mm-
mm-

===
Subject: Re: Division by zero. Go ahead and laugh.
> I believe that the lim 1/x as x->0 is inŽnity. Am I wrong ?
Yes. There is no such real number as inŽnity.
If you had said you can make |1/x| larger than any
arbitrarily large
(Žnite) number by choosing x small enough, that would be
correct.
Bob Kolker
===
Subject: Re: Division by zero. Go ahead and laugh.
>> I believe that the lim 1/x as x->0 is inŽnity. Am I wrong ?
> Yes. There is no such real number as inŽnity.
> If you had said you can make |1/x| larger than any
arbitrarily large
> (Žnite) number by choosing x small enough, that would be
correct.
Hah. x is still allways <>0. Then for every x>0 there is
Žnite number that is 1/x, so 1/x can not make larger than any
arbitrarily large Žnite number, because 1/x is Žnite number,
if x>0.
And math today is incorrect nonsense.
--
amount and biggest
1+1+1+...= inŽnite and Žnite in math today
Petri Keckman
===
Subject: Re: Division by zero. Go ahead and laugh.
> And math today is incorrect nonsense.
No it isn¹t. No one has found a contradiction in the theory
of real or
complex variables.
Bob Kolker
===
Subject: Re: Division by zero. Go ahead and laugh.
>> I believe that the lim 1/x as x->0 is inŽnity. Am I wrong ?
> Yes. There is no such real number as inŽnity.
> If you had said you can make |1/x| larger than any
arbitrarily large
> (Žnite) number by choosing x small enough, that would be
correct.
> Hah. x is still allways <>0. Then for every x>0 there is
> Žnite number that is 1/x, so 1/x can not make larger than
any
> arbitrarily large Žnite number, because 1/x is Žnite number,
> if x>0.
Perhaps it is the quantiŽers that confuse you.
Pick an arbitrary Žnite number, say 10^100. Make |1/x|
larger than your choice means that I can Žnd x such
that 1/x > 10^100. Do you believe that?
Pick another Žnite number. I can again Žnd 1/x
larger than that.
Nobody is claiming that EVERY 1/x is greater than
EVERY Žnite number, or however you¹re interpreting that.
Merely that that any particular speciŽed, Žxed Žnite
value is exceeded by inŽnitely many choices of 1/x.
Again, I don¹t know whether it is the English or the
mathematical reasoning that is confusing you.
- Randy
===
Subject: Re: Division by zero. Go ahead and laugh.
>> If you had said you can make |1/x| larger than any
arbitrarily large
>> (Žnite) number by choosing x small enough, that would be
correct.
>> Hah. x is still allways <>0. Then for every x>0 there is
>> Žnite number that is 1/x, so 1/x can not make larger than
any
>> arbitrarily large Žnite number, because 1/x is Žnite
number,
>> if x>0.
> Perhaps it is the quantiŽers that confuse you.
> Pick an arbitrary Žnite number, say 10^100. Make |1/x|
> larger than your choice means that I can Žnd x such
> that 1/x > 10^100. Do you believe that?
Yeah. You right. I was stupidly wrong.
--
amount and biggest
1+1+1+...= inŽnite and Žnite in math today
Petri Keckman
===
Subject: Re: Division by zero. Go ahead and laugh.
>> I believe that the lim 1/x as x->0 is inŽnity. Am I wrong ?
> Yes. There is no such real number as inŽnity.
> If you had said you can make |1/x| larger than any
arbitrarily large
> (Žnite) number by choosing x small enough, that would be
correct.
> Hah. x is still allways <>0. Then for every x>0 there is
> Žnite number that is 1/x, so 1/x can not make larger than
any
> arbitrarily large Žnite number, because 1/x is Žnite number,
> if x>0.
> And math today is incorrect nonsense.
I¹m going to take that and run with it.
===
Subject: Re: Division by zero. Go ahead and laugh.
>> Hah. x is still allways <>0. Then for every x>0 there is
>> Žnite number that is 1/x, so 1/x can not make larger than
any
>> arbitrarily large Žnite number, because 1/x is Žnite
number,
>> if x>0.
>> And math today is incorrect nonsense.
> I¹m going to take that and run with it.
Error, error, should be:
so 1/x can not make larger than any arbitrarily large Žnite
number,
because
(1/x)+1 is Žnite number that is larger.
--
amount and biggest
1+1+1+...= inŽnite and Žnite in math today
Petri Keckman
===
Subject: Re: Division by zero. Go ahead and laugh.
> so 1/x can not make larger than any arbitrarily large Žnite
number,
> because
> (1/x)+1 is Žnite number that is larger.
How to make 1/x larger than N. Set pick x so that |x| < 1/N.
For any large N there exists an x for which |1/x| > N.
Obviously the x
you choose is related to the given N.
Bob Kolker
===
Subject: Re: Division by zero. Go ahead and laugh.
> Error, error, should be:
> so 1/x can not make larger than any arbitrarily large Žnite
number,
> because (1/x)+1 is Žnite number that is larger.
An error again. Damned.
--
amount and biggest
1+1+1+...= inŽnite and Žnite in math today
Petri Keckman
===
Subject: Re: Division by zero. Go ahead and laugh.
>
> universe of real objects does seem difŽcult to imagine how
it might
> happen.
> Dividing by 0 where multiplication and division have their
usual
> meanings is impossible in the mathematical universe. It
implies that
1=0.
> Bob Kolker
>
> But, it¹s easy to see that the limit lim 1/x as x->0 is
everywhere
in
> nature. This implies that while nature might not do 1/0, it
is
certainly
> doing lim 1/x as x->0. Nature is doing this.
> Lim 1/x as x => 0 does not exist.
> Bob Kolker
> I believe that the lim 1/x as x->0 is inŽnity. Am I wrong ?
There is no limit to 1/x as x increases without bound.
> Really guys, I¹m rusty, so bear with me a little.
> In the above, we¹re talking about 2 different things
however.
> 1) Mathematical operations which exist as abstractions, and
> 2) Manipulations in the physical world which resemble
mathematical
things.
> It would be nice to bridge these solidly. Never been done I
think,
===
Subject: Re: Division by zero. Go ahead and laugh.
> There are no paradoxes or inconsistencies in mathematics,
unless you
> remove
> all the ad hoc stipulations and band aids which hold it all
together.
> Confess.
> Have you met James Harris, by any chance?
OK now that¹s hitting below the x-axis !
I come up with some very loony ideas, I will admit, and I
order erasers by
the case load, but sometimes my explorations take in some
interesting
directions, and since I have no time to do any real math
anymore I thought
I¹d look at some phylosophical issues, which are generally
garbage, so
maybe
this will all wind up in the sci-Ž section, but I wont hang
around in here
for 5 years trying to prove my point because I dont have one.
All just
questions, and seeking the eternal wisdom of my little beige
box.
===
Subject: Re: Division by zero. Go ahead and laugh.
> Brian, he¹s a troll.
> Probably. But I prefer not to leave nonsense unanswered in
> alt.math.undergrad (or alt.algebra.help, though that¹s not
> relevant this time).
> Brian
You guys kill me. OK, I¹m not going to turn this into a proof
of FLT like
others might, but I¹m not trying to troll the place either.
I also dont want to waste anyone¹s time. So I¹ll be as clear
as possible to
explain what I¹m trying to understand.
If you let 1/0 = inŽnity, then any result can be obtained
algebraically
and
arithmetic becomes inconsistent. I am saying that this may
have something
to
do with the physical properties of the universe in which you
are located.
Here¹s the motivation. You are a human being. You understand
math in your
head and everything is logical, but the entire apparatus of
mathematics was
constructed in this universe, and therefore logic itself may
be succeptible
to physics related inžuences. This creates yet another
chicken and the egg
type of paradox, and I do not believe that this question is
even solvable,
but, physics may be responsible for the fact that you cannot
divide by
zero.
Lastly, I am asking a serious academic question and not
trolling. If you
want to blast this idea to pieces then I welcome you into the
debate.
===
Subject: Re: Division by zero. Go ahead and laugh.
> If you let 1/0 = inŽnity,
meaningless clap trap. In the Želd of rational, real or
complex numbers
there is no number inŽnity.
If you let my grandma have balls, she would be my granpa.
Bob Kolker
===
Subject: Re: Division by zero. Go ahead and laugh.
> If you let 1/0 = inŽnity,
> meaningless clap trap. In the Želd of rational, real or
complex numbers
> there is no number inŽnity.
> If you let my grandma have balls, she would be my granpa.
> Bob Kolker
===
Subject: Re: Division by zero. Go ahead and laugh.
>The inconsistency which I refer to is division by zero,
itself. It has
been
>deŽned to not exist, out of laziness.
> Lefty: you don¹t know what you are talking about. Not here,
not about
> paradoxes.
> Simply put: you are talking and making assertions about
what is,
> based on misinformation, confusion, and ignorance. I do not
know if it
> is simply that you do not want to learn, or that you are
too lazy to
> do so, because even when corrected, and pointed to the
accurate facts,
> you continue to restate your igorance over and over.
> There is no laziness involved in saying that division by
zero is
> undeŽned. The only laziness is in your insistence that
repeating the
> same stupidity over and over is preferable to learning
something.
You did not even address my assertion, all you did was attack
me
personally.
Let me just restate something for you and I¹m asking you to
stick to the
subject matter.
Here¹s my claim. You say 1/0 is undeŽned. I say, 1/0 =
inconsistency. You
do not need to deŽne it out of existence to understand that
1/0 is
something you dont want to do. It does seem peculiar that
you¹d be picking
on poor zero, because no other number in all of R gets
treated that way.
Got a beef with zero or something ?
Can you reply to this ?
===
Subject: Re: Division by zero. Go ahead and laugh.
days. My association with the Department is that of an
alumnus.
>>The inconsistency which I refer to is division by zero,
itself. It has
>been
>>deŽned to not exist, out of laziness.
>> Lefty: you don¹t know what you are talking about. Not
here, not about
>> paradoxes.
>> Simply put: you are talking and making assertions about
what is,
>> based on misinformation, confusion, and ignorance. I do
not know if it
>> is simply that you do not want to learn, or that you are
too lazy to
>> do so, because even when corrected, and pointed to the
accurate facts,
>> you continue to restate your igorance over and over.
>> There is no laziness involved in saying that division by
zero is
>> undeŽned. The only laziness is in your insistence that
repeating the
>> same stupidity over and over is preferable to learning
something.
>You did not even address my assertion, all you did was
attack me
personally.
Didn¹t seem worth my time, seeing how you have already been
answered,
but ignored the reply in order to repeat your inane and
ignorant
assertion. And here you did it yet again.
>Let me just restate something for you and I¹m asking you to
stick to the
>subject matter.
>Here¹s my claim. You say 1/0 is undeŽned. I say, 1/0 =
>inconsistency.
You are saying words that do not mean anything.
>You
>do not need to deŽne it out of existence to understand that
1/0 is
>something you dont want to do.
The expression a/b=c means one thing, and one thing only: it
means
that b*c = a. The reason 1/0 is undeŽned is that the equation
0*c = 1 has no solutions. Simple. Nothing to do with
laziness. Nothing
to do with things you don¹t want to do, nothing to do with
deŽning
out of existence, nothing to do with inconsistency, nothing
to do
with any of the big words you insist on using without knowing
what
they mean.
>It does seem peculiar that you¹d be picking
>on poor zero, because no other number in all of R gets
treated that way.
0 is the additive identity of R. It is the ONLY additive
identity of
R. In any ring, the additive identity has the peculiar
property that,
multiplied by anything, it yields itself. So in any ring, the
equation
0*c = a has no solutions for a different from 0.
In the integers, there are plenty of other divisions that are
undeŽned: 1/2 is not deŽned in the integers.
>Got a beef with zero or something ?
No.
>Can you reply to this ?
Of course I can. Are you capable of learning something from
the
responses, or are you just interested in making noise by
saying stupid
things?
--
It¹s not denial. I¹m just very selective about
what I accept as reality.
--- Calvin (Calvin and Hobbes)
Arturo Magidin
magidin@math.berkeley.edu
===
Subject: Re: Division by zero. Go ahead and laugh.
> Here¹s my claim. You say 1/0 is undeŽned.
Pay attention. a/b = c if and only if a = bc. 1/0 cannot be
deŽned.
Why? if 1/0 = b then (by deŽnition) 1 = b*0 = 0. In short if
1/0 had a
value then 1 = 0 which is impossible so 1/0 is not deŽned.
Another way
of saying the same thing is that a/b is deŽned if and only if
a = bx
has a solution x. 1/0 is a meaningless combination of marks
and is not
deŽned.
Bob Kolker
===
Subject: Re: Division by zero. Go ahead and laugh.
> Here¹s my claim. You say 1/0 is undeŽned.
> Pay attention. a/b = c if and only if a = bc. 1/0 cannot be
deŽned.
> Why? if 1/0 = b then (by deŽnition) 1 = b*0 = 0. In short
if 1/0 had a
> value then 1 = 0 which is impossible so 1/0 is not deŽned.
Another way
> of saying the same thing is that a/b is deŽned if and only
if a = bx
> has a solution x. 1/0 is a meaningless combination of marks
and is not
> deŽned.
> Bob Kolker
I understand what you are saying exactly.And, at the risk of
sounding like
an argumentative imbecile, I continue to press -
- I am not saying that division by zero should be allowed.
What I am saying
is that if it were allowed, then you would have problems.
Again, this is
known. Division by zero must be isolated somehow so that
arithmetic will
remain intact. BUT - deŽning it to not exist is incorrect.
When you
divide by zero you are _not_ creating a non-existence. What
you are
creating is something else, possibly a contradiction,
paradox, or
inconsistency. I think that this is due to topology of
space/time where
mathematician lives.
There is currently no algebraic symbology to express this,
and if there
were
it might lead to new understandings.
===
Subject: Re: Division by zero. Go ahead and laugh.
> - I am not saying that division by zero should be allowed.
What I am
saying
> is that if it were allowed, then you would have problems.
If it were allowed the system would be -inconsistent-
therefore useless.
In an inconsistent mathematical system -anything- can be
proved.
Again, this is
> known. Division by zero must be isolated somehow so that
arithmetic will
> remain intact.
Impossible. When you allow something you also allow the
consequence.
> BUT - deŽning it to not exist is incorrect. When you
> divide by zero you are _not_ creating a non-existence.
Dividing by zero is meaningless.
> What you are
> creating is something else, possibly a contradiction,
paradox, or
> inconsistency. I think that this is due to topology of
space/time where
> mathematician lives.
You are pissing up a rope. There is no way of making a
meaningless
comvibnation of symbols mean something in the context the
theory.
If you want an arithmetic in which zero division is allowed
you must
give up all the other things you like about arithmentic in
order to be
consistent.
> There is currently no algebraic symbology to express this,
and if there
were
> it might lead to new understandings.
No it wouldn¹t. It would still be nonsens.
This is not like trying to Žnd the square root of minus one.
This can
be done by Žnding a broader algebraic structure in which
negative
numbers have square roots. There is no such extension to
allow division
by zero.
Bob Kolker
===
Subject: Re: Division by zero. Go ahead and laugh.
>>
> - I am not saying that division by zero should be allowed.
What I am
saying
> is that if it were allowed, then you would have problems.
> If it were allowed the system would be -inconsistent-
therefore useless.
> In an inconsistent mathematical system -anything- can be
proved.
You did¹nt even bother to read what I said. Or if you did,
you certainly
did¹nt understand what I was saying.
Where did I say that one must allow division by zero ? I
did¹nt. I said
that
it needs to be understood in the proper context instead of
burying your
head
in the sand and pretending that it does not exist. Clearly it
DOES
exist,
otherwise we would not be having this conversation. Dont you
agree ?
> Again, this is
> known. Division by zero must be isolated somehow so that
arithmetic
will
> remain intact.
> Impossible. When you allow something you also allow the
consequence.
Undoubtedly. And I also qualiŽed that by stating that it must
be
isolated
to prevent your consequences.
Seriously, this conversation is just a little bit more than
highschool
level
stuff.
You can, if you wish, deŽne an operator, number of other
object which
could
be added to R such that arithmetic could be consistent even
if you allow
division by zero.
For example, let @ be the object which represents
inconsistency. Maybe @ is
a number. Maybe it¹s not. I dont know. Maybe you should get
creative,
Žgure
it out, then YOU can tell ME what it is.
If @ was contained in R, or maybe an R¹ or something, then
you could say
that division is wellbehaved everywhere except that 1/0 = @.
This can all
be
done by deŽnition. Sure seems more realistic that just
stating that things
do not exist - as if -.
> BUT - deŽning it to not exist is incorrect. When you
> divide by zero you are _not_ creating a non-existence.
> Dividing by zero is meaningless.
> What you are
> creating is something else, possibly a contradiction,
paradox, or
> inconsistency. I think that this is due to topology of
space/time where
> mathematician lives.
> You are pissing up a rope. There is no way of making a
meaningless
> comvibnation of symbols mean something in the context the
theory.
> If you want an arithmetic in which zero division is allowed
you must
> give up all the other things you like about arithmentic in
order to be
> consistent.
> There is currently no algebraic symbology to express this,
and if there
were
> it might lead to new understandings.
> No it wouldn¹t. It would still be nonsens.
You are being very closed minded to this. I am not going to
pursue this for
the next 5 years. I need quality feedback. If you want to
criticize
professionally then please continue. If you want to ridicule
then please
move on.
===
Subject: Re: Division by zero. Go ahead and laugh.
> You are being very closed minded to this. I am not going to
pursue this
for
> the next 5 years. I need quality feedback. If you want to
criticize
> professionally then please continue. If you want to
ridicule then please
> move on.
You are right. I reject inconsistency root and branch. If tha
makes me
closed minded, then so be it. Logical inconsistency is the
death of
science and mathematics, both of which I care for a great
deal.
I am being both professional and charitable. Serious advocacy
of
inconsistency is either a bad joke (on your part) or a clear
symptom of
inanity. Which is it?
Bob Kolker
===
Subject: Re: Division by zero. Go ahead and laugh.
> Undoubtedly. And I also qualiŽed that by stating that it
must be
isolated
> to prevent your consequences.
In order to do that you have to eliminated logical
implication. Which
would render mathematics totally useless. Do you really want
to get rid
of implication and inference? Mathematics is all about
implication and
infernce. It is all about drawing conclusions from the axioms
of the
system. What you propose is the elimination of mathematics in
the entirety.
> Seriously, this conversation is just a little bit more than
highschool
level
> stuff.
> You can, if you wish, deŽne an operator, number of other
object which
could
> be added to R such that arithmetic could be consistent even
if you allow
> division by zero.
I have told you why that is not possible if you want to do any
mathematics. Why bother with axioms if you can¹t draw
conclusions from
the. Just write down isolated facts and forget any sort of
logically
hierarchical system.
> For example, let @ be the object which represents
inconsistency. Maybe @
is
> a number. Maybe it¹s not. I dont know. Maybe you should get
creative,
Žgure
> it out, then YOU can tell ME what it is.
It isn¹t a number. It is isn¹t anything that makes logical
sense. You
want a symbolic expression of inconsistency? Here it is: A
and not-A
where A is any propositon. There is your @.
Bob Kolker
===
Subject: Re: Division by zero. Go ahead and laugh.
>Here¹s my claim. You say 1/0 is undeŽned.
>>Pay attention. a/b = c if and only if a = bc. 1/0 cannot be
deŽned.
>>Why? if 1/0 = b then (by deŽnition) 1 = b*0 = 0. In short
if 1/0 had a
>>value then 1 = 0 which is impossible so 1/0 is not deŽned.
Another way
>>of saying the same thing is that a/b is deŽned if and only
if a = bx
>>has a solution x. 1/0 is a meaningless combination of marks
and is not
>>deŽned.
>>Bob Kolker
> I understand what you are saying exactly.And, at the risk
of sounding
like
> an argumentative imbecile, I continue to press -
> - I am not saying that division by zero should be allowed.
What I am
saying
> is that if it were allowed, then you would have problems.
Again, this is
> known. Division by zero must be isolated somehow so that
arithmetic will
> remain intact. BUT - deŽning it to not exist is incorrect.
When you
> divide by zero you are _not_ creating a non-existence. What
you are
> creating is something else, possibly a contradiction,
paradox, or
> inconsistency. I think that this is due to topology of
space/time where
> mathematician lives.
> There is currently no algebraic symbology to express this,
and if there
were
> it might lead to new understandings.
You might enjoy taking a look at:
Wheels [CapitalEth] On Division by Zero (thesis for the
licentiate degree),

