think I have responded to Herman.
Since your initial challenge of my use of the
phrase mathematical objects was in text directed
to Herman, one would think you would understand the
context.
I really don't care what we call those things:
(1, 2, 3, N, P(N)), whatever they are. To the
extent that I need to communicate with Herman
I must identify them in some way, so I did.
I wasn't trying to make a philosophical point.
>... Things like N and P(N)
> certainly exist in ZF, but that doesn't mean they have to
exist in other
> systems.
>I thought mathematical objects was a harmless phrase. Yet
>after you objected to its use, seeking to accommodate you
further,
>I chose what I felt was an even more general phrase:
something
>expressible in mathematics but apparently you don't like that
>one either. So you must have some phrase in mind that
>appropriately characterizes objects such objects.
>What phrase do we use, you tell me?
> Does a nontrivial zero of the Riemann zeta function qualify
as
> something expressible in mathematics?
> Does it exist?
I don't know.
This continues in the same spirit as nitpicking to death
my use of mathematical objects or things expressible
in mathematics.
1, 2, 3, N, and P(N), are all mathematical objects, and
also expressible in mathematics, I have said no more
than this.
However, I'm trying to accommodate you. Whatever phrase
you feel correctly describes 1, 2, 3, N, P(N) then just
give me that phrase and I'll use it. Then: problem
solved.
> I think you are
> just obfuscating the issue by substituting different phrases
without ever
> getting at what existence means.
Your definition of exists is something like:
S exists if it's a sentence in some model. Well,
if that's how you want to define exists, great, but...
it fails to capture Herman's student's concept of
exist, so it's of no relevance my original
question.
>Is a prime composite something that can be expressed in
mathematics?
>It's obvious that no such thing exists, but the
specification makes
>perfectly good sense. Your two classes of things that can be
expressed
>in mathematics seems rather uninteresting, considering that
the second
>class is necessarily empty.
>Hard to respond to this. It's incorrect to describe these
classifications
>as if they were mine or some theory of mine, that's false.
The full
>quote was:
>My point was, the word exist was being used to differentiate
between
>two classes of things that can be expressed in mathematics;
I wasn't
>the one making the distinction.
>Since I wasn't making a distinction between things that
>exist and don't exist, they are not my two classes, but
rather
>inferred. I've implicitly made the point that I don't
distinguish
>between the existence or non-existence of 1, 2, N, P(N),
>and so on, they all exist enough (to be imagined) as far as
>I'm concerned.
> Is it your opinion that measurable cardinals exist?
My opinion is: I don't know.
I don't know what messurable means here.
> Does this mean that
> if we have a proof of the form If a measurable cardinal
exists, then
X,
> then we can conclude X unconditionally without the
hypothesis?
>If one is going to say some things do exist and yet others
don't
>exist
>or might not exist then surely I'm entitled to know what
the
>definition
>of exist means and why it's important to me.
> Then why do you keep ignoring the explanation?
Because I was asking Herman for his definition,
not yours. Just as one might ask JSH for his
definition of factor.
Also, I'm not ignoring your definition, I just
find it very ho hum.... very so what?, so
to speak.
>As I said, it's something that follows from the axioms.
>I see your point, but this isn't Herman's opinion, is it?
> I'm not debating Herman.
Wow...
We're we debating? is that what that was?
>Because if it is, and existence follows from axioms
>(such as ZF) then the existence of N and P(N) are
>beyond any questioning. ZF axioms exist, and that's the end
>of the story. I doubt that's what Herman had in mind.
>I know you're a very knowledgeable guy, and I thank you
>for your input, but I posted to this thread to question
>Herman... which isn't to say I don't find your views useful
>or interesting, but my intention was more along the lines
>of testing the consistency of some of my opinions of
>mathematics against views which I am not so familiar
>with, so I'm genuinely curious to know what he thinks.
>However, given that you say that existence is something
>that follows from axioms. In your opinion did natural
>numbers exist prior to axiomatizations of them?
> Maybe the axioms were:
> 1 exists.
> 2 exists.
> 3 exists.
> .
> .
> .
And maybe they weren't?
===
Subject: Re: The basic idea behind my great forthcoming proof
>|It occurred to me recently that I have almost no instinct
for what can
>|be proved without using AC.
>I'm guessing that it's fairly common among research
mathematicians
>not to have a good sense for when they're using AC,
> This is not so clear to me. I had no idea for a _long_ time
that when
> I said that a countable union of countable sets was
countable I was
> using AC - I see this given as a theorem in the introductory
material
> in one of my favorite books on real analysis, without
mention of AC,
> although the author _seems_ to intend to say so when he's
using
> AC because he does so elsewhere; I've had many colleagues
> react with disbelief when I told them this needed AC, etc.
I was quite surprised the other day when a friend pointed out
that the
proof of Caratheodory's extension theorem that we'd been given
in
lectures used AC - specifically in the part where we showed the
would-be outer measure is countably subadditive. I spent a
long time
trying to remove AC from the proof (e.g. by tweaking the
definition of
the outer measure) but then realised there was no point: it is
easy to
see Caratheodory implies that an uncountable totally ordered
set X
cannot be written as a countable union of countable sets, and
that
apparently cannot be proved in ZF (e.g. for X = w_1).
Michael
===
Subject: Re: The basic idea behind my great forthcoming proof
> Dave Seaman says...
>I haven't been debating Herman, but I am not aware of
anything he has
>said that is inconsistent with what I said about existence.
> I think that he rejects ZFC *because* he doesn't believe in
the
> existence of infinite sets.
Yes, that's what I thought, too.
>That contradicts the idea that
> existence means provable existence inside some theory.
It does? I didn't say anything about which came first, the
axioms or the
intuition.
> People choose their axioms because they believe they capture
> some intuitions about mathematical objects. So there is some
> notion of mathematical object that is prior to formalization,
> and helps guide the formalization.
I agree. But the fact remains that Herman chooses his axioms,
just like
everyone else. He certainly is not saying that every
conceivable
mathematical object exists.
> Of course, there is feedback that goes on, as well. You come
> up with axioms to capture your intuitions, and then working
> with the axioms (or working with models of those axioms)
> can cause you to change your intuitions. (which might suggest
> different axioms).
Yes.
--
Dave Seaman
Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling.
<http://www.commoncouragepress.com/index.cfm?action=book&
bookid=228>
===
Subject: Re: The basic idea behind my great forthcoming proof
>I've not yet seen a careful definition that
>establishes exactly what existence means in
>this context (yet some people talk about it
>often), and more importantly why it should
>matter to me. In the most general sense, all
>mathematical objects exist because they
>can be imagined. Right?
>Perhaps so, but then the question becomes, What is a
mathematical
>object?
> Something doesn't add up here.
> Yes, I don't think you're putting much effort into
> comprehending what I've written, and I don't mean
> to be confrontational but I've no idea how to put
> it another way.
You are saying that all mathematical objects exist because
they can be
imagined. I disagree. I can imagine a prime composite, even
though it
is obvious from the definitions that no such thing can exist.
Is it a
mathematical object?
> You are the one who keeps claiming that
> you don't understand what it means for something to exist in
> mathematics, and when someone endeavors to explain to you
what the word
> means, you accuse him of not knowing what a mathematical
object is.
> Where did I say that?
> If I claimed anything it was that I didn't understand what
Herman's
> definition of exist was.
> Let's go back and compare answers:
> Your answer. An object exists if it is a mathematical
object.
> End of story.
> I never said this.
You said all mathematical objects exist because they can be
imagined.
> Can you try to make a better effort understanding what
> I have written, please.
If you insist that saying all mathematical objects exist
because they
can be imagined does not mean an object exists if it is a
mathematical
object, then I don't see what difference it makes how much
effort I
put into trying to understand what you say.
>If you've been reading this thread you're aware that my
>questions are directed at Herman, who has raised issue,
>indirectly that perhaps N and P(N) don't exist.
> In some particular system.
> No - regardless of the system.
> The actual quote (that I responded to) from Herman was:
> But, suppose here we have someone who thinks that
> infinite sets do not exist at all, let alone P(N)
> Herman's student, refuses to accept that N exists.
> He didn't say the student doesn't accept P(N) in ZF
> only, and accepts P(N) in some other system...
Of course he didn't say that. Neither would I. It would be
logically
inconsistent to say that, because P(N) provably exists in ZF.
> He didn't say that his student rejects the notion
> of P(N) in all known theories but reserves the
> possibility that he might be wrong, and that
> P(N) may exist after all...
But it's implicit from this that he rejects all theories in
which the
existence of P(N) is a theorem (including ZF).
> His student, as a matter of fact asserts that N,
> P(N) (any infinite set) doesn't exist (and no
> axioms shall sway him) which implies his usage
> of exist is more along the lines of a natural
> language statement, which also seems (to me) to
> fit the constructive mindset that has been
> expressed previously in this thread.
Yes, he chooses his axioms to fit his intuition, but that
doesn't
invalidate what I said. For him, as for everyone else, an
object exists
if it can be deduced from the axioms. He simply chooses not to
consider
axioms that contradict his intuition.
> Since your initial challenge of my use of the
> phrase mathematical objects was in text directed
> to Herman, one would think you would understand the
> context.
What made you think I didn't understand the context?
> However, I'm trying to accommodate you. Whatever phrase
> you feel correctly describes 1, 2, 3, N, P(N) then just
> give me that phrase and I'll use it. Then: problem
> solved.
My answer has not changed.
> I think you are
> just obfuscating the issue by substituting different
phrases without
ever
> getting at what existence means.
> Your definition of exists is something like:
> S exists if it's a sentence in some model.
Not just some model, but the particular axiom system that one
chooses
to study. The choice may be guided to some extent by
intuition. Your
proposed alternative, that every conceivable mathematical
object
exists, simply is not workable, and is contrary to the way the
word is
actually used.
>Well,
> if that's how you want to define exists, great, but...
> it fails to capture Herman's student's concept of
> exist, so it's of no relevance my original
> question.
I disagree. Even Herman has axioms.
These are the truths that we consider to be self-evident...
--
Dave Seaman
Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling.
<http://www.commoncouragepress.com/index.cfm?action=book&
bookid=228>
===
Subject: Re: The basic idea behind my great forthcoming proof
> Well, we do have someone who doesn't believe infinite sets
exist, and
goes
> so far as to say sets don't exist at all! That person is
me.
> I don't believe in the Platonic realm or whatever, but I
do assign
greater
> degrees of realness to mathematical objects. Sets have my
lowest
> realness rating.
> But you do talk and think about them, right? Perhaps only in
disguised
form.
> And when you do, you probably think about them *as if* they
are fixed,
> completed totalities? In that case, you are assuming lots of
properties
> about them....
Not necessarily. It doesn't really matter how you think of
them as
long as the final result (of whatever thought process you go
through)
is a valid proof of whatever assertion you're claiming.
Michael
===
Subject: Re: The basic idea behind my great forthcoming proof
Dave Seaman says...
> Yes, I don't think you're putting much effort into
> comprehending what I've written, and I don't mean
> to be confrontational but I've no idea how to put
> it another way.
>You are saying that all mathematical objects exist because
they can be
>imagined.
No, he's not saying that. He's *asking* what meaning of exists
is being used by Herman.
> His student, as a matter of fact asserts that N,
> P(N) (any infinite set) doesn't exist (and no
> axioms shall sway him) which implies his usage
> of exist is more along the lines of a natural
> language statement, which also seems (to me) to
> fit the constructive mindset that has been
> expressed previously in this thread.
>Yes, he chooses his axioms to fit his intuition, but that
doesn't
>invalidate what I said.
It certainly does. If I say
The power set of N does not exist.
Therefore, I reject ZFC.
The first line is a notion of existence that *precedes*
axiomatization.
>For him, as for everyone else, an object exists
>if it can be deduced from the axioms.
No, that's *not* what Herman's student means.
>He simply chooses not to consider
>axioms that contradict his intuition.
Which means that existence doesn't follow from the axioms,
it precedes axioms.
> Since your initial challenge of my use of the
> phrase mathematical objects was in text directed
> to Herman, one would think you would understand the
> context.
>What made you think I didn't understand the context?
Well, it's hard to understand why you would be saying
what you're saying if you understood. You seem to be
arguing against fpluser's notion of existence when
fpluser is not trying to put forth a notion of existence. He's
trying to find out what Herman's notion is.
--
Daryl McCullough
Ithaca, NY
===
Subject: Re: The basic idea behind my great forthcoming proof
Dave Seaman says...
> Dave Seaman says...
>I haven't been debating Herman, but I am not aware of
anything he has
>said that is inconsistent with what I said about existence.
> I think that he rejects ZFC *because* he doesn't believe in
the
> existence of infinite sets.
>Yes, that's what I thought, too.
>That contradicts the idea that
> existence means provable existence inside some theory.
>It does? I didn't say anything about which came first, the
axioms or the
>intuition.
Well, if you have not yet come up with any axioms, and you
already have a belief in the existence (or nonexistence) of
certain mathematical objects, then how can it possibly be
that existence means provable existence from axioms?
--
Daryl McCullough
Ithaca, NY
===
Subject: Re: The basic idea behind my great forthcoming proof
> Dave Seaman says...
> Yes, I don't think you're putting much effort into
> comprehending what I've written, and I don't mean
> to be confrontational but I've no idea how to put
> it another way.
>You are saying that all mathematical objects exist because
they can be
>imagined.
> No, he's not saying that. He's *asking* what meaning of
exists
> is being used by Herman.
> This has been bothering me lately.
> What difference does the existence of P(N) make?
> For that matter what difference does the existence
> of 7 make?
> I've not yet seen a careful definition that
> establishes exactly what existence means in
> this context (yet some people talk about it
> often), and more importantly why it should
> matter to me. In the most general sense, all
> mathematical objects exist because they
> can be imagined. Right?
***************************************************
Seems to me he is putting forth his own notion of existence
here. He is
claiming that mathematical objects exist because they can be
imagined.
That's what I was responding to.
> His student, as a matter of fact asserts that N,
> P(N) (any infinite set) doesn't exist (and no
> axioms shall sway him) which implies his usage
> of exist is more along the lines of a natural
> language statement, which also seems (to me) to
> fit the constructive mindset that has been
> expressed previously in this thread.
>Yes, he chooses his axioms to fit his intuition, but that
doesn't
>invalidate what I said.
> It certainly does. If I say
> The power set of N does not exist.
> Therefore, I reject ZFC.
> The first line is a notion of existence that *precedes*
> axiomatization.
I have already explained that I am not arguing that point. I
didn't say
which came first. The question that fpluser asked was, what
difference
does it make whether something exists or not? The difference
lies in
what can be deduced from that existence.
>For him, as for everyone else, an object exists
>if it can be deduced from the axioms.
> No, that's *not* what Herman's student means.
I didn't say anything about a causal relationship there. It's
a material
implication.
>He simply chooses not to consider
>axioms that contradict his intuition.
> Which means that existence doesn't follow from the axioms,
> it precedes axioms.
Which leaves the question, what difference does intuition make
in a
mathematical discussion? In order for intuition to have any
mathematical
relevance, it must be included in some logical system. In
other words,
in order for mathematical existence to make a difference to
fpluser or to
anyone else, we need axioms.
> Since your initial challenge of my use of the
> phrase mathematical objects was in text directed
> to Herman, one would think you would understand the
> context.
>What made you think I didn't understand the context?
> Well, it's hard to understand why you would be saying
> what you're saying if you understood. You seem to be
> arguing against fpluser's notion of existence when
> fpluser is not trying to put forth a notion of existence.
He's
> trying to find out what Herman's notion is.
I have quoted the exact question. That's what I was responding
to.
--
Dave Seaman
Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling.
<http://www.commoncouragepress.com/index.cfm?action=book&
bookid=228>
===
Subject: Re: The basic idea behind my great forthcoming proof
> Dave Seaman says...
> Dave Seaman says...
>I haven't been debating Herman, but I am not aware of
anything he has
>said that is inconsistent with what I said about existence.
> I think that he rejects ZFC *because* he doesn't believe
in the
> existence of infinite sets.
>Yes, that's what I thought, too.
>That contradicts the idea that
> existence means provable existence inside some theory.
>It does? I didn't say anything about which came first, the
axioms or the
>intuition.
> Well, if you have not yet come up with any axioms, and you
> already have a belief in the existence (or nonexistence) of
> certain mathematical objects, then how can it possibly be
> that existence means provable existence from axioms?
What is your answer to fpluser's question concerning what
difference
existence makes, if you are not allowed to consider axioms or
deduction? This is a math group, right?
--
Dave Seaman
Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling.
<http://www.commoncouragepress.com/index.cfm?action=book&
bookid=228>
===
Subject: Re: The basic idea behind my great forthcoming proof
> Dave Seaman says...
> Yes, I don't think you're putting much effort into
> comprehending what I've written, and I don't mean
> to be confrontational but I've no idea how to put
> it another way.
>You are saying that all mathematical objects exist because
they can be
>imagined.
> No, he's not saying that. He's *asking* what meaning of
exists
> is being used by Herman.
Wow, amazing Dave, NO CONTEXT left in from Herman. Why am I not
surprised? That was a part of my initial post, wasn't it? So
why did you
remove it?
Let's add that context:
Jurjus
> |And, be fair: can anyone become a mathematician, nowadays,
> |without being turned into a ZFC-talker?
> As often as people on usenet refer to ZFC, I think the
typical
> mathematician knows very little about it, nor does it make
> much difference to them.
> Keith Ramsay
> You're right.
> But, suppose here we have someone who thinks that infinite
sets do <<<
> not exist at all, let alone P(N). Would such a person be
likely to <<<
> graduate, or even to start studying math and persevere? <<<
> Still, such people may still be good mathematicians. <<<
> Herman Jurjus
This has been bothering me lately.
What difference does the existence of P(N) make?
For that matter what difference does the existence
of 7 make?
I've not yet seen a careful definition that
establishes exactly what existence means in
this context (yet some people talk about it
often), and more importantly why it should
matter to me. In the most general sense, all
mathematical objects exist because they
can be imagined. Right?
Does 10^10^10^10 exist ? In what sense,
and how is its existence distinct from N, or
P(N)?
===
Subject: Re: The basic idea behind my great forthcoming proof
>I've not yet seen a careful definition that
>establishes exactly what existence means in
>this context (yet some people talk about it
>often), and more importantly why it should
>matter to me. In the most general sense, all
>mathematical objects exist because they
>can be imagined. Right?
>Perhaps so, but then the question becomes, What is a
mathematical
>object?
> Something doesn't add up here.
> Yes, I don't think you're putting much effort into
> comprehending what I've written, and I don't mean
> to be confrontational but I've no idea how to put
> it another way.
> You are saying that all mathematical objects exist because
they can be
> imagined.
I didn't say that.
In fact, I'm near certain the actual quote from which you are
drawing
doesn't use the phrase mathematical objects, certainly not all
and
instead explicitly lists, 1, 2, 3, N, P(N). I did use the
phrase
and so on to indicate there were perhaps other objects like
the ones
I listed which are similar (4, 5, 6..., P(P(N)),
P(P(P(N)))...) which
I regard in the same way but that is by no means *all*.
> I disagree. I can imagine a prime composite, even though it
> is obvious from the definitions that no such thing can
exist. Is it a
> mathematical object?
I can't imagine a prime composite. I can certainly say prime
composite in English, and also I can type the phrase but I
can't
imagine the object itself. Sorry.
I would not think a prime composite is a mathematical object.
When
Robin Chapman said, Mathematics is the study of mathematical
objects, I don't think he had in mind that prime composites
were such
objects (I could be wrong but, this is my read). Although, I
think he
felt that 1, 2, 3, N, and P(N) were such objects.... Why are
prime
composites of any interest here? This is your invention, I
don't care
what you call those things.
This infatuation of yours with these concepts doesn't seem in
any way
to address my original question to Herman, so I regard them as
diversionary.
If I discuss these things enough with you, sure it will
eventually bloom
into it's own topic, and it has already to some extent but
it's not my
interest or original intent.
> You are the one who keeps claiming that
> you don't understand what it means for something to exist
in
> mathematics, and when someone endeavors to explain to you
what the
word
> means, you accuse him of not knowing what a mathematical
object
is.
> Where did I say that?
> If I claimed anything it was that I didn't understand what
Herman's
> definition of exist was.
> Let's go back and compare answers:
> Your answer. An object exists if it is a mathematical
object.
> End of story.
> I never said this.
> You said all mathematical objects exist because they can be
imagined.
Nope.
> Can you try to make a better effort understanding what
> I have written, please.
> If you insist that saying all mathematical objects exist
because they
> can be imagined does not mean an object exists if it is a
mathematical
> object, then I don't see what difference it makes how much
effort I
> put into trying to understand what you say.
Am I insisting this?
Apparently you don't need me to carry out this debate.
>If you've been reading this thread you're aware that my
>questions are directed at Herman, who has raised issue,
>indirectly that perhaps N and P(N) don't exist.
> In some particular system.
> No - regardless of the system.
> The actual quote (that I responded to) from Herman was:
> But, suppose here we have someone who thinks that
> infinite sets do not exist at all, let alone P(N)
> Herman's student, refuses to accept that N exists.
> He didn't say the student doesn't accept P(N) in ZF
> only, and accepts P(N) in some other system...
> Of course he didn't say that. Neither would I. It would be
logically
> inconsistent to say that, because P(N) provably exists in ZF.
You said that his denial of N and P(N) was relative to a
particular
system. [It's in the context above] I showed, after citing
Herman's
original text, that Herman's student refuses to believe in any
infinity,
his non belief of P(N) isn't relative to any system. It's
universal.
> He didn't say that his student rejects the notion
> of P(N) in all known theories but reserves the
> possibility that he might be wrong, and that
> P(N) may exist after all...
> But it's implicit from this that he rejects all theories in
which the
> existence of P(N) is a theorem (including ZF).
Yes, that seems reasonable (obvious even)....
> His student, as a matter of fact asserts that N,
> P(N) (any infinite set) doesn't exist (and no
> axioms shall sway him) which implies his usage
> of exist is more along the lines of a natural
> language statement, which also seems (to me) to
> fit the constructive mindset that has been
> expressed previously in this thread.
> Yes, he chooses his axioms to fit his intuition, but that
doesn't
> invalidate what I said. For him, as for everyone else, an
object exists
> if it can be deduced from the axioms. He simply chooses not
to consider
> axioms that contradict his intuition.
Ok, but Herman's student doesn't think N, P(N) exists,
so obviously he's not using your definition, which means
your definition of exists, while beautiful, thought provoking,
noble, and all those other wonderful things, fails to address
my question.
> Since your initial challenge of my use of the
> phrase mathematical objects was in text directed
> to Herman, one would think you would understand the
> context.
> What made you think I didn't understand the context?
Your failure to articulate, and reflect my viewpoints. Repeated
statements that don't characterize my opinions, but most
importantly the nagging way in which you've shifted the focus
of my sub-thread away from it's intended purpose.
> However, I'm trying to accommodate you. Whatever phrase
> you feel correctly describes 1, 2, 3, N, P(N) then just
> give me that phrase and I'll use it. Then: problem
> solved.
> My answer has not changed.
Which was what?
a. <No answer>?
Or
b. I'm not debating with Herman ?
Both are unacceptable.
> I think you are
> just obfuscating the issue by substituting different
phrases without
ever
> getting at what existence means.
> Your definition of exists is something like:
> S exists if it's a sentence in some model.
> Not just some model, but the particular axiom system that
one chooses
> to study. The choice may be guided to some extent by
intuition. Your
> proposed alternative,
You must have me confused with someone else, I don't have a
proposed
alternative.
> that every conceivable mathematical object
> exists, simply is not workable, and is contrary to the way
the word
is
> actually used.
Did I say that? surely not.
>Well,
> if that's how you want to define exists, great, but...
> it fails to capture Herman's student's concept of
> exist, so it's of no relevance my original
> question.
> I disagree. Even Herman has axioms.
Oh, and I said Herman didn't have axioms?! Jeez...
Herman doesn't have axioms! Herman doesn't have axioms!
> These are the truths that we consider to be self-evident...
> --
> Dave Seaman
> Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling.
>
<http://www.commoncouragepress.com/index.cfm?action=book&
bookid=228>
===
Subject: Re: The basic idea behind my great forthcoming proof
> Dave Seaman says...
> Yes, I don't think you're putting much effort into
> comprehending what I've written, and I don't mean
> to be confrontational but I've no idea how to put
> it another way.
>You are saying that all mathematical objects exist because
they can be
>imagined.
> No, he's not saying that. He's *asking* what meaning of
exists
> is being used by Herman.
> This has been bothering me lately.
> What difference does the existence of P(N) make?
> For that matter what difference does the existence
> of 7 make?
> I've not yet seen a careful definition that
> establishes exactly what existence means in
> this context (yet some people talk about it
> often), and more importantly why it should
> matter to me. In the most general sense, all
> mathematical objects exist because they
> can be imagined. Right?
> ***************************************************
> Seems to me he is putting forth his own notion of existence
here. He is
> claiming that mathematical objects exist because they can be
imagined.
> That's what I was responding to.
This lacks the original context which does help if you're
honestly
interested to know why I asked the question.
I don't think my most general notion of exists is as
controversial
as you make it out to be. Certainly, things which have no
physical
presence, can't be said to exist if they can't be thought of.
If one can not think of something, and one can not physically
contact
(through any sense) something then, it's beyond consciousness,
I don't
think such a thing exists, if there is such a thing, certainly
no counter
examples can be provided.
After it was obvious you didn't like my use of mathematical
objects,
I dropped it and specifically focused on things expressed by
mathematics then 1, 2, 3, N, P(N) so one would think, since I
altered my position to avoid your objections, you would
recognize
that perhaps I wasn't postulating an existence scheme, that it
wasn't of great importance for me to characterize it, whatever
it
was. Take it as a retraction if you will, I was never
interested in
defining some rigorous concept of exists, but rather just
establishing
an absolute lower bound for conditions of existence.