by Jesper Carlstr.9am
===
Subject: Re: Division by zero. Go ahead and laugh.
> Here¹s my claim. You say 1/0 is undeŽned.
> Pay attention. a/b = c if and only if a = bc. 1/0 cannot be
deŽned.
Gosh, it¹s very curious then that complex analysis texts
which discuss C*,
the one-point compactiŽcation of C, deŽne 1/0. It¹s truly
amazing that
they do what cannot be done! What a mistake! Will you perhaps
take it
upon yourself to inform the authors of such texts of their
horrid error?
Of course, instead of doing that, it would actually be much
more proŽtable
for you to think about why you are incorrect in claiming,
bluntly, that 1/0
cannot be deŽned. (Hint: In systems allowing division of
nonzero
quantities by zero, division is not deŽned as you have
indicated.)
David Cantrell
> Why? if 1/0 = b then (by deŽnition) 1 = b*0 = 0. In short
if 1/0 had a
> value then 1 = 0 which is impossible so 1/0 is not deŽned.
Another way
> of saying the same thing is that a/b is deŽned if and only
if a = bx
> has a solution x. 1/0 is a meaningless combination of marks
and is not
> deŽned.
===
Subject: Re: Division by zero. Go ahead and laugh.
> Gosh, it¹s very curious then that complex analysis texts
which discuss
C*,
> the one-point compactiŽcation of C, deŽne 1/0.
What are the algebraic properties of the compactiŽcation
point. Can you
do arithmentic on them. Is the compactiŽcation point a
memeber of the
additive group of complex numbers? Does the compactiŽer
satisfy any
cancellation laws? Is it a member of the multiplicative group
of the
complex number - {0}? Tell us how to do complex arithmetic
with this
added compactiŽcation point. The compactiŽer is a topological
artifact
to assure the convergence of all sequences. It does not Žt in
with the
complex -Želd-.
It¹s truly amazing that
> they do what cannot be done! What a mistake! Will you
perhaps take it
> upon yourself to inform the authors of such texts of their
horrid error?
Learn the difference between algebraic properties and
topological
properties.
Bob Kolker
===
Subject: Re: Division by zero. Go ahead and laugh.
> Gosh, it¹s very curious then that complex analysis texts
which discuss
> C*, the one-point compactiŽcation of C, deŽne 1/0.
> What are the algebraic properties of the compactiŽcation
point. Can you
> do arithmentic on them.
I suspect those are merely rhetorical questions. (Indeed, I
suspect
that
you already knew that 1/0 is deŽned in C* and _chose_ to
ignore that fact
when you said, without qualiŽcation, that 1/0 cannot be
deŽned.) For the
answers to those questions, you may look at texts which
discuss C*. You
should Žnd something like
-(oo) = oo
oo + z = z + oo = oo for all z in C
oo * z = z * oo = oo for all nonzero z in C*
z/oo = 0 for all z in C
z/0 = oo for all nonzero z in C*
[Note that oo+oo, 0*oo and oo*0 are often undeŽned in C*.]
> Is the compactiŽcation point a memeber of the additive
group of
> complex numbers?
Of course not, because oo is not a member of C itself. But I
think that
what you intended to ask is whether C* is a group under
addition. Well,
of course it isn¹t, just as C is not a group under
multiplication.
> Does the compactiŽer satisfy any cancellation laws?
No. Similarly, if z*0 = w*0, we cannot conclude that z = w.
> Is it a member of the multiplicative group of the complex
number - {0}?
Like C itself, neither C* nor C* - {0} is a group under
multiplication.
But of course C* - {0, oo} is.
> Tell us how to do complex arithmetic with this added
compactiŽcation
> point.
I¹ve already covered that. See above.
If it disturbs you that some operations are often undeŽned in
C*, so be
it. The important thing for the purpose of this thread is
that z/0 _is_
deŽned for nonzero z. And if you knew that already but chose
to ignore it
when proclaiming, without qulaiŽcation, that 1/0 cannot be
deŽned, then
that tells us a good bit about you. (Are you perhaps thinking
about a new
career in politics?)
> The compactiŽer is a topological artifact
> to assure the convergence of all sequences.
Granted, I was the one who used the term compactiŽcation Žrst
in this
thread. But topological properties need not be discussed at
all. Perhaps
you would have been happier if I had called C* merely the
one-point
extension of C.
> It does not Žt in with the complex -Želd-.
Right, in the same sense that 0 does not Žt in with a
multiplicative
group.
> It¹s truly amazing that
> they do what cannot be done! What a mistake! Will you
perhaps take it
> upon yourself to inform the authors of such texts of their
horrid
> error?
> Learn the difference between algebraic properties and
topological
> properties.
I already know the difference.
David Cantrell
===
Subject: Re: Division by zero. Go ahead and laugh.
> If it disturbs you that some operations are often undeŽned
in C*, so be
> it. The important thing for the purpose of this thread is
that z/0 _is_
> deŽned for nonzero z. And if you knew that already but
chose to ignore
it
> when proclaiming, without qulaiŽcation, that 1/0 cannot be
deŽned, then
> that tells us a good bit about you. (Are you perhaps
thinking about a new
> career in politics?)
If 1/0 is deŽned a logical contradiction follows, to wit, 1 =
0. If you
like your systems to be consistent then forget about dividing
by zero.
The only place were dividing by zero makes any sense is in an
algebra
who only element is 0. Which is kind of useless.
Bob Kolker
===
Subject: Re: Division by zero. Go ahead and laugh.
> If 1/0 is deŽned a logical contradiction follows, to wit, 1
= 0. If you
> like your systems to be consistent then forget about
dividing by zero.
> The only place were dividing by zero makes any sense is in
an algebra
> who only element is 0. Which is kind of useless.
Precisely.
So what do we do ? We declare it to be undeŽned, or does not
exist.
What is really happening here is some type of disintegration
of logical
consistency at the singularity which is created by attempting
to divide
1/0.
Dont you Žnd this amazing ? I am perplexed by this. I Žnd it
really
incredible. Yes, you can deŽne things so that the whole
question just goes
away. But, if you dare to attack it, there could be something
to learn from
it.
===
Subject: Re: Division by zero. Go ahead and laugh.
>>If 1/0 is deŽned a logical contradiction follows, to wit, 1
= 0. If you
>>like your systems to be consistent then forget about
dividing by zero.
>>The only place were dividing by zero makes any sense is in
an algebra
>>who only element is 0. Which is kind of useless.
> Precisely.
> So what do we do ? We declare it to be undeŽned, or does not
exist.
We declare -division by 0- as undeŽned. 0 itself is a
perfectly nice
number and it is special being the identity of the additive
group.
Bob Kolker
> What is really happening here is some type of
disintegration of logical
> consistency at the singularity which is created by
attempting to divide
1/0.
There is no inconsistency. If division by 0 is allowed then 1
= 0 which
means everything is equal to 0.
> Dont you Žnd this amazing ? I am perplexed by this. I Žnd
it really
> incredible. Yes, you can deŽne things so that the whole
question just
goes
> away. But, if you dare to attack it, there could be
something to learn
from
> it.
The lesson to be learned is not to permit the rules of the
system to
lead to inconsistency. That is not the least bit astounding.
It is just
plain good sense.
Bob Kolker
===
Subject: Re: Division by zero. Go ahead and laugh.
> If it disturbs you that some operations are often undeŽned
in C*, so
> be it. The important thing for the purpose of this thread
is that z/0
> _is_ deŽned for nonzero z. And if you knew that already but
chose to
> ignore it when proclaiming, without qulaiŽcation, that 1/0
cannot be
> deŽned, then that tells us a good bit about you. (Are you
perhaps
> thinking about a new career in politics?)
> If 1/0 is deŽned a logical contradiction follows, to wit, 1
= 0.
No contradiction follows. But please, by all means, feel free
to exhibit
here how you think that 1 = 0 could be deduced after deŽning
1/0 in, say,
C*. If you have in mind what I suspect you do, then you will
have made
an unwarranted assumption in order to get your contradiction.
Namely, you
will have incorrectly assumed at some point that 0 has a
multiplicative
inverse, thereby allowing factors of 0 to be cancelled. But
of course
that¹s nonsense; 0 cannot have a multiplicative inverse, and
so cancelling
factors of 0 would be unjustiŽable.
> If you like your systems to be consistent then forget about
dividing by
> zero.
Wrong.
> The only place were dividing by zero makes any sense is in
an algebra
> who only element is 0.
Wrong.
David Cantrell
===
Subject: Re: Division by zero. Go ahead and laugh.
> No contradiction follows. But please, by all means, feel
free to exhibit
> here how you think that 1 = 0 could be deduced after
deŽning 1/0 in,
say,
> C*.
DeŽnition a/b = c if and only a = b*c. That is what division
is.
assume 1/0 = x. This means 1 = 0*x (deŽnition of division).
0*x = 0 (well known theorem for rings)
hence 1 = 0.
QED
If you have in mind what I suspect you do, then you will have
made
> an unwarranted assumption in order to get your
contradiction. Namely, you
> will have incorrectly assumed at some point that 0 has a
multiplicative
> inverse,
Division means precisely multiplicative inverse. That is what
it is.
My proof stands.
Bob Kolker
===
Subject: Re: Division by zero. Go ahead and laugh.
> No contradiction follows. But please, by all means, feel
free to
> exhibit here how you think that 1 = 0 could be deduced
after deŽning
> 1/0 in, say, C*.
> DeŽnition a/b = c if and only a = b*c. That is what
division is.
Fascinating. So according to _your_ deŽnition:
Since 0 = 0*0, we must have 0/0 = 0. And
since 0 = 0*1, we must also have 0/0 = 1.
Therefore, since 0/0 equals both 0 and 1,
by transitivity, we have 0 = 1.
Seriously now: Of course I know the deŽnition you intended to
give. And
there¹s clearly nothing wrong with using that deŽnition of
division in
some systems. But you were responding to a statement
concerning C*, for
heaven¹s sake! The deŽnition you intended to give is therefore
inappropriate, not only to C* but to any system in which
division of
nonzero quantities by 0 is deŽned.
> assume 1/0 = x. This means 1 = 0*x (deŽnition of division).
No, that¹s obviously _not_ what it means in C*.
> 0*x = 0 (well known theorem for rings)
> hence 1 = 0.
> QED
> If you have in mind what I suspect you do, then you will
have made
> an unwarranted assumption in order to get your
contradiction. Namely,
> you will have incorrectly assumed at some point that 0 has a
> multiplicative inverse,
> Division means precisely multiplicative inverse. That is
what it is.
No, in C* and other systems in which division of nonzero
quantities by 0 is
deŽned, division does not mean that.
> My proof stands.
It never even got off the žoor!
David Cantrell
===
Subject: Re: Division by zero. Go ahead and laugh.
>No contradiction follows. But please, by all means, feel
free to
>exhibit here how you think that 1 = 0 could be deduced after
deŽning
>1/0 in, say, C*.
>>DeŽnition a/b = c if and only a = b*c. That is what
division is.
> Fascinating. So according to _your_ deŽnition:
> Since 0 = 0*0, we must have 0/0 = 0. And
> since 0 = 0*1, we must also have 0/0 = 1.
> Therefore, since 0/0 equals both 0 and 1,
> by transitivity, we have 0 = 1.
FOr that reason 0/0 is not permitted either. It is not equal
to any
particular value therefore is not deŽned. Get it straight.
Dividing by
zero in any circumstance leads to inconsistency which is why
we don¹t
divide by zero.
0/0 is not a number.
Bob Kolker
===
Subject: Re: Division by zero. Go ahead and laugh.
> No, I do not divide by zero. But it did occur to me to
mention it here
as
> an illustration of mathematical culture.
> I have said before that mathematicians do not like
paradoxes and
> inconsistencies. It¹s all quite natural that they wouldn¹t,
and I don¹t
> blame them. But there are many who have voiced their
disagreement with
this
> opinion. Please let me qualify my belief.
> Division by zero creates an inconsistency in arithmetic,
and any result
can
> be derived if such a thing is allowed. Now then, instead of
always
> stipulating for example, 1/x, x not 0, we simply state that
1/0 is
> undeŽned, and avoid it completely. We avoid it like the
plague. The
only
> place I could Žnd these things being treated seriously is
in singularity
> theory. But in general, people do not even acknowledge that
this is a
> singularity. They just state žatly that 1/0 does not exist.
And
other
> indeterminate forms are treated pretty much the same.
> The fact that arithmetic falls apart when you let 1/0 is
fascinating and
> amazing. There is no reason to relegate this operation to
the trash can.
It
> would be better to acknowledge the inconsistency, the fact
that we don¹t
> really understand why it exists, and to simply work around
it.
> The problem of division by zero is that it creates many
paradoxes.
> Mathematicians hate this, and therefore they have deŽned
division by
zero
> to be non-existent. Why didn¹t you guys just throw away zero
altogether ?
> I¹ll tell you why. Because you want to keep zero around
because it¹s a
handy
> thing to have, but those damned paradoxes have got to go,
so we¹ll just
> deŽne them away.
> It¹s the biggest cover up since Watergate.
> There are no paradoxes or inconsistencies in mathematics,
unless you
remove
> all the ad hoc stipulations and band aids which hold it all
together.
> Confess.
Žnd a lot of answers about the work done on such problems.
Note that
1/0 in the usual sense is still undeŽned not because it¹s
being
supressed, simply because doing so producing interesting and
intuitive
results. Nothing prevents studying systems where division by
zero
works yet those objects are recognized as being distinct from
the
common length and area understanding of division.
Thinking in those terms, it¹s easy to see why division by 0 is
undeŽned in a geometric sense. What would 1/0 mean? It would
answer
the question Œif the area is 1, and the length of one side is
0, what
is the length of the other side¹ and we realize there is no
answer. We
see that for our rectangle to have a meaningful area it can¹t
have a
side of length 0. Even if the other side was inŽnite in
length the
area would still not be 1, and so we chose to leave it
undeŽned and
in that speciŽc context since one dimensional rectangles
don¹t make
sense. Other contexts where a zero divisor might arise exist,
yet
addition and division when seen in geometric terms clearly
doesn¹t
allow 1/0 to be meaningful or useful. I hope that gives you
some
insight.
===
Subject: Re: Division by zero. Go ahead and laugh.
> No, I do not divide by zero. But it did occur to me to
mention it
here
as
> an illustration of mathematical culture.
> I have said before that mathematicians do not like
paradoxes and
> inconsistencies. It¹s all quite natural that they wouldn¹t,
and I don¹t
> blame them. But there are many who have voiced their
disagreement with
this
> opinion. Please let me qualify my belief.
> Division by zero creates an inconsistency in arithmetic,
and any result
can
> be derived if such a thing is allowed. Now then, instead of
always
> stipulating for example, 1/x, x not 0, we simply state that
1/0 is
> undeŽned, and avoid it completely. We avoid it like the
plague.
The
only
> place I could Žnd these things being treated seriously is in
singularity
> theory. But in general, people do not even acknowledge that
this is a
> singularity. They just state žatly that 1/0 does not exist.
And
other
> indeterminate forms are treated pretty much the same.
> The fact that arithmetic falls apart when you let 1/0 is
fascinating
and
> amazing. There is no reason to relegate this operation to
the trash
can.
It
> would be better to acknowledge the inconsistency, the fact
that we
don¹t
> really understand why it exists, and to simply work around
it.
> The problem of division by zero is that it creates many
paradoxes.
> Mathematicians hate this, and therefore they have deŽned
division by
zero
> to be non-existent. Why didn¹t you guys just throw away zero
altogether ?
> I¹ll tell you why. Because you want to keep zero around
because it¹s a
handy
> thing to have, but those damned paradoxes have got to go,
so we¹ll just
> deŽne them away.
> It¹s the biggest cover up since Watergate.
> There are no paradoxes or inconsistencies in mathematics,
unless you
remove
> all the ad hoc stipulations and band aids which hold it all
together.
> Confess.
> Žnd a lot of answers about the work done on such problems.
Note that
> 1/0 in the usual sense is still undeŽned not because it¹s
being
> supressed, simply because doing so producing interesting
and intuitive
> results. Nothing prevents studying systems where division
by zero
> works yet those objects are recognized as being distinct
from the
> common length and area understanding of division.
> Thinking in those terms, it¹s easy to see why division by 0
is
> undeŽned in a geometric sense. What would 1/0 mean? It
would answer
> the question Œif the area is 1, and the length of one side
is 0, what
> is the length of the other side¹ and we realize there is no
answer. We
> see that for our rectangle to have a meaningful area it
can¹t have a
> side of length 0. Even if the other side was inŽnite in
length the
> area would still not be 1, and so we chose to leave it
undeŽned and
> in that speciŽc context since one dimensional rectangles
don¹t make
> sense. Other contexts where a zero divisor might arise
exist, yet
> addition and division when seen in geometric terms clearly
doesn¹t
> allow 1/0 to be meaningful or useful. I hope that gives you
some
> insight.
It has given me much appreciated insight.
Numbers are metaphysical entities. They are abstractions. I
am not sure if
these quantities can really be said to exist in the physical
universe.
Obviously they seem to, but this would make an enormous
argument.
However, there are two numbers which seem to Žt the physical
world almost
perfectly. Zero and 1. I wont bore you with my musings except
to say that
when considering these numbers, it is almost as if the
abstract world is
bridged to the physical universe by the existence of zero and
1. Zero and 1
must exist in the real world. It almost resembles the
dichotomy of
existence, true or false.
Certainly you can have 1 banana. And certainly you can have
zero bananas.
The other numbers are all quite dubious in the real world.
Might have something to do with why there are singularities.
Or maybe it¹s
the whiskey talkin¹ ?
Who knows.
===
Subject: Re: Division by zero. Go ahead and laugh.
> However, there are two numbers which seem to Žt the
physical world
almost
> perfectly. Zero and 1. I wont bore you with my musings
except to say that
> when considering these numbers, it is almost as if the
abstract world is
> bridged to the physical universe by the existence of zero
and 1. Zero and
1
> must exist in the real world. It almost resembles the
dichotomy of
> existence, true or false.
Just when we thought Plato was dead and buried, you shoed up.
Go away.
Bob Kolker
===
Subject: Re: Division by zero. Go ahead and laugh.
>> However, there are two numbers which seem to Žt the
physical world
>> almost perfectly. Zero and 1. I wont bore you with my
musings except
>> to say that when considering these numbers, it is almost
as if the
>> abstract world is bridged to the physical universe by the
existenceof
>> zero and 1. Zero and 1 must exist in the real world. It
almostresembles

>> the dichotomy of existence, true or false.
> Just when we thought Plato was dead and buried, you shoed
up. Go away.
No no. That sounds interesting. Please Lefty tell us more.
--
amount and biggest
1+1+1+...= inŽnite and Žnite in math today
Petri Keckman
===
Subject: Re: Division by zero. Go ahead and laugh.
>> However, there are two numbers which seem to Žt the
physical world
>> almost perfectly. Zero and 1. I wont bore you with my
musings except
>> to say that when considering these numbers, it is almost
as if the
>> abstract world is bridged to the physical universe by the
existenceof
>> zero and 1. Zero and 1 must exist in the real world. It
almostresembles
>> the dichotomy of existence, true or false.
> Just when we thought Plato was dead and buried, you shoed
up. Go away.
> No no. That sounds interesting. Please Lefty tell us more.
I promise that I will not argue about this for the next 5
years. I imagine
it will probably show up on one of those usenet crank
websites, who
cares.
I reserve the right to imagine whatever the hell I please.
Let freedom
reign.
===
Subject: attn : Keckman : Division by zero. Go ahead and
laugh.
Lets say that you wanted to make a fundamental mathematical
statement about
the physical universe. Please take a look at this axiom. I
think that it
says something about numbers and theri relationship to the
physical
universe.
Please critique this -
Maybe I lost my mind, I dont care. Please tell me what you
think.
----------------------------------------------
Axiom
No two objects in the physical universe can ever be
identical. Even if they
are identical in every respect, they still occupy two
separate locations,
again making them different. You can have one of something,
and you
can have zero of something, but that¹s all, because no two
objects can
ever be identical.
Proof.
suppose not,
contradiction,
qed,
bla bla bla
===
Subject: Re: attn : Keckman : Division by zero. Go ahead and
laugh.
> You can have one of something, and you can have zero of
something,but
> that¹s all, because no two objects can
> ever be identical.
Hey. I¹m clad i found friends here because i too like to
wonder things

like that.
I do not agree you. I think what comes to amounts there can
only be one.
That there isn¹t any something is nothing. It is nonsense.
Zero is not a
number.
That¹s why i observe the Naturals starting from number 1, not
from zero,
like
Peano did.
But what i think there is: there is true and false, on/off.
And sometimes
those are meaned like numbers one and zero, but in deeper
understanding
they are not.
In universum something exist. And it¹s amount is 1. We can¹t
count those
things that does not
exist.
Let¹s go ahead...If you have any agrees/not agrees, tell.
--
amount and biggest
1+1+1+...= inŽnite and Žnite in math today
Petri Keckman
===
Subject: Re: attn : Keckman : Division by zero. Go ahead and
laugh.
> You can have one of something, and you can have zero of
something,but
> that¹s all, because no two objects can
> ever be identical.
> Hey. I¹m clad i found friends here because i too like to
wonder
things
> like that.
> I do not agree you. I think what comes to amounts there can
only be one.
> That there isn¹t any something is nothing. It is nonsense.
Zero is not a
> number.
> That¹s why i observe the Naturals starting from number 1,
not from zero,
> like
> Peano did.
> But what i think there is: there is true and false, on/off.
And sometimes
> those are meaned like numbers one and zero, but in deeper
understanding
> they are not.
> In universum something exist. And it¹s amount is 1. We
can¹t count those
> things that does not
> exist.
> Let¹s go ahead...If you have any agrees/not agrees, tell.
> --
> amount and biggest
> 1+1+1+...= inŽnite and Žnite in math today
> Petri Keckman
I¹m irritated by a physics which consists of hodgepodge
mathematical models
which are basically things which are just pasted together
from algebra.
You have two separate worlds. The abstract world of math, and
the physical
universe. Math is traditionally exported from the abstract
world into the
real world in the form of models. I think that physics is
suffering from
this. Physics should be a mathematics in it¹s own right. And
in thinking
about this, it seems that the best place to start is some
type of
understanding of what a number is in the physical universe.
Are there
numbers which really exist and are not merely abstractions ?
I think so.
And
I think that those numbers would be 1 and 0. You can have
something, and
then you can have the abscence of something. You must be able
to deŽne
non-existence otherwise everything would exist. Clearly there
are
things
which do not exist.
If something exists you have 1 of that thing. You cannot have
two, because
no two things can be identical. You can only have one of any
item.
You can have zero of an item. This merely means that
something does not
exist in the physical universe.
It is problematic to say that you have 1.5 inches, or 1/2
mile, or .75
liters. These things are quite impossible. Maybe you can come
close, but
it¹s always just an illusion. You will never be able to
obtain 2.5 inches
of
material in the physical universe, it is physically
impossible to obtain
_exactly_ 2.5 inches of material, for example.
So you have 1 and zero. Two numbers which seem to exist very
precisely
in
both the physical world and the abstract world. And this
whole thing hinges
on the fact that no two physical objects can ever be
identical.
If you acknowledge this as a fundamental axiom, you may be
able to
reconstruct physics as an axiomatized system instead of a
hodge podge of
algebraic contraptions.
===
Subject: Re: attn : Keckman : Division by zero. Go ahead and
laugh.
> You can have one of something, and you can have zero of
something,but
> that¹s all, because no two objects can
> ever be identical.
> Hey. I¹m clad i found friends here because i too like to
wonder things

> like that.
Great. Now you two can start a new group (maybe
alt.math.loons) and
===
Subject: Re: attn : Keckman : Division by zero. Go ahead and
laugh.
> You can have one of something, and you can have zero of
something,but
> that¹s all, because no two objects can
> ever be identical.
>
> Hey. I¹m clad i found friends here because i too like to
wonder
things
> like that.
> Great. Now you two can start a new group (maybe
alt.math.loons) and
I¹m sorry you feel that way Fenix. I never gave much of a
damn about
philosophy, but we are discussing philosophical mathematics
and it is on
topic.
And, unless you understand the universe which you are sitting
inside of,
you
will never know if your logic is well behaved. You must
understand
space/time if you want to consider your logical calculus to
be absolutely
bulletproof. It is far from being bullshit.
It seems that paradoxes may exist in the general mathematical
model,
possibly because of the topology of space/time.
Fenix - this is not a waste of time.
I do not have any other tools to attack this problem and must
fall back on
philosophy. Do you know of a better tool to use in answering
this question
?
===
Subject: Re: attn : Keckman : Division by zero. Go ahead and
laugh.
> Great. Now you two can start a new group (maybe
alt.math.loons) and
We could do it here too, no need to read these.
There are Žlters in newsreaders. Why don¹t you use it?
If you still try somehow to disturb this and answer this,
i wonder what is your meaning. To read or not to read
this math bullshit.
--
amount and biggest
1+1+1+...= inŽnite and Žnite in math today
Petri Keckman
===
Subject: Re: Would NASA impress?
<4162f4a9$40$fuzhry+tra$mr2ice@news.patriot.net>
<cjvi08$t79$1@nntp.itservices.ubc.ca>
X-CompuServe-Customer: Yes
X-Coriate: interspeed.co.nz
X-Ecrate: tanandtanlawyers.com
X-Pose: George Cox
X-Punge: Micro$oft
X-Sanguinate: The MVS Guy
X-Terminate: SPA(GIS)
X-Tinguish: Mark GrifŽth
X-Treme: C&C,DWS
>The word is written in 1 Kings 7:23 as kuf-vav-hei
(numerical value
>100 + 6 + 5 = 111), but is read (and written in 2 Chronicles
4:2) as
>kuf-vav (numerical value 100 + 6 = 111).
Okay. Then he gave the wrong citation but was otherwise
correct[1]. I
have no idea how that hei got in there, but am reluctant to
ascribe it
to a scribal error.
[1] Torah means teaching, and is sometimes applied to more
than the
Chumash.
--
Shmuel (Seymour J.) Metz, SysProg and JOAT
<http://patriot.net/~shmuel>
Unsolicited bulk E-mail 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: Would NASA impress?
<cjf77t$9hn$1@news.LF.net>
<874qlg93zn.fsf@phiwumbda.org>
<9NM6d.97746$9Y5.48176@fed1read02>
<87acv8l2aj.fsf@phiwumbda.org>
<x5is9w83p3.fsf@lola.goethe.zz>
<873c0ykghf.fsf@phiwumbda.org>
X-CompuServe-Customer: Yes
X-Coriate: interspeed.co.nz
X-Ecrate: tanandtanlawyers.com
X-Pose: George Cox
X-Punge: Micro$oft
X-Sanguinate: The MVS Guy
X-Terminate: SPA(GIS)
X-Tinguish: Mark GrifŽth
X-Treme: C&C,DWS
>Leviticus 11:22-23
There¹s a sticky linguistic point there, although a literal
reading
would give that impression. However, even the people that
claim to
read the Bible literally will insist that, e.g., Shir
Hashirim (The
Song of Song¹s that is Shlomoh¹s), is a metaphor.
--
Shmuel (Seymour J.) Metz, SysProg and JOAT
<http://patriot.net/~shmuel>
Unsolicited bulk E-mail 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: Would NASA impress?
<874qlg93zn.fsf@phiwumbda.org>
<9NM6d.97746$9Y5.48176@fed1read02>
<87acv8l2aj.fsf@phiwumbda.org> <x5is9w83p3.fsf@lola.goethe.zz>
<873c0ykghf.fsf@phiwumbda.org>
<41658bfe$17$fuzhry+tra$mr2ice@news.patriot.net>
Discussion, linux)
>>Leviticus 11:22-23
> There¹s a sticky linguistic point there,
What *is* the sticky linguistic point?
> although a literal reading would give that impression.
However, even
> the people that claim to read the Bible literally will
insist that,
> e.g., Shir Hashirim (The Song of Song¹s that is Shlomoh¹s),
is a
> metaphor.
Of course, this passage is less likely metaphor, right?
,----
| 20 All fowls that creep, going upon all four, shall be an
abomination
| unto you. 21 Yet these may ye eat of every žying creeping
thing that
| goeth upon all four, which have legs above their feet, to
leap withal
| upon the earth; 22 Even these of them ye may eat; the
locust after his
| kind, and the bald locust after his kind, and the beetle
after his
| kind, and the grasshopper after his kind. 23 But all other
žying
| creeping things, which have four feet, shall be an
abomination unto
| you.
`----
I don¹t think it¹s particularly important or interesting that
the
count is wrong (I¹ve never been much for literal inerrancy),
but it
does seem to be straightforwardly wrong.
--
Jesse F. Hughes
Well, you know as soon as you have a new number I will be
happy to
add it to the list. Don¹t try those childish tit-for-tat
games with
me. -- Ross Finlayson on Cantor¹s theorem.
===
Subject: Re: Would NASA impress?
>Leviticus 11:22-23
>> There¹s a sticky linguistic point there,
>What *is* the sticky linguistic point?
...
>| 23 But all other žying
>| creeping things, which have four feet
...
>I don¹t think it¹s particularly important or interesting
that the
>count is wrong (I¹ve never been much for literal inerrancy),
but it
>does seem to be straightforwardly wrong.
Perhaps Shmuel¹s sticky linguistic point is that, *literally*,
things which have six feet also *do* have four feet (in 15
different ways, at that). Of course Gricean presuppositions,
and
all that (as expounded, for instance, in McCawley¹s book
`Everything that Linguists have Always Wanted to Know about
Logic (but were afraid to ask)¹), mean that this particular
version of literal interpretation is (quite rightly) taken
to be a fraud when presented as part of everyday talk. But
theological exegesis is not, thank G-d, everyday talk.
Lee Rudolph (eisegesis done cheap!)
===
Subject: Re: measurable function