Whatever I think of 1, 2, 3, N, P(N) is irrelevant, and
whatever
I think of mathematical objects also is irrelevant... because
that
never was an issue. What was an issue was: What does Herman's
student mean by mathematical object? I still want to know, but
I don't think I will.... Oh well.
> His student, as a matter of fact asserts that N,
> P(N) (any infinite set) doesn't exist (and no
> axioms shall sway him) which implies his usage
> of exist is more along the lines of a natural
> language statement, which also seems (to me) to
> fit the constructive mindset that has been
> expressed previously in this thread.
>Yes, he chooses his axioms to fit his intuition, but that
doesn't
>invalidate what I said.
> It certainly does. If I say
> The power set of N does not exist.
> Therefore, I reject ZFC.
> The first line is a notion of existence that *precedes*
> axiomatization.
> I have already explained that I am not arguing that point. I
didn't say
> which came first. The question that fpluser asked was, what
difference
> does it make whether something exists or not? The difference
lies in
> what can be deduced from that existence.
>For him, as for everyone else, an object exists
>if it can be deduced from the axioms.
> No, that's *not* what Herman's student means.
> I didn't say anything about a causal relationship there.
It's a material
> implication.
>He simply chooses not to consider
>axioms that contradict his intuition.
> Which means that existence doesn't follow from the axioms,
> it precedes axioms.
> Which leaves the question, what difference does intuition
make in a
> mathematical discussion? In order for intuition to have any
mathematical
> relevance, it must be included in some logical system. In
other words,
> in order for mathematical existence to make a difference to
fpluser or to
> anyone else, we need axioms.
> Since your initial challenge of my use of the
> phrase mathematical objects was in text directed
> to Herman, one would think you would understand the
> context.
>What made you think I didn't understand the context?
> Well, it's hard to understand why you would be saying
> what you're saying if you understood. You seem to be
> arguing against fpluser's notion of existence when
> fpluser is not trying to put forth a notion of existence.
He's
> trying to find out what Herman's notion is.
> I have quoted the exact question. That's what I was
responding to.
> --
> Dave Seaman
> Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling.
>
<http://www.commoncouragepress.com/index.cfm?action=book&
bookid=228>
===
Subject: Re: The basic idea behind my great forthcoming proof
Dave Seaman says...
[Quoting fpluser]
> What difference does the existence of P(N) make?
> For that matter what difference does the existence
> of 7 make?
> I've not yet seen a careful definition that
> establishes exactly what existence means in
> this context (yet some people talk about it
> often), and more importantly why it should
> matter to me. In the most general sense, all
> mathematical objects exist because they
> can be imagined. Right?
>***************************************************
>Seems to me he is putting forth his own notion of existence
here.
No, I think that's a misinterpretation on your part. He's
*asking* about the meaning of exists (more specifically,
Herman's meaning) not expounding on his own definition.
Yes, the way that he goes about asking what people mean
by exists is by suggesting a possible meaning, but it
seems clear to me that he doesn't care about that
definition---it's just a straw man for getting the
conversation started.
His subsequent posts to me bear out my interpretation.
--
Daryl McCullough
Ithaca, NY
===
Subject: Re: The basic idea behind my great forthcoming proof
>I've not yet seen a careful definition that
>establishes exactly what existence means in
>this context (yet some people talk about it
>often), and more importantly why it should
>matter to me. In the most general sense, all
>mathematical objects exist because they
>can be imagined. Right?
I suggest you read carefully the last quoted sentence in that
paragraph
above. Do you deny having written that?
> You are saying that all mathematical objects exist because
they can be
> imagined.
> I didn't say that.
Yes, you did. The exact quote is in the last full sentence of
the first
> In fact, I'm near certain the actual quote from which you
are drawing
> doesn't use the phrase mathematical objects, certainly not
all
and
> instead explicitly lists, 1, 2, 3, N, P(N).
The exact quote is in the last full sentence of the first
quoted
> Your answer. An object exists if it is a mathematical
object.
> End of story.
> I never said this.
> You said all mathematical objects exist because they can be
imagined.
> Nope.
The exact quote is in the last full sentence of the first
quoted
> Your definition of exists is something like:
> S exists if it's a sentence in some model.
> Not just some model, but the particular axiom system that
one
chooses
> to study. The choice may be guided to some extent by
intuition. Your
> proposed alternative,
> You must have me confused with someone else, I don't have a
proposed
> alternative.
> that every conceivable mathematical object
> exists, simply is not workable, and is contrary to the way
the word
is
> actually used.
> Did I say that? surely not.
The exact quote is in the last full sentence of the first
quoted
--
Dave Seaman
Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling.
<http://www.commoncouragepress.com/index.cfm?action=book&
bookid=228>
===
Subject: Re: The basic idea behind my great forthcoming proof
> Dave Seaman says...
> Yes, I don't think you're putting much effort into
> comprehending what I've written, and I don't mean
> to be confrontational but I've no idea how to put
> it another way.
>You are saying that all mathematical objects exist
because they can be
>imagined.
> No, he's not saying that. He's *asking* what meaning of
exists
> is being used by Herman.
> This has been bothering me lately.
> What difference does the existence of P(N) make?
> For that matter what difference does the existence
> of 7 make?
> I've not yet seen a careful definition that
> establishes exactly what existence means in
> this context (yet some people talk about it
> often), and more importantly why it should
> matter to me. In the most general sense, all
> mathematical objects exist because they
> can be imagined. Right?
> ***************************************************
> Seems to me he is putting forth his own notion of existence
here. He is
> claiming that mathematical objects exist because they can
be imagined.
> That's what I was responding to.
> This lacks the original context which does help if you're
honestly
> interested to know why I asked the question.
I know the context. The question was about what you said, and
that's why
I quoted you.
> I don't think my most general notion of exists is as
controversial
> as you make it out to be. Certainly, things which have no
physical
> presence, can't be said to exist if they can't be thought of.
Who mentioned things that can't be thought of? Are you
claiming that
things that don't exist are necessarily unthinkable? That
sounds like
the contrapositive of what you said in the first place, that
everything
that can be imagined necessarily exists. That is your
position, not
mine.
> If one can not think of something, and one can not
physically contact
You are off in a fairy land, talking about things that can't
be imagined.
Counterexamples to FLT can certainly be imagined, but we now
know that
they do not exist. Existence and imagination are not the same
thing, no
matter how many times you claim otherwise.
--
Dave Seaman
Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling.
<http://www.commoncouragepress.com/index.cfm?action=book&
bookid=228>
===
Subject: Re: The basic idea behind my great forthcoming proof
Dave Seaman says...
>I've not yet seen a careful definition that
>establishes exactly what existence means in
>this context (yet some people talk about it
>often), and more importantly why it should
>matter to me. In the most general sense, all
>mathematical objects exist because they
>can be imagined. Right?
>I suggest you read carefully the last quoted sentence
>in that paragraph above. Do you deny having written that?
Objection! Badgering the witness!
--
Daryl McCullough
===
Subject: Re: The basic idea behind my great forthcoming proof
>I've not yet seen a careful definition that
>establishes exactly what existence means in
>this context (yet some people talk about it
>often), and more importantly why it should
>matter to me. In the most general sense, all
>mathematical objects exist because they
>can be imagined. Right?
> I suggest you read carefully the last quoted sentence in
that paragraph
> above. Do you deny having written that?
No, but I do recall distancing myself from mathematical
objects, don't
you?
I had actually overlooked the possibility that you considered
anything
related
to that phrase after I invested so much energy into renouncing
it early in
our exchange, if that was your original beef, you must be over
it now, no?
> You are saying that all mathematical objects exist
because they can be
> imagined.
> I didn't say that.
> Yes, you did. The exact quote is in the last full sentence
of the first
Yes, I did, and as a result we had this exchange:
You said:
Perhaps so, but then the question becomes, What is a
mathematical
object?
I responded:
I don't think the phrase mathematical object in this context
warrants much investigation, whatever type of objects Herman
meant to be discussing are the kind I meant. I thought
mathematical object was generic enough, perhaps not.
My point was, the word exist was being used to differentiate
between
two classes of things that can be expressed in mathematics; I
wasn't
the one making the distinction.
If one is going to say some things do exist and yet others
don't
exist
or might not exist then surely I'm entitled to know what the
definition
of exist means and why it's important to me.
Obviously I wasn't so hung up on the statement which you
fixated on,
to the extent that I was unable to abandon it. After this
exchange,
I thought a rational person would no longer bother to
investigate it....
Then we had another exchange and I distance myself further from
that statement:
So, please tell me. When you respond to Herman, and
talk of things like N and P(N) or any X in the context of
mathematics that may or may not exist, generically, what phrase
would you use to characterize those things?
I thought mathematical objects was a harmless phrase. Yet
after you objected to its use, seeking to accommodate you
further,
I chose what I felt was an even more general phrase: something
expressible in mathematics but apparently you don't like that
one either. So you must have some phrase in mind that
appropriately characterizes objects such objects.
What phrase do we use, you tell me?
Here, I'm appealing to you to just construct the appropriate
phrase
for me, so you don't have to fixate on the original...
Shouldn't it have been obvious to you, or any rational person,
that
after all these efforts to renounce that term, that my
original claim
was more or less retracted? If you don't think so - I'll
retract it now.
Done.
At least one (and I suspect most) Other(s) reading this thread
recognized my lack of interest to publish a strong concept of
what exists means. That it is not an objective of mine, and
requests for me to do such a thing are an annoyance, yet you
seem incapable of understanding this.
If you want to pester me for my definition of exists, sure...
eventually I'll give one, and one sort of leaked out in the
middle
of our continued discussion, but that doesn't reflect a desire
on my part to come up with some methodical definition of what
it means for mathematical objects to exist. What it reflects is
that, when people put specific questions to me, I make an
attempt to answer them, even if I think they're a bit loaded.
and not relevant to the point at hand (in this case, Herman's
student).
I've made myself clear, haven't I? There should at this point
be
no ambiguity about my unwillingness to publish such a
definition
even if previously you thought that was an objective of mine.
Now, about your definition of exists, even though I didn't
really
want to get into it, at this point I may as well embrace
whatever
argument you're making.
The last thing in the world I wanted to do was to get side
tracked
by your concept of exist, but now it's the main topic.. so...
if we must:
Your definition of exists is insufficient.
For example, I've seen a few mathematicians on sci.math talk
of N, and potential axiomatizations of N in such a way that
suggests
that N exists prior to axiomatization of it, that is to say
axioms are
chosen which best characterize our intuitions about what N is.
Examples follow:
Tim Chow said:
The Peano postulates could mean several things. They could
refer to
Peano's original postulates, which by modern standards are
slightly vague.
They could refer to what modern mathematicians call
first-order Peano
Arithmetic, or they could refer to what they call second-order
Peano
Arithmetic. Neither of these suffices to prove, or even state,
everything
that mathematicians have proved in number theory, and
first-order PA does
not even suffice to *characterize* the natural
numbers---meaning that there
are other things (nonstandard integers) besides the natural
numbers
that
satisfy those axioms.
Notice (in the last 3 lines) that knowledge of how we expect
axioms
chosen to represent N behave is known prior to N's
axiomatization.
After some axiomatization of N, we could infer results, which
make us
say, oops... maybe this axiom set doesn't properly
characterize what
we had in mind (N).
There is some concept of what N *should* be, prior to it's
axiomatization.
Also...
We cannot expect that the natural numbers sequence can be
defined in
terms of anything essentially more primitive than itself, but
we can
elaborate on what our conception of it comprises in terms of
notions
already developed, with the goal of clarifying our reasoning
with it.
Stoll indicates that the natural number sequence is itself a
primative
thing prior to axiomatization of it.
Also...
David Ullrich said:
On the one hand it seems clear that the second-order PA
axioms characterize N. On the other hand it seems clear
that we can do second-order arithmetic in first-order
set theory... (my guess is that the point is that we're talking
about two different notions of second-order arithmetic
here, but when I assume that then I get all confused
over exactly what notion the first sentence refers to.)
If second order PA characterizes N (and implicitly
FOPA doesn't) then doesn't it suggest that we have a
notion of what N should be prior to axiomatizing it?
Also...
Mike Oliver has said
The key thing which Hofstaedter elides about G is that we
know perfectly well whether it should be true or false, and
the answer is true. It just can't be proved within Peano
Arithmetic. It's true that if you add not G to Peano Arithmetic
you get a consistent theory. However that theory does not
correctly describe the natural numbers.
Also...
My last and only question to you about your notion of exists
was:
Did natural numbers exist prior to their axiomatization?
Your response was:
Maybe the axioms were:
1 exists.
2 exists.
3 exists.
.
.
.
Is that your argument? Maybe? That things like numbers can
have implicit axiomatizations? What other kinds of things can
have implicit axiomatizations?
That seems like a slippery slope... are you certain that you
don't want to revise the notion of exists that you keep
insisting upon?
[snip as we shift to new topic]
===
Subject: Re: The basic idea behind my great forthcoming proof
> Dave Seaman says...
> Yes, I don't think you're putting much effort into
> comprehending what I've written, and I don't mean
> to be confrontational but I've no idea how to put
> it another way.
>You are saying that all mathematical objects exist
because they can
be
>imagined.
> No, he's not saying that. He's *asking* what meaning of
exists
> is being used by Herman.
> This has been bothering me lately.
> What difference does the existence of P(N) make?
> For that matter what difference does the existence
> of 7 make?
> I've not yet seen a careful definition that
> establishes exactly what existence means in
> this context (yet some people talk about it
> often), and more importantly why it should
> matter to me. In the most general sense, all
> mathematical objects exist because they
> can be imagined. Right?
> ***************************************************
> Seems to me he is putting forth his own notion of
existence here. He
is
> claiming that mathematical objects exist because they can
be imagined.
> That's what I was responding to.
> This lacks the original context which does help if you're
honestly
> interested to know why I asked the question.
> I know the context. The question was about what you said,
and that's why
> I quoted you.
Yes, but didn't I renounce use of that phrase? So, why 4 or 5
posts later
would you gotcha me with something we already dispensed with?
> I don't think my most general notion of exists is as
controversial
> as you make it out to be. Certainly, things which have no
physical
> presence, can't be said to exist if they can't be thought
of.
> Who mentioned things that can't be thought of?
Me? Did I break the cookie jar again? And here I was just
trying
to answer your question.... oops.
>Are you claiming that
> things that don't exist are necessarily unthinkable?
Nope.
Jeez, just tell me what you want to say and I'll repeat it like
a robot if it makes you happy. Ok ?
It's as though your approach toward me is intentionally
belligerent.
Honestly, I wonder (really I do), if you have me confused with
some other guy who you didn't get along with. It's like I'm
getting
this unending stream of disagreement and contrarian attitude
and I don't know what causes it, where it's coming from, and
most importantly how to shut it off (short of ceasing
communication
with you).
> That sounds like
> the contrapositive of what you said in the first place, that
everything
> that can be imagined necessarily exists. That is your
position, not
> mine.
> If one can not think of something, and one can not
physically contact
> You are off in a fairy land, talking about things that can't
be imagined.
> Counterexamples to FLT can certainly be imagined
I don't think so. I think one could imagine things that seem
like
counter examples of FLT though.
>, but we now know that
> they do not exist. Existence and imagination are not the
same thing, no
> matter how many times you claim otherwise.
And I didn't assert that they were the same thing no matter
how many times you claim otherwise, I described it as a
prerequisite for abstract objects to exist, that they must be
able to be thought of, imagined, conceived, whatever phrase
along those lines that you would like.
Since I'm doubtful that continuing this conversation will be of
any benefit to either of us, I'm bowing out of this thread and
will leave you with the last word.
> --
> Dave Seaman
> Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling.
>
<http://www.commoncouragepress.com/index.cfm?action=book&
bookid=228>
===
Subject: Re: The basic idea behind my great forthcoming proof
> Dave Seaman says...
>I've not yet seen a careful definition that
>establishes exactly what existence means in
>this context (yet some people talk about it
>often), and more importantly why it should
>matter to me. In the most general sense, all
>mathematical objects exist because they
>can be imagined. Right?
>I suggest you read carefully the last quoted sentence
>in that paragraph above. Do you deny having written that?
> Objection! Badgering the witness!
lol, yes I think I needed some comic relief.
===
Subject: Re: The basic idea behind my great forthcoming proof
>I've not yet seen a careful definition that
>establishes exactly what existence means in
>this context (yet some people talk about it
>often), and more importantly why it should
>matter to me. In the most general sense, all
>mathematical objects exist because they
>can be imagined. Right?
> I suggest you read carefully the last quoted sentence in
that paragraph
> above. Do you deny having written that?
> No, but I do recall distancing myself from mathematical
objects,
don't
> you?
> I had actually overlooked the possibility that you
considered anything
> related
> to that phrase after I invested so much energy into
renouncing it early
in
> our exchange, if that was your original beef, you must be
over it now,
no?
I thought the important words in that sentence were exist
because they
can be imagined, not mathematical objects. Especially in light
of
your recent reaffirmation of the contrapositive of that
statement, when
you claimed that things that don't exist are things that can't
be
imagined. Sorry, but your St. Anselm act is not working.
> Shouldn't it have been obvious to you, or any rational
person, that
> after all these efforts to renounce that term, that my
original claim
> was more or less retracted? If you don't think so - I'll
retract it now.
> Done.
Does your retraction also apply to these statements from your
previous
post?
> I don't think my most general notion of exists is as
controversial
> as you make it out to be. Certainly, things which have no
physical
> presence, can't be said to exist if they can't be thought of.
> If one can not think of something, and one can not
physically contact...
You are still arguing, even as recently as a few hours ago, as
if
existence is indistinguishable from being imaginable. It isn't.
--
Dave Seaman
Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling.
<http://www.commoncouragepress.com/index.cfm?action=book&
bookid=228>
===
Subject: Re: The basic idea behind my great forthcoming proof
> Counterexamples to FLT can certainly be imagined
> I don't think so. I think one could imagine things that seem
like
> counter examples of FLT though.
Then maybe how you can explain how Ribet proved his theorem?
I was under the impression that he started by imagining a
counterexample
to FLT, and from that hypothesis he reached the conclusion
that there was
also a counterexample to the Taniyama-Shimura conjecture. Are
you saying
what he did is impossible? Then why did Wiles get so excited
when he
heard the news?
--
Dave Seaman
Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling.
<http://www.commoncouragepress.com/index.cfm?action=book&
bookid=228>
===
Subject: Re: The basic idea behind my great forthcoming proof
> Counterexamples to FLT can certainly be imagined
> I don't think so. I think one could imagine things that
seem like
> counter examples of FLT though.
> Then maybe how you can explain how Ribet proved his theorem?
> I was under the impression that he started by imagining a
counterexample
> to FLT, and from that hypothesis he reached the conclusion
that there was
> also a counterexample to the Taniyama-Shimura conjecture.
Are you saying
> what he did is impossible? Then why did Wiles get so excited
when he
> heard the news?
Well these things are beyond my knowledge. I assumed FLT was
proven.
If counter example means something that proves FLT is false,
and yet
experts
say FLT is true, then I assume no such counter example can
exist, but as I
said
this is beyond me so there is no point in me trying to
answer.. I simply
don't know (one way or the other).
> --
> Dave Seaman
> Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling.
>
<http://www.commoncouragepress.com/index.cfm?action=book&
bookid=228>
===
Subject: Re: The Bird-in-the-Hand Game
|Consider a game with the following rules: You are given an
initial
|score and a time limit. We'll denote the time remaining by t,
and your
|score at that time by x[t]. Your score changes over time as a
random
|walk with a fudge factor that attracts it to zero:
If it weren't for the fact that the score tends upward for
negative values,
the optimal strategy would I believe be to quit immediately.
Given the upward bias, however, the optimal strategy is
instead to
quit when the score is >= some slow-growing function f(t). The
probability of the score exceeding any given bound sometime
during time T approaches 1, just because the score over the
long
run wanders around 0, and has a positive probability of going
up
beyond the given limit starting from 0. So f(t) tends to
+infinity as t
goes to infinity.
But I would assume it grows rather slowly. It sounds hard to
figure
out just what f is, and harder to figure out the expected
payoff
for someone starting out with a given score s at a given time
t, where
s < f(t). But perhaps it can be done.
Keith Ramsay
===
Subject: Re: The Bird-in-the-Hand Game
> Given the upward bias, however, the optimal strategy is
instead to
> quit when the score is >= some slow-growing function f(t).
Yes, I had come to that same conclusion myself. The question
is, what
is f?
> It sounds hard to figure
> out just what f is
I have certainly found it very difficult.
The problem statement looks so natural that I would expect it
to lend
itself to an elegant solution. But that's just a feeling with
no
analytical basis.
===
Subject: Re: The Definition of a Manifold
.
> _Pace_ Wigner, what's unreasonable isn't that correct
mathematics
> correctly understood and applied can be effective in
understanding
> the world, it's that Cargo Cult Mathematics seems to actually
> work sometimes (whereas Cargo Cult Science hardly ever does).
> Lee Rudolph
You were thinking of string theory maybe?
===
Subject: Re: The paternity of Christ
>So are you saying that the Y-chromosome that God made for
Jesus would
>somehow be different than the one he made for Adam?
> Why would you ask such a silly question? There isn't nearly
enough
> information in scripture to draw that sort of conclusion.
Odds favor
> the conjecture that they were different, of course, unless
you propose
> that Jesus and Adam were identical twins.
Never mind all that. Do you propose that Adam existed?
===
Subject: Re: The paternity of Christ
>So are you saying that the Y-chromosome that God made for
Jesus would
>somehow be different than the one he made for Adam?
> Why would you ask such a silly question? There isn't
nearly enough
> information in scripture to draw that sort of conclusion.
Odds favor
> the conjecture that they were different, of course, unless
you propose
> that Jesus and Adam were identical twins.
> Never mind all that. Do you propose that Adam existed?
Well, assuming it would make the thread even funnier. ;+)
Some claim that Jesus = God. Which means Mother of Jesus =
Mother of God.
So, Jesus' Y chomosome would be God's. However, AFAICT, Nobody
says Adam
had God's Chromosomes. God likely made Adam's genes like he
made
everything else's -- complete with a billion-year history.
Jerry
===
Subject: Re: The paternity of Christ (was: something
disrespectful if not
blasphemous)
> So are you saying that the Y-chromosome that God made for
Jesus would
> somehow be different than the one he made for Adam?
> Why would you ask such a silly question? There isn't nearly
enough
> information in scripture to draw that sort of conclusion.
Odds favor
> the conjecture that they were different, of course, unless
you propose
> that Jesus and Adam were identical twins.
They wouldn't need to be identical twins to have identical Y
chromosomes.
OTOH, this is all moot, as we know that Adam and Jesus are
*not* identical
twins, but rather two of the many identities of the same
person. He also
goes by the name Herc in this ng. He is constantly being
tortured by
the
government and Big Media, he's never had a thought in his head
(they're all
stolen from him), and a restraining order (or something like
that) is
keeping him away from his Eve. If you don't believe this, call
someone in
Villagopolis, New Zealand (or something like that) and ask.
===
Subject: Re: The paternity of Christ (was: something
disrespectful if not
blasphemous)
>The virgin birth is indeed treated as a special sort of
>sonship in scripture.
>Len.
Yes, of course. The son of man.
duke, American-American
*****
The existance of God makes perfectly good
sense based on the evidence at hand.
*****
===
Subject: Re: The paternity of Christ is OT in sci.math
> The subject says it.
I'm sure the good folks in alt.politics.democrats and
alt.business thank
you
for your helpful post.
===
Subject: Re: The paternity of Christ is OT in sci.math
> The subject says it.
How? Isn't genetics science and math? ;=)
> I'm sure the good folks in alt.politics.democrats and
alt.business thank
you
> for your helpful post.
As for alt.politics.democrats and alt.business, point taken;
those groups
demoved.
===
Subject: Re: The two envelope paradox
> The paradox is no paradox at all, simply an error in
calculating
> probabilities introduced by treating the values as
independent
> choices.
> Well, I think you're right, but I don't know if you explaned
it as
> well as you could. But trying to explain it better I too am
left with
> nothing much better than what you came up with.
Wow, this problem is really tough. It's easy to see that there
is no reason
to switch envelopes, but what is hard is to directly refute
the argument
that you should switch envelopes. I've got these two arguments
in my head,
the valid one and the invalid one, and while I am convinced of
one of them
(the valid one), I have a real tough time directly critiquing
the other one
(the invalid but very persuasive one), except by pointing out
that it
contradicts the valid argument and is therefore wrong (for
some reason).
I think what it comes down to is that you aren't given prior
probabilities
for all the relevant events. There are four events of interest
here, and
their intersection is what characterizes the puzzle:
A) One of the envelopes contains $10 and the other contains
$20.
B) One of the envelopes contains $20 and the other contains
$40.
C) You pick the envelope with the lower amount.
D) You pick the envelope with the higher amount.
It is implicit that the prior probabilities P(C) and P(D) are
1/2 each. So
you are in effect given those. What you are not given is P(A)
and P(B). If
you were given those then there would be no puzzle and the
calculation of
expected value would be straightforward (assuming A&B are
independent of
C&D).
Another observation is that the puzzle cannot be modelled in
terms of
frequencies. If we knew P(A) and P(B), then we could model the
whole thing,
e.g., we can derive P(A&C), P(A&D), etc. We can represent this
graphically,
with an a region for each possibility whose area corresponds
to the
probability of the possibility. Then the conditional
probability P(A|[A&D]
^ [B&C]) can be seen without any problem in the graphical
representation.