>Let T:W->(W¹,A¹), where A¹ is a sigma algebra in W¹. Show
that
>T^-1(A¹) is a sigma algebra and it is a smallest sigma
algebra such
>that T is measurable.
>Please help me with this problem.
Hint: apply the deŽnition of a sigma algebra to T^{-1}A¹.
What can
you say about unions and intersections?
--
Shmuel (Seymour J.) Metz, SysProg and JOAT
<http://patriot.net/~shmuel>
Unsolicited bulk E-mail 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: Monte Carlo experiment
> I¹ll try again, although I am not sure what you mean by your
> question: [ ... ]
Your explanation demonstrates that you have understood quite
well what I mean.
Han de Bruijn
===
Subject: Re: Help me ,Intermediate Chebyshev polynomials T(r
,x) ,
abs(x) =< 1
>
> Chebyshev polynomials are very well known.
> Here x in(-1,1)
> T2=2x^2-1, T3=4*x^3-3*x, T4=8*x^4-8*x^3+1 ...
> Since T2=cos(2*cos^[-1](x)) ,I believe we may compute Tr
for all real
values
> of r.
> That the reason why I use the word INTERMEDIATE,
>
> Well, Chebyshev polynomial T_n(x) is a special case of
Jacobi
> polynomial P_n^{(alpha,beta)}(x) which can be deŽned using
a 2F1
> hypergeometric function. That, in turn, makes sense for
non-integer n,
> replacing factorials by gamma functions.
> But is that result any better than cos(n*arccos(x)) for
non-integer n ?
Here is what Maple help says:
If the Žrst parameter is a non-negative integer, the
ChebyshevT(n, x)
function computes the nth Chebyshev polynomial of the Žrst
kind
evaluated at x.
[...]
For n not equal to a non-negative integer, the analytic
extension of
the Chebyshev polynomial of the Žrst kind is speciŽed by:
ChebyshevT(n,x) = hypergeom([-n,n],[1/2],(1-x)/2)
Here are some samples...
> ChebyshevT(1/2,x);
/1
ChebyshevT|-, x|
2 /
> simplify(%);
1 (1/2)
- (2 x + 2)
2
> ChebyshevT(sqrt(2),x);
/ (1/2)
ChebyshevT2 , x/
> simplify(%);
/ (1/2)
ChebyshevT2 , x/
> series(%,x=1);
1 2 2 3 1 4 7 5
1 + 2 (x - 1) + - (x - 1) - -- (x - 1) + -- (x - 1) - ---- (x
- 1)
3 45 90 2025
/ 6
+ O(x - 1) /
> ChebyshevT(sqrt(-1),x);
ChebyshevT(I, x)
> simplify(%);
ChebyshevT(I, x)
> series(%,x=1);
1 2 1 3 5 4 17 5
1 - (x - 1) + - (x - 1) - - (x - 1) + --- (x - 1) - ---- (x -
1) +
3 9 126 1134
/ 6
O(x - 1) /
--
G. A. Edgar
http://www.math.ohio-state.edu/~edgar/
===
Subject: Re: Probability of rolling snake eyes
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id i98EPF810132;
>My answer and my son¹s answer is 1 in 36. My son¹s teacher
says that
>is not the correct answer. Help?
Assuming two fair independent dice and one roll, the teacher
is wrong.
===
Subject: Re: Spherical geometry
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id i98ERaC10316;
>Hi all,
>I¹m doing a project in my Fourier Analysis class on
spherical geometry,
does
>anyone know of any good books or reference material? I am an
undergrad, so
I
>would need information at that level please :)
>John
John,
Go to Amazon.com and search on Spherical Astronomy. Astronomy
is the
science that uses spherical trig the most, so all the good
books on spherical
trig are astronomy books.
- MO
===
Subject: Re: Spherical Geometry
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id i98EPG910156;
><pre>
>Hello I have to create a report by comparing and contrasting
plane and
>spherical geometry. COuld someone help me with some examples
please?
>It would be much appreciated thanx!
>GSMILES@worldnet.att.net
></pre>
Derrick,
Here are a couple of nice things to compare:
1) Sum of Anlges: Let sigma = sum of the angles. Then in the
plane sigma =
pi, whereas on the sphere sigma > pi. Extra: what is the
upper bound on
sigma on the sphere?
2) Area of the triangle: In the plane it is A = (1/2)bh,
where b is the
base and h the height. On the sphere it is A = sigma - pi,
where sigma is
the sum of the angles.
3) Law of sines: On the plane it is a/sin(alpha) =
b/sin(beta) =
c/sin(gamma), where alpha (beta, gamma) is the angle opposite
the side a (b,
c). On the sphere it is sin(a)/sin(alpha) = sin(b)/sin(beta) =
sin(c)/sin(gamma).
4) Pythagorean Theorem: On the plane it is a^2 + b^2 = c^2,
on the sphere it
is cos(a)cos(b) = cos(c).
5) Law of cosines (generalization of Pythagoream Theorem): On
the plane it
is c^2 = a^2 + b^2 - 2(ab)cos(gamma), where a, b, c are the
sides of the
triangle and gamma is the anlge opposite c. On the sphere it
is cos(c) =
cos(a)cos(b) + sin(a)sin(b)cos(gamma). Extra: the law of
cosines for angles
on a sphere is
cos(gamma) = -cos(alpha)cos(beta) + sin(alpha)sin(beta)cos(c)
Hope these little tid-bits help.
- MO
===
Subject: Re: Spherical Geometry
><pre>
>Hello I have to create a report by comparing and contrasting
plane and
>spherical geometry. Could someone help me with some examples
please?
>It would be much appreciated thanx!
>GSMILES@worldnet.att.net
> Derrick,
> Here are a couple of nice things to compare:
> 1) Sum of Angles: Let sigma = sum of the angles. Then in
the plane sigma
= pi, whereas on the sphere sigma > pi. Extra: what is the
upper bound on
sigma on the sphere?
> 2) Area of the triangle: In the plane it is A = (1/2)bh,
where b is the
base and h the height. On the sphere it is A = sigma - pi,
where sigma is
the sum of the angles.
A = (sigma - pi) * R^2, R sphere radius, R is taken unity
here.
> 3) Law of sines: On the plane it is a/sin(alpha) =
b/sin(beta) =
c/sin(gamma), where alpha (beta, gamma) is the angle opposite
the side a (b,
c).
Permit me to add:
In the plane it is the circum-diameter. BUT,
> On the sphere it is sin(a)/sin(alpha) = sin(b)/sin(beta) =
sin(c)/sin(gamma).
what this geometrical parameter is, has never been identiŽed
so far
AFAIK .I had posted twice in this NG earlier. I am curious to
know,
not its expression but the non-dimensional geometrical
parameter it
represents. You had given all corrosponding quantities in
plane and
sphereical geometries, but this one is unidentiŽed, to me it
appears
as a blank in our knowledge about the Sine Rule. It has
relevance to
non-euclidean geometries, or so I believe.
> 4) Pythagorean Theorem: On the plane it is a^2 + b^2 = c^2,
on the sphere
it is cos(a)cos(b) = cos(c).
> 5) Law of cosines (generalization of Pythagoream Theorem):
On the plane it
is c^2 = a^2 + b^2 - 2(ab)cos(gamma), where a, b, c are the
sides of the
triangle and gamma is the anlge opposite c. On the sphere it
is cos(c) =
cos(a)cos(b) + sin(a)sin(b)cos(gamma). Extra: the law of
cosines for angles
on a sphere is
> cos(gamma) = -cos(alpha)cos(beta) +
sin(alpha)sin(beta)cos(c)
> Hope these little tid-bits help.
> - MO
===
Subject: More about the sum of odd numbers
Starting from the ridiculous supposition that the sum of any
two odd
numbers is always even we quickly contradict ourselves. First
let¹s
talk about what odd means. Odd and even are just special
cases of
divisibility. An odd number mod 2 will be 1.. or in other
words
non-zero. An even number mod 2 will be 0. You could just as
easily
extend it to higher numbers. Any number mod 3 that doesn¹t
equate zero
is also odd. In general, any number mod some positive integer
n not
being zero is going to be odd. This is nothing new. This is
the
generally accepted and understood deŽnition about the general
nature
of evenness and oddness.
even means evenly divisible by
2 mod 2 = 0 = even
3 mod 2 = 1 = odd
etc.
2 mod 3 = 2 = odd
3 mod 3 = 0 = even
Well which are we to deal with? a number clearly can¹t be
both odd and
even.
Take two odd numbers: 2j+1 and 2k+1, allowing j and k to
differ by an
odd amount we have 2j+1 and 2(j+(2r+1)) +1.
The sum being: 4j + 4r + 4 = 4(j + r + 1) = 2(2(j+r+1))
2 mod 3 = 2 = odd
odd * odd * (j + r + 1)
Sure, j+r+1 can be anything we wish. In the case where it¹s
odd we
have that the sum of two odds equates to the product of three
odds.
In the case where it¹s even we have that the sum of two odds
equates
to the product of an even with two odds. In one case, the sum
of two
odds is an odd. In the other, it is indeed even.
Some here tend to claim that in general the sum of two odds
will
equate to an odd sum. This is hypocricy when they attack
fallacious
arguments on the grounds of claiming to apply to the general
case when
they only apply to a speciŽc case. Here we have a perfect
example of
a generalization which turns out to be nothing more than a
special
case.
===
Subject: Re: More about the sum of odd numbers
X-RFC2646: Format=Flowed; Original
> Starting from the ridiculous supposition that the sum of
any two odd
> numbers is always even we quickly contradict ourselves.
First let¹s
> talk about what odd means. Odd and even are just special
cases of
> divisibility. An odd number mod 2 will be 1.. or in other
words
> non-zero. An even number mod 2 will be 0. You could just as
easily
> extend it to higher numbers. Any number mod 3 that doesn¹t
equate zero
> is also odd. In general, any number mod some positive
integer n not
> being zero is going to be odd. This is nothing new. This is
the
> generally accepted and understood deŽnition about the
general nature
> of evenness and oddness.
> even means evenly divisible by
Even means either divisible by 2 or congruent to 0 mod 2.
Even does
not
mean evenly divisible by anything other than 2. For example,
3 is odd
but
evenly divisible by three.
HTH.
Michael
===
Subject: Re: More about the sum of odd numbers
> Starting from the ridiculous supposition that the sum of
any two odd
> numbers is always even we quickly contradict ourselves.
First let¹s
> talk about what odd means. Odd and even are just special
cases of
> divisibility. An odd number mod 2 will be 1.. or in other
words
> non-zero. An even number mod 2 will be 0. You could just as
easily
> extend it to higher numbers. Any number mod 3 that doesn¹t
equate zero
> is also odd. In general, any number mod some positive
integer n not
> being zero is going to be odd. This is nothing new. This is
the
> generally accepted and understood deŽnition about the
general nature
> of evenness and oddness.
> even means evenly divisible by
> 2 mod 2 = 0 = even
> 3 mod 2 = 1 = odd
> etc.
> 2 mod 3 = 2 = odd
> 3 mod 3 = 0 = even
> Well which are we to deal with? a number clearly can¹t be
both odd and
> even.
Bravo, Mr. Bandel. You have clearly outdone yourself in
the clearest most possible way of stating the most ridiculous
proof. Ok, maybe not the most ridiculous, but certainly up
there.
Should I?
By the general Œeven¹ and Œodd¹, you do have the deŽnition
right
but you need to state the context of odd and eveness. IIRC,
the general context of Œoddness¹ and Œeven-ness¹ is based on
mod 2.
By throwing the deŽnition out the window and replacing it
with what you want (ie. mod 3), you have entered the realm of
Œwhatever the hell makes you happy¹.
You are proving the sum of two odds equal to an odd.
Your implied(should I even say this here?) context is
mod 2. Which is your initial starting point 2 odd #s
2j+1 and 2k+1, with the two differing by an odd #.. ie.
j = k + (2r+1) for some r in Z (I really HOPE you are
talking about the Integers, and not some complex imaginary
realm).
Now, in the proof, you state mod 3.
At that point, there is little point in continuing.
By stating a different context within your proof, you have
commited a fallacy; in that, you are not playing by the
rules and by doing that, you get whatever results you want.
That, as far as I can remember from my math courses, is
going to get you a F.
If you want a decent discussion/proof session, play
by the damn rules. Not by what you Žniky feel like.
Why don¹t you just be a man and admit you don¹t understand
what you are talking about.
===
Subject: Re: More about the sum of odd numbers
Trivial, Troll
Teapot Tempest
Filled with ridiculous, hypocricy, + fallacious arguments
===
Subject: Re: Proposal For New Newsgroup
sci.math:
>How about alt.math.james-harris?
It¹s easy enough to create, but do you think he would conŽne
himself to it?
--
Stan Brown, Oak Road Systems, Tompkins 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: Proposal For New Newsgroup
X-RFC2646: Format=Flowed; Original
> sci.math:
>>How about alt.math.james-harris?
> It¹s easy enough to create, but do you think he would conŽne
> himself to it?
IF we only posted replies to James Harris on
alt.math.james-harris, he
would.
===
Subject: Re: Proposal For New Newsgroup
<Wvy9d.5$Mh7.2@trnddc04>
Discussion, linux)
>> sci.math:
>How about alt.math.james-harris?
>> It¹s easy enough to create, but do you think he would
conŽne
>> himself to it?
> IF we only posted replies to James Harris on
alt.math.james-harris, he
> would.
A long time ago, James cared about replies. Nowadays, his
caring
comes and goes, but he mostly seems to think of himself as a
genius
broadcasting to the masses. He¹s not so interested in
conversation as
monologue.
That¹s not to say he doesn¹t participate in an occasional
conversation, but I doubt he¹d go look for one. I¹d guess
that he¹s
much more interested in audience size than in an exchange of
ideas.
Sorry, but I predict that alt.math.james-harris will fail
utterly.
--
This page contains information of a type (text/html) that can
only be
viewed with the appropriate Plug-in. Click OK to download
Plugin.
--- Netscape 4.7 error message
===
Subject: Re: Proposal For New Newsgroup
X-RFC2646: Format=Flowed; Response
> Hello Frank,
> That would be a great idea, why not just do it?
Well, I have taken the Žrst step: I submitted a proposal to
the alt.conŽg

newsgroup. After allowing at least a week for feedback on
this proposal, I
will then send a newgroup message. Feel free to go to
alt.conŽg and
critique the proposal. I promise to pay close attention to
any advice on
this proposal, for I would like this new group to be
available on as many
sites as possible.
===
Subject: Re: Proposal For New Newsgroup
>How about alt.math.james-harris?
That¹s an oxymoron. How about misc.Žction. jh ??
===
Subject: Re: Proposal For New Newsgroup
> How about alt.math.james-harris?
I wonder if he may be related to Bernard Harris, author of a
1966
book, Theory of Probability.
David Ames
===
Subject: Re: Proposal For New Newsgroup
>> How about alt.math.james-harris?
>I wonder if he may be related to Bernard Harris, author of a
1966
>book, Theory of Probability.
No.
--
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: Derivatives...
>Just when I thought I was beginning to Žgure it out.....
>If the tangent line to y = f(x) at (4,3) passes through the
point
>(0,2), Žnd f(4) and f¹(4).
>Looking through the book, I can¹t even see how to set this
one up.
What does f(4) mean? It means the value of the function f at
point
x=4. That¹s given in the problem.
What does f¹(4) mean? It means the slope of the tangent line
to the
function at x=4. You have two points in that line, so you have
enough information to calculate its slope.
Very often when you don¹t see how to work a problem, the best
advice
is to go back to the deŽnitions.
--
Stan Brown, Oak Road Systems, Tompkins 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: Derivatives...
>>Just when I thought I was beginning to Žgure it out.....
>>
>>If the tangent line to y = f(x) at (4,3) passes through the
point
>>(0,2), Žnd f(4) and f¹(4).
>>Looking through the book, I can¹t even see how to set this
one up.
> If f(x) passes through the point (4,3) this means f(4) = 3.
The
derivative
> of f(x) at (4,3) is equal to the slope of the tangent line
to f(x) at
(4,3).
> We know this tangent line passes through the points (4,3)
and (0,2).
> this one can calculate its slope m (Œrise over run¹) using
the formula
m
=
> (y1 - y2)/(x1 - x2). So we have m = (3-2)/(4-0) = 1/4. So
f¹(4) =
1/4.
> You know, it¹s not considered proper procedure to solve
other
> people¹s homework, even if you¹re offering a
detailed/helpful
> explanation as opposed to simply give the answer (though in
> this case, since a valid answer requires an explanation of
> the steps, this is arguably simply giving the answer).
> This is clearly a homework or practice exercise -- it is
much
> more helpful to guide the OP on how to solve it than to
simply
> solve it and show him.
ŒProper procedure¹? Anyways, it seemed to me that hints would
not have
been
beneŽcial to the OP and he¹d be better off with a concrete
example.
Especially considering that some of the answers are pretty
much Œby
deŽnition¹.
l8r, Mike N. Christoff
===
Subject: Compact vs Quasicompact
Just a while ago I was discussing with my prof. about
something in a
book. We came to term Quasicompactness, which I thought as
kinda wierd
coz it is just the deŽnition of compactness as I knew since
ages
ie. every open covering has a Žnite subcovering (that was the
deŽnition of quasicompactness in the book .. Hartshorne again
*sweat*)
Well my prof. just told me that I had the wrong deŽnition,
which just
shoked me. He said that compactness also includes
Hausdorfness. I was
totally, uhm speechless... I don¹t know whether I have been
having the
wrong deŽnition of compactness all this while or was it that
my
professor is just having a nonstandard deŽnition of
compactness. As far
as I know, or at least since the time I attended my Žrst
topology
class.. compactness is just what here my professor calls
quasicompact.
Would anyone please enlighten me.
Jose Capco
===
Subject: Re: Compact vs Quasicompact
> Just a while ago I was discussing with my prof. about
something in a
> book. We came to term Quasicompactness, which I thought as
kinda wierd

> coz it is just the deŽnition of compactness as I knew since
ages
> ie. every open covering has a Žnite subcovering (that was
the
> deŽnition of quasicompactness in the book .. Hartshorne
again *sweat*)
> Well my prof. just told me that I had the wrong deŽnition,
which just
> shoked me. He said that compactness also includes
Hausdorfness. I was
> totally, uhm speechless... I don¹t know whether I have been
having the
> wrong deŽnition of compactness all this while or was it
that my
> professor is just having a nonstandard deŽnition of
compactness. As far
> as I know, or at least since the time I attended my Žrst
topology
> class.. compactness is just what here my professor calls
quasicompact.
> Would anyone please enlighten me.
In my view, the professors comment that your deŽnition is
just wrong
is a bit harsh; perhaps he is not that familiar with the
topological
literature?
Some people prefer the pair compact and compact Hausdorff,
others
prefer the pair quasicompact and compact. The former seems to
be universal in english texts, while the latter can still be
found
in Bourbaki and older continental topological texts.
Perhaps you have already met the similar situation with
ring and ring with 1 vs. ring without one and ring.
I guess (and this is purely guessing, I do not have
information on this)
that the notion of compactness was Žrst introduced in times
when
only Hausdorff topological spaces were considered; when
non-Hausdorff
spaces found their way into mainstream topology it was not
clear, how
the notion of compactness should be transferred to the new
setting.
Personally, I prefer your version, which also seems to become
more
and more dominant. So, time is on your side.
However, as long as a text makes clear, which terms are used,
it should
not pose any problems to accept the (perhaps old-fashioned
:-) conventions
in algebraic geometry.
Marc
===
Subject: Re: Compact vs Quasicompact
> Well my prof. just told me that I had the wrong deŽnition,
which just
> shoked me. He said that compactness also includes
Hausdorfness.
Surely your professor was not that scholar in topology ...
===
Subject: Re: Compact vs Quasicompact
> Just a while ago I was discussing with my prof. about
something in a
> book. We came to term Quasicompactness, which I thought as
kinda wierd