But this whole approach is undoable if we don't know P(A) and
P(B). We
simply do not know P(A&C), etc. In fact if we plug in
different values for
P(A) and P(B) then we get different answers for P(A|[A&D] ^
[B&C]).
But even with all that, the reasoning sticks, it's not
directly addressed.
The weighted sum approach seems to very appropriate to this
problem.
There's 1/2 chance I picked the lower envelope, therefore
there's 1/2
chance the other envelope has $40, and similarly for $10,
therefore the
weighted sum approach must work. There has to be a direct way
of addressing
this argument that isn't abstruse, and I don't see it. I'm
sure it's
sitting right in front of me in this thread, but I am not
seeing it.
> It's not an error in calculating probabilities, it's only an
error in
> calculating expected gain (as you first said).
> The probability that you will get a higher value if you
switch is
> still 50% and the probability of getting a lower value is
still 50% if
> you switch.
> The error is in the expected value formula. Calculating
expected
> value for switching as:
> .5 * x/2 + .5 * 2x
> Is only valid if the probability is independent of x (the
amount of
> money you find in the first envelope). But it's not
independent, so
> the formula is invalid. When you pick the high value, you
don't have
> an equal probability of getting twice that. And when you
pick the low
> value, you don't have an equal probablity of getting half
that.
> Because we don't know if the value is high or low when we
see the
> amount of money in the first envelope, it seems like a valid
> assumption to make, but it's not.
> This is a lot like the 3 door game-show problem but in the
opposit
> direction.
Actually I find this problem comparatively easy. It's a really
nice
problem, though.
===
Subject: Re: The two envelope paradox
>
be
> very grateful.
> The two-envelope paradox', Analysis, 55 (1995), pp. 6-11.
> Emanuel Rutten
>Two envelopes
>I looked this up on the web. Didn't find exactly the
requested paper,
but
>dozens of write-ups. Amusing little puzzle, though I'm not
sure why it
>seems to have attracted so much interest. Maybe I'm missing
something,
but
>here's my off-the-cuff analysis.
>You are given two envelopes, and told that each contains
money.
>You are also told that the amount in one envelope is twice
that in the
>other.
>You open one envelope, discover that it contained $20
>You are offered the chance to switch for the other unopened
envelope.
>You reason that the other may contain $10 or $40 with equal
probability,
so
>that your average return for switching is $25... better
than the $20 you
>have, so you switch.
> The standard mathematical rebuttal of this paradox is to say
that there
is
> no uniform probability distribution over the integers, and
so you cannot
> assume that the envelope contains any integral number of
dollars with
equal
> probability. In other words, the assumption that is contains
$10 and $40
> with equal probabilities is false.
> This does not really resolve the paradox, however, because
of the
following
> more sophisticated version.
> For n >= 1, with probability 1/2^n, I place 3^(n-1) and 3^n
dollars in
> the two envelopes, and you know that I have followed this
rule. Now you
> play the game as before.
> You open one envelope and observe see x = 3^m dollars. Of
course, if x=1,
then
> you definitely swap.
> Otherwise, you reason, that the n that I used must have been
m-1 or m.
> Since it is twice as likely to have been m-1 as m, the other
envelope
> contains x/3 with probability 2/3, and 3x with probability
1/3.
> So your expected gain from swapping is 2x/9 + x = 11x/9, and
you should
> always swap.
> Derek Holt.
Having looked a bit farther into other discussions of this
problem
available
on the web, I'm properly awed by the depths of analysis to
which it has
been
submitted, and also quite aware that my naive thoughts on the
matter are
unlikely to be of interest to a real mathematician.
However, I was amused to find that a straightforward numerical
examination
of this alternative formulation did reveal a bit of a surprise.
Obviously it cannot make a difference to swap envelopes
without looking at
the content of one of them, no matter how seductive the
reasoning. Making
a choice and changing your mind with no additional information
to go on
can't matter: the envelopes don't know that you might have
chosen another
one first.
Similarly, looking at one and then unconditionally switching
every time
cannot be advantageous. If the information from opening the
envelope is
ignored, this is the same as not opening it. If you compare
the total
gain
from a strategy of staying with your first choice to a
strategy of always
switching, one expects them to be the same. And indeed the
totals are
identical if you examine a series in which all observed
relative
frequencies
are assumed to exactly match the probabilities defining the
distribution
from which they are drawn. This is fact. The problem is to
reconcile this
indisputable equality of actual overall expectation with the
seemingly
inarguable calculation of a better expected gain for switching
than for
sticking with ones first choice.
The surprise I mentioned was that the calculation of
expectation does prove
to be correct in almost every case, so that choosing to switch
does in fact
almost always yield a greater return that choosing to stick
with ones
original choice.
For example, if you consider a long series in which you happen
to observe a
3 in the first envelope 90 times, your return is 270 if you
stick with that
choice, but if you always switch you will get 60*1 + 30*9, for
a total
return of 330, just as the expected gain calculation would
predict.
This continues for almost every possible initial observation,
except those
involving an initial selection of the largest number actually
included in
the sample under examination. In this one case the strategy of
always
switching ensures that you will switch to a lower valued
envelope, while
sticking with the first choice allows you to keep this highest
value.
And it turns out that this difference in this one boundary
case (it can be
only one case if the observed frequencies are to exactly match
the
theoretical distribution) is precisely enough to balance out
all the
consistent differences in all the other cases.
In other words, despite the accurate projection of a higher
expected gain
for switching in all but the boundary cases, when boundary
cases are
included the overall expectation from either fixed strategy is
the same:
always stay with your first choice, or always switch, in
either case your
total expectation is identical.
I've ignored the other boundary at 1 until now. Of course it
is rational
to switch when you observe a minimal bounding value. So a
strategy of
switching when you see a 1 and staying with your initial
choice in all
other
cases is best if we consider the high end boundary to be
completely
unknowable.
The seeming paradox is revealed as a problem in calculating
expectation
without taking into account the effects of a finite bound on
the values to
be seen in any finite series of trials. A properly calculated
expectation
is not paradoxical.
I must respectfully disagree with the seeming consensus of the
mathematicians whose analyses I scanned. The seeming paradox
has little to
do with the specific distributions assumed for the source of
the pairs. It
is resolved in all cases by inclusion of the boundary cases in
the
calculation of expected gain over any finite series.
I haven't thought it through completely, but it would seem
that a winning
strategy would be to switch in every case in which you observe
a value less
than the maximum value you have previously observed, otherwise
to stick
with
your first choice.
Of course if you are told a maximum bound ahead of time there
is no
problem.
Always switch unless you see that value.
Bill Modlin
===
Subject: Re: Three-signed arithmetic : T space
There is another question about this mathematical system:
I introduced the terms MONAL and BINAL elsewhere - from which
TERNAL &c. follow. This was in another context.
BINAL would be a term for REQUIRES TWO numbers, where BINARY
means
HAVE TWO.
Consider a NEGATION FUNCTION in Three-signed arithmetic.
If -, +, and * are mapped onto a piece of paper (for our ease
of
understanding), they produce - as you showed - three
NUMBER-LINES from
one ORIGIN.
Thus, a position on a number-line is MONAL (requires ONE
specification
from the origin).
The NEGATIVE of that position will be in the SPACE between the
OTHER
two signa, and will be BINAL (requires TWO specifications).
As - is not longer the negation operator, we have to use some
jargon
such as NEG.
Perhaps NEG(+5) gives -5,*5
Perhaps NEG(+5) gives -2.5,*2.5
Perhaps NEG(+5) gives -2.5(root3),*2.5(root3)
YOU, Timothy Golden, are making the rules - and I am not
fussing over
the exact rules. Instead, I am showing that there is a ONE
parameter/TWO parameter transmutation going on when negating.
Odd!
Charles Douglas Wehner
===
Subject: Re: Three-signed arithmetic : T space
Hi
Yes. I agree with what you are saying and I would choose your
first
supposition for NEG().
I do not like the word negative as a concept since the
construction
is already sign based and the negative sign has taken on its
own form
of negation.
I would think that the NEG function you are thinking of
returns a
value in Y such that the sum of the original and the result
are zero:
Sum( y, NEG(y) ) = 0.
Or:
y * NEG( y ) = 0.
If y is + 5, then NEG(y) is - 5 * 5, since - 5 + 5 * 5 = 0.
You are definitely getting the concept.
As you say:
> I am showing that there is a ONE
> parameter/TWO parameter transmutation going on when negating.
If you consider summation of the simplest values( like *1 and
+1 )
you will find that they do not condense to one simple value in
T.
The expression * 1 + 1 is not reducible.
However any three termed sum is always reducible:
- 2 + 3 * 4 = + 1 * 2.
+ 4 + 5 - 2 = + 9 - 2.
- 8.12 + 3.10 * 5.14 = - 5.02 * 2.04.
+ 2 + 3 + 6 = + 11.
This then leads to the Y space, which is at most a pair of
three-signed values in T summed.
This has been defined much earlier in this thread.
So your NEG() function also needs Y to work in.
> There is another question about this mathematical system:
> I introduced the terms MONAL and BINAL elsewhere - from which
> TERNAL &c. follow. This was in another context.
> BINAL would be a term for REQUIRES TWO numbers, where BINARY
means
> HAVE TWO.
> Consider a NEGATION FUNCTION in Three-signed arithmetic.
> If -, +, and * are mapped onto a piece of paper (for our
ease of
> understanding), they produce - as you showed - three
NUMBER-LINES from
> one ORIGIN.
> Thus, a position on a number-line is MONAL (requires ONE
specification
> from the origin).
> The NEGATIVE of that position will be in the SPACE between
the OTHER
> two signa, and will be BINAL (requires TWO specifications).
> As - is not longer the negation operator, we have to use
some jargon
> such as NEG.
> Perhaps NEG(+5) gives -5,*5
> Perhaps NEG(+5) gives -2.5,*2.5
> Perhaps NEG(+5) gives -2.5(root3),*2.5(root3)
> YOU, Timothy Golden, are making the rules - and I am not
fussing over
> the exact rules. Instead, I am showing that there is a ONE
> parameter/TWO parameter transmutation going on when negating.
> Odd!
> Charles Douglas Wehner
===
Subject: to keep you warm on those long winter nights
_.---._
./ _.-._`-._
/ .' _`. `-.
| | _ //' /|
: | //'`-. / |
( ` .' ;/ /
/ | /_.
` `. ' .' ./ `-._.-.
| .-'' / .-' .---. .--.
/ |` :_ /_/ /__...----...' `.___...__
.n
-' . : : _ __ .__.-' ``.' /` .' `.
`````--. .' /
`' | ` | .' `. / / -. .' `.
```-- .' /
/ ' c ; .' / . .' `
``---.-'' /
`-. / '|/ / / / ' ` -' : `.-._
:
`' |/| `-.-..-'__ / / / `. .
``--..-----..__.-._ |
| / ``| /'. / `-. `.
`-'
| / __ / ,' `-.
:F_P:
| / ``--.._________-' `-. .' `- `.
`-. |/ `--._.-' `. `.
`' ( __
`-._
`-'
`--..o
--
===
Subject: Re: to keep you warm on those long winter nights
You clearly have too much idle time on your hands.
Bob Kolker
===
Subject: Re: to keep you warm on those long winter nights
It's summer in this hemishpere.
--
Mensanator
2 of Clubs
http://members.aol.com/mensanator666/2ofclubs/2ofclubs.htm
===
Subject: Re: to keep you warm on those long winter nights
> You clearly have too much idle time on your hands.
> Bob Kolker
LOL!! I was thinking the same thing!
===
Subject: Re: to keep you warm on those long winter nights
> <snip>
Are you ten years old or something?
---
John Hattan Grand High UberPope - First Church of Shatnerology
john@thecodezone.com http://www.shatnerology.com
===
Subject: Re: to keep you warm on those long winter nights
posted in alt.atheism:
> <snipAre you ten years old or something?
Not yet.
--
To assume the existence of an unperceivable being ... does not
facilitate
understanding
the orderliness we find in the perceivable world.
- Letter to an Iowa student who asked, What is God? July,
1953; Einstein
Archive 59-085
(random sig, produced by SigChanger)
rukbat at optonline dot net
===
Subject: Torsion submodule
If R is a noncomutative integral domain and M is a R-module.
How do you
prove that Tor(M) = { m in M | r m = 0 for some r in R} is a
submodule of
M??
The proof is trivial if R is commutative, but I can't see how
Tor(M) is a
submodule when R is noncommuative?! For example if m, n both
belong to
Tor(M) such that rm = 0 and sn = 0 for r, s in R, then what is
the
annihilator of m + n. Trying rs as the annihilator of m + n we
get: (rs)m +
(rs)n = rsm + r(sn) = rsm. Also if m belongs to Tor(M) then
what is the
annihilator of ym for all y belonging to R??