> coz it is just the deŽnition of compactness as I knew since
ages
> ie. every open covering has a Žnite subcovering (that was
the
> deŽnition of quasicompactness in the book .. Hartshorne
again *sweat*)
> Well my prof. just told me that I had the wrong deŽnition,
which just
> shoked me. He said that compactness also includes
Hausdorfness. I was
> totally, uhm speechless... I don¹t know whether I have been
having the
> wrong deŽnition of compactness all this while or was it
that my
> professor is just having a nonstandard deŽnition of
compactness. As far
> as I know, or at least since the time I attended my Žrst
topology
> class.. compactness is just what here my professor calls
quasicompact.
> Would anyone please enlighten me.
> Jose Capco
That a Bourbakiism. In English we say:
(a) compact (b) compact Hausdorff
for the concepts Bourbaki called
(a) quasicompact (b) compact
the idea being (I guess) that the more common concept
deserves the
shorter name.
--
G. A. Edgar
http://www.math.ohio-state.edu/~edgar/
===
Subject: Re: Compact vs Quasicompact
> Just a while ago I was discussing with my prof. about
something in a
> book. We came to term Quasicompactness, which I thought as
kinda
wierd
> coz it is just the deŽnition of compactness as I knew since
ages
> ie. every open covering has a Žnite subcovering (that was
the
> deŽnition of quasicompactness in the book .. Hartshorne
again *sweat*)
> Well my prof. just told me that I had the wrong deŽnition,
which just
> shoked me. He said that compactness also includes
Hausdorfness. I was
> totally, uhm speechless... I don¹t know whether I have been
having the
> wrong deŽnition of compactness all this while or was it
that my
> professor is just having a nonstandard deŽnition of
compactness.
Topologists don¹t assume that compact spaces are Hausdorff,
but
algebraic geometers assume (or used to assume) that
topologists
assume that compact spaces are Hausdorff, so they call
compact spaces
quasicompact spaces.
Does that make sense ? :-)
--
Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html
Lacan, Jacques, 79, 91-92; mistakes his penis for a square
root, 88-9
Francis Wheen, _How Mumbo-Jumbo Conquered the World_
===
Subject: Re: The beast, again.
> The bestial primes:
> 61, 661, 6661, 6666666661, 666666666666666661,
6666666666666661
> 6666666666666666666661, 6666666666666666666666666661
As the Bestial Primes are of the form : 2*(10^n -1)/3 -5
The diabolical exponents n are:
2,3,4,10,18,21,22,28,43,66,121,133,178 ...
The Conjecture of the Beast : There are inŽnitely many
Bestial Primes
and are more numerous than the Mersenne Primes.
===
Subject: Re: The beast, again.
posting-account=c1-XewwAAAD2xIEFOZsEHyTOzSJpZ-h-
> The bestial primes:
> 61, 661, 6661, 6666666661, 666666666666666661,
6666666666666661
> 6666666666666666666661, 6666666666666666666666666661
> As the Bestial Primes are of the form : 2*(10^n -1)/3 -5
> The diabolical exponents n are:
2,3,4,10,18,21,22,28,43,66,121,133,178 ...
> The Conjecture of the Beast : There are inŽnitely many
Bestial
Primes
> and are more numerous than the Mersenne Primes.
a beastly magic square...
3 107 5 131 109 311
7 331 193 11 83 41
103 53 71 89 151 199
113 61 97 197 167 31
367 13 173 59 17 37
73 101 127 179 139 47
all entries are prime numbers, and each row, column, and
diagonal adds up to 666.
===
Subject: Re: The beast, again.
> We all know that Rev. 13:18 says 666 is the number of the
Beast,
> but did you know that...
> 670 - Approximate number of the Beast
> DCLXVI - Roman numeral of the Beast
> 666.0000000 - Number of the High Precision Beast
> 665.9999954 - Number of the Pentium Beast
> [...]
surprised no one has mentioned
666 sigma - Number of the quality beast
(see http://www.isixsigma.com/)
===
Subject: Re: Complex problems/roots...
>Hae everybody,
>(c + di) / (a + bi) = (ac + bd)/(a+b) +
(ad-bc)*i/(a+b)
>can somebody explain me why this is right?
Rationalize the denominator by multiplying top and bottom of
your
Žrst fraction by (a-bi).
--
Stan Brown, Oak Road Systems, Tompkins 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: need help
>was wondering if you knew how to enter the intergral of
dx/dt=x/t into a
ti-89 calculator i have asked everyone i know adn they can
not seem to Žgure
it out. thank you for your time
Can you do some transformation so that the left side has only
x and
the right side has only t?
--
Stan Brown, Oak Road Systems, Tompkins 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: ((.99999.... == 1) == 1) = 1
In sci.math, S. Enterprize Company
<smart1234@aol.com>
>>In sci.math, S. Enterprize Company
>><smart1234@aol.com>
>.99999... = sigma 9 * 10^-i from i = 1 to inŽnity
>
>10 * .9999999999.... = 10 * sigma 9 * 10^-i from i = 1 to
inŽnity
>
>10 * .9999999999.... = sigma 9 * 10^(1-i) from i = 1 to
inŽnity
>
> = 9*10^(1 - 1) + sigma 9 * 10^(1-i) from i = 2 to inŽnity
> = 9 + sigma 9 * 10^(1-i) from i = 2 to inŽnity
> = 9 + sigma 9 * 10^-i from i = 1 to inŽnity
> = 9 + .9999....
> = 9.9999....
>
>so now we know that 10 * .9999..... = 9.9999.....
>
>that¹s sufŽcient to show that .9999.... = 1.
>
>because 9 * .9999.... = 10 * .9999.... - .9999.... = 9
>
>and if at any point you doubt my rightness in doing
arithmetic like
>thi with an inŽnitely repeating decimal. just replace it
with the
>sigma and you will see that the algebra works out Žne.
>
>if you disprove this, then you¹ve made a mistake.
> Not one math function supports that .99999... = 1. This
shows you
> that there is more proof against .99999... = 1 than for it.
>>And why would a math function particularly care about
whether
>>two representations are equal...?
>>But ponder this for a moment.
>>Let x = .999... = 1 - d, for some weird non-zero d.
>>Contemplate the surrealism of the underlying metaphor, and
>>then contemplate (x + 9) / 10 and (x * 10) - 9.
>>x .999... 1 - d
>>(x + 9) / 10 .999... 1 - d/10 (right shift)
>>(x * 10) - 9 .999... 1 - 10*d (left shift)
>>Are they all the same number, or not?
>>For the Žnite case, where d = 10^(-n), one gets:
>>x .999...990 1 - d
>>(x + 9) / 10 .999...999 1 - d/10 (right shift)
>>(x * 10) - 9 .999...900 1 - 10*d (left shift)
>>but the last three digits vanish as one takes the limit.
>>Or one can attempt (.999...)^2, yielding 1 - 2d + d^2.
>>If we assume (.999...)^2 = .999... , then we have a solution
>>to the equation x^2 - x = 0, which has two roots, 0 and 1,
>>and .999... clearly isn¹t 0. If we do not assume (.999...)^2
>>= .999..., we are left with Yet Another Value for the
>>inŽnite expansion .999... , namely 1 - 2d + d^2.
>>I¹d have to do some work but presumably one can prove
>>that, if d_i = 9 for every integer i > 0, then
>>(sum (i = 1, +oo) (d_i * 10^(-i)))^2 = sum(i = 1, +oo) (p_i
* 10^(-i))
>>is such that p_i = 9 for every digit as well. One might
represent
>>in shorthand what¹s going on:
>>.9 * .999... = .8999999...
>>.09 * .999... = .0899999...
>>.009 * .999... = .0089999...
>>...
>>.0...09 * .999... = .0...089999...
>>If one adds the Žrst two, one gets .9899999... .
>>The Žrst three gives .9989999... .
>>The Žrst four gives .9998999... . This probably
>>can be used for the basis of an inductive proof, with
>>some care; i = 1 might correspond to the number
>>.[9]8, i = 2 to the addition .8[9]99 + .0[8]99, i = 3 to the
>>addition .89[9]999 + .08[9]999 + .00[8]999 , and so on.
>>Of course such a proof would be pointless Žddlywork for
most. :-)
>>If d is in fact non-zero, it¹s very weird, exhibiting many
of
>>the properties of inŽnities (and since inŽnity is not a real
>>number, well...). If d is zero, of course, .999... = 1, and
>>1 - d = 1 - d/10 = 1 - 10*d = 1 - 2d + d^2 = 1.
> Nope
> 1 - 2(0) + 0^2 =1
> 1 -2(1) + 1^2 = 0
> This doesn¹t at all prove .9999... = 1
An interesting point. However, what is .9999... - 1, then?
I¹ve been assuming .9999... - 1 = d (since .9999... = 1 - d,
1 - .9999... = 1 - (1 - d) = d) but you can perhaps explain
your logic here.
Also, if d = 10d = d/10 = 2d - d^2, then one has an
interesting
non-zero number d. After all, oo = 10 * oo = oo / 10 ( 2*oo -
oo^2
is undeŽned).
>>You may be convinced by this, or not -- but it¹s clear that,
>>if .999... != 1, .999... represents a whole bunch of very
near
>>numbers on or around 1 (and can be taken as near as desired
to
>>1, without actually reaching it; if x = .999... is not near
>>enough, take (x + 9) / 10 to get a little nearer).
>>So choose.
>>[1] .999... = 1.
>>[2] .999... is an entire new class of numbers which are of
little or no
>> utility in contemporary mathematics.
>>[3] .999... is a meaningless set of symbols.
>>[rest snipped]
>>--
>>#191, ewill3@earthlink.net
>>It¹s still legal to go .sigless.
> Smart¹s Alt. Physics News Group
>
http://pub39.bravenet.com/forum/show.php?usernum=3320272813&
cpv=1
> S. Enterprize (Science Journal)
> http://smart1234.s-enterprize.com/
--
#191, ewill3@earthlink.net
It¹s still legal to go .sigless.
===
Subject: Re: The coefŽcient of the power of the polynomials
m is a very small number, say 3,4,5, while n is generally
greater than
1000.
Can you give me some details on that algorithm?
Yes, there¹s a straightforward algorithm which runs in
O((m+1)^n). How
large are your (m,n)?
The solution can be derived from an (m+1)-nomial number
triangle. You
might
be able to write something quicker using that.
--Keith Lewis klewis {at} mitre.org
The above may not (yet) represent the opinions of my employer.
===
Subject: Re: The coefŽcient of the power of the polynomials
<ck29vl$qaj$1@newslocal.mitre.org>
>m is a very small number, say 3,4,5, while n is generally
greater than
1000.
Well then (3+1)^1000 would be an unacceptable runtime.
>Can you give me some details on that algorithm?
The brute force approach -- Anything equivalent to n nested
loops going
from
0 to m. You could use recursion or an OO language to avoid
having to code
all n loops. You¹d end up multiplying at each stage of the
loop, and
adding
at the center.
exponent[0] = 0;
product[0] = 1;
for (i1 = 0; i1 <= m; i1++) {
product[1] = product[0] * a[i1];
exponent[1] = exponent[0] + i1;
for (i2 = 0; i2 <= m; i2++) {
product[2] = product[1] * a[i2];
exponent[2] = exponent[1] + i2;
for (i3 = 0; i3 <= m; i3++) {
product[3] = product[2] * a[i3];
exponent[3] = exponent[2] + i3;
c[exponent[3]] += product[3];
}
}
}
I¹m sure there¹s a faster way.
--Keith Lewis klewis {at} mitre.org
The above may not (yet) represent the opinions of my employer.
===
Subject: Re: The coefŽcient of the power of the polynomials
<ck29vl$qaj$1@newslocal.mitre.org>
<ck6omu$ac9$1@newslocal.mitre.org>
posting-account=s20ORQ0AAAAAbo26p41slsJ5qq5G69eX
coefŽcients by making use of the convolution, i.e.
The coefŽcients of (a0 + a1*x + a2*x^2)^2 can be easily
obtained by
taking the convolution of (a0 a1 a2) with itself. For higher
order,
we can simply repeat this process,
( ( (a0 + a1*x + a2*x^2) ^2) ^2 .... )^2
However, if we can express these coefŽcients in an explicit
form, it
will be more preferred. A certain kind of bound will also be
much
helpful.
===
Subject: Re: The coefŽcient of the power of the polynomials
|> m is a very small number, say 3,4,5, while n is generally
greater than
1000.
| Well then (3+1)^1000 would be an unacceptable runtime.
|> Can you give me some details on that algorithm?
| The brute force approach -- Anything equivalent to n nested
loops going
from
| 0 to m. You could use recursion or an OO language to avoid
having to
code
| all n loops. You¹d end up multiplying at each stage of the
loop, and
adding
| at the center.
|
| exponent[0] = 0;
| product[0] = 1;
| for (i1 = 0; i1 <= m; i1++) {
| product[1] = product[0] * a[i1];
| exponent[1] = exponent[0] + i1;
| for (i2 = 0; i2 <= m; i2++) {
| product[2] = product[1] * a[i2];
| exponent[2] = exponent[1] + i2;
| for (i3 = 0; i3 <= m; i3++) {
| product[3] = product[2] * a[i3];
| exponent[3] = exponent[2] + i3;
|
| c[exponent[3]] += product[3];
| }
| }
| }
|
| I¹m sure there¹s a faster way.
(By the way, 4^1,000 is a 603 digit number.)
There are at least three faster ways, one of the easiest ones
to explain
is the binary method:
express the exponent in base two: 1000 (base ten) =
1111101000 (base 2)
then, discard the high order bit (it¹s always a 1): 111101000
Then, going from left to right, perform the following:
if the digit is a one, square it and multiply that by the
original
number
if the digit is a zero, square the number
That¹s it.
So, we:
square 4 and then multiply that times 4,
square the above number, and multiply that times 4,
square the above number, and multiply that times 4,
square the above number, and multiply that times 4,
square the above number,
square the above number, and multiply that times 4,
square the above number,
square the above number,
square the above number.
The other two methods use a power tree, and another method
factors the
exponent to facilitate mulitplying.
________________________________Gerard
S.
===
Subject: Re: The coefŽcient of the power of the polynomials
> m is a very small number, say 3,4,5, while n is generally
greater than
1000.
In this case, you may be able to use a Chinese remaindering
trick:
compute this modulo enough primes that, using bounds on the
coefŽcients of the nth power of your polynomial, you can
reconstruct
the coefŽcients unambiguously.
As for the coefŽcient bounds, one good bet is the Theorem of
Vieta
expressing the coefŽcients of the monicization (the unique
scalar
multiple of the polynomial which is monic) of the polynomial
in terms
of the roots. Remember to repeat them n-fold when you do the
nth-powering.
I believe that Chinese remaindering is the technique used to
compute
the determinants of integer matrices while avoiding coefŽcient
explosions that can happen in the rationals and numerical
instability
that can happen in the reals.
---- David
To send me email, move the r from the beginning to the end of
the part
before the @ and insert alum. at the beginning of the part
after the
@.
===
Subject: Re: The coefŽcient of the power of the polynomials
<ck29vl$qaj$1@newslocal.mitre.org>
posting-account=s20ORQ0AAAAAbo26p41slsJ5qq5G69eX
>> In this case, you may be able to use a Chinese
remaindering trick:
>> compute this modulo enough primes that, using bounds on the
>> coefŽcients of the nth power of your polynomial, you can
reconstruct
>> the coefŽcients unambiguously.
I tried to read some references on the Chinese remainder
theorem.
However, I have no idea how to solve my problem by using this
theorem.
Could you please give me some more hints?
===
Subject: is sphere in cone
X-RFC2646: Format=Flowed; Original
I have a cone with starting point, direction and opening
angle. I can
very easily test if apoint is inside the cone (closed end).
.x
--- Želd of view
---
---
.e--------> .l
---
---
---
e-> = Point of eye-position
l-> = Point the eye is looking at
x-> = Any point - check if it¹s visible
fov = Angle of frustum (view angle of eye)
phi = acos[(l-> - e->) dot (x-> - e->)]
if phi > fov then point is outside. Easy.
Now, assume the point is a sphere with radius R. How can I
check this?
You help is very much appreciated. This is no homework, I¹m
out of
homework age.
--
-Gernot
int main(int argc, char** argv) {printf
(%silto%c%cf%cgl%ssic%ccom%c, ma, 58, Œg¹, 64, ba, 46, 10);}
________________________________________
Looking for a good game? Do it yourself!
GLBasic - you can do
www.GLBasic.com
===
Subject: Re: is sphere in cone
X-newsreader: xrn 9.02
>I have a cone with starting point, direction and opening
angle. I can
>very easily test if apoint is inside the cone (closed end).
> .x
> --- Želd of view
> ---
> ---
>.e--------> .l
> ---
> ---
> ---
>e-> = Point of eye-position
>l-> = Point the eye is looking at
>x-> = Any point - check if it¹s visible
>fov = Angle of frustum (view angle of eye)
>phi = acos[(l-> - e->) dot (x-> - e->)]
>if phi > fov then point is outside. Easy.
>Now, assume the point is a sphere with radius R. How can I
check this?
You Žrst check whether the sphere center is inside the cone.
If it
is not, the case is clear. If the sphere center is inside the
cone,
you compare the sphere radius with the closest distance
between
the sphere center and the cone wall. This closest distance
is found by looking at the rectangular triangle with one side
given
by the line from the cone center e to the sphere center (of
known
length
by Pythagoras), another line from the cone center to a point
on
the cone closest to the sphere center (tangential point), and
a
third line from the sphere center to this tangetial point.
This triangle has three known parameters: the side mentioned,
the right angle, and another angle seen from the point e
between the cone wall and the sphere center.
The shortest distance follows by planar trigonometry from
these
values. For simplicity in 3D one should write the three side
vectors
of the triangle as a sum and form the dot product of this
equation in
turn with any of the three side vectors..
If this distance is smaller than R, you are again done.
If it is not, you have to check the distance between the
sphere center and the lid on the cone, which is of similar
complexity.
mathar@mpia.de
===
Subject: Re: 5 - 8¹s`
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id i98F24U15209;
>Does any one know the trick arrange 5 8¹s so they equal nine
without
>using + or -??
what about 8 + 88/88 ?
:-))
highegg
===
Subject: Re: 5 - 8¹s`
>>Does any one know the trick arrange 5 8¹s so they equal
nine without
>>using + or -??
> what about 8 + 88/88 ?
>:-))
[sqrt(88)] needs only two 8¹s, plus the greatest integer
function ([]) and
square root, but no + or -.
--
Dave Seaman
Judge Yohn¹s mistakes revealed in Mumia Abu-Jamal ruling.
<http://www.commoncouragepress.com/index.cfm?action=book&
bookid=228>
===
Subject: Re: 5 - 8¹s`
|>>Does any one know the trick arrange 5 8¹s so they equal
nine without
|>>using + or -??
|> what about 8 + 88/88 ?
| [sqrt(88)] needs only two 8¹s, plus the greatest integer
function ([])
and
| square root, but no + or -.
How about just one eight: addone(8) where the addone
function adds one the the argument.
Or, how about no eights: nine() where the nine
function returns the value 9.
It¹s a matter of deŽning the rules on what functions (if any)
are
allowed.
_____________________________________________________Gerard S.
===
Subject: Re: 5 - 8¹s`
> |>>Does any one know the trick arrange 5 8¹s so they equal
nine without
> |>>using + or -??
> |> what about 8 + 88/88 ?
> | [sqrt(88)] needs only two 8¹s, plus the greatest integer
function ([])
and
> | square root, but no + or -.
> How about just one eight: addone(8) where the addone
> function adds one the the argument.
> Or, how about no eights: nine() where the nine
> function returns the value 9.
> It¹s a matter of deŽning the rules on what functions (if
any) are
> allowed.
_____________________________________________________Gerard S.
I don¹t want to get into ASCII art, so rather than using a
line or some
dots over some digits to represent repeats, I¹ll bracket
them. In other
words, 0.4{56} would represent 0.4565656565656... . Then we
could
write 9 as (8 * 8) / (8 * .{88}).
Rick
===
Subject: Re: 5 - 8¹s`
> |>>Does any one know the trick arrange 5 8¹s so they equal
nine without
> |>>using + or -??
> |> what about 8 + 88/88 ?
> | [sqrt(88)] needs only two 8¹s, plus the greatest integer
function ([])
and
> | square root, but no + or -.
> How about just one eight: addone(8) where the addone
> function adds one the the argument.
> Or, how about no eights: nine() where the nine
> function returns the value 9.
> It¹s a matter of deŽning the rules on what functions (if
any) are
> allowed.
_____________________________________________________Gerard S.
How about 8/(0.8 repeated)*sqrt(8*8)/8?
--
Mark Thornquist
===
Subject: Re: A (not so ?) complex pb
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id i98F24Z15204;
>Hello all,
>In order to complete a proof, I need to prove the following
fact:
>If a and b are complex numbers with Re(a) and Re(b) > 0, and
if z is
>such that Re(z) > 0 and z^2 = a.b, then
>Re(z) >= Min( Re(a) , Re(b) ).
>I¹m pretty sure that this is true (and that equality holds
iff a = b,
>but I don¹t need this).
>Any help or clue welcome!
>(If you are wondering, this is not for homework, and I can
even explain
>the context should you be interested...)
>Dede
hi Dede,
let¹s write z = z1 + i*z2, a = a1 + i*a2, b = b1 + i*b2
these must satisfy the system of equations
z1^2 - z2^2 = a1*b1 - a2*b2
2*z1*z2 = a1*b2 + a2*b1
now, suppose for a moment a2, b2 have different signs (or any
of them is
zero). The 1st equation gives us then z1^2 >= a1*b1, which
yields the desired
inequality z1>=min(a1,b1)
If a2,b2<0 then the 2nd equation gives us z2<0 and we can
multiply a2,b2,z2
by -1 while the equations remain unchanged. Therefore,
without loss of
generality we can assume
a2,b2,z2 >0
moreover, without loss of generality, we assume a1>=b1
Suppose now z1 < b1 and match it with the equations:
b1^2 + a2*b2 > z2^2 + a1*b1
2*b1*z2 > a1*b2+a2*b1
Since a1>=b1, the 2nd inequality implies: 2*b1*z2 >
b1*b2+a2*b1 =>
z2 >= (a2+b2)/2 (I)
and since a1*b1 >= b1^2, the Žrst inequality gives us
z2^2 < a2*b2 (II)
but (I), (II) together give a contradiction to
arithmetic-geometric mean
inequality. Contradiction, therefore z1 >= b1, which is what
we wanted to
prove.
highegg
===
Subject: statistics~~.
hello....doctor~
probability function of X is P(X = sqrt{(2^n)/n}) = 1/(2^n)
(n=1,2,....)
show the existence of the mean and variance of X.
----------------------------------------------------
i am unfamiliar with form P(X = sqrt{(2^n)/n}) = 1/(2^n).
i view this form for the Žrst time.
is this right problem ?
i need your advice.
thank you very much for your advice.
===
Subject: Re: I can trisect an angle
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id i98FBE216563;
> Mathematician: I need to trisect this angle, but I know it
can¹t
>be done.
>>Even in classical Greek times they did trisections. Just
not with
>>straightedge & compasses.
>> <<a
>href=http://mathworld.wolfram.com/Trisectrix.html>http://
mathworld.wolf
ram.com/Trisectrix.html</a>
>>--
>>G. A. Edgar <a
>href=http://math.ohio-state.edu/~edgar/>http://
math.ohio-state.edu/~edg
ar/</a>
>I can trisect an angle. but I have a different way of
trisecting an
>angle for an obtuse, right, and acute angle. ( with only a
compass and
>a straight edge )
marvellous!
But can you put your leg arround your neck? That¹s deŽnitely
a better
skill to improve.
highegg
===
Subject: Re: no comments?
[talking about Petry Keckman¹s confusion]
> Limitless: not bounded, without limit. Therefor not Žnite,
therefor
> inŽnite.
> Note: limited and limitless are not mathematical terms. You
have
> introduced them because you want something that means
between Žnite and
> inŽnite, even though there is no middle ground.
There is in fact a subtle distinction between unbounded
(which *is* a
mathematical term) and inŽnite, even when restricted to
countable
inŽnities (countable ordinals, to use a mathematical term) --
and this
distinction is what Keckman feels in his bones but can¹t get
quite get
into his head. For example, the binary expansions of dyadic
rationals
(numbers of the form a/2^b) have unbounded length, but are
not inŽnite
(i.e. they are always Žnite) -- yet there also exist
rationals with an
inŽnite (literally: never ending) binary expansion.
In the case of natural numbers, they *also* have unbounded
value, and an
unbounded number of digits (or bits), but *unlike rationals*
there are NO
natural numbers with an inŽnite number of digits (or bits):
every sequence
of digits representing a natural number has a beginning and
an end.
Keckman¹s biggest problem is that he believes in dynamic set
theory,
where sets can acquire and lose elements while staying the
same set,
which is quite contrary to every mathematician¹s view of sets.
Another issue is that, strictly speaking, the Peano Axioms
(as I know them)
say nothing at all about sets -- they only talk about natural
numbers.
One can speak about the *class* of natural numbers satisfying
a certain
property (and that is a convenient way to talk about
induction), but
technically Induction only claims that if the Žrst natural
number has a
certain property, and if that property is inherited through
the successor
operation, then every natural has that property. This axiom
makes it
possible to have Žnite proofs about inŽnite collections of
natural
numbers, e.g. every even number is the successor of an odd
number (if we
start counting at 1). Without induction, we could still prove
this for
every even number -- but each one would need a different
proof! This is
of course in the technical sense of a proof being a sequence
of theorems
each derived by an inference rule from axioms or earlier
theorems.
Whether the class of natural numbers as deŽned by the Peano
axioms is a
set or not is a question for Set Theory -- and needs an
explicit axiom,
namely the axiom of inŽnity. Without that axiom, we would
still have all
the same natural numbers (as a class of hereditarily-Žnite
sets, as for
example the Žnite Von Neumann ordinals), but we couldn¹t
claim that this
class was a set.
Language clariŽcation: a class is a collection of objects
(members)
that share a certain proprty. Technically it is shorthand for
a formula
that asserts this property in certain contexts. A set is an
object of
Set Theory that satisŽes the axioms of the membership
relation. It is
(informally speaking) the collection of its members as deŽned
by that
membership relation (which is *the* primitive of the theory).
Michel.
P.S. This thread was getting on my nerves, and I had to get my
peace/piece!
===
Subject: Re: no comments?
> Without induction, we could still prove this for
> every even number -- but each one would need a different
proof!
Shut up. Of course there is induction allthough N is not
inŽnite.
Induction is as valid as before.
loop
f(n)=true
forever
Can run forever.
But in any situation we use the set of ns, we needed to
observe the limit
of that construction when n->oo. For example what happens to
the
goal sets and start sets in Cantor¹s table tricks.
--
amount and biggest
1+1+1+...= inŽnite and Žnite in math today
Petri Keckman
===
Subject: Re: no comments?
> Keckman¹s biggest problem is that he believes in dynamic
set theory,
> where sets can acquire and lose elements while staying the
same set,
> which is quite contrary to every mathematician¹s view of
sets.
My biggest problem is beneath. So your problem is that
there is no set N containing inŽnite n. Something we have
to use instead. And it could be dynamic.
My/your problem
*************
Talking about natural numbers:
There must be number atleast 6 in set that has 6 item.
6 {1,2,3,4,5,6}
Could be bigger of course
8 {1,2,3,4,5,8}
but atleast there must be 6.
There must number atleast n in set that has n item.
n {1,2,3,...,n}
Could be bigger but atleast there must be number that is n.
So what number _atleast_ should be in set that has oo item?
Atleast there should be number that is oo.
*************
--
amount and biggest
1+1+1+...= inŽnite and Žnite in math today
Petri Keckman
===
Subject: Re: no comments?
: ...it has been made conclusion that N is inŽnite, but it is
wrong.
: The right word should be limitless. And this is really not
just semantic.
Ah - right.
: If set is limitless it has as many items as ever, but
allways Žnite
: number of them. Same as Paxiomn=n+1 keeps the bigness of
number allways
: Žnite, so is the amount of numbers.
You¹re blathering - but I believe you know that you have no
idea what
you¹re talking about. What are you smoking and do you have
some for
sale? This doesn¹t even make good Žction, sad to say.
: And this is not just semantic,
No, it¹s also pure crap.
: Now lets try once again. Set N is said to be inŽnite.
InŽnete is said to

: be bigger than any number in N.
Who said that?
Here¹s a suggestion - buy a decent book on set theory,
sequester yourself
up in your bedroom for a year, then come back and post.
Justin
===
Subject: Re: no comments?
> InŽnete is said to be bigger than any number in N.
> Who said that?
Because all naturals are Žnite. InŽnite is bigger.
Or are you trying to say tha inŽnite is smaller than Žnite?
Buah buah buah!!! And you call yourself a mathematician and
that i should
learn the basics. Buah buah.
--
amount and biggest
1+1+1+...= inŽnite and Žnite in math today
Petri Keckman
===
Subject: Re: no comments?
:> InŽnete is said to be bigger than any number in N.
:> Who said that?
: Because all naturals are Žnite. InŽnite is bigger.
: Or are you trying to say tha inŽnite is smaller than Žnite?
The deŽnition of inŽnity is not something bigger than all
naturals,
which is, as I mentioned, pseudomathematical. You can deŽne
various
cardinalities of sets and power sets and then deŽne bijective
correspondances, and from there you get the deŽnitions of
Žnite and the
various inŽnities. Everything¹s pretty straightforward.
Maybe you should rephrase your questions within the language
of set
theory. Then I¹m sure you¹d see your errors.
: Buah buah buah!!! And you call yourself a mathematician and
that i should
: learn the basics. Buah buah.
I didn¹t call myself a mathematician, but I am, and you
should.
Justin
===
Subject: Re: no comments?
> You can deŽne various
> cardinalities of sets and power sets and then deŽne
bijective
> correspondances, and from there you get the deŽnitions of
Žnite and the
> various inŽnities. Everything¹s pretty straightforward.
You cannot do those deŽnitions before you somehow just
suppose that there

is
inŽnite set. And after that you get those deŽnitations of
inŽnite set.
And again, it is nothing. It is pseudo lgical nonsense.
> The deŽnition of inŽnity is not something bigger than all
naturals,
The concept inŽnity was in philoshophy and math before you
tried to
exactly
deŽne it. And it meaned just that.
n->oo means that n grows without limit. It never reach but it
tries.
Allthough
it never can get one step further.
If it is not deŽned that oo>n for every Žnite n, how the hell
you know
where that n
should run?
--
amount and biggest
1+1+1+...= inŽnite and Žnite in math today
Petri Keckman
===
Subject: Re: no comments?
:> You can deŽne various
:> cardinalities of sets and power sets and then deŽne
bijective
:> correspondences, and from there you get the deŽnitions of
Žnite and
the
:> various inŽnities. Everything¹s pretty straightforward.
: You cannot do those deŽnitions before you somehow just
suppose that there

: is inŽnite set.
Let N={0,1,2,3,...}. Rough, eh?
: And after that you get those deŽnitations of inŽnite
: set. And again, it is nothing. It is pseudo lgical nonsense.
Your grammar and spelling are atrocious.
deŽnitations -> deŽnitions?
pseudo lgical -> pseudological?
:> The deŽnition of inŽnity is not something bigger than all
naturals,
: The concept inŽnity was in philoshophy and math before you
tried to
: exactly deŽne it. And it meaned just that.
meaned -> meant?
philoshophy -> philosophy?
Before mathematics gave these things concrete deŽnitions they
*were*
poorly deŽned and understood concepts. Your problem is that
you¹re still
trying to deal with them on that poor level. Mathematics has
since taken
a Žrm grasp on those things. Now it¹s your turn.
: n->oo means that n grows without limit. It never reach but
it tries.

What do you mean it tries? It grows without limit, what¹s so
hard to
understand about that?
: Allthough it never can get one step further.
allthough -> although?
What on earth do you mean? One step further than what? n
grows without
bound, why is that so difŽcult for you to grasp?
: If it is not deŽned that oo>n for every Žnite n, how the
hell you know

: where that n should run?
It¹s not.
What do you mean, where it should run? It just grows without
bound. It
doesn¹t run anywhere. It¹s not an olympic sprinter, for
goodness¹
sakes.
Justin
===
Subject: Re: no comments?
> Let N={0,1,2,3,...}. Rough, eh?
Why it is so hard to understand this:
*************
Talking about natural numbers:
There must be number atleast 6 in set that has 6 item.
6 {1,2,3,4,5,6}
Could be bigger of course
8 {1,2,3,4,5,8}
but atleast there must be 6.
There must number atleast n in set that has n item.
n {1,2,3,...,n}
Could be bigger but atleast there must be number that is n.
So what number _atleast_ should be in set that has oo item?
Atleast there should be number that is oo.
*************
So there cannot be set N of inŽnite n which are all Žnite.
--
amount and biggest
1+1+1+...= inŽnite and Žnite in math today
Petri Keckman
===
Subject: Re: no comments?
: Why it is so hard to understand this:
: *************
: Talking about natural numbers:
: There must be number atleast 6 in set that has 6 item.
: 6 {1,2,3,4,5,6}
: Could be bigger of course
: 8 {1,2,3,4,5,8}
: but atleast there must be 6.
: There must number atleast n in set that has n item.
: n {1,2,3,...,n}
: Could be bigger but atleast there must be number that is n.
: So what number _atleast_ should be in set that has oo item?
: Atleast there should be number that is oo.
: *************
: So there cannot be set N of inŽnite n which are all Žnite.
Here you have precisely indicated your comprehension issue.
You are using
nonstandard vague deŽnitions for things, and *your deŽnitions
do not*
extend to the things that *you* do not understand. This is
not the fault
of mathematics, it is the fault of your žawed deŽnitions.
This is precisely why I suggested you go and study the
basics. If you
knew the basics, you wouln¹t be spouting off things like the
above
nonsense. You can¹t claim to understand them if you¹re not
using them.
The set N={1,2,...} is the set of all natural numbers. This
is not a
difŽcult concept for most people. What you seem to be trying
to do above
is deŽne size by the number of consecutive naturals starting
with 1
that must be in a set in order to contain that particular
natural number
given by size. Clearly this deŽnition does not apply to
inŽnity, but
this is *not* the correct deŽnition of cardinality of *any*
set.
In proper ordinal arithmetic, the natural numbers (and zero)
are merely
(and a priori meaningless) symbols used to denote sets, and
all of those
sets are constructed beginning with only the empty set {}.
The jump to
the various inŽnities follows naturally from the construction.
But of course, if you¹d studied the basics, you¹d know this.
Justin
===
Subject: Re: no comments?
> Clearly this deŽnition does not apply to inŽnity, but
> this is *not* the correct deŽnition of cardinality of *any*
set.
Do you know how those cardinalities came to math, which you
you have
studied
from books?
Galilei wondered N->2N bijection. And discovered that 2N is a
subset of N
but still
there is a bijection from N->2N.
Cantor studied Galilei¹s thoughts and make deŽnation. Set
that has subset

that
is as big (i.e there is that bijection) as the set itself is
inŽnite.
If i say and proof, that N can not be inŽnite, then there is
no meaning
about those
cardinals, aleph_0 etc... They are nonsense.
There is not even the Žrst inŽnite set. That conception is
paradoxical.
And so
there can not be different aleph_s or what ever..
--
amount and biggest
1+1+1+...= inŽnite and Žnite in math today
Petri Keckman
===
Subject: Re: no comments?
: Do you know how those cardinalities came to math, which you
you have
: studied from books?
Sigh. Okay, you¹re pretty dense, so let me spell it out for
you:
Before the proper deŽnitions of these things were developed,
they *were*
poorly deŽned entities. Now that things have been formalized,
they are
all very clear and understandable.
It doesn¹t matter one iota the origins for purposes of
understanding how
the mathematics *is*. Zeno¹s paradoxes are interesting, of
course, but
pointless for any formal understanding of limits.
You, as I said, are still mumbling at the level of ignorance
and are
complaining that *your* understanding is not sufŽcient.
Until you speak the language, there is nothing to discuss.
It¹s been a blast!
Justin
===
Subject: Re: no comments?
>> InŽnete is said to be bigger than any number in N.
>> Who said that?
> Because all naturals are Žnite. InŽnite is bigger.
> Or are you trying to say tha inŽnite is smaller than Žnite?
> Buah buah buah!!! And you call yourself a mathematician and
that i should
> learn the basics. Buah buah.
No, he¹s saying that it isn¹t necessarily a number.
===
Subject: Re: no comments?
> No, he¹s saying that it isn¹t necessarily a number.
And he is rigth in it. Allthough that necessarily is not
needed.
Nothing paradoxical nonsense is not number.
But if there is any hint of sense in your Žnite and inŽnite
language i hope you somehow see that inŽnite is bigger than
Žnite.
--
amount and biggest
1+1+1+...= inŽnite and Žnite in math today
Petri Keckman
===
Subject: Re: no comments?
> You¹re blathering - but I believe you know that you have no
idea what
> you¹re talking about.
Then i guess you do not have no i idea what i¹m talking about.
Then you can not say is it right or wrong.
--
amount and atleast
1+1+1+...= inŽnite and Žnite in math today
Petri Keckman
===
Subject: Re: no comments?
> You¹re blathering - but I believe you know that you have no
idea what
> you¹re talking about.
: Then i guess you do not have no i idea what i¹m talking
about.
: Then you can not say is it right or wrong.
If you come over and take a crap on my doorstep, I may not be
certain of
what you had for dinner last night, but I know it¹s s**t and
I don¹t want
it there.
Look, most of the people here, myself included, have a
Žrm-enough grasp
of set theory, cardinality, and ordinality to know that your
sentences
don¹t contain any coherent statements about mathematics. You
don¹t know
the basics, and hence your attempt to construct any
statements at any
slightly higher level falls žat. You¹re basically doing
pseudomath -
stringing big words together in a way that might make sense
to you given
what little you know but in reality is just mumbling. This is
why I
recommend you go and learn those basics before you post.
Justin
===
Subject: Re: no comments?
> of set theory, cardinality, and ordinality to know that
your sentences
> don¹t contain any coherent statements about mathematics.
Quite easy concepts to learn. I know them.
> You don¹t know
> the basics,
Knowing and learning basics is other thing than really
understand them
in holeness of math, logic and philosophy and science history
and
structure,
which is other thing to accept them - those other peoples
concepts -
and it is other thing trying to change them.
Learning and understanding math <> accepting all
> and hence your attempt to construct any statements at any
> slightly higher level falls žat.
That is a problem of language. When something change in one
conception,
then all
changes. That is called change of paradigma.
--
amount and biggest
1+1+1+...= inŽnite and Žnite in math today
Petri Keckman
===
Subject: Re: no comments?
> Limitless: not bounded, without limit. Therefor not Žnite,
therefor
> inŽnite.
An other words. There is no conception in english language
for our Žnnish
word rajoittamaton.
It is like not Žxed Žnite. But something that can allways be
bigger but
not inŽnite.
It is like natural number. Not the size of N but the size of
number in N.
That is Žnite but it can allways be bigger (in other
situation).
--
amount and biggest
1+1+1+...= inŽnite and Žnite in math today
Petri Keckman
===
Subject: Re: no comments?
>>An other words. There is no conception in english language
for our
Žnnish
>>word rajoittamaton.
> You mean unbounded.
I interpreted limitless as unbounded. A bound is a limit.
===
Subject: Re: no comments?
>> An other words. There is no conception in english language
for our
>> Žnnish word rajoittamaton.
> You mean unbounded.
I would not like to use other word than that. Word limitless
has another
poin too.
It tells that 1+1/2+1/4+1/8+... actually never is same as two.
Only limit(1+1/2+1/4+1/8+...)=2 but that is another thing.
1+1/2+1/4+1/8+...<>limit(1+1/2+1/4+1/8+...)
And i would also like to use only two point - not three
1+1/2+1/4+1/8+..
--
amount and biggest
1+1+1+...= inŽnite and Žnite in math today
Petri Keckman
===
Subject: Re: no comments?
>> An other words. There is no conception in english language
for our
>> Žnnish word rajoittamaton.
> You mean unbounded.
> I would not like to use other word than that. Word
limitless has another

> poin too.
> It tells that 1+1/2+1/4+1/8+... actually never is same as
two.
Since the three dots imply limit of the sequence of
sums, actually it is the same as 2.
> Only limit(1+1/2+1/4+1/8+...)=2 but that is another thing.
No, it¹s the same thing.
> 1+1/2+1/4+1/8+...<>limit(1+1/2+1/4+1/8+...)
Really? What do you think the thing on the left is,
if not a limit?
> And i would also like to use only two point - not three
> 1+1/2+1/4+1/8+..
Why? Does it have a different meaning you are going
to introduce?
- Randy
===
Subject: Re: no comments?
>> 1+1/2+1/4+1/8+...<>limit(1+1/2+1/4+1/8+...)
> Really? What do you think the thing on the left is,
> if not a limit?
Left is that Acilles will never actually reach the tortoise.
Those ... tree dots that means inŽnity. And because everything
is relative it does¹n matter how far you are, it is still the
same you have not been taken any step to inŽnity.
But we can mark limit(1+1/2^n+...)=2 where n->oo
--
amount and biggest
1+1+1+...= inŽnite and Žnite in math today
Petri Keckman
===
Subject: Re: no comments?
> An other words. There is no conception in english language
for our
> Žnnish word rajoittamaton.
>> You mean unbounded.
> I would not like to use other word than that.
I mean i use word limitless
--
amount and biggest
1+1+1+...= inŽnite and Žnite in math today
Petri Keckman
===
Subject: Re: no comments?
> Assuming, as you seem to do, that only positive integers
can be in the
> set, then an inŽnite set does not have a biggest element.
That¹s what
> it means to be an inŽnite set.
I don¹t need to have word biggest.
Talking about natural numbers:
There must be number atleast 6 in set that has 6 item?
6 {1,2,3,4,5,6}
Could be bigger of course
8 {1,2,3,4,5,8}
but atleast it was 6.
there must number atleast n in set that has n item?
n {1,2,3,...,n}
Could be bigger but atleast there is n.
So what number atleast must be in set that has oo item?
Atleast there must be number that is oo.
There is no enough Žnite natural numbers for inŽnite set of
Natural
numbers.
InŽnite is bigger than any Žnite natural number.
--
amount and biggest
1+1+1+...= inŽnite and Žnite in math today
Petri Keckman
===
Subject: Re: no comments?
> The right word should be limitless. And this is really not
just semantic.
Yes it is. It is a word you have substituted because you
don¹t accept what we say when we use the word inŽnite
but you seem able to accept the same concepts with a
different word. Shall we use the word glxylq to deŽne
such sets? Would that help? The name is just a name.
> If set is limitless it has as many items as ever, but
allways Žnite
> number of them.
What does always mean here? You act as if the cardinality
of the set changes in time. Time does not enter into it. There
are only two possibilities:
1. The size of N is Žnite. If it is Žnite, it is always
Žnite since the size is Žxed.
2. The size of N is not Žnite. If it is not Žnite, we
shall use the term glxylq.
So, do you think that N is Žnite? Finite means there is
a largest element.
> Let¹s mark every item in N by label so that 1->1,
2->2,...,n->n,... How
> many
> labels? Answer: InŽnite, because N is said to be inŽnite.
Correct.
> What is the biggest number?
There isn¹t one.
> Should be inŽnite
Only if it exists. It doesn¹t exist.
> because there is one label for every number
> and one
> number for every label, but all n in N are Žnite.
cotradiction.
What two statements are in contradiction? Where does all n
in N are Žnite conžict with there is no largest n in N?
- Randy
===
Subject: Re: no comments?
>> The right word should be limitless. And this is really not
just
>> semantic.
> Yes it is. It is a word you have substituted because you
> don¹t accept what we say when we use the word inŽnite
> but you seem able to accept the same concepts with a
> different word. Shall we use the word glxylq to deŽne
> such sets? Would that help? The name is just a name.
If you deŽne that glxylq->oo but never glxylq=oo
So that it just converge but never reach. And even that word
is stupid
because we are in the Žrst step forever.
Amount=1
loop
Amount=Amout*2 ŒŽrst step
forever
>> If set is limitless it has as many items as ever, but
allways Žnite
>> number of them.
> What does always mean here? You act as if the cardinality
> of the set changes in time.
Yes. I think there is no Žxed set called N. In world there is
just
actual situations where we use natural numbers.
> Time does not enter into it. There
> are only two possibilities:
> 1. The size of N is Žnite. If it is Žnite, it is always
> Žnite since the size is Žxed.
> 2. The size of N is not Žnite. If it is not Žnite, we
> shall use the term glxylq.
and third 3. The size of N must allways handle as a Žnite set.
Every algorithm must sometimes stop.
> So, do you think that N is Žnite? Finite means there is
> a largest element.
It is not Žxed Žnite. Next time we need it we can produce
bigger set.
In our mind, or in algorithm.
>> Let¹s mark every item in N by label so that 1->1,
2->2,...,n->n,... How
>> many
>> labels? Answer: InŽnite, because N is said to be inŽnite.
> Correct.
>> What is the biggest number?
> There isn¹t one.
I agree. But there must be number that is atleast inŽnite.
>> Should be inŽnite
> Only if it exists. It doesn¹t exist.
You only say that it does not exist because it does not exist.
But it should.
>> because there is one label for every number
>> and one
>> number for every label, but all n in N are Žnite.
cotradiction.
> What two statements are in contradiction? Where does all n
> in N are Žnite conžict with there is no largest n in N?
copy paste from other place:
I don¹t need to have word biggest.
Talking about natural numbers:
There must be number atleast 6 in set that has 6 item?
6 {1,2,3,4,5,6}
Could be bigger of course
8 {1,2,3,4,5,8}
but atleast there is 6.
there must number atleast n in set that has n item?
n {1,2,3,...,n}
Could be bigger but atleast there is n.
So what number atleast must be in set that has oo item?
Atleast there must be number that is oo.
--
amount and biggest
1+1+1+...= inŽnite and Žnite in math today
Petri Keckman
===
Subject: Re: no comments?
[snippage]
> Time does not enter into it. There
> are only two possibilities:
> 1. The size of N is Žnite. If it is Žnite, it is always
> Žnite since the size is Žxed.
> 2. The size of N is not Žnite. If it is not Žnite, we
> shall use the term glxylq.
> and third 3. The size of N must allways handle as a Žnite
set.
> Every algorithm must sometimes stop.
The limitations on our *abilities* to perform calculations
are not
limitations on the underlying mathematical *concepts*.
Yes, every algorithm must either 1) Stop, or 2) Be terminated
in an
uncomplete state, but that has no bearing on the underlying
*concepts*
with which we work.
===
Subject: Re: no comments?
> [snippage]
If you snip, you lost something.
I don¹t need to have word biggest.
Talking about natural numbers:
There must be number atleast 6 in set that has 6 item?
6 {1,2,3,4,5,6}
Could be bigger of course
8 {1,2,3,4,5,8}
but atleast there is 6.
there must number atleast n in set that has n item?
n {1,2,3,...,n}
Could be bigger but atleast there is n.
So what number atleast must be in set that has oo item?
Atleast there must be number that is oo.
--
amount and biggest
1+1+1+...= inŽnite and Žnite in math today
Petri Keckman
===
Subject: Re: no comments?
> Yes, every algorithm must either 1) Stop, or 2) Be
terminated in an
> uncomplete state, but that has no bearing on the underlying
*concepts*
> with which we work.
In a sense mathematics has Žrst deŽned number 1. Then from
that number 1

they have deŽned
number 2 and so on. In these days they have done it useing
set therory.
I think there is that order. That concept Naturals doesn¹t
have meaning
without deŽnition.
It can not been take as given. We have done it.
How human beings learn those concepts? What comes to a little
chirld or
ape men in the past?
First they learn some concept. Maybe mother. Maybe food was
nearly

only concept for early ape men.
Then they learn more concepts but they are all deŽned somehow
form
earliers concept.
not(food)-> hungry or
something like that.
If there doesn¹t exist 1 there can not exist 2 either.
--
amount and biggest
1+1+1+...= inŽnite and Žnite in math today
Petri Keckman
===
Subject: Re: no comments?
> there
> is error in structure and language called mathematichs
Wow. Deja vu all over again.
V.
--
email: lastname at cs utk edu
homepage: www cs utk edu tilde lastname
===
Subject: Re: no comments?
>> there is error in structure and language called
mathematichs
> Wow. Deja vu all over again.
And again.
> Assuming, as you seem to do, that only positive integers
can be in the
> set, then an inŽnite set does not have a biggest element.
That¹s what
> it means to be an inŽnite set.
I don¹t need to have word biggest.
Talking about natural numbers:
There must be number atleast 6 in set that has 6 item?
6 {1,2,3,4,5,6}
Could be bigger of course
8 {1,2,3,4,5,8}
but atleast it was 6.
there must number atleast n in set that has n item?
n {1,2,3,...,n}
Could be bigger but atleast there is n.
So what number atleast must be in set that has oo item?
Atleast there must be number that is oo.
There is no enough Žnite natural numbers for inŽnite set of
Natural
numbers.
InŽnite is bigger than any Žnite natural number.
--
amount and biggest
1+1+1+...= inŽnite and Žnite in math today
Petri Keckman
===
Subject: Re: no comments?
>> Then these:
>> f(1) = 1
>> f(2) = 2
>> f(3) = 3
>> f(4) = 4
>> f(5) = 5
>> f(6) = 6
>> f(7) = 7
>> f(8) = 8
>> f(9) = 9
>> f(10) = 10
>> .
>> .
>> .
> With these i mean those rows.
> Why math could not just say that Amount of Naturals->oo but
never
> Amount of Naturals=oo
Because the Amount of Naturals is not *approaching* anything.
That
would assume that it is somehow changing. n(N)=oo.
> Same as in limit(1+1/2+1/4^2+...1/n^2) is just a limit.
> It never actually si the same.
You are talking about two completely different Želds of
mathematics.
Limits do not apply to the cardinality of sets. However,
inŽnite sums
are *deŽned* to be the limit of Žnite sums.
> I would feel so much easier then. If math does not change
his attitude
> i¹m gonna be like James Harris for rest of my life. And you
will
> see how many daily post it will mean. Just wait.
Math is not a person that can have an attitude. If you want
to prove
yourself less capable of learning than James Harris, I will
be happy to
toss you into my kill Žle. James learns. If you will not,
that is
your choice.
--
Will Twentyman
email: wtwentyman at copper dot net
===
Subject: Re: no comments?
> n(N)=oo.
I just can¹t believe that person who has learn so much about
math
Like Will Twentyman seems can not follow this simpple proof
that
there is contradiction saying that N has inŽnite member and
they are all Žnite.
*************
Talking about natural numbers:
There must be number atleast 6 in set that has 6 item.
6 {1,2,3,4,5,6}
Could be bigger of course
8 {1,2,3,4,5,8}
but atleast there must be 6.
There must number atleast n in set that has n item.
n {1,2,3,...,n}
Could be bigger but atleast there must be number that is n.
So what number _atleast_ should be in set that has oo item?
Atleast there should be number that is oo.
*************
--
amount and biggest
1+1+1+...= inŽnite and Žnite in math today
Petri Keckman
===
Subject: Re: no comments?
> I just can¹t believe that person who has learn so much
about math
> Like Will Twentyman seems can not follow this simpple proof
that
> there is contradiction saying that N has inŽnite member and
> they are all Žnite.
N is inŽnite set and all n are Žnite of course...
> *************
> Talking about natural numbers:
> There must be number atleast 6 in set that has 6 item.
> 6 {1,2,3,4,5,6}
> Could be bigger of course
> 8 {1,2,3,4,5,8}
> but atleast there must be 6.
> There must number atleast n in set that has n item.
> n {1,2,3,...,n}
> Could be bigger but atleast there must be number that is n.
> So what number _atleast_ should be in set that has oo item?
> Atleast there should be number that is oo.
> *************
--
amount and biggest
1+1+1+...= inŽnite and Žnite in math today
Petri Keckman
===
Subject: Re: no comments?
> Because the Amount of Naturals is not *approaching*
anything. That
> would assume that it is somehow changing. n(N)=oo.
We have to be satisŽed for that. In every actual situation we
use set
Naturals in computers, in our minds or whatever, it has to be
stopped
somewhere
Algorithm have to stop.
Amount=1
loop
Amount=Amount*n
forever
Where n is any n in N.
It is so hopeless even try to catcj the inŽnite. Do you hear
INFINITE!
Not just something that you are used to use and manipulate
handly in
mathemathics marking it oo and so on, but I N F I N I T E.
Maybe not lately, but as a child, you perhaps has tried to
think inŽnite
big. And no matter how big ig big you it is allways the same
as it have
the size of set {1}. We can not take the Žrst step an a way
to inŽnite.
>> Same as in limit(1+1/2+1/4^2+...1/n^2) is just a limit.
>> It never actually si the same.
> You are talking about two completely different Želds of
mathematics.
No no. Zeno Akhilles. Finite sums of inŽnite many litlle
numbers.
They are just the same inŽte we are handling here everywhere.
> Limits do not apply to the cardinality of sets. However,
inŽnite sums
> are *deŽned* to be the limit of Žnite sums.
And there must be inŽnite of those sums in 1+1/2+1/4+1/8...
Not Žnite.
--
amount and biggest
1+1+1+...= inŽnite and Žnite in math today
Petri Keckman
===
Subject: Re: no comments?
> What must be the biggest number in set that has n item?
>> It depends on whether the set is Žnite or inŽnite.
> Of course that that n belongs to N.
> n {1,2,3,...,n}
> Could be bigger but atleast it is n.
> So what must be the biggest item atleast in set that has oo
item?
>> Assuming, as you seem to do, that only positive integers
can be in the
> Oh yeah. You realize that we are talking here about
Naturals by Peano¹n
> axioms.
> Expet i don¹t like 0. I prefer 1.
>> set, then an inŽnite set does not have a biggest element.
That¹s what
>> it means to be an inŽnite set. So your question is
meaningless.
> Where do we need that inŽnite set. Why not any n in N is
enough
> for amount.
> There must be a bijection from N to N.
> And if there is inŽnite item in N.
> Then these:
> f(1) = 1
> f(2) = 2
> f(3) = 3
> f(4) = 4
> f(5) = 5
> f(6) = 6
> f(7) = 7
> f(8) = 8
> f(9) = 9
> f(10) = 10
> Must be inŽnite. For every n in N...But there could not be
that f(oo)=oo
> InŽnite set does not exist.
InŽnite sets can, and do, exist that contain only Žnite
elements. Not
Žnitely many elements, but Žnite elements.
> Or it is paradoxical conception.
It¹s not paradoxical, you¹re just wrong.
--
Will Twentyman
email: wtwentyman at copper dot net
===
Subject: Re: no comments?
> Let¹s mark every item in N by label so that 1->1,
2->2,...,n->n,...
> How many labels? Answer: InŽnite, because N is said to be
> inŽnite. What is the biggest number?
>> Answer: there isn¹t one. Nor does there have to be one.
> Of course there isn¹t. For every n in n+1 etc...
> Talking about natural numbers:
> What must be the biggest item _atleast_ in set that has 6
item?
> 6 {1,2,3,4,5,6}
> Could be bigger of course
> 8 {1,2,3,4,5,8}
> but atleast it was 6.
That isn¹t what you asked. What you meant to ask is: what is
the
inŽmum of the supremums of the 6-element subsets of N?
> What must be the biggest number in set that has n item?
> n {1,2,3,...,n}
> Could be bigger but atleast it is n.
Which means: it depends on the set.
> So what must be the biggest item atleast in set that has oo
item?
Again, you are assuming there is a meaningful answer that
somehow
corresponds to a number. The correct answer is: There isn¹t
one.
Even if you ask it precisely the answer is still: There isn¹t
one.
> Atleast it must be oo.
oo is not in the set of natural numbers. Therefor it¹s not
the biggest
item at least in the set that has oo items.
> There is no enough Žnite natural numbers for inŽnite set of
Natural
> numbers.
> InŽnite is bigger than any Žnite natural number.
Here¹s what you don¹t seem to understand: *inŽnity is not a
natural
number*.
--
Will Twentyman
email: wtwentyman at copper dot net
===
Subject: Re: no comments?
> Here¹s what you don¹t seem to understand: *inŽnity is not a
natural
> number*.
I agree. it is not. It is nonsense too what comes to the
amount of n in N.
You do not seem to understand what i have tryied to say.
--
amount and biggest
1+1+1+...= inŽnite and Žnite in math today
Petri Keckman
===
Subject: Projection of a Borel set in R^2
I need to see an example of a Borel set of R^2 whose
projection into one of
the coordinate axis is not Borel.
Are there any?
===
Subject: Re: Construct compact set of R whose limit points
form a countable
set
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id i98FmXG19993;
James <ADemetris@gmail.com>
http://mathforum.org/discuss/sci.math/m/641335/641335
> Construct a compact set of real numbers whose limit points
> form a countable set.
> How should I start?
In the case where the countable set is inŽnite, consider
the closure of the set {x : sin(1 / sin(1/x) } or the closure
of the set {1/m + 1/n : m,n = 1, 2, 3, ...}.
Dave L. Renfro
===
Subject: Coloring a map with 3 colors
Hi all,
It¹s just a random question about the coloring problem. You
know, on a map
where countries are separated by boundaries and you want to
assign a color
to each country so that no neighboring countries have the
same color.
(a) Transform the thing as a graph. Each country would be a
vertex and a
boundary would be a line joining the two.
(b) So we have polygons drawn on a plane. For every polygon
that has 4 or
more edges, add extra edges so that we only have triangles
left. For
example
if we have 4 points A, B, C, and D with lines joining A-B,
B-C, C-D, and
D-A, add an extra line A-C. I believe it¹s always possible to
do that,
because if we have n points A_1, ..., A_n, where A_i and
A_i+1 are
connected
(mod n for the indices), then connect A_1 to A_2, A_3,...
A_n-1.
(c) Pick two adjacent vertices. Assign colors Red and Blue.
These two
vertices are part of two triangles, so we pick the color
Green to complete
the triangles. We repeat this thing until each vertex has
been assigned a
color. We simply grow a region of non-colored by carefully
picking the
color
of the non assigned color in a triangle.
(d) Remove the lines added in step (b) and you have your
coloring.
I know it¹s supposed to be a hard problem, so there must be
something
wrong.
Can somebody point me to the real problem or the mistake in
the algorithm
described above?
Tony Bruguier
--
http://antoine.caltech.edu
===
Subject: Re: Coloring a map with 3 colors
>I know it¹s supposed to be a hard problem, so there must be
something
wrong.
>Can somebody point me to the real problem or the mistake in
the algorithm
>described above?
How would your algorithm work on this example?
(best viewed with a Žxed-width font)
--------------------------------------------------
| |
| Unopia |
| |
| ----------------------------------|
| | | |
| | | |
| | | |
| | Dosocia | Tressippia |
| | | |
| | | |
| |---------------------------------|
| | |
| | |
| | Quatronia |
| | |
| | |
--------------------------------------------------
Since all 4 countries touch the other 3, you need 4 colors.
--Keith Lewis klewis {at} mitre.org
The above may not (yet) represent the opinions of my employer.
===
Subject: Re: Coloring a map with 3 colors
>I know it¹s supposed to be a hard problem, so there must be
something
wrong.
>Can somebody point me to the real problem or the mistake in
the algorithm
>described above?
What happens when you apply it to the graph {1<->2, 2<->3,
3<->1
1<->4, 2<->4, 3<->4}?
--
I¹m not interested in mathematics that might have anything
to do with reality. -- Russell Easterly, in sci.math
===
Subject: Re: Construct compact set of R whose limit points
form a countable
set
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id i98FvH720911;
James <ADemetris@gmail.com>
http://mathforum.org/discuss/sci.math/m/641335/641335
> Construct a compact set of real numbers whose limit points
> form a countable set.
> How should I start?
A minor correction to my previous reply ...
In the case where the countable set is inŽnite, consider
the closure of the set {x : sin(1 / sin(1/x) = 0 } or the
closure of the set {1/m + 1/n : m,n = 1, 2, 3, ...}.
Dave L. Renfro
===
Subject: normal vectors
I have a problem ;(
http://members.lycos.co.uk/tatry1023/wektory.png
Data:
vector [A,B,C] (vector length: 1)
vector [a,b,c] (vector length: 1)
angle Alfa <0 , 2*Pi)
Search:
vector [x,y,z].
vector length: 1
===
Subject: Re: normal vectors
lunter@interia.pl (lunter) dixit:
>I have a problem ;(
>http://members.lycos.co.uk/tatry1023/wektory.png
>Data:
>vector [A,B,C] (vector length: 1)
>vector [a,b,c] (vector length: 1)
>angle Alfa <0 , 2*Pi)
>Search:
>vector [x,y,z].
>vector length: 1
Try:
u = vector [A,B,C]
v = vector [a,b,c]
w = vector [x,y,z]
t = u cross-product v
w = v*cos(alpha) + u*sin(alpha)
===
Subject: Re: normal vectors
>lunter@interia.pl (lunter) dixit:
>>I have a problem ;(
>>http://members.lycos.co.uk/tatry1023/wektory.png
>>Data:
>>vector [A,B,C] (vector length: 1)
>>vector [a,b,c] (vector length: 1)
>>angle Alfa <0 , 2*Pi)
>>Search:
>>vector [x,y,z].
>>vector length: 1
>Try:
>u = vector [A,B,C]
>v = vector [a,b,c]
>w = vector [x,y,z]
>t = u cross-product v
>w = v*cos(alpha) + u*sin(alpha)
Sorry the last 2 lines should be:
t = v cross-product u
w = v*cos(alpha) + t*sin(alpha)
If I understood your question correctly, vectors u and v are
perpendicular, and you want vector w such that w is
perpendicular to u
and at an angle of alpha with vector v. This will do it.
===
Subject: Re: normal vectors
> I have a problem ;(
> http://members.lycos.co.uk/tatry1023/wektory.png
> Data:
> vector [A,B,C] (vector length: 1)
> vector [a,b,c] (vector length: 1)
> angle Alfa <0 , 2*Pi)
> Search:
> vector [x,y,z].
> vector length: 1
You want the cross-product vector of the given vectors
divided by its
length.
There are two equal but opposite results, depending on the
order of the
given vectors in the cross product.
The cross product of [A,B,C] with [a,b,c], in that order is
[B*c -C*b, C*a-A*c, A*b - B*a]
===
Subject: Re: 3^k + 2^k revisited
>Now, what can we say about 3^k + 2^k being a cube?
For k = 0 1 2 3 4 5 (mod 6) we have
3^k + 2^k = 2 5 6 0 6 2 (mod 7).
So k = 6n+3. But 3^k + 2^k then lies between two consecutive
cubes
(3^(2n+1))^3 < 3^(6n+3) + 2^(6n+3) < (3^(2n+1)+1)^3
Mike Guy
===
Subject: Cardinality of a sigma-algebra
Let A1, A2, ..., An be n subset of Omega.
Then the smallest sigma-algebra generated by {A1, A2, ...,
An} has at
most 2^2^n elements.
How to prove it?
And if A1, A2, ..., An make a Žnite partition of Omega, what
about the
cardinality of the sigma-A. generated?
P.S. Please also correct my English :-)
Josh.
===
Subject: Approximating Inverse of Analytic Function
I have an analytic function: f(z)=z*c^(-e^z), c in
C{x:x<=0}U{1}
for which I need to approximate the inverse, numerically. The
inverse
function is multivalued, but I am interested basically on the
principal
branch.
Burmann¹s Theorem (which is a generalization of the Lagrange
Inversion
Theorem)
http://mathworld.wolfram.com/BuermannsTheorem.html
allows me to do this about the origin, giving the coefŽcients
of the
series expansion of the inverse as:
a_n=1/n!*d^(n-1)/dz^(n-1){[z/f(z)]^n}|_{z=0}
so I¹ve constructed the following Maple code, which works
nicely:
a:=proc(n,c)#coefŽcients of series
local expr;
if n=1 then
expr:=1/n!*1/c^(-exp(z));
else
expr:=1/n!*(diff(1/c^(-n*exp(z)),z$(n-1)))
Ž;
simplify(subs(z=0,expr));
end:
IF:=proc(w,c,k)#inverse function series
sum(Œa(n,c)¹*w^n,n=1..k);
end:
Now, the Žrst branch point of the inverse, occurs at:
solve(diff(z*c^(-exp(z)),z)=0,z);
W(1/log(c))
Therefore, the radius of convergence of the above series
cannot exceed:
|W(1/log(c))|
However for the values of c I am interested in, this series
has a really
small radius of convergence and I am interested in calculating
numerically the inverse around other points, say, past the
region of
convergence around 0, above.
Can one suggest a modiŽcation to Burmann¹s Theorem that would
allow me
to calculate elsewhere?
Perhaps if I expanded around another point z_0=/=branch point
and z_0
away from the branch cut?
Is there any way to do this in Maple? How can I pick the z_0
(reliably)
to expand around there, making sure I don¹t fall on the
branch cut?
--
I. N. Galidakis --- http://users.forthnet.gr/ath/jgal/
===
Subject: Re: Approximating Inverse of Analytic Function
> I have an analytic function: f(z)=z*c^(-e^z), c in
C{x:x<=0}U{1}
> for which I need to approximate the inverse, numerically.
The inverse
> function is multivalued, but I am interested basically on
the principal
> branch.
> Burmann¹s Theorem (which is a generalization of the
Lagrange Inversion
> Theorem)
> http://mathworld.wolfram.com/BuermannsTheorem.html
> allows me to do this about the origin, giving the
coefŽcients of the
> series expansion of the inverse as:
> a_n=1/n!*d^(n-1)/dz^(n-1){[z/f(z)]^n}|_{z=0}
> so I¹ve constructed the following Maple code, which works
nicely:
> a:=proc(n,c)#coefŽcients of series
> local expr;
> if n=1 then
> expr:=1/n!*1/c^(-exp(z));
> else
> expr:=1/n!*(diff(1/c^(-n*exp(z)),z$(n-1)))
> Ž;
> simplify(subs(z=0,expr));
> end:
> IF:=proc(w,c,k)#inverse function series
> sum(Œa(n,c)¹*w^n,n=1..k);
> end:
> Now, the Žrst branch point of the inverse, occurs at:
> solve(diff(z*c^(-exp(z)),z)=0,z);
> W(1/log(c))
> Therefore, the radius of convergence of the above series
cannot exceed:
> |W(1/log(c))|
> However for the values of c I am interested in, this series
has a really
> small radius of convergence and I am interested in
calculating
> numerically the inverse around other points, say, past the
region of
> convergence around 0, above.
> Can one suggest a modiŽcation to Burmann¹s Theorem that
would allow me
> to calculate elsewhere?
> Perhaps if I expanded around another point z_0=/=branch
point and z_0
> away from the branch cut?
> Is there any way to do this in Maple? How can I pick the
z_0 (reliably)
> to expand around there, making sure I don¹t fall on the
branch cut?
> --
> I. N. Galidakis --- http://users.forthnet.gr/ath/jgal/
There is a useful technique for rearranging a Taylor¹s
series. Steve
Weinberg discussed it in connection with scattering from
well-behaved,
but strongly repulsive potentials. It was published in
Physical Review
sometime in the 1960¹s.
Suppose you have a function f(z) whose nearest singularity is
at z=-a,
that is, on the negative real axis. You know the coefŽcients
of the
Taylor¹s series about z=0, but do not know the function in
closed
form. Suppose you want to evaluate f(z=b), say, where b is
real and
positive, and b>>a. You can do something useful by writing
z = W(w)
where w is a new complex variable, and W is an analytic
function that maps
|w|=1, say, into a curve that excludes the point z=-a (for
example,
W(-1) = -a/2 ) but includes z=b (e.g. W(+1) = 2b ) and maps
the inside
of the unit circle (|w| < 1 ) to the interior of the closed
curve in the
z-plane. It is convenient if W(0) = 0.
Now compute the Taylor¹s series expansion in w:
f(W(w)) = f(0) + f¹(0) W¹(0) w + [f(0){W¹(0)}^2 + f¹(0) W(0)]
w^2 /2
+ ...
Since you know W analyticially and you know the coefŽcients
f¹(0), f(0),
...
you have a rearranged series that is guaranteed to converge
at W=b.
Alternatively, since you have an integral representation of
the inverse
function in terms of the function you are inverting, you
might want
to substitute the (conformal) change of variable in that and
use any
knowledge you have of the location of singularities to
distort the con-
tour of integration appropriately. Then you might be able to
evaluate
the contour integral numerically.

--
Julian V. Noble
Professor Emeritus of Physics
jvn@lessspamformother.virginia.edu
^^^^^^^^^^^^^^^^^^
http://galileo.phys.virginia.edu/~jvn/
For there was never yet philosopher that could endure the
toothache
patiently. -- Wm. Shakespeare, Much Ado about Nothing. Act v.
Sc.
1.
===
Subject: Re: The seventeengon, 17th roots of 1, Galois Želds
and groups
> *what* other thread?
>>I found this at MIT and added some notes, and posted it in
the thread
>>I am doing on cyclotomic polynomials for primes p = 3, ...,
17.
>>I thought it was interesting so posted this notice for
those who missed
>>the other thread.
> --ils duces d¹Enron!
> http://larouchepub.com
> The pyramid is opening!
> Which one?
> The one with the ever-widening hole in it!
> -- The Firesign Theatre
Dale
===
Subject: Re: The seventeengon, 17th roots of 1, Galois Želds
and groups
<tTz9d.983$6q2.376@newssvr14.news.prodigy.com>
posting-account=jcZk7AwAAADXpPEyHtVyWC264SxtppRB
> *what* other thread?
>>I found this at MIT and added some notes, and posted it in
the
thread
>>I am doing on cyclotomic polynomials for primes p = 3, ...,
17.
>>I thought it was interesting so posted this notice for
those who
missed
>>the other thread.
> --ils duces d¹Enron!
> http://larouchepub.com
> The pyramid is opening!
> Which one?
> The one with the ever-widening hole in it!
> -- The Firesign Theatre
> Dale
Or search for seventeengon, or cyclotomic. The URL didn¹t
come out as I
wanted.
Maybe this will do it.
===
Subject: Re: The seventeengon, 17th roots of 1, Galois Želds
and groups
posting-account=jcZk7AwAAADXpPEyHtVyWC264SxtppRB
<font size=+0><a
href=/group/sci.math/browse_frm/thread/135a90cafcb27c4a/
5d9f9691314fbfd4#5
d9f9691314fbfd4>
This one, titled cyclotomic ...
===
Subject: Re: The seventeengon, 17th roots of 1, Galois Želds
and groups
I don¹t know HTML. alas.
> <font size=+0><a
>
href=/group/sci.math/browse_frm/thread/135a90cafcb27c4a/
5d9f9691314fbfd4#5d
9f9691314fbfd4>
> This one, titled cyclotomic ...
===
Subject: Prime numbers and the RSA algorithm
I have been reading about this and I have a question. It is
probably
naive or simplistic, but I would like to understand.
The question is not about the RSA algorithm itself but the
nature of
prime numbers and factorization.
For RSA encryption, you choose two large prime numbers p and
q and the
public key N is the product of p and q. The strength of the
encryption
depends upon the impracticality of factoring N back into p
and q.
The question is how to Žnd a large prime number p. Say you
randomly
choose p. You either check if it is prime by factoring it or
by
dividing it by every number between 2 and the square root of
p (with
the understanding that there may be more efŽcient algorithms
for
checking primes.)
If it is easy to check p for primeness, then it must be
relatively
easy to check N for its factors. I.e., by factorization or by
dividing
it by every number between 2 and the square root of N.
As a mathematical layman, I must be missing something. Or is
it the
case that what I said is true, but the point is that it might
take a
minute or so to check p but would take much longer for a
value which
is roughly the square of p.
Or is the answer that factoring large numbers is impractical.
But
testing large numbers for primeness IS practical. This must
be the
answer, but I haven¹t found any information on how this is
done.
I thank anyone for clearing up my confusion for me!
===
Subject: Re: Prime numbers and the RSA algorithm
> I have been reading about this and I have a question. It is
probably
> naive or simplistic, but I would like to understand.
> The question is not about the RSA algorithm itself but the
nature of
> prime numbers and factorization.
> For RSA encryption, you choose two large prime numbers p
and q and the
> public key N is the product of p and q. The strength of the
encryption
> depends upon the impracticality of factoring N back into p
and q.
> The question is how to Žnd a large prime number p. Say you
randomly
> choose p. You either check if it is prime by factoring it
or by
> dividing it by every number between 2 and the square root
of p (with
> the understanding that there may be more efŽcient
algorithms for
> checking primes.)
> If it is easy to check p for primeness, then it must be
relatively
> easy to check N for its factors. I.e., by factorization or
by dividing
> it by every number between 2 and the square root of N.
> As a mathematical layman, I must be missing something. Or
is it the
> case that what I said is true, but the point is that it
might take a
> minute or so to check p but would take much longer for a
value which
> is roughly the square of p.
> Or is the answer that factoring large numbers is
impractical. But
> testing large numbers for primeness IS practical. This must
be the
> answer, but I haven¹t found any information on how this is
done.
> I thank anyone for clearing up my confusion for me!
There are tests for primeness that are quicker than searching
for
factors, at least when applied to very large prime candidates.
In particular, a lot of composites can be quickly eliminated
withoutactually factoring them, though testing for a few of
the smallest
small primes as factors also eliminates a lot of candidates
quickly.
Were it to for things like Carmichael numbers, these tests
would be
very simple indeed.
For a large number that is the product of two primes of
nearly equal
size, the process of actual factoring can be lengthy enough
to be
impractical, requiring years for the numbers used in less
critical
coding to centuries for the most secure.
===
Subject: Re: Prime numbers and the RSA algorithm
> I have been reading about this and I have a question. It is
probably
> naive or simplistic, but I would like to understand.
> The question is not about the RSA algorithm itself but the
nature of
> prime numbers and factorization.
You want Professor Caldwell¹s Prime Pages:
http://primepages.org/
In particular the section on proving primality.
Phil
--
They no longer do my traditional winks tournament lunch -
liver and bacon.
It¹s just what you need during a winks tournament lunchtime
to replace lost

===
Subject: Re: Prime numbers and the RSA algorithm
> The question is how to Žnd a large prime number p. Say you
randomly
> choose p. You either check if it is prime by factoring it
or by
> dividing it by every number between 2 and the square root
of p (with
> the understanding that there may be more efŽcient
algorithms for
> checking primes.)
There *are* more efŽcient ways of checking primality. Much
more
effcient if you allow the slight possibility that a composite
is
falsely stated as prime, e.g., the Miller-Rabin test.
> Or is the answer that factoring large numbers is
impractical. But
> testing large numbers for primeness IS practical. This must
be the
> answer, but I haven¹t found any information on how this is
done.
Yes; at least the best that has been achieved is fast testing
but not fast factorization.
On the web you might look at Chris Caldwell¹s prime pages
http://www.utm.edu/research/primes/prove/index.html
But if you are after serious mathematics, see the book
of Crandall and Pomerance
http://www.amazon.co.uk/exec/obidos/ASIN/0387947779/
--
Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html
Lacan, Jacques, 79, 91-92; mistakes his penis for a square
root, 88-9
Francis Wheen, _How Mumbo-Jumbo Conquered the World_
===
Subject: Re: Prime numbers and the RSA algorithm
>
> The question is how to Žnd a large prime number p. Say you
randomly
> choose p. You either check if it is prime by factoring it
or by
> dividing it by every number between 2 and the square root
of p (with
> the understanding that there may be more efŽcient
algorithms for
> checking primes.)
> There *are* more efŽcient ways of checking primality. Much
more
> effcient if you allow the slight possibility that a
composite is
> falsely stated as prime, e.g., the Miller-Rabin test.
And if you do not allow that possibility you can always do a
full
primality test on numbers that are probably prime by
Miller-Rabin.
BTW, this is how primality provers work. First trial division
by
small primes. Next Miller-Rabin, or suchlike. Next a full
proof.
And as for more efŽcient, indeed *far* more efŽcient. I think
that proving prime a probable prime of about 80 digits would
take
0.1 seconds on current hardware (at least it was less than 0.5
seconds on hardware some 5 years ago).
--
dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam,
nederland,
+31205924131
home: bovenover 215, 1025 jn amsterdam, nederland;
http://www.cwi.nl/~dik/
===
Subject: Re: Derivative of exponential function
> I¹m in the process of home-schooling my son in calculus.
We¹re about to
> Žnish up on limits, so that we can start differentiation.
I¹ve relearned
> enough that I¹ve been able to derive, from the deŽnition of
derivative,
> the derivatives of c, x, cx, f(x)+g(x), f(x)*g(x), f(g(x)),
sin(x),
> (f(x))^n, (f(x))^(-n), tan(x). A pretty solid beginning.
But there¹s
> one big hole: I can¹t prove that the derivative of e^x is
e^x.
> I can see that if Lim(h->0) ((e^h-1)/h) == 1, then I¹m set.
Numerically,
> I can crank out values for ((e^h-1)/h) with h getting very
small, and
> see that they get really close to 1. But, that¹s not proof.
> I was going to base a proof on the Taylor series for e^x,
but that
> depends upon already knowing the nth derivatives of e^x, so
that was
> out. Then I was going to try using my knowledge of the
behavior of
> e^x, but upon examination, that was all based on knowing
things about
> its derivative.
> Is there some simple (or subtle) trick that I¹m
overlooking? Is the
> proof of this limit actually incredibly hard?
> I can¹t look in the book, because it¹s with him (he lives
with his
> mother). I¹ve even tried typing ((e^h-1)/h) into Google,
but that
> just turned up a bunch of PDF Žles. Any help, or am I going
to need
> to do some serious hand-waving?
How about this - deŽne a function f(x) to be the same as it¹s
derivative. Assume you can do a series expansion of f(x).
Explicitly
do the series expansion for f(x) and d/dx (f(x)) and set them
equal to
each other (for all x). This should deŽne the coefŽcients in
the
series expansion. Now call this function f(x) the exponential
function, exp(x).
I have seen this done before and thought it was quite neat. I
don¹t
know if it can mathematically justiŽed, as I am only a
physicist, but
it looks quite neat to me, and you can avoid all the
handwaving.
Ian Taylor
===
Subject: Re: Derivative of exponential function
>How about this - deŽne a function f(x) to be the same as it¹s
>derivative. Assume you can do a series expansion of f(x).
Explicitly
>do the series expansion for f(x) and d/dx (f(x)) and set
them equal to
>each other (for all x). This should deŽne the coefŽcients in
the
>series expansion. Now call this function f(x) the exponential
>function, exp(x).
>I have seen this done before and thought it was quite neat.
I don¹t
>know if it can mathematically justiŽed, as I am only a
physicist, but
>it looks quite neat to me, and you can avoid all the
handwaving.
Actually, that is my favorite deŽnition of the exponential
function:
The unique solution to the differential equation
f¹ = f
f(0) = 1.
That is, the exponential function is the eigenfuction of the
differential operator corresponding to the eigen value 1 with
f(0) = 1.
--
Stephen J. Herschkorn herschko@rutcor.rutgers.edu
===
Subject: Re: Derivative of exponential function
> How about this - deŽne a function f(x) to be the same as
it¹s
> derivative. Assume you can do a series expansion of f(x).
You don¹t have to assume this --- it follows.
--
Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html
Lacan, Jacques, 79, 91-92; mistakes his penis for a square
root, 88-9
Francis Wheen, _How Mumbo-Jumbo Conquered the World_
===
Subject: Re: Derivative of exponential function
Robin,
> How about this - deŽne a function f(x) to be the same as
it¹s
> derivative. Assume you can do a series expansion of f(x).
> You don¹t have to assume this --- it follows.
Well, there is another function which satisŽes the condition;
its series
expansion is Sigma a_ix^i where all a_i are 0. :)
===
Subject: Re: Derivative of exponential function
A solution to your problem given two standard deŽnitions of
e^x have
already been given.
Here is a further possibility: if e^x is deŽned to be
lim (1+x/n)^n,
as it sometimes is in school (in connection with cumulative
interest,
for example), then you can argue as follows: the derivative
of each
term in the limit is
(1+x/n)^{n-1},
and
lim (1+x/n)^{n-1} = lim 1/(1+x/n) * lim (1+x/n)^n = 1 * e^x.
Since the convergence of the derivatives is, in fact,
(locally)
uniform in x, the original function is differentiable and the
derivative of the limit is the limit of the derivatives. Of
course you
wouldn¹t really try to say the latter to a high school
student, but
this might give a reasonable justiŽcation.
Hope this helps,
Lasse
---
(lasse@remove.for.spam.maths.warwick.ac.uk)
===
Subject: Re: Derivative of exponential function
>(f(x))^n, (f(x))^(-n), tan(x). A pretty solid beginning. But
there¹s
>one big hole: I can¹t prove that the derivative of e^x is
e^x.
>>How are you _deŽning_ e^x? You need to tell us that Žrst...
>Well, that seems to be the unanimous opinion, and something
that I¹d
>completely ignored. I was deŽning it kind of intuitively, I
guess.
>e^0 == 1
>e^n == e*e^(n-1) (n>0, n in Z)
>e^(-n) == 1/(e^n) (n>0, n in Z)
>(e^(1/q))^q == e (q!=0, q in Z)
>e^(p/q) == (e^(1/q))^p (q!=0, p,q in Z)
>Then, I get real hand-wavy and talk about how any a in R is
the
>limit of a bunch of p_i/q_i, and so there¹s e^a.
>Kind of weak, I guess. The real žaw (that I overlooked) is
in the
>second line above, since I¹ve never deŽned e. Or, are there
bigger
>žaws in my intuitionistic deŽnition (aside from my not
knowing anything
>about Cauchy sequences)?
It¹s possible to deŽne things this way, but although it may
be the
approach that¹s going to make the most sense to some people
it¹s
actually not the approach that makes things easiest to prove
- as
you point out you have a certain amount of work to do before
you
can even say the deŽnition makes sense.
>Is there some simple (or subtle) trick that I¹m overlooking?
There are at least two standard ways to go about all this; at
least one has already been mentioned:
(i) _DeŽne_ log(x) = int_1^x dt/t for x > 0. Then show log has
various properties (for example log(xy) = log(x) + log(y)
follows
easily, it¹s not hard to show that log is a bijection from
(0, inŽnity) onto R, etc.) Then deŽne exp to be the inverse
of log.
(ii) DeŽne exp(x) by the power series.
>DeŽning my terms, appears to be the consensus view. I guess
that I¹ll
>need to wait until the weekend and see what deŽnition Fadell
& Fadell
>use.
>Thaks to *all* who responded.
************************
David C. Ullrich
===
Subject: Re: Derivative of exponential function
> (ii) DeŽne exp(x) by the power series.
Even an idiot like me was able to prove the essential
properties
e^{x+y}=e^x*e^y and e^x*e^-x=1 last night using the binomial
theorem and
other properties of Pascal¹s Triangle (as well as continuity
and
differentiability, of course).
Fun to do Analysis I again after all these years.
===
Subject: Re: Derivative of exponential function
> I can see that if Lim(h->0) ((e^h-1)/h) == 1, then I¹m set.
Numerically,
> I can crank out values for ((e^h-1)/h) with h getting very
small, and
> see that they get really close to 1. But, that¹s not proof.
But e is the number especially chosen so that lim_{h->0} (e^h
- 1)/h = 1.
Exercise:
(i) prove that for each positive a, lim_{h->0} (e^h - 1)/h
exists.
(ii) if we call the limit L(a) prove that L is continuous and
increasing.
(iii) Prove that L(ab) = L(a) + L(b)
(iv) that there is a unique e with L(e) = 1.
--
Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html
Lacan, Jacques, 79, 91-92; mistakes his penis for a square
root, 88-9
Francis Wheen, _How Mumbo-Jumbo Conquered the World_
===
Subject: Re: Derivative of exponential function
boundary=------------050706010405010202000802
-------------------------------------------------------------
--------
>>
>(f(x))^n, (f(x))^(-n), tan(x). A pretty solid beginning. But
there¹s
>one big hole: I can¹t prove that the derivative of e^x is
e^x.
>
>>How are you _deŽning_ e^x? You need to tell us that Žrst...
>>
>Well, that seems to be the unanimous opinion, and something
that I¹d
>completely ignored. I was deŽning it kind of intuitively, I
guess.
>e^0 == 1
>e^n == e*e^(n-1) (n>0, n in Z)
>e^(-n) == 1/(e^n) (n>0, n in Z)
>(e^(1/q))^q == e (q!=0, q in Z)
>e^(p/q) == (e^(1/q))^p (q!=0, p,q in Z)
>Then, I get real hand-wavy and talk about how any a in R is
the
>limit of a bunch of p_i/q_i, and so there¹s e^a.
If you are going to work from exponentiation this direct way,
you can
show easily show the following:
For Žxed positive a, let f(x) = a^x. If f is differentiable
at 0,
then f is differentialble everywhere, and
f¹(x) = f¹(0) f(x).
The hard part then is to show that f is differentiable at 0,
and
that if a = e (however you deŽne it), then f¹(0) = 1.
--
Stephen J. Herschkorn herschko@rutcor.rutgers.edu
===
Subject: Re: Derivative of exponential function
>>(f(x))^n, (f(x))^(-n), tan(x). A pretty solid beginning.
But there¹s
>>one big hole: I can¹t prove that the derivative of e^x is
e^x.
>How are you _deŽning_ e^x? You need to tell us that Žrst...
> Well, that seems to be the unanimous opinion, and something
that I¹d
> completely ignored. I was deŽning it kind of intuitively, I
guess.
> e^0 == 1
> e^n == e*e^(n-1) (n>0, n in Z)
> e^(-n) == 1/(e^n) (n>0, n in Z)
> (e^(1/q))^q == e (q!=0, q in Z)
> e^(p/q) == (e^(1/q))^p (q!=0, p,q in Z)
> Then, I get real hand-wavy and talk about how any a in R is
the
> limit of a bunch of p_i/q_i, and so there¹s e^a.
> Kind of weak, I guess. The real žaw (that I overlooked) is
in the
> second line above, since I¹ve never deŽned e.
I think that¹s the problem. You can probably show that
lim (2^h - 1) / h < 1, and lim (3^h - 1) / h > 1,
then make a continuity argument that there has to be some a,
2 < a < 3, such that lim (a^h - 1) / h = 1,
then announce that you¹re going to call that number e.
--
Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email)
===
Subject: (mod 2^4096) on a 4096 bit MAA?
Hello. I need to perform the 4097-bit modular reduction
a (mod 2^4096)
with only 4096-bit (or less) modular operations, but haven¹t
a clue if
such a thing is possible.
I think this might be equivalent to simulating a group with
multiple
smaller groups? Is this the right terminology?
Any suggestions would be much appreciated!
--
Jay Miller
PGP Fingerprint | 5f7adbbe bfc60727 96dca94c 616d5080 09e3e846
===
Subject: Re: (mod 2^4096) on a 4096 bit MAA?
> Hello. I need to perform the 4097-bit modular reduction
> a (mod 2^4096)
> with only 4096-bit (or less) modular operations, but
haven¹t a clue if
> such a thing is possible.
> I think this might be equivalent to simulating a group with
multiple
> smaller groups? Is this the right terminology?
> Any suggestions would be much appreciated!
download GP/Pari.
For example, if you want to know what 3^123456789 (mod
2^4096) is, you would
do
Mod(3,2^4096)^123456789
and you¹d get the answer back in a fraction of a second.
Phil
--
They no longer do my traditional winks tournament lunch -
liver and bacon.
It¹s just what you need during a winks tournament lunchtime
to replace lost