I'm trying to follow a proof in a book regarding properties of
modules over
PIDs. In the book, there is the following argument: Let M be a
finitely
generated R-module wher R is a PID. Also for any r in R let
E(r) = {m in M
|
rm = 0}. Now, suppose w annihilates M and suppose w = bc with
(b, c) = (1)
where w, b, c are in R. Let x, y in R be such that 1 = xb +
yc. Now if m is
in M then, m = xbm + ycm. Then xbm belongs to E(c) since cxbm
= xwm = 0.
How
is this true?? Can cxbm = xwm, if R is noncommuatative?
Also, just to make sure I understand this right if rsm = srm
for r, s in R
and m in M then r and s must commute right? Also rm = mr only
if R is
commuataive right?
I'm very confused : ( Hope someone can clear this up for me.
Bill
===
Subject: Re: Torsion submodule
> If R is a noncomutative integral domain and M is a R-module.
How do you
> prove that Tor(M) = { m in M | r m = 0 for some r in R} is a
submodule of
^^^^^^
nonzero! :-)
> M??
I don't think you do :-(
> The proof is trivial if R is commutative, but I can't see
how Tor(M) is a
> submodule when R is noncommuative?! For example if m, n both
belong to
> Tor(M) such that rm = 0 and sn = 0 for r, s in R, then what
is the
> annihilator of m + n. Trying rs as the annihilator of m + n
we get: (rs)m
> + (rs)n = rsm + r(sn) = rsm. Also if m belongs to Tor(M)
then what is the
> annihilator of ym for all y belonging to R??
Let's try to disprove this result. As a simple case let's look
at the
direct sum of two cyclic modules, say M = R/I (+) R/J where I
and
J are left ideals of R. Then (1,0) and (0,1) are certainly
torsion.
What about (1,1). That is annihilated by r iff r is in I
intersect J.
Can we find a domain R with nonzero left ideals I and J
meeting trivially?
I think we can. Let R = Z<X,Y>, the polynomial ring generated
by noncommuting indeterminates X and Y (that is a domain isn't
it?)
and I and J be the left ideals generated by X and Y.
> I'm trying to follow a proof in a book regarding properties
of modules
> over PIDs.
They are generally assumed to be commutative ....
> In the book, there is the following argument: Let M be a
> finitely generated R-module wher R is a PID. Also for any r
in R let E(r)
> = {m in M | rm = 0}. Now, suppose w annihilates M and
suppose w = bc with
> (b, c) = (1) where w, b, c are in R. Let x, y in R be such
that 1 = xb +
> yc. Now if m is in M then, m = xbm + ycm. Then xbm belongs
to E(c) since
> cxbm = xwm = 0. How is this true?? Can cxbm = xwm, if R is
> noncommuatative?
It could, but it doesn't have to.
> Also, just to make sure I understand this right if rsm = srm
for r, s in
R
> and m in M then r and s must commute right?
That's the only way to guarantee rsm = srm (but in some
instances
rsm = srm might hold even if rs = sr doesn't).
> Also rm = mr only if R is
> commuataive right?
Normally we write the action of a ring on a module on one side
(i.e., we would write rm or mr but not both).
> I'm very confused : ( Hope someone can clear this up for me.
Are you telling us all the properties the author is assuming?
--
Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html
His mind has been corrupted by colours, sounds and shapes.
The League of Gentlemen
===
Subject: Re: Torsion submodule
> I'm trying to follow a proof in a book regarding properties
of modules
over
> PIDs. In the book, there is the following argument: Let M be
a finitely
> generated R-module wher R is a PID. Also for any r in R let
E(r) = {m in M
|
> rm = 0}. Now, suppose w annihilates M and suppose w = bc
with (b, c) =
(1)
> where w, b, c are in R. Let x, y in R be such that 1 = xb +
yc. Now if m
is
> in M then, m = xbm + ycm. Then xbm belongs to E(c) since
cxbm = xwm = 0.
How
> is this true?? Can cxbm = xwm, if R is noncommuatative?
But does one not assume commutativity for a ring to be
principal?
Otherwise you would have to distinguish between left and right
ideals.
> Also, just to make sure I understand this right if rsm = srm
for r, s in
R
> and m in M then r and s must commute right?
No, the equation is equivalent to (rs-sr)m=0. So the condition
is that
(rs-sr) annihilates m, eg. when s and r commute.
> Also rm = mr only if R is
> commuataive right?
This has nothing to do with commutativity of R; the expression
mr
which is element of module times element of ring is not
defined in
general; in the common definition of a module we only have
scalar
multiplication from the left, no matter whether R is
commutative or
not.
===
Subject: Re: Ullrich says probability 1 does not imply
certainty
>...I have to say that your comments in this
> thread have all been very illuminating. Statements
> about things contradicting the axioms when you
> have no idea what axioms you're talking about,
> then statements about applied mathematics that
> indicate you have no idea what's done in applied
> mathematics. Anything else you'd like to tell us
> about?
Hmm; there is. This is not a private seminar nor is it
anyone's classroom. Since you, especially, are not
giving grades here, your judgments are worth about
as much as the ink, used to pen them.
===
Subject: Re: Ullrich says probability 1 does not imply
certainty
>...I have to say that your comments in this
> thread have all been very illuminating. Statements
> about things contradicting the axioms when you
> have no idea what axioms you're talking about,
> then statements about applied mathematics that
> indicate you have no idea what's done in applied
> mathematics. Anything else you'd like to tell us
> about?
>Hmm; there is. This is not a private seminar nor is it
>anyone's classroom. Since you, especially, are not
>giving grades here, your judgments are worth about
>as much as the ink, used to pen them.
things wrong in this thread.
************************
David C. Ullrich
===
Subject: Re: Ullrich says probability 1 does not imply
certainty
>...I have to say that your comments in this
>thread have all been very illuminating. Statements
>about things contradicting the axioms when you
>have no idea what axioms you're talking about,
>then statements about applied mathematics that
>indicate you have no idea what's done in applied
>mathematics. Anything else you'd like to tell us
>about?
> Hmm; there is. This is not a private seminar nor is it
> anyone's classroom. Since you, especially, are not
> giving grades here, your judgments are worth about
> as much as the ink, used to pen them.
So, if I read this correctly, you're saying that every
potentially mathematical comment you've made to date,
in this thread, is to be evaluated by that criterion?
Specifically, you made a comment that, on its surface,
appeared to be a comment regarding your belief regarding
some particular event (infinite sequence from {H,T}):
Such a sequence, violates the axioms of probability.
Others, Ullrich among them, have tried to get you to
explain your reasons for making such a claim, and in
particular, have asked you to give some mention of these
axioms that you claim are violated.
Your response has been to retreat from that aspect
of the discussion. I'm sure you haven't broken any
laws, either of the city/state/nation you reside in,
or of netiquette, per se.
If I'm reading your above comment (This is not a private
seminar ...) correctly, it appears that this retreat was
simply due to the facts that
(1) this is not a private seminar
(2) this is not anyone's classroom
and (3) no one is being graded on sci.math.
In short, since you have no academic stake in the discussion,
you are free to toss out as many nonsensical statements as
you like, and no one should call you to explain your reasoning
for any of them.
Note that I don't intend this as any form of castigation, and
I apologize if that sort of tone comes across: such would be
unintentional. However, civility suggests that people engaged
in a discussion deserve to know who is actually engaged in the
topic, and who is merely using the discussion as a handy source
of amusement.
If you're simply a bored soul hoping to stir up a little
chatter
in the outfield, that's fine. If so, a consistent clue to that
effect would help keep people from taking your comments
seriously.
...............STANDARD DISCLAIMER (short form)...........
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without postage. List was current at time of printing.
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===
Subject: Re: Ullrich says probability 1 does not imply
certainty
<hdn2jvkc6ri4aelkinpictb8aattadja8a@4ax.com>
<mjdYa.87601$3o3.6052020@bgtnsc05-news.ops.worldnet.att.net>
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X-Treme: C&C,DWS
In
<mjdYa.87601$3o3.6052020@bgtnsc05-news.ops.worldnet.att.net>,
on
at 08:13 PM, Nat Silver <mathelp@worldnet.att.net> said:
>There are limits. And don't
>prentend to be a Philistine.
--
Shmuel (Seymour J.) Metz, SysProg and JOAT
Any unsolicited bulk E-mail will be subject to legal action. I
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===
Subject: Re: Ullrich says probability 1 does not imply
certainty
<uo32jvs5ebghqbfv5hmdgh68o7ads1emqk@4ax.com>
<ab9Ya.90015$0v4.6092403@bgtnsc04-news.ops.worldnet.att.net>
<tr82jv8bbfdtqcbss90ouqetf7u32asecu@4ax.com>
<YxaYa.87389$3o3.6038231@bgtnsc05-news.ops.worldnet.att.net>
<c5n2jvgjpmh8g8egdbk52tp6q8mc5gpqmf@4ax.com>
<4hdYa.87598$3o3.6052171@bgtnsc05-news.ops.worldnet.att.net>
<oiu2jvgk8sn13mnkrucdguvd2oi2d82rm8@4ax.com>
<RcfYa.90345$0v4.6121163@bgtnsc04-news.ops.worldnet.att.net>
X-Cise: tanbanso@iinet.net.au
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In
<RcfYa.90345$0v4.6121163@bgtnsc04-news.ops.worldnet.att.net>,
on
at 10:22 PM, Nat Silver <mathelp@worldnet.att.net> said:
>To me, its obvious that an infinite sequence of random
>coin flips cannot be periodic.
To every complex question there is an answer that is simple,
obvious
and wrong.
--
Shmuel (Seymour J.) Metz, SysProg and JOAT
Any unsolicited bulk E-mail will be subject to legal action. I
reserve the
right to publicly post or ridicule any abusive E-mail.
Reply to domain Patriot dot net user shmuel+news to contact
me. Do not
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===
Subject: Re: Ullrich says probability 1 does not imply
certainty
<hdn2jvkc6ri4aelkinpictb8aattadja8a@4ax.com>
<mjdYa.87601$3o3.6052020@bgtnsc05-news.ops.worldnet.att.net>
<lpu2jv0hos5hmakqm79dks2aiom4gfudhb@4ax.com>
<FgfYa.90348$0v4.6121324@bgtnsc04-news.ops.worldnet.att.net>
X-Cise: tanbanso@iinet.net.au
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In
<FgfYa.90348$0v4.6121324@bgtnsc04-news.ops.worldnet.att.net>,
on
at 10:26 PM, Nat Silver <mathelp@worldnet.att.net> said:
>Let's not pretend the study of infinite sequences
>can directly be applied mathematics.
Indeed, only a Taylor would think that they had any relevance.
--
Shmuel (Seymour J.) Metz, SysProg and JOAT
Any unsolicited bulk E-mail will be subject to legal action. I
reserve the
right to publicly post or ridicule any abusive E-mail.
Reply to domain Patriot dot net user shmuel+news to contact
me. Do not
reply
to spamtrap@library.lspace.org
===
Subject: Re: Ullrich says probability 1 does not imply
certainty
> Specifically, you made a comment that, on its surface,
> appeared to be a comment regarding your belief regarding
> some particular event (infinite sequence from {H,T}):
> Such a sequence, violates the axioms of probability.
Axiom 1 (Positiveness)
The probability, assigned to each event in a sample space,
is a non-negative number in the closed interval [0, 1].
Axiom 2 (Certainty)
The probability of the entire space is 1.
Axiom 3. (Unions)
If A and B are mutually exclusive events,
then P(A U B) = P(A) + P(B)
We have not defined random. But let's continue anyway.
Consider the interval [0, 1] itself as our space.
Suppose we want to choose a number at random
from the space. We want all numbers to be equally likely.
If you claim the probability of choosing such a number is 0,
then the sum of the probabilities is not 1, violating Axiom 2.
If you claim that the probability of choosing such a number is
greater than zero, again you would be violating Axiom 2, since
the sum of the probabilities, then, is infinite. An
alternative is
to concede that choosing a number at random in this interval
is not a well-defined operation.
Taking another tack, how would you propose to do it?
One answer: Maybe generating decimal expansions as
sequences. That would be valid (from the axioms) for only
finitely many steps, giving a nice non-zero probability in the
range of those numbers whose decimal expansions begin
with the generated digits.
If we want to extend our probability measure to points,
we just define their probability to be zero, which seems
to work well in practice. However, we lose Axioms 2 and 3.
===
Subject: Re: Ullrich says probability 1 does not imply
certainty
>Axiom 1 (Positiveness)
>The probability, assigned to each event in a sample space,
>is a non-negative number in the closed interval [0, 1].
>Axiom 2 (Certainty)
>The probability of the entire space is 1.
>Axiom 3. (Unions)
>If A and B are mutually exclusive events,
>then P(A U B) = P(A) + P(B)
This axiom refers to two (2) events. A corollary will talk
about an
arbitrary finite number of events. It says nothing about
infinitely
many events. There is actually another axiom called countable
additivity that you left out, which talks about countably many
events.
But not uncountably many. There is _no_ axiom that says that
the
probability of the union of uncountably many mutually exclusive
events must be the sum of their probabilities.
>Consider the interval [0, 1] itself as our space.
>Suppose we want to choose a number at random
>from the space. We want all numbers to be equally likely.
>If you claim the probability of choosing such a number is 0,
>then the sum of the probabilities is not 1, violating Axiom 2.
No, because the interval is not a finite set (and in fact not
a countable set).
Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia
Vancouver, BC, Canada V6T 1Z2
===
Subject: Re: Ullrich says probability 1 does not imply
certainty
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In
<z_wYa.89088$3o3.6148142@bgtnsc05-news.ops.worldnet.att.net>,
on
at 06:36 PM, Nat Silver <mathelp@worldnet.att.net> said:
>We don't know.
Well, one of you doesn't.
In
<cHzYa.89240$3o3.6159693@bgtnsc05-news.ops.worldnet.att.net>,
on
at 09:40 PM, Nat Silver <mathelp@worldnet.att.net> said:
>Hmm; there is. This is not a private seminar nor is it
>anyone's classroom. Since you, especially, are not
>giving grades here, your judgments are worth about
>as much as the ink, used to pen them.
His opinions are worth no more in the classroom than on
Usenet. The
value of his opinions is not determined by where he expresses
them,
but by how informed they are. If he weren't such a jolly
Feller he
would not have been so gentle with your ignorance.
--
Shmuel (Seymour J.) Metz, SysProg and JOAT
Any unsolicited bulk E-mail will be subject to legal action. I
reserve the
right to publicly post or ridicule any abusive E-mail.
Reply to domain Patriot dot net user shmuel+news to contact
me. Do not
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to spamtrap@library.lspace.org
===
Subject: Re: Ullrich says probability 1 does not imply
certainty
> His opinions are worth no more in the classroom than on
Usenet. The
> value of his opinions is not determined by where he
expresses them,
> but by how informed they are. If he weren't such a jolly
Feller he
> would not have been so gentle with your ignorance.
One wonders if the original poster has been smoking a few too
many Doobs.
===
Subject: Re: Ullrich says probability 1 does not imply
certainty
> Specifically, you made a comment that, on its surface,
> appeared to be a comment regarding your belief regarding
> some particular event (infinite sequence from {H,T}):
> Such a sequence, violates the axioms of probability.
>Axiom 1 (Positiveness)
>The probability, assigned to each event in a sample space,
>is a non-negative number in the closed interval [0, 1].
>Axiom 2 (Certainty)
>The probability of the entire space is 1.
>Axiom 3. (Unions)
>If A and B are mutually exclusive events,
>then P(A U B) = P(A) + P(B)
>We have not defined random. But let's continue anyway.
>Consider the interval [0, 1] itself as our space.
>Suppose we want to choose a number at random
>from the space. We want all numbers to be equally likely.
>If you claim the probability of choosing such a number is 0,
>then the sum of the probabilities is not 1, violating Axiom 2.
>If you claim that the probability of choosing such a number is
>greater than zero, again you would be violating Axiom 2, since
>the sum of the probabilities, then, is infinite.
No it doesn't violate Axiom 2, because Axiom 3 talks about the
union
of _two_ events; it doesn't say anything about the union
of uncountably many events.
In the standard axioms for probability theory there are
technical details about measure theory, and then the
thing corresponding to Axiom 3 refers to unions of
_countably many_ events. Nothing about uncountable
unions in the axioms you give here nor in the standard
axioms (there's a reason for that).
>An alternative is
>to concede that choosing a number at random in this interval
>is not a well-defined operation.
Not needed - all we have to do is understand what Axiom 2
actually says, and not attempt to deduce things from it that
it simply does not say.
>Taking another tack, how would you propose to do it?
>One answer: Maybe generating decimal expansions as
>sequences. That would be valid (from the axioms) for only
>finitely many steps, giving a nice non-zero probability in the
>range of those numbers whose decimal expansions begin
>with the generated digits.
>If we want to extend our probability measure to points,
>we just define their probability to be zero, which seems
>to work well in practice. However, we lose Axioms 2 and 3.
Find a book on probability theory. (A book on grownup
actual probability theory, ie one that assumes a
knowledge of measure theory and bases probability
on that). In the standard setup (with a uniform probability
on [0,1] as above) points do indeed have probability zero.
And it follows that _countable_ sets have probability
zero. Perfectly consistent with Axiom 2 and the
countably-many-sets version of Axiom 3.
************************
David C. Ullrich
===
Subject: Re: Ullrich says probability 1 does not imply
certainty
> Specifically, you made a comment that, on its surface,
> appeared to be a comment regarding your belief regarding
> some particular event (infinite sequence from {H,T}):
> Such a sequence, violates the axioms of probability.
> Axiom 1 (Positiveness)
> The probability, assigned to each event in a sample space,
> is a non-negative number in the closed interval [0, 1].
> Axiom 2 (Certainty)
> The probability of the entire space is 1.
> Axiom 3. (Unions)
> If A and B are mutually exclusive events,
> then P(A U B) = P(A) + P(B)
> We have not defined random. But let's continue anyway.
> Consider the interval [0, 1] itself as our space.
> Suppose we want to choose a number at random
> from the space. We want all numbers to be equally likely.
> If you claim the probability of choosing such a number is 0,
> then the sum of the probabilities is not 1, violating Axiom
2.
Really? Could you demonstrate how you evaluated that sum and
PROVE that it's not equal to 1?
The entire space in this case is [0,1]. The probability of
that space is 1. That does not violate axiom 2.
The probability of every interval of size 0.1 is 0.1. There
are 10 such intervals, for a total probability of 10*0.1 = 1.
The probability of every interval of size 1/n is 1/n for any
natural number n. There are n mutually exclusive such
intervals,
for a total probability of 1.
Now show me your sum.
- Randy
===
Subject: Re: Ullrich says probability 1 does not imply
certainty
> Axiom 1 (Positiveness)
> The probability, assigned to each event in a sample space,
> is a non-negative number in the closed interval [0, 1].
> Axiom 2 (Certainty)
> The probability of the entire space is 1.
> Axiom 3. (Unions)
> If A and B are mutually exclusive events,
> then P(A U B) = P(A) + P(B)
> We have not defined random. But let's continue anyway.
> Consider the interval [0, 1] itself as our space.
> Suppose we want to choose a number at random
> from the space. We want all numbers to be equally likely.
> If you claim the probability of choosing such a number is 0,
> then the sum of the probabilities is not 1, violating Axiom
2.
Nonsense, just take the probability of an event to be its
Lebesgue measure.
Try some post 1850 mathematics for a change. Game over.
===
Subject: Re: Ullrich says probability 1 does not imply
certainty
<hdn2jvkc6ri4aelkinpictb8aattadja8a@4ax.com>
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<1mp4jv07p25bbb49vvs2sa9p7o15l3s0a7@4ax.com>
<z_wYa.89088$3o3.6148142@bgtnsc05-news.ops.worldnet.att.net>
<l7d5jvc4tahb3d0qbh2e9eogu84c1jir7m@4ax.com>
<cHzYa.89240$3o3.6159693@bgtnsc05-news.ops.worldnet.att.net>
<3F32CD97.6010708@farir.com>
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In
<_WDYa.89488$3o3.6184376@bgtnsc05-news.ops.worldnet.att.net>,
on
>Consider the interval [0, 1] itself as our space.
>Suppose we want to choose a number at random
>from the space. We want all numbers to be equally likely. If
you
>claim the probability of choosing such a number is 0, then
the sum of
>the probabilities is not 1,
Wrong. If x is in [0,1], then then {x} has probability 0,
[0,x) U
(x,1] has probability 1 and the sum of the probabilities is 1.
>violating Axiom 2.
No, the probability of [0,1] is 1, in accordance with your
axiom of
certainty.
--
Shmuel (Seymour J.) Metz, SysProg and JOAT
Any unsolicited bulk E-mail will be subject to legal action. I
reserve the
right to publicly post or ridicule any abusive E-mail.
Reply to domain Patriot dot net user shmuel+news to contact
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===
Subject: Re: Unbelievable!
>
http://mr-31238.mr.valuehost.co.uk/assets/Flash/psychic.swf
> SOLUTION . --
> Let x:= ab:=10a+b where a,bin {0,1,...,9}, a>=1.
> Then Result:= x-(a+b)= 9a where a in {1,2,3,4,5,6,7,8,9}.
Therefore
> please the same symbol for 9,18,27,36,45,54,63,72,81 .
> In other words , give the same relevant symbol to numbers :
> 81 , 72 , 63 , 54 , 45 , 36 , 27 , 18 , 9 .
> According to the table (see below) you find the desired
result.
> This solution/time: 15 minutes/ belongs to my son Tudor
Lupas,
> student by Computer Sciences.
> Your son may enjoy learning about Casting Out Nines -- the
basis
> of this result. Many of my prior posts discuss related
topics, see
> -Bill Dubuque
===
Subject: very very simple question
X-URL:
http://mygate.mailgate.org/mynews/sci/sci.math/
a79bbdddcbbd2278bf1ae05e62be20
b8.35661%40mygate.mailgate.org
This is the simplest question in the world about sigma
algebras.
I'm trying to read a book which treats measure theory (and
then applies
it to quantum mechanics - the book is David W. Jordan's An
Introduction
to Hilbert Space and Quantum Logic, but that's not
particularly relevant
as this definition he gives is pretty standard, I think).
Definition:
Let X be a set, and let A be a collection of subsets of X
satisfying
i). X is an element of A
ii). if S is an element of A, then X S is an element of A,
iii). if S is a countable subset of A, then U S is an element
of A.
Okay, can you see the error in George Boole's set-up?
How can A be a collection of subsets of X and, at the same
time, X be an
element of A ? Surely the only way this can happen is if X = A.
How about I draw a circle? I label it X. Then, inside this
circle I draw
some more circles,and label these A_1, A_2, etc. So the A are
subsets of
X. But the X is not a subset of A and is therefore not an
element of A,
thus proving that i) is false.
Does anybody think this realisation is worth publishing, or
have I
missed something?
Appreciate your help,
David.
PS. Can anyone detect that I'm starting to get a bit
frustrated with set
theory?
--
===
Subject: Re: very very simple question
> Definition:
> Let X be a set, and let A be a collection of subsets of X
satisfying
> i). X is an element of A
> ii). if S is an element of A, then X S is an element of A,
> iii). if S is a countable subset of A, then U S is an
element of A.
> Okay, can you see the error in George Boole's set-up?
> How can A be a collection of subsets of X and, at the same
time, X be an
> element of A ? Surely the only way this can happen is if X =
A.
No.
Let X = {1,2,3}
Let A = {{}, {1}, {2.3}, {1,2,3}}.
Then A is a collection of subsets of X. And X is an element of
A.
No worries!
--
Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html
His mind has been corrupted by colours, sounds and shapes.
The League of Gentlemen
===
Subject: Re: very very simple question
X-URL:
http://mygate.mailgate.org/mynews/sci/sci.math/
2c9ba176a98ad7f297656495904d98
c7.35661%40mygate.mailgate.org
> How can A be a collection of subsets of X and, at the same
time, X be
an
> element of A ? Surely the only way this can happen is if X
= A.
> No.
> Let X = {1,2,3}
> Let A = {{}, {1}, {2.3}, {1,2,3}}.
> Then A is a collection of subsets of X. And X is an element
of A.
> No worries!
bit as you are saying that {1,2,3} is a subset of {1,2,3}
which is how I
came to deduce that X = A in the first place - there cannot be
an
example whereby A doesn't contain X, can there?
David.
--
===
Subject: Re: very very simple question
> How can A be a collection of subsets of X and, at the
same time, X be
> an element of A ? Surely the only way this can happen is
if X = A.
> No.
> Let X = {1,2,3}
> Let A = {{}, {1}, {2.3}, {1,2,3}}.
> Then A is a collection of subsets of X. And X is an element
of A.
> No worries!
> bit as you are saying that {1,2,3} is a subset of {1,2,3}
I haven't cheated at all.
--
Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html
His mind has been corrupted by colours, sounds and shapes.
The League of Gentlemen
===
Subject: Re: very very simple question
X-URL:
http://mygate.mailgate.org/mynews/sci/sci.math/
295ddce92436860949fc48d84e747c
c3.35661%40mygate.mailgate.org
> I haven't cheated at all.
Not even a tiny bit?
--
===
Subject: Re: very very simple question
> How can A be a collection of subsets of X and, at the
same time, X be
an
> element of A ? Surely the only way this can happen is if
X = A.
> No.
> Let X = {1,2,3}
> Let A = {{}, {1}, {2.3}, {1,2,3}}.
> Then A is a collection of subsets of X. And X is an element
of A.
> No worries!
>bit as you are saying that {1,2,3} is a subset of {1,2,3}
What's the definition of the word subset? Not what you think
it is, I suspect.
>which is how I
>came to deduce that X = A in the first place
It doesn't follow that X = A. It follows that X is the _union_
of
the elements of A...
> - there cannot be an
>example whereby A doesn't contain X, can there?