===
Subject: Re: (mod 2^4096) on a 4096 bit MAA?
Words by Phil Carmody <thefatphil_demunged@yahoo.co.uk>:
>download GP/Pari.
>For example, if you want to know what 3^123456789 (mod
2^4096) is, you
would do
>Mod(3,2^4096)^123456789
>and you¹d get the answer back in a fraction of a second.
Heh. I very much appreciate the response, though I¹m afraid
it isn¹t
quite that simple in my case. I should have supplied a bit of
background.
I¹m implementing the Arazi Inversion formula -
d = ( 1 + o ( -o^{e-2} mod e ) ) / e,
- on an 8-bit uC that happens to have a 4096-bit modular
arithmetic
accelerator (MAA) attached to it. To avoid the expensive
division,
I¹m hoping to use only modular operations to perform this
calculation.
It turns out that when |o| is the size in bits of o, the
following
formula is equivalent:
d = ( ( 1 + o ( -o^{e-2} mod e ) ) * e^{-1} ) mod 2^{|o|},
where Œe^{-1}¹ indicates the modular inverse of e. The latter
forumla
is much faster for |o| < 4096 but when |o| is equal to 4096,
2^{|o|}
exceeds the capacity of my MAA!
Hopefully my question now makes more sense: is there any way
to
compute
a (mod 2^|o|)
with less than |o| modular operations only?
--
Jay Miller
PGP Fingerprint | 5f7adbbe bfc60727 96dca94c 616d5080 09e3e846
===
Subject: Re: Necessary condition for a matrix to have real
eigenvalues
>Would anyone tell me what is the necessary conditions for a
>non-hermitian matrix to have real eigenvalues ?
As others have stated, there is no particularly good answer
here:
you haven¹t really told us anything about the matrix except an
implication that it¹s square and has entries in R (or maybe C
?).
Statistically speaking, most such matrices do not have real
eigenvalues. You¹d have to start by mentioning some clue you
have
that makes you hope they have real eigenvalues.
>I have some matrices which have real eigenvalues. I need to
reduce
>their dimension. How can I make sure the eigenvalues are
still real
>after I reduce their dimension.
I don¹t think others have commented on this. How exactly are
you
reducing the dimension?
I prefer to think of matrices as (representing) linear maps:
I guess
you are starting off with a linear map L : R^n --> R^n for
some n.
When you say the matrix has real eigenvalues, does that mean
_all_
its eigenvalues are real? If so, then there is a basis for R^n
consisting of real eigenvectors v_i. When you want to reduce
the
dimension, does that mean you have a subspace V of R^n which
is invariant under L ? (Or perhaps you have also a map R^n
--> V
which you compose with L and with the inclusion V --> R^n ?)
The natural kind of subspace to choose would be one which is
spanned
by some of the eigenvectors. In that case, V would indeed be
invariant under L, and (L|V) : V --> V would obviously still
have
real eigenvalues, namely the same eigenvectors you started
with.
On the theoretical level there¹s nothing to prove!
As a numerical issue I guess there would potentially be
round-off
problems which might mean that the matrix representing L|V
might
not have real eigenvalues. (This would require that at least
one
pair of the real eigenvalues were equal or at least close.)
Whether this kind of failure actually occurs or not would
depend on
the algorithm you are using to construct the matrix for L|V .
So I guess I¹ve come back to my original question: what does
it mean
when you say you are reducing the dimension?
dave
===
Subject: re:Necessary condition for a matrix to have real
eigenvalues
How can I know the characteristic polynomial has no complex
root?
The matrix is quite large, about 100x100
if I reduce the dimension of the matrix to eg. 90x90 by
deleting 10
rows and columns, is there any method to make sure the
characteristic
polynomial does not have complex roots?
----------------------------------------------------------
** SPEED ** RETENTION ** COMPLETION ** ANONYMITY **
----------------------------------------------------------
http://www.usenet.com
===
Subject: Re: Necessary condition for a matrix to have real
eigenvalues
>How can I know the characteristic polynomial has no complex
root?
>The matrix is quite large, about 100x100
>if I reduce the dimension of the matrix to eg. 90x90 by
deleting 10
>rows and columns, is there any method to make sure the
characteristic
>polynomial does not have complex roots?
If the matrix is symmetric (real) or Hermitian (complex),
this is the case. There are cases which can be reduced to
this; the problem of obtaining the roots is almost always
subject to errors.
Otherwise it is very difŽcult to decide. For a large matrix,
all the methods I can think of are numerically unstable.
This is why obtaining roots and vectors together is a good
idea, as reconstructions from the combinations are often
much more accurate than doing them separately.
--
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: Question about induction argument
>>The question is more clearly stated as: Let X be a Žnite
collection
>>of open intervals that cover the closed interval I = [a0,
b0]. Show
>>that there exists a Žnite subcollection Y = {(a_i, b_i): i
= 1...m}
>>of X such that Y covers I , a1 < a0 < b1, a_m < b0 < b_m,
and
>>a_(k+1) < b_k < b_(k+1) for 1 <= k <= m-1.
>>
>Whew, it¹s understandable.
>>Not so hard. Use induction on n = |X|.
>>
>Then let¹s make X inŽnite.
That was not the initial question. However, since [a0,b0] is
compact, X must have a Žnite subcover.
--
Stephen J. Herschkorn herschko@rutcor.rutgers.edu
===
Subject: Re: Question about induction argument
>Consider interval [a_0, b_0]. There is a collection {(a_j,
b_j):
>1<=j<=n} of (open) intervals such that
>
>Union (j=1 -> n) (a_j, b_j) contains [a_0, b_0].
>
>It is intuitively clear that there exists a subset of {(a_j,
b_j)},
>{(a_k, b_k): 1<= k <= m <= n} such that :
>
>a_1 < a_0 < b_1, a_m < b_0 < b_m and
>
>for m>1, a_(k+1) < b_k < b_(k+1), 1 <= k <= m-1
>
>How does one construct a formal induction argument to prove
this?
>
> Not so hard. Use induction on n = |X|.
<n we assume that they are also met. I¹m having a little
difŽculty
proving that if this is the case, then if |X|=n they are also
met.
Any hints would be appreciated.
I have also wondered about using induction in this way: Since
the
only variable is n, as long as its elements are bounded, open
intervals whose union contains [a_0, b_0], the speciŽc
composition of
X does not matter. That is the different X¹s for |X|<n can
have
totally unrelated elements, and the induction is valid to any
X such
that |X|=n. This is not intuitively clear to me, and I¹m sure
there
===
Subject: Re: Question about induction argument
>Consider interval [a_0, b_0]. There is a collection {(a_j,
b_j):
>1<=j<=n} of (open) intervals such that
>
>Union (j=1 -> n) (a_j, b_j) contains [a_0, b_0].
>
>It is intuitively clear that there exists a subset of {(a_j,
b_j)},
>{(a_k, b_k): 1<= k <= m <= n} such that :
>
>a_1 < a_0 < b_1, a_m < b_0 < b_m and
>
>for m>1, a_(k+1) < b_k < b_(k+1), 1 <= k <= m-1
>
>How does one construct a formal induction argument to prove
this?
>
>
>
>>Not so hard. Use induction on n = |X|.
>>
><n we assume that they are also met. I¹m having a little
difŽculty
>proving that if this is the case, then if |X|=n they are
also met.
>Any hints would be appreciated.
>I have also wondered about using induction in this way:
Since the
>only variable is n, as long as its elements are bounded, open
>intervals whose union contains [a_0, b_0], the speciŽc
composition of
>X does not matter. That is the different X¹s for |X|<n can
have
>totally unrelated elements, and the induction is valid to
any X such
>that |X|=n. This is not intuitively clear to me, and I¹m
sure there
The proposition:
Let X be a Žnite collection of open intervals that cover a
closed
interval I = [a0, b0]. Show that there exists a Žnite
subcollection
Y = {(a_i, b_i): i = 1...m} of X such that Y covers I , a1 <
a0
< b1, a_m < b0 < b_m, and a_(k+1) < b_k < b_(k+1) for 1 <= k
<= m-1.
Outline of the induction step: Assume true for any interval
closed by a
cover with n intervals and that |X| = n+1. Let a be the least
value of lower bounds of intervals in X and let (a1, b1) be
any
interval in X where a1 = a. Then X {(a1,b1)} covers [b1, b0].
--
Stephen J. Herschkorn herschko@rutcor.rutgers.edu
===
Subject: Re: Question about induction argument
> Consider interval [a_0, b_0]. There is a collection {(a_j,
b_j):
> 1<=j<=n} of (open) intervals such that
>
> Union (j=1 -> n) (a_j, b_j) contains [a_0, b_0].
>
> It is intuitively clear that there exists a subset of
{(a_j, b_j)},
> {(a_k, b_k): 1<= k <= m <= n} such that :
>
> a_1 < a_0 < b_1, a_m < b_0 < b_m and
>
> for m>1, a_(k+1) < b_k < b_(k+1), 1 <= k <= m-1
>
> How does one construct a formal induction argument to prove
this?
> The following is a rough proof ignoring the induction idea.
You may
> need to Žll in a few details to make it rigorous. You know
that
> [a_0,b_0] is a subset of Union (j=1,n) of (a_j,b_j). There
are two
> possibilities:
> 1) [a_0,b_0] is a subset of one of the (a_j,b_j). This
trivially
> satisŽes the conditions.
> 2) [a_0,b_0] is not a subset of any of the (a_j,b_j). In
this case,
> consider a minimal collection (with renumbering as
necessary) of
> (a_k,b_k) where k=1 to m, a_1<a_2<a_3<...<a_m and Union
(k=1,m) of
> (a_k,b_k) contains [a_0,b_0]. Now, to be a minimal
collection,
> b_1<b_2<b_3<...<b_m, otherwise you have an i where
(a_k,b_k) contains
> (a_(k+1),b_(k+1)). Now, suppose a_1<a_0<b_1 is not true,
then either
> a_0 < a_1, or b_1<a_0. In the Žrst case, a_0 is not in the
union,
> violating the fact that the union is a cover. In the second
case, the
> (a_k,b_k) where k=2 to m must be a cover, violating the
minimality.
> Similarly, a_m < b_0 < b_m. Now, suppose there is a k,
1<=k<=m-1,
> where a_(k+1) < b_k < b_(k+1) is not true. We know from
above that
> b_k<b_(k+1), so we must have b_k<a_(k+1). In that case, the
union does
> not cover [b_k,a_(k+1)]. Since it does cover [a_0,b_0],
this must mean
> that the intersection of [b_k,a_(k+1)] and [a_0,b_0] is
empty. So,
> either b_1<b_k<a_(k+1)<a_0 -> b_1<a_0, a contradiction with
the above,
> or b_0<b_k<a_(k+1)<a_m -> b_0 < a_m, a contradiction with
the above.
> QED
> I think the only detail that could be strengthened is the
existence of
> the minimal covering of [a_0,b_0], but since there is a
Žnite cover of
> open sets to begin with, getting a minimal cover by
throwing out the
> excess sets shouldn¹t be too hard.
more stringent than that of the original problem: In the
minimal
collection, as you indicate, both a_j¹s and b_j¹s are ordered,
whereas in the original only the b_j¹s are ordered. I guess
one would
still require some kind of induction to establish its
existence. That
is, one can somehow use induction to perform the throwing out
and
===
Subject: Expected duration of pass the buck game?
Originator: tchow@lagrange.mit.edu.mit.edu (Timothy Chow)
At time t = 0 there are N+1 people, each of which has M
bucks. At each
time step, each person passes exactly one of his bucks to
another person
(each victim is chosen independently and uniformly at random
from the N
candidates).
(1) How long does it take until the Žrst time someone has no
bucks?
(2) If people are removed as soon as they have no bucks, how
long does
it take for 2 people to acquire all the bucks?
I don¹t expect an exact answer, just an approximation to the
expected
value.
--
Tim Chow tchow-at-alum-dot-mit-dot-edu
The range of our projectiles---even ... the
artillery---however great, will
never exceed four of those miles of which as many thousand
separate us from
the center of the earth. ---Galileo, Dialogues Concerning Two
New Sciences
===
Subject: Re: Skolem¹s Paradox and why is math the way it is?
> Read this:
> Does Goedel submarine higher-order logic?
When I follow that link I get a comment about a footnote, and
all it
says is how did Godel know results that were proved by other
people
years later?. Did you perhaps give the wrong link?
> Why do we care about uncountably many real numbers when in
reality
> there are only a countable number that we can prove
theorems
about?
>
> Actually, that¹s not true. Although there can be only a
countable
> number of theorems, they can still address an uncountable
number
of
> reals. As a simple example, there are an uncountable
number of
reals
> in the Cantor set, and a simple procedure by which to
determine
given
> any real if it is a member (namely, can its decimal
expansion,
when
> convereted to base 3, be written without using the digit
1).
>
> Jonathan Hoyle
>
> question more clearly. There are many sets, we agree on
that.
Some
> sets have bijections with the set of naturals, others
don¹t, we
agree
> on that.
>
> I don¹t. You can apply a canonical ordering operator onto
any set
that is
> not ordering-sensitive.
> Which statement were you disagreeing with? Are you saying
that all
> sets have bijections with the set of naturals, because all
I said was
> that some do and others don¹t?
> No, only inŽnite sets are equivalent, in my theory. There
are also
> alternatives where that is not so.
> This has to do with deŽning, or rather, acknowledging, that
P(X) = X +
1.

I reference for that would be nice, it seems like some formal
symbols
to me, and I don¹t understand what the intended
interpretation is
supposed to be.
> ... But each time you use an axiom to prove a theorem, the
axiom
> produces, at most, a Žnite number of sets, and you can use
each
axiom
> only a Žnite number of times in a single proof. Doesn¹t it
seem
that
> each proof generates only a Žnite number of sets, even
though some
of
> these sets might be considered large themselves? And if the
number
> of theorems is somewhat countable, then doesn¹t it seem like
there
> is only a somewhat countable number of sets that we can
prove
> theorems about?
>
>
> I eschew axioms, but if you call inŽnity an axiom then it
generates
> inŽnitely many sets and all the proofs about those sets.
> The axiom of inŽnite I know asserts the existance of ONE
set (that
> either is, or at least contains omega). And then the axiom
of
> speciŽcation creates a countable number of subsets, and one
can
> invoke the other axioms a Žnite number of times, and there
is still
> just an countable number of sets you¹ve made.
> I just say there is at least nothing and then that through
excluded middle
there
> is everything else, and that the axiom is actually a
theorem.
> In the above reference, there is discussion about Goedel
essentially
agreeing
> with Skolem and Loewnheim. Also, the reals are the reals
are the reals.
I didn¹t see that in the reference.
> You, for instance, cited a theorem about ONE set (the
cantor set).
So
> there is at least one set in the universe. But you want be
to
believe
> that you just proved a theorem about many sets. However,
all you
> proved is that there is a certain subset of another set,
namely the
> cantor set is a subset of the reals. That is a theorem
about ONE
set.
> How am I supposed to tell if there are as MANY sets in the
universe
>
> Easily, put them in a line.
> The axioms don¹t put anything into a line, the whole point
is that the
> axioms don¹t generate ALL the points on the line, only a
countable
> number of them. Consistently one can add more axioms to
have more
> numbers on the line, that¹s what the diagonal arguement
says, but if
> one doesn¹t add more axioms, then you only have a countable
number of
> points that the original ZF(C) axioms talk about directly.
And sadly,
> even after adding more axioms you still only have a
countable number.
> That¹s not sad. There are just countably many, inŽnite sets
are
equivalent.
I think I believe that too, that¹s why I don¹t like axioms
like ZF
that make it appear otherwise.
> Are second order logic and countable models of the
uncountable
not
> slippery slopes? (They are.)
> I don¹t see how it is a slippery slope to be clear what you
are
> tlaking about and what you are not.
> The issue I hope to make clear to you is that pushing off
the resolution
into
> high-order logics, in an admitted way that there is no
resolution in
resort to
> some metalogic, is handwaving that does not offer a
resolution, and that
if a
> resolution is to exist it exists in Žrst order logic, and
that all
high-order
> and meta logic is enframed or emposed within Žrst order
logic.

I don¹t know what you mean by emframed and emposed. And what
about
something between Žrst order and second order logic?
> It¹s a theory about one set: all of them.
> I lost you here. There is no set of all sets.
> Why not?
We need some way of saying what is and is not a set, what is
your
deŽntion and/or what axioms do you use to assert the
existance of the
sets you claim exist?
===
Subject: Re: Skolem¹s Paradox and why is math the way it is?
>rather than adjust Einstein¹s equations to meet the large
scale
>observations that we make to be consistent with the matter
that we DO
>see.
> Physicists have been doing that for decades. They simply
haven¹t been
> able to devise anything that works better than GR. The most
promising
> work right now is in an area that you probably don¹t like,
because it
> is absolutely ladden with unobservable phenomena.
Your comments are very vague. People are introducing
unobservable
phenomena to make theories where observations are consistent
with the
matter we do see? Why? What is the point? As for not doing
better
than GR, I¹ve seen claims of better results from people whose
work I
don¹t understand. So I can¹t say, but I¹d be interested in
your
opinion if did successfully understand their models. What
went wrong?
===
Subject: Re: Skolem¹s Paradox and why is math the way it is?
> The countable model isn¹t the intended model because
there are
some
> collections of its elements of rank smaller than some Žxed
ordinal
> which don¹t get counted as sets. Why would you want to Žx
the
axioms?
> There¹s nothing wrong with them.
>
> I was interested in Žxing the axioms because I don¹t see
the
point of
> numbers that we can¹t talk about. If we had a bijection
that
said,
> these are the sets and the other axioms don¹t apply to
this
particular
> bijection, then we¹d know what the model was conŽned too,
no
> mysterious other elements whose properties are based on
what
order I
> same as your model, how can you tell from the inside that
it is
> defective, and what makes it defective in your opinion.
>
> The fact that it¹s missing a bijection.
>
> Why do you keep saying that *I* am missing a bijection,
when I say
> that *you* are missing a logical correspondance that
satisŽes the
> same logical properties that the bijections of your model
do. You
are
> the one that refuses to formalize a true statement with the
properties
> of a bijection just so you can make some type theory on
your class
of
> sets, that has NO application that I¹ve ever seen.
>
>
> The fact that you say your model is countable means you
must admit
> there is a bijection not in the model. This is presumably
what you
are
> calling a logical correspondence. I freely admit that the
bijection
> exists, and I say this shows your model is incomplete. I¹m
asking you
> to consider a model that has *all* of the bijections in it.
>
> I think I understand what you are saying, and it¹s really
CLOSE to
> answering my questions. Let me try to ask it again more
carefully.
> Apparantly the Žrst order ZF axioms don¹t adequately
describe the
> universe of sets because they allow countable models. You
tell me to
> Žx this by considering a different model. Wouldn¹t it be
better
to
> change the axioms so that the objectional model was no
longer a model?
> It seems better than assuming that things exist that are
independant
> of the axioms. And it sounds like you are saying that you
need second
> order axioms to avoid the countable model. Since IF-logic
is rather
> new, is it possible that there are IF axioms that are
sufŽcient to
> remove substandard models? Is it just going to be the case
that all
> axioms of any order are substandard?
>
> Second-order axioms can Žx the model up to isomorphism. I
don¹t know
> whether IF axioms will do the job.
Don¹t second order axioms assume the existance of all the
functions as
a basis? That means you basically already have to have the
reals to
use it, don¹t you?
===
Subject: Re: Skolem¹s Paradox and why is math the way it is?
> Eray Ozkural exa says...
>> OC we haven¹t, (and as in a thread some time ago),
>> *can¹t* precisely deŽne the deŽnability of reals...
>Why, I thought deŽnability was sufŽciently well deŽned. Do
you have
>a counter-example or a proof for these second and third
sentences of
>yours?
> A notion of deŽnable gives rise to a function f(s) from
strings
> to reals in [0,1]: namely, if s is an unambiguous deŽnition
of a
> real r in [0,1] then f(s)=r, otherwise f(s) = 0. Let s_n be
the nth
> string in a complete enumeration of strings. Let r_n be the
real
> f(s_n). Then r_0, r_1, etc. is a deŽnable enumeration of
all the
> deŽnable reals in [0,1].
> But using Cantor¹s proof, we can come up with a new real d
in [0,1]
> that is unequal to any of the reals in the enumeration r_0,
r_1, etc.
> Furthermore, d is *deŽnable* in terms of the function f(s).
So,
> the notion of deŽnable used in the construction of f did not
> actually include all deŽnable reals.
> So any deŽnable collection of deŽnable reals is incomplete,
in the
> sense that there is a deŽnable real not in that set.
I disagree with the interpretation of your theorem. I claim
that such
a function SHOULD exist (because it is a consistent relation
from
strings to the interval), but that what you REALLY proved if
that the
set of functions from strings to reals is INcomplete because
you
proved that the assumption of the existance of that function
(AS A
SET) leads to a contradiction. That just shows that the set of
functions (in ZF) is TOO SMALL, nothing more. SpeciŽcally it
does
NOT prove the existance of a real that is NOT on the list
GIVEN that
the relation IS NOT A SET in ZF. I agree that if I added
another
axiom to ZF to add that set, THEN there would be another real
in ZF
too, but we could still do the process again to have a NEW
relation
that includes the NEW real. This just proves that ZF is
incomplete.
And we already knew that, so you really aren¹t showing
anything new at
all.
===
Subject: Re: Skolem¹s Paradox and why is math the way it is?
> J.E. says...
> Well, even if you don¹t assume Platonism, it¹s still a
theorem that
> there are uncountably many real numbers. So surely that¹s a
good
> reason to care about uncountably many real numbers?
>>
>> I disagree, there is a theorem, that says for any given
model, and
any
>> injection in that model, there exists another number that
is in the
>> reals that has no preimage. It only proves ONE more number,
>I¹ve SEEN the diagnal arguement!!!!! I asked a question, I
didn¹t
>make a statement
> Yes, you did. You said you disagreed that there were
uncountably
> many real numbers. The paragraph starting with I
disagree... and
> ending with It only proves ONE more number was not a
question,
> it was a number of statements.
The proof just demonstrates the lack of a bijection. It
doesn¹t
demonstrate that many elements exist in the set of reals. I
haven¹t
seen any proof of that, and my question is where people get
that idea,
since it¹s not what the theorem says.
>I¹m asking WHY mathematicians DEFINE real numbers so broadly
as to
>include numbers that can be neither described NOR
constructed.
> Because it is the most straight-forward way to deŽne the
real
> numbers: It is the smallest set containing the rational
numbers
> which is closed, in the sense that every Cauchy sequence
converges.
> That isn¹t true for the deŽnable reals or the constructible
> reals. That is, if you have an inŽnite convergent sequence
of
> deŽnable reals
> r_0, r_1, ...
> the limit isn¹t necessarily deŽnable.
You may, in reality, be correct, but your statement wasn¹t
very
convincing (or detailed or complete). Can you prove (in ZF)
that the
sequence exists? Can you prove (in ZF) that the sequence is
cauchy?
If so, then can¹t you deŽne the limit to be the equivalence
class of
cauchy sequences that have the same limit? What is undeŽnable
about
that? In fact if you care about completeness, I thought that
that WAS
the deŽnition of real numbers you use. Since there are only a
countable class of sequences that you can PROVE are cauchy,
there is
NO proof that there are MORE real numbers required to be
limits of
cauchy sequences. How do you get around that, your example
didn¹t
show very many details.

> To work exclusively with constructible, or deŽnable reals
is much
> more difŽcult, with no actual beneŽt for physics or science
in
> general. You say that you are motivated by physics and
science, but
> I certainly don¹t see any motivation from *physics* to
restrict our
> attention to constructible reals, or to assume that all
reals are
> deŽnable.
How is it more difŽcult? How can you even work with things
you do
NOT deŽne? THAT seems hard.
===
Subject: Re: Skolem¹s Paradox and why is math the way it is?
J.E. says...
>The proof just demonstrates the lack of a bijection.
And that¹s what it means for a set to be uncountable: that
there
is no bijection between that set and the set of naturals.
>You may, in reality, be correct, but your statement wasn¹t
very
>convincing (or detailed or complete). Can you prove (in ZF)
that the
>sequence exists?
Who cares? Why is it important (for applications to physics,
for
example) that a sequence be deŽnable in ZF?
>> To work exclusively with constructible, or deŽnable reals
is much
>> more difŽcult, with no actual beneŽt for physics or
science in
>> general. You say that you are motivated by physics and
science, but
>> I certainly don¹t see any motivation from *physics* to
restrict our
>> attention to constructible reals, or to assume that all
reals are
>> deŽnable.
>How is it more difŽcult?
Try it and see.
>How can you even work with things you do
>NOT deŽne? THAT seems hard.
Well, it¹s not.
--
Daryl McCullough
Ithaca, NY
===
Subject: Re: Skolem¹s Paradox and why is math the way it is?
>> Well, even if you don¹t assume >>
>> Do you think constructable reals are a better class? For
once, all
>> analysis theorem are false in them... (for instance, it is
possible
>> to get functions strictly monotonous with derivative
always 0).
> I deŽned constructable reals (the class of sets that are
subsets
of
> the reals and that can be asserted by the axioms of ZF to
exist
> through recursive calls to Žnite strings and the ZF
axioms), and it
> is clearly a subclass of any other class of any model of
the reals,
> right?
> If you are trying to make those things work, you must be
*very* precise
with
> your terms (and your arguments). Your *deŽnition* is
hopelessly mangled.
> But say a real x is constructible is there is a theorem in
ZF of the
shape
> there exists one and only one real x such that P(x), where
P is a
> Žrst-order formula.
> Then, it is, indeed , isomorphic to a subset of R in any
model of ZF.
That is what I intended. I thought the word Žrst order formula
could be vague (since some people use different signatures,
and
because I¹ve seen the axiom of speciŽcation worded
differently with
different provisos about free variables), so I avoided it,
but that is
what I was trying to say something equivalent to. The reals
that ZF
proves exist.
> So why would any theorem based on the ZF axioms we share in
> common be true in one model be false in my model? I simply
don¹t get
> what you are saying.
> Why, indeed? But the fact is true nevertheless, as the
thing you¹ve got
is
> not a *real* model of R. For instance, the set of your
reals *is*
> denumerable, and R is not... And worse, I insist there
exists a strictly
> increasing fonction from your set to itself, derivable, and
of derivative
0
> everywhere (look at Besicovitch is you dont believe me)
I looked up Besicovitch, and there must be some more speciŽc
term
that gets more than biographies. A theorem¹s name, a set¹s
name, a
date of discovery/invention, something. Because I don¹t see
HOW
someone proves that introducing unprovable reals makes another
function stop existing, so I¹m DEFINATELY not going to be
able to
reproduce this result on my own. If my model doesn¹t have a
real
model of R, then ZF is to blame, not me.

> The only point
> SEEMS to be so that one someone says, Look here¹s a model
of your
> axioms and it has countably many members, that one can
introduce
> another axiom to assert an additional real number (with no
> particular property) that wasn¹t in THAT model.
>>
>> No axiom here : hard proof. You cannot show a model *and*
prove
>> from the inside there are countably many members : in
those models,
>> the required bijections don¹t exist
> We can prove from the OUTSIDE that it is. Being uncountable
(on
the
> inside) just means that your class of bijections (on the
inside) is
> small, not that the range of any bijection (like the set of
reals
> onthe inside) on the outside is large. This IS the skolem
paradox.
> You didn¹t answer WHY we choose a model that is not
countable. Why is
> this an advantage.
> But what make you believe I choose a model at all? When I
do maths in ZF,
I
> dont use models, for godsake (and in ZF, there is only one
set having
the
> properties of R, modulo some isomorphism)

Modulo isomorphism uses the fact that some bijections that
should
exist can¹t be proven to exist. So it¹s a lousy equivalence
class.

> But
> neutrinos have mass (or energy) and momentum, and lepton
number as
> well as spin and so on.
> Not when they were introduced (as for mass, it is *very*
recent)
Neutrinos ALWAYS had energy and momentum and lepton number
and spin,
they were introduced to CONSERVE those quantities. And mass
is just a
word for some energy that we don¹t know where it comes from.
Most of
the mass of a nucleus is from potential energy and gluons in
the
nucleus.
> If someone introduced a nothino that had no
> mass, no energy, no momentum, no fermion numbers of any
kind, no spin,
> and didn¹t interact with anything ever or decay into
anything ever,
> then they would be laughed at.
> You know about leptinos, I presume...
observing something new? It¹s possible, but I wasn¹t aware of
it.

>> There is no point. I want to know WHY mathematicians PREFER
> an interpretation of the axioms that is not countable, and
what
> difference it makes to mathematics. Please, please, please
tell me
> why, and what difference it makes!
>>
>>
>> Coherence, that¹s all. The Lowenheim-sSkolem paradox is
just that :
>> fromthe inside, it vanishes. Compare with non-standard
analysis :
>> there are models with inŽnitely small reals, and you can
never
>> prove they don¹t exist in your favorite model. But there
is the same
>> trouble with integers already : we have the same integers
0,1,2,...,
>> 10^10^10,... and we agree on recursion and Peano axioms,
and yet you
>> cannot be sure my integers are the same as yours....
> The Lowenheim-Skolem paradox is what? There is a theorem
about a
> countable model. That model is not isomorphic to some other
model,
> that I¹ve never seen deŽned.
> ???? There is a standard model; in ZF, it is called R....
How do you PROVE that that standard intended model has more
elements
than the countable model? And calling something R doesn¹t
DEFINE it,
so my statement about never seeing it deŽned still stands.

> As for non-standard analysis. I think
> it¹s Žne, it¹s just as consistent as standard analysis, I
can see the
> appeal of standard analysis if the class of sets is smaller
than in
> nonstandard analysis, since then there is.
> ???????? My question was : what makes you sure you are the
one doing
> standard analysis? We have no way to compare our
non-stansdard integers
(if
> any)...
Standard analysis uses fewer axioms and doesn¹t have
nonstandard sets
or refer to nonstandard elements. In non-standard analysis
you can
introduce nonstandard sets and elements in the proof of a
theorem that
doesn¹t mention nonstandard sets or elements, and know that
there
corresponds a standard proof for the same theorem in standard
anaylsis. So unless you actually WANT to prove results about
nonstandard sets and and/or nonstandard elements, then it is
just a
convenience for proving theorems about standard analysis. I
don¹t
understand your claims about me not using standard analysis
AT ALL.
> Finally you said Coherence. What do you mean by that? Why
does
> anyone care about LACK of a bijection enough to give it a
name
> (uncountable)? There are only countably many that we have
to deal
> with in real life.
> Now, you are suddenly speaking nonsense. Why should we care
about
anything
> (like integers > 10^10^10^10 ?)

Integers could be used to represent many things, and maybe
some
representations are sparse and so could have some very large
integers
that we still care about. You still haven¹t said why an
artiŽcial
LACK of a bijection is useful. Just because you choose not to
count
the things you care about doesn¹t make you have MORE of them.
I think
you¹re being a bit silly.

> and since
> the set of all reals in ONE model is actually countable
(from the
> outside) then it seems that the uncountably many reals are
> ghosts in the wind.
>>
>> No, for this works also for models of the whole set
theory. You
>> must admit theory and meta-theory dont speak of the same
thing.
>> What do you make of Berry paradox, then (the smallest
integer not
>> describable in less than forty english words)?
>
> The Berry Paradox, is not well-deŽned in my book, since
> describable in less than forty english words is not well
formed.
>>
>> Why? If you suddenly say you are only interested in
formalized
>> theories, what are those models considerations doing here?
in *my*
>> book, as you say, I simply don¹t know what is a model of
IR (let
>> alone a denumerable one) ; the sentence P(IN) is
nondenumerable is
>> a theorem, and the sentence there is a bijection between
IR and
>> P(IN) is another. Worse, the sentence x is a constructible
real
>> is as badly formed as the sentence n can be described in
less than
>> 40 words
>>
>> There is no
> such number, and there is no proof that such a number
exists.
>>
>> ??? What do you do of the obvious usual (meta) proof
(there is only
>> a Žnite number of 40-words sentences, so there is a
smallest
>> integer not described by those)
> There are a Žnite number of 40-word sentances. But until we
have
> enough axioms to interpret each 40-word sentance as
describing a
> unique integer,
> ???? How do you interpret more formalized sentences like
(the smallest
> integer x such that there all integers between x and
x+10^1000 are non
> primes) You dont need an axiom for that... and you can
restrict yourself
to
> sentences of the shape the smallest x such that P(x) if
there is one,
and
> else 0, which, by Peano, are enough for such a guarantee...
The sentance the smallest integer x such that there all
integers
between x and x+10^1000 are non primes doesn¹t look
formalized to me.
And it¹s hard to tell just from looking at it, if the
formalized
version would be more or less than 40 words long. The barry
paradox
seems rather silly to me since people just use UNformalized
words to
sneak ever MORE words under the 40 word limit. If you started
with
the set of naturals and then restricted yourself to the Žrst
order
sentances that singled out just one natural number that were
generated
from the Žrst 40 interations of the a Žxed inductive
construction of
the Žrst order sentances, and considered the smallest integer
that
didn¹t belong to any of the singletons so produced, you¹d get
an
integer. If you paid me enough money and gave me a Žxed
inductive
construction, then I could tell you want that number would be,
assuming incompleteness of ZF didn¹t rear it¹s head within
the Žrst
40 iterations. But there is clearly no paradox involved, and I
seriously doubt anyone wants to pay me enough money to Žnd
out that
number. And since ZF is incomplete I can¹t in good faith
garantee
that I can compute the number anyway without adding more
axioms, and
I¹m sure you would get mad if I added more axioms. What is
the point
of using sloppy language and pretending there is a paradox?
> then we won¹t know what integer ANY of those sentances
> describe. And once we did, then the axioms themselves would
assign
> meaning to the word described (internal to the intended
> interpretation of the sentances acording to the new axioms)
that would
> preclude the naive interpretation of the sentance outside
any models.
> I have
> NO idea what you mean by You must admit theory and
meta-theory dont
> speak of the same thing,, can you explain and/or give an
example?
>>
>> Well we are in it from the beginning : Skolem theorem ,
saying, for
>> instance that there exists denumerable models of ZFC (if
ZFC is
>> consistent, of course) is not really a theorem of ZFC
(well, it
>> is, but the sets of the model and the sets of ZFC they
represent are
>> not at the same level, for instance, in the model, there
is a set of
>> all sets which are not members of themselves,
notwithstanding the
>> fact that such a thing is logically impossible in ZFC (or
in any
>> consistent set theory))
>>
>>
> What am I supposed to admit, and why (what is the basis of
this
> compulsion)?
>>
>> see above. The reals in the model are not the real reals.
> What are the real reals? How do you tell which model is the
intended
> model? I haven¹t seen a deŽnition of the intended model.
> The reals of ZF are the intended model (Dedekind cuts, say)
I¹ve seen those deŽnitions, but there are descriptions of
cuts that
cannot be proven in ZF to correspond to sets. So how am I
supposed to
know if that real number is included in your model or not? If
it¹s
existance is independant of your axioms, why do you act like
you KNOW
it is there?

>> Seems like fancy talk to say that the set of bijections
> between the reals is incomplete because our axioms made it
so.
> Why not Žx the axioms?
>>
>> Alas, no Žx is possible...
>
> Theorem? Proof? You¹ve investigated all axioms possible?
>>
>> No, you don¹t understand. No axioms are necessary here,
just
>> deŽnitions. The set of axioms (ZF, say) is reasonable, and
the
>> problem of cardinality is not limited to IR. If you really
mean to
>> restrict your reals to constructible ones, you must give a
new
>> deŽnition for IR. Good luck with Žnding one as useful as
the ones
>> (Cauchy sequences, Dedekind cuts, or even digits
sequences) everyone
>> else use.
> What do you mean no axioms?
> The axiom set is ZF. We dont need new axioms for R, but
only a sensible
> deŽnition.
But the axioms of ZF are incomplete, so you can¹t tell what
sets are
captured by the deŽnition withIN ZF. Hence you don¹t know
what are
the real numbers and what aren¹t.

> Axioms are assumptions and without
> assumptions we have nothing (except logic itself). I¹m
assuming IR
> are irrational numbers.
> Oh, Q is no longer included in R ?
Where did you get that? I didn¹t say that. YOU claimed you
didn¹t
need axioms, and YOU mentioned IR. You didn¹t deŽne IR, so I
told
you that I had assumed you meant irrational numbers. I wasn¹t
TRYING
to exclude Q for any reason, I was TRYING to understand what
YOU were
saying. Are you trying to say anything at all or are you just
making
silly questions that make you feel smarter than other people.
I for
one am trying to have a conversation. If you want to use
terms that
I¹m not familiar with and make fun of me for trying to guess
what you
are saying, you are free to do so, but I don¹t see how it¹s
supposed
to help advance any conversation(s).

> The point is that only SOME IR numbers come
> up while doing physics, and particularly only countably
many come up,
> so we can just give them names, like pi and e.
> Sure, but if you go this way, dont bother with irrationals
at all :
> 314159265358979/100000000000000 is a very good name for pi,
for all
> practical purposes (and if not, just plug a few more
decimals)
YOU can do that if you want. I¹d prefer to use whatever is
easiest
but STILL doesn¹t create problems. And pi can easily be added
to the
number system. In fact all reals that ZF proves exist seem to
be
easily added to the number system. There are only countably
many of
them, but the others appear to be useless since for MANY MANY
years NO
ONE has proven theorems about them (since I haven¹t seen new
axioms
added to set theory to prove more subsets of the naturals
exist than
formerly could be proven to exist).
> Who said ZF is
> reasonable?
> We (the math community) did. Love it or leave it.
I¹m thinking of leaving, but as a scientist it seems only
responsible
to take as much as I can or should with me when I leave. Why
do you
like pretending that real numbers exist that you can¹t prove
exist? I
want to know if I¹ll have problems if I leave them behind. For
instance earlier you said that if I left them behind that
weird
functions would appear in analysis. I couldn¹t Žnd your
reference,
but that sounds like EXACTLY the kind of information I wanted
to know.
Just saying it¹s pretty isn¹t a response to someone who asks
what it
does and does not do or prove.

> Jaako Hintikka had a principle that generalized the axiom
> of choice (in Principles of Mathematics Revisited), and I
haven¹t
> seen any mathematician incorperate it into ZF, instead they
seem to
> love ZFC, and I don¹t know why, just like I don¹t know
which model
> they like (not the countable one) or why. These other
(uncounted)
> reals seem to only come up when someone outside the model
hands us a
> list of reals, why can¹t we ignore them since we don¹t need
to go
> outside the model to do math or science?
> ??????????
In ZFC sometime people talk like there are more reals than
what they
can prove exist, but they don¹t actually prove this ever (and
they
can¹t). The closest they come is demonstrating the
INCOMPLETENESS of
the set of bijections in ZF. These other reals only seem to
come up
for the SOCIAL purpose of making holier than thou statements
to
other people, and never appear in theorems except in the
MIDDLE of
proofs by contradict that make invalid assumptions, and
ANYTHING could
appear in the MIDDLE of such a proof. So why does anyone
pretend
these numbers exist?
> Why don¹t
> we use the class of all constructable (made from the empty
set using
> the axioms of set theory) sets as a universe? Is there
something
> wrong with that?
>>
>> Yes : how do you deŽne them? (see Berry)
> How were they deŽned in the countable model of set theory?
If others
> can do it, why do I get the third degree from you about
saying I could
> do it too?
> Because you still seem not to understand : they were deŽned
*outside*
ZF.
> Anyway, there is nothing wrong with using that class
(G.9adel did that)
in
> metamathematics,but mathematicians dont usually use models,
especially
not
> when speaking of R, or doing real analysis
How do YOU deŽne the reals that you can¹t prove exist? If we
both
have to step outside of ZF to make our deŽnitions where does
this
holier than thou arguement come from. If I use the class of
reals
that can be proven in ZF to exist, then it becomes PAINFULLY
obvious
that the class is DECIFIENT, and this is a problem with *ZF*,
not with
my choice to make this deŽciency OBVIOUS. Just because some
other
choice makes is LESS obvious doesn¹t make the problem go
away. If a
car is hurtling towards you at high speed, closing your eyes
doesnt
protect you.

>> If so, what problems does it cause? Isn¹t that
> class countable? Doesn¹t it have as many theorems about it?
> No, it has no theorems at all, as it cannot be described in
the langage
in
> which we prove theorems. You should really study
non-standard analysis.
I *have* studied non-standard analysis. It doesn¹t allow me
to prove
the existance of any more standard real numbers, so I didn¹t
see the
point in using it. Standard analysis based on ZF is
incomplete, so
why add more axioms that don¹t move it toward completeness?
As for
having NO theorems, earlier you said that the model had a
strictly
increasing function from itself to to itself, derivable, and
of
derivative 0 everywhere. But now you claim that there are bi
theorems
about this class, so your claim isn¹t a theorem. If there is
no such
theorem, WHAT did you base your claim ON?

> Doesn¹t it have only a countable many members? Doesn¹t it
just
> lack a member that is a bijection from the member that
represents
> the naturals to the member that represents the reals
> And WHAT
> happens when a logician considers the universe of
constructable
> sets AND the bijection between the two members. Obviously
the old
> axioms can¹t (consistently) apply to this newest member of
the
> universe, but why does no one want to consider that
universe at
> all? Wouldn¹t it be nice, say, to represent irrational real
> numbers on a computer in an exact way by having every real
number
> we CARE about be assigned a distinct integer?
>>
>> No, this is mere folly. After all, we do represent reals
already, by
>> approximate values. And for exact symbolic form, this is
of no use,
>> as the problem of deciding if two things like cos
(sqrt(2+e^pi))
>> ... (much more complicated forms, of course) are equal is
>> undecidable (which means it is quite hard in practice)
> This is the Žrst thing you¹ve said that I think I
understand, so I¹ll
> try to repeat it in my own words to see if I did. There are
only
> countably many reals that we can describe with a Žxed set
of axioms
> and Žnite strings.
> Right
> Each integer can be recursivley assigned a
> strings and calls to previous sets and calls to axioms in
such a way
> that every real number, function, and set in general
> (in this model)
> is the image of
> an integer,
> Right
> but the fact of determining which integers are mapped to
> sets (corresponding to well formed strings and appropriate
calls to
> earlier sets and appropriate calls to the Žnite axioms) is
> undecidable
> ??? No, in this direction, all works perfectly well
> and the correspondance is not one-one so different
> integers could be assigned to the same set and Žnding the
class of
> integers assigned to a particular set is also undecidable.
> Yes
> Is that
> what you were saying? That seems like conveniance, not
coherence.
> No : if you cannot say if a string is 0, it is not a mere
inconvenience,
it
> means your system of representation of reals is žawed.
Decimal notations
> (inŽnite, of course) dont have this problem...
You say this representation is žawed because multiple
integers can
represent the same real, but your representation has the
appearance of
more reals than can be proven to exist, which makes IT žawed.
Is
there no hybrid where we include only the reals that had
integers
corresponding to them? I thought that that WAS the countable
model.
If it isn¹t, what is the difference? It if IS, what is wrong
with it?

> And as a physicist, it seems like maybe it is more convient
to use
> other axioms than ZF that are more convient. For instance
David
> Hestenes and others have proposed axioms for Geometric
Calculus which
> seems large enough to do all of physics. Maybe a countable
model of
> that is easiest to use.
> I stil dont understand what you mean, but I am sure you are
confusing
> theories (axioms and theorems) and models (objects
satisfying these
axioms,
> and so satisfying all the theorems they imply)
I¹m saying that if the ZF axioms are incomplete but assumed
to be
consistent, then why not use axioms whose consistency follows
from the
consistency of ZF, but whose properties might not have the
same
objectional parts as ZF.

> As for approximate values. Spacetime is four dimensional,
and it
> seems silly to be forced to choose a representation (like
the Dirac
> matrixes) or coordinates on an arbitrary basis, because
then linear
> error on each coordinate becomes really weird shaped error
> parallellopipeds in spacetime. A more exact system for the
model
> seems better to use until you compute the exact
observables, and then
> you have compare the exact results to the approximate
measured
> results. Does that make any sense?
> No, not at all. Nobody has ever measured a real, and nobody
ever will.
I didn¹t claim that anyone measured a real. I don¹t appreciate
mathematicians obsessing about the numberline, when it is more
truthfully vector spaces that are used in physics. If one
approximates the reals FIRST and then builds a vector space
up from
numbers, then you get basis dependant results. I¹d prefer an
isotropic approximation of the whole vector space, or if
that¹s too
hard, one that can be adjusted to suit the Žnite measurements
we end
up making at the end and is consistent with the anisotropies
of the
Žnal measurements, but doesn¹t introduce OTHER anisotropies.
>> As a physicist, that seems appealing. What¹s wrong
> with using that universe? It seems big enough to do
physics. Why
> should I care about whatever number is missing, after all it
> couldn¹t have come up in any calculation or predicition I
made for
> an experiment, so it¹s seems to have rather nothing to do
with hard
> science.
>>
>> Strange, I didn¹t know physicists used reals for their
experiments;
>> I could have sworn they used decimals (or intervals of the
form a+/-
>> b, with a decimal and b power of 10)
> Physicists make models of the real world. They make models
that are
> testable. They perform experiments and make observations
and compare
> them to the models. They then revise the models as
necissary. I
> think that many physicists are sloppy with statistics
(still better
> than many other scientists), and I wanted to seperately
cleaning the
> model making parts and the making observation parts and the
comparison
> parts. I prefer statistics closer the Bayesian method of
comparing
> models to the observed data, than the classical approach of
cut-off
> values for one (decimal) model. It¹s the only honest way to
compare
> one model too all data ever taken. The classical approach
assumes one
> makes a theory and then starts taking data and then
compares the
> results to the theory and then rejects or doesn¹t reject
the theory
> and then throws the data away to never be used again. It¹s
very
> wasteful to do that, but to do otherwise and use classical
> statistics is dishonest and unfair.
> ???????????? Anyway, this is OT
YOU mentioned decimals, and that isn¹t an accurate
description of what
physicists do because there are many computations performed,
frequencies, differences, radius of curvature measurements.
And there
are different ways to measure each. I don¹t know why YOU are
bringing
this up, but it can¹t have been OT, because you must have had
a reason
to ask. I still don¹t know what that reason is myself, and I
guess
that if you forgot, then it might NOW be OT.
===
Subject: Re: Skolem¹s Paradox and why is math the way it is?
>Someone else claimed that I brought up a collapsing
wavefuntion
> No.
I didn¹t try to bring it up. It wasn¹t in my original post. I
don¹t
believe in collapsing wavefunctions, yet out of nowhere
people were
claiming I did.
>either some people here aren¹t trying to understand me
> They understand you only too well. You¹re wrong.
I¹ve asked to be shown where I¹m wrong. Some people have said
I¹m not
wrong. Others have said that everything is Žxed by second
order
axioms, and the others just accuse be of being inconsistent.
If you
thought something was inconsistent, then how can you
understand it.
Why trying to understand someone isn¹t the goal to Žnd the
most
probable consistent interpretation of what they are saying?

>or that I¹m not being clear
> You¹re being clear, but inconsistent.
If I¹ve being clear in asking why people like the ZF axioms,
then why
aren¹t people answering my question? If I¹m being clear about
why I
think they fail to do what we expect, then why don¹t people
correct my
expectations or show me how the axioms do meet my
expectations?
===
Subject: Re: Skolem¹s Paradox and why is math the way it is?
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Hash: SHA1
>If I¹ve being clear in asking why people like the ZF axioms,
then why
>aren¹t people answering my question?
Are we supposed to *like* the ZF axioms? For most
mathematicians, it
is sufŽcient that they do what is needed. They don¹t have to
also
be likeable.
> If I¹m being clear about why I
>think they fail to do what we expect, then why don¹t people
correct my
>expectations or show me how the axioms do meet my
expectations?
Several people have done that. You appear to not be listening.
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--
vote for regime change in Washington, Nov 02.
===
Subject: Re: Skolem¹s Paradox and why is math the way it is?
> It¹s its own powerset, Eray.
> By one of Cantor¹s theorems, |A| < |P(A)| for any A. Is not
that right?
Well, think of it this way, maybe: every set is an ordinal,
and the
order type is the powerset is the successor is the
cardinality.
and the the correct statement would be |U| <= |U|. That¹s true
because any set equals itself.
Something like the empty set or ur-element shares some
properties with
other things that are the proper class.
I admit that the powerset result is a strong result, but
don¹t let
that stop me from considering it meaningless.
The set of all sets does not exist, per se, in ZF, it is
disqualiŽed from existence primarily by the axiom of
regularity.
Basically its existence is denied because there are a variety
of
antinomies or paradoxes associated with it. One of these is
called
Cantor¹s paradox, that the set of all sets would be its own
powerset.
Another is the Burali-Forti paradox, that the order type of
all
ordinals would be itself an ordinal. These are clearly
associated
with basic problems of considering any inŽnite number, and
thus their
consideration leads to either: a)ultraŽnitism, or b)
combining each
of them into basically a superclass of conundrums, and trying
to solve
the problem at once, instead of just restating it.
One approach I use to consider this is that each set
represents an
ordinal, and that a given ordinal can be represented by any
of a
variety of sets the order type of which is the ordinal, and
as well
any of those sets very žexibly represents the ordinal.
Then, that leads into zero vis-a-vis inŽnity. Apply the
predecessor
operation to any Žnite ordinal, where that returns the
predecessor or
the value unchanged for a limit ordinal for predecession. For
each of
to predecession, and in some arguable sense: the same thing.
Another notion with zero vis-a-vis inŽnity is that inŽnity is
negative one. That¹s about Ord, or for that matter inŽnity,
being
less than nothing.
is conŽrmator. That¹s something along the lines of saying
that if
you apply the successor operation to U, the universal set
once, it¹s
still the same set, and again, it¹s zero. Yet, where that¹s
so then
if it is still the same set, then there is no way to
determine that
the successor function was applied to it already, unless you
apply it
twice at once. That¹s similar to the consideration that the
successor
of any Žnite ordinal is a Žnite ordinal, yet there exists an
inŽnite ordinal.
There exists an inŽnite ordinal because the set of all Žnite
ordinals is an inŽnite ordinal.
There are notions where there is exactly one inŽnite ordinal
for each
Žnite ordinal.
Plainly, there is a lot of room left within inŽnity that the
foundations of mathematics have yet to explain.
Consider something long the lines of epsilon-delta style
limits, they
make something useful like integral calculus unambiguous.
That¹s a
very good tool and it¹s wise to only deal with the Žnite.
Yet, while
that¹s so, and the limit of the sum of the inŽnite series
1/2^x for
positive integer x, in ascending order, is 1, that sum also
is one
over all of the positive integers, because you¹re reading
this.
References to Zeno aside, it¹s not enough that the
foundations of
mathematics puts off resolution of these plain problems in
the study
of the science and mathematics of the inŽnite, and I don¹t
want to be
guilty of this myself: it¹s meaningless.
The state of the art of set theory is useful for many things,
as long
as you don¹t address the inŽnite. The idea is to advance the
state
of the art because these complications need to be addressed
at once
and not never. Explaining half, and then half again, and then
again,
forever, leads to too much work for mathematicians.
Look at something like EF, it makes sense, although some
aspects are
not initially intuitive. Nobody ever told me about the
one-sidedness of points that are contiguous members of a real
number
line, I¹ve had to rationalize that myself, and I share these
observations wih you because I want you to later be willing
to thank
me. Anything that can be explained to another is not a unique
insight.
I¹ve spent years studying this exotic Želd, then I posted to
sci.math.
There are more things to tell to you.
Ross F.
--
He don¹t know me very well, do he? - Bugs Bunny
===
Subject: Re: Skolem¹s Paradox and why is math the way it is?
>I¹m saying that it does land everywhere.
> OK, so you adhere to Many Worlds[1]. It¹s viable, but it
also has
> issues.
No I don¹t adhere to Many Worlds, or to be more precise I¹ve
never
agreed with anyone I¹ve met that claimed to agree with Many
Worlds. I
have similar views in that I think the ONE wavefunction that
exists in
conŽguration space has geographically different parts that
correspond
to different versions of me that observe each of the results,
but I
don¹t talk about many worlds or splitting or interfering or
creating
of worlds, just smoot continuous wavefunction evolution like
any
no-collapse quantum person would use. I follow Mermin more
than
Everrett (if that¹s even how you spell his name).
>the >wave-function splits
> Please describe the apparatus that measures the splitting.
No apparatus measures it perfectly, but any hamiltonian that
causes
correlations between the wavefunction part corresponding to
the
and the wavefunction part corresponding to that cross section
would be
adequate enough to see a coarse approximation to the pattern
after
many trials. Any Bohmian model should describe this to you in
whatever language you like, since I¹m unlike to happen to
pick word
you already understand.
> Actually, I haven¹t. How do you know that I don¹t prefer
GBN?
Well I¹ve been asking why people like ZF axioms, and you
haven¹t been
disparaging them in favor of GBN axioms. You could have any
opinion
at all, I will only know the opinions you choose to share.

>So it is interesting to you that there is no bijection from
>the naturals to the power set of the naturals, right?
> Yes.
Can you describe how this is anything other than an ambiguous
result
to have?
>Are you concerned if that is bothersome to other people?
> No. Are you concerned that SR[2] is bothersome to other
people?
I am concerned that SR bothers some people, but that is
probably
because in my experience it comes from having bad teacher of
SR, and I
disapprove of bad teaching. It is possible that I had bad
teachers
for math. An over-reliance on the Moore method by my
instructors
might be the culprit, but I don¹t like to place blame on my
own
teachers, seems too easy and cheap. As a teacher I¹ll happily
blame
myself for my students¹ misunderstandings, but as a study it
seems
unjust to blame my teachers.
>It is bothersome to the people who are not making further
axioms to
>add to ZF and instead are merely using ZF as is, because
there
>doesn¹t seem to be an uncountable number of things described
by
>ZF, given the faithful countable model by Skolem.
> What do you mean by be? Are you taking a Platonic position
that
> every time you prove (Exists x)P(x), that there is a real
object
> described by that x? Platonism is not a necessary part of
Mathematics.
The ZF axioms fail to describe an uncountable number of
things,
because they are incapable of proving that that many distinct
things
exist.
>I want to get rid of unobservables from math because I think
it
>would *help* to remove them from math Žrst and *then* from
physics.
> It would make more sense to Žrst remove unobservables from
Physics.
If I new how to do that Žrst, I would. It seems hard when
physics
uses math that already has elements that are independant of
proof to
exist.
>I was trying to say that the fact one can prove theorems
with axioms
>doesn¹t say anything about the axioms themselves.
> Why would you want to say anything about the axioms beyond
stating the
> theorems that can be derived from them?
Incompleteness is a problem, because I want to use
mathematics to
describe things, and the mathematical language of ZF is
descriptively
incomplete (which means among other things (I¹d bet), that I
can
describe things that I cannot prove whether or not they
exist).
>I still don¹t understand these other criteria,
> That¹s the problem; Mathematics is an art form. One of the
key terms
> is elegant. The criteria for judging Mathematics are more
intuitive
> to a musician or a sculptor than to a Physicist.
Descriptive completeness isn¹t beautiful or elegant to
mathematicians?
>I¹m willing to consider that, but then I
>want to know why you think the power set is interesting.
> Because it generalizes. Because it leads into other
results. Because
> the proof is so simple.
It seems vague, like if someone claimed there was some deep
symmetry
on the žoor of a room that was mostly covered with a bland
rug. If
the deep symmetries are forever unobservable, then what is
the point.
In ZF, all the big parts of the big sets cannot be proven to
exist.
So why consider them? Why not stick to deŽnite results about
things
we have observed and statistical claims about the things we
haven¹t?

>I understand talking about the collection of all subsets
that exist
>in a model of a particular set,
> But you seem to be equating the model with the theory that
it is a
> model of. They are not the same.
A theory can only describe things that exist in all it¹s
models,
right? So if consistency is independant of the theory, then
it can¹t
be described in the theory. If truth is independent of the
theory,
then it can¹t be described in the theory.
>but to talk about there being more subsets when you
>haven¹t yet introduced any further axioms is weird.
> What further axioms do you need? ZF already includes the
requisite
> axioms. ZF is not a constructive theory; proving (Exists
x)P(x) does
> not mean that we have show a construction for such an x,
but only that
> the statement is a theorem of the system.
I don¹t know what your deŽnition of construct is, I¹m talking
about
sets that you can prove exist from the axioms. The class of
such
subsets of the naturals is clearly not different in kind than
the
class of naturals themselves. The fact that no bijection
exists from
the two classes, is an inadequacy of the thoery to describe
it¹s
provability, it says NOTHING about the alleged larger class
of subsets
that COULD have been proved with MORE axioms.

>But from physics we want a Žnished mathematical theory
> That¹s impossible, if you want it to be consistent. It will
always be
> incomplete.
Then a theory that is clear about the incompleteness being in
the
parts we don¹t use should be good enough then. ZF doesn¹t
seem set up
to make it clear where descriptive completeness starts and
stops,
that¹s a failing in my opinion.

>it¹s cumbersome to add another mathematical
>axiom and then go do all our physics again,
> It¹s also unnecessary. ZF is perfectly adequate for the
tasks to which
> physicists have applied it to date.
That statement is not a theorem of ZF. How can we tell that
the
incomplete parts of the theory of ZF don¹t interest with the
physically interesting parts of our physical models based on
ZF.

>I wanted to know why uncountability is desirable