>David.
************************
David C. Ullrich
===
Subject: Re: very very simple question
Supersedes: <bh0bi0$ju8$3@news8.svr.pol.co.uk> I haven't cheated at all.
> Not even a tiny bit?
Not even a tiny bit.
--
Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html
His mind has been corrupted by colours, sounds and shapes.
The League of Gentlemen
===
Subject: Re: very very simple question
> This is the simplest question in the world about sigma
algebras.
> I'm trying to read a book which treats measure theory (and
then applies
> it to quantum mechanics - the book is David W. Jordan's An
Introduction
> to Hilbert Space and Quantum Logic, but that's not
particularly relevant
> as this definition he gives is pretty standard, I think).
> Definition:
> Let X be a set, and let A be a collection of subsets of X
satisfying
> i). X is an element of A
> ii). if S is an element of A, then X S is an element of A,
> iii). if S is a countable subset of A, then U S is an
element of A.
> Okay, can you see the error in George Boole's set-up?
> How can A be a collection of subsets of X and, at the same
time, X be an
> element of A ?
A is set whose elements are certain subsets of X.
X is a subset of X.
Hence X may well appear as an element of A.
> element of A ? Surely the only way this can happen is if X =
A.
> How about I draw a circle? I label it X. Then, inside this
circle I draw
> some more circles,and label these A_1, A_2, etc. So the A
are subsets of
> X. But the X is not a subset of A and is therefore not an
element of A,
> thus proving that i) is false.
You confused subset of A with element of A.
Of course, in your example with the disks, X is also not an
elemement of A.
But this is a property of that example.
Marc
===
Subject: Re: very very simple question
> How can A be a collection of subsets of X and, at the
same time, X be
an
> element of A ? Surely the only way this can happen is if
X = A.
 No.
 Let X = {1,2,3}
 Let A = {{}, {1}, {2.3}, {1,2,3}}.
 Then A is a collection of subsets of X. And X is an
element of A.
 No worries!
> bit as you are saying that {1,2,3} is a subset of {1,2,3}
which is how I
> came to deduce that X = A in the first place - there cannot
be an
> example whereby A doesn't contain X, can there?
> David.
X and A are different creatures.
X = {1,2,3} so its elements are numbers,
A = {{}, {1}, ... , {1,2,3}} has sets as its elements.
In probability theory, the elements of A correspond to events
occuring.
===
Subject: Re: very very simple question
> I haven't cheated at all.
> Not even a tiny bit?
No - here is an explanation:
In any piece of mathematics, you have to go back to the
definitions
for all the little details to make sense. What is the
definition
of subset? A is a subset of B if every element of A is an
element
of B. So, is every element of X an element of X? Yes. Thus,
by definition, X is a subset of X. We use the term proper
subset
to mean a subset which is not all of X (proper subset probably
corresponds more naturally to the English meaning of subset).
The one people usually have the hardest time with is the fact
that
the empty set is a subset of X (in fact its a subset of any
set).
In this case, it is best to translate every element of A is an
element of B into something slightly more formal like given
any x, if x is in A, then x is in B. Then, if you trace back to
the logical definition of if ... then you find that it can only
be false if the if part is true and the then part is false.
But, for the empty set, the if part is never true (since no
element
is in the empty set), so the whole if....then statement is
true,
regardless of what x is. Thus the empty set is a subset of X
(or
of any other set, including itself). Notice that the complement
of X (with respect to X) is the empty set, so every
sigma-algebra
also contains the empty set.
Note that in Robin's example, the elements of the sigma algebra
A are {}, {1}, {2,3}, {1,2,3}. These elements are subsets of
{1,2,3} (and they happen to be closed under complementation and
union). Perhaps the first confusing thing to get over when
first
seeing sigma algebras and topologies is that the elements of a
set can be other sets (the fact that those sets must be subsets
of another set can also add consusion).
Good luck,
Hugh
===
Subject: what courses usually make up a 4-year degree in math?
math is my main passion, however unfortunately i dunno whether
i have
the deductive reasoning needed to go all the way to phd. ahh
life,
God seldom matches what you like most with what you're most
talented
at. anyway, im practically entirely self-taught and am just
now after
many years getting to a point where college is an option
financially.
but having very little college classtime im utterly unfamiliar
with
what the general requirements for a 4-year degree are. i was
always
under the impression a 4 year degree required some extremely
high
maths, but recently had a brief opportunity to talk to a
coworker who
has one, and from what i gathered she didn't do much beyond
your basic
intro theory of algebra/theory of calculus!.. i asked what
about
galois theory and she didnt even know what it was... so now im
thinking maybe i have a chance for a 4 year after all (in the
subject
where my heart lies)
(in lieu of a math degree i was thinking of going for
something like
phd in computer programming with a pure math minor, but after
that
conversation now i dunno WHAT to do... computer programming is
where
i seem to have the most innate talent, though my heart is
already
stolen by maths...)
===
Subject: Re: what courses usually make up a 4-year degree in
math?
> math is my main passion, however unfortunately i dunno
whether i have
> the deductive reasoning needed to go all the way to phd. ahh
life,
> God seldom matches what you like most with what you're most
talented
> at. anyway, im practically entirely self-taught and am just
now after
> many years getting to a point where college is an option
financially.
> but having very little college classtime im utterly
unfamiliar with
> what the general requirements for a 4-year degree are. i was
always
> under the impression a 4 year degree required some extremely
high
> maths, but recently had a brief opportunity to talk to a
coworker who
> has one, and from what i gathered she didn't do much beyond
your basic
> intro theory of algebra/theory of calculus!.. i asked what
about
> galois theory and she didnt even know what it was... so now
im
> thinking maybe i have a chance for a 4 year after all (in
the subject
> where my heart lies)
> (in lieu of a math degree i was thinking of going for
something like
> phd in computer programming with a pure math minor, but
after that
> conversation now i dunno WHAT to do... computer programming
is where
> i seem to have the most innate talent, though my heart is
already
> stolen by maths...)
Depending on the abstract algebra course, it may not include
Galois
theory, leaving that to graduate school.
A rough list for the 4-year degree in the US...
calculus, differential eqations, basic analysis, complex
variables
linear algebra, abstract algebra
combinatorics, probability, statistics
===
Subject: Re: what courses usually make up a 4-year degree in
math?
> math is my main passion, however unfortunately i dunno
whether i have
> the deductive reasoning needed to go all the way to phd. ahh
life,
> God seldom matches what you like most with what you're most
talented
> at. anyway, im practically entirely self-taught and am just
now after
> many years getting to a point where college is an option
financially.
> but having very little college classtime im utterly
unfamiliar with
> what the general requirements for a 4-year degree are. i was
always
> under the impression a 4 year degree required some extremely
high
> maths, but recently had a brief opportunity to talk to a
coworker who
> has one, and from what i gathered she didn't do much beyond
your basic
> intro theory of algebra/theory of calculus!.. i asked what
about
> galois theory and she didnt even know what it was... so now
im
> thinking maybe i have a chance for a 4 year after all (in
the subject
> where my heart lies)
The requirements vary a lot from department to department. I
would guess at
a
bare minimum these would generally include calculus,
differential
equations,
linear algebra, and introductory analysis and/or abstract
algebra. Good
students in good departments who want to do graduate work in
mathematics
typically take all of these and more, in particular the entire
sequences in
analysis and algebra (honors versions, if available) and then
graduate
courses
if they have the time. It's possible to get into good
departments with
considerably less background than this, however.
If you are not interested in pursuing a graduate degree in
mathematics, you
can
surely get away with less than the above, and you may be well
advised to
take at
least the main courses that a CS student would take (I don't
know what
those
are) if you want to do graduate work in CS. But if you like
mathematics
and
your school offers the courses, I cannot imagine that you
would ever regret
taking the entire analysis and algebra sequences (including
Galois theory,
which
is quite nice).
===
Subject: Re: what courses usually make up a 4-year degree in
math?
> math is my main passion, however unfortunately i dunno
whether i have
> the deductive reasoning needed to go all the way to phd. ahh
life,
> God seldom matches what you like most with what you're most
talented
> at. anyway, im practically entirely self-taught and am just
now after
> many years getting to a point where college is an option
financially.
> but having very little college classtime im utterly
unfamiliar with
> what the general requirements for a 4-year degree are. i was
always
> under the impression a 4 year degree required some extremely
high
> maths, but recently had a brief opportunity to talk to a
coworker who
> has one, and from what i gathered she didn't do much beyond
your basic
> intro theory of algebra/theory of calculus!.. i asked what
about
> galois theory and she didnt even know what it was... so now
im
> thinking maybe i have a chance for a 4 year after all (in
the subject
> where my heart lies)
> (in lieu of a math degree i was thinking of going for
something like
> phd in computer programming with a pure math minor, but
after that
> conversation now i dunno WHAT to do... computer programming
is where
> i seem to have the most innate talent, though my heart is
already
> stolen by maths...)
I think that if you self taught yourself Galois Theory, then
you definitely
have
what it takes to do a math degree at any major state
university. (To keep
your job options open, how about the minor in computer
programming.) If I
were
you, I would definitely call around some universities, and
talk with the
director of undergraduate studies. I think that they would be
very
interested
in talking with you.
--
Stephen Montgomery-Smith
stephen@math.missouri.edu
http://www.math.missouri.edu/~stephen
===
Subject: Re: what courses usually make up a 4-year degree in
math?
>Depending on the abstract algebra course, it may not include
Galois
>theory, leaving that to graduate school.
>A rough list for the 4-year degree in the US...
>calculus, differential eqations, basic analysis, complex
variables
>linear algebra, abstract algebra
>combinatorics, probability, statistics
Of course, a more rigorous program, such as those taught at
better
schools internationally, would include, in addition to the
topics
above, other topics, such as game theory, topology,
approximation
theory, and touch on other advanced topics. They also usually
only
take 3 years, but do not include any liberal arts curriculum.
===
Subject: Where can I found an Erdos' proof?
Hi!
In reading the book The Man Who Loved Only Numbers about Paul
Erdos, by
Paul Hoffman, I learned that Erdos presented a elementary
proof that for
all
natural number n, exists a prime number between n and 2n.
Can anyone tell me where can I find Erdos' proof?
Jaime Gaspar
______________________________
Homepage: www.jaimegaspar.com
E-mail: e-mail@jaimegaspar.com
===
Subject: Re: Where can I found an Erdos' proof?
> In reading the book The Man Who Loved Only Numbers about Paul
> Erdos, by Paul Hoffman, I learned that Erdos presented a
elementary
> proof that for all natural number n, exists a prime number
between n
> and 2n.
> Can anyone tell me where can I find Erdos' proof?
Chapman with the title Bertrand's Postulate.
===
Subject: Re: Yet another choice question (the other yacc)
> Can one prove the following (a variation on
Schrder-Bernstein)
> without AC?
No.
> If there exist an injection and a surjection from one set to
> another, then there exists a bijection between these two
sets.
Let |A| = 2^{aleph 0}, and let |B| = aleph 1 + 2^{aleph 0}.
Without
AC, you can get an injection and a surjection from A to B, but
you
can't get an injection from B to A.
===
Subject: Re: Yet another choice question (the other yacc)
> Can one prove the following (a variation on
Schr.9ader-Bernstein) without
AC?
> If there exist an injection and a surjection from one set to
another,
> then there exists a bijection between these two sets.
No because this statement implies that there exists an
uncountable
well-orderable subset of R. Consider the sets R and R union
w_1 where
w_1 is the first uncountable ordinal. There is an obvious
injection
one way, and since every countable ordinal is order isomorphic
to a
countable subset of R, there is a surjection from R to w_1, so
also
from R to R union w_1. So if your statement holds then R
bijects with
R union w_1; then the inverse image of w_1 under the bijection
is an
uncountable well orderable subset of R. And apparently that
can't be
proved without AC...
Michael
===
Subject: Re: Yet another choice question (the other yacc)
> Can one prove the following (a variation on
Schr.9ader-Bernstein) without
AC?
> If there exist an injection and a surjection from one set to
another,
> then there exists a bijection between these two sets.
Apologies - Fred Galvin posted more or less the same thing as
me a day
earlier, but somehow I missed his message.
Michael
===
Subject: Re: Yet another choice question (the other yacc)
> Can one prove the following (a variation on
Schr=F6der-Bernstein)
> without AC?
>No.
> If there exist an injection and a surjection from one set to
> another, then there exists a bijection between these two
sets.
>Let |A| =3D 2^{aleph_0}, and let |B| =3D aleph_1 +
2^{aleph_0}. Without
>AC, you can get an injection and a surjection from A to B,
but you
>can't get an injection from B to A.
This same proof extends to showing that Herschkorn's
proposition implies that, if k is a cardinal with k=k^2,
the Hartogs function k* (the smallest aleph not less than
or equal to k) satisfies k* <= 2^k.
--
This address is for information only. I do not claim that
these views
are those of the Statistics Department or of Purdue University.
Herman Rubin, Department of Statistics, Purdue University
hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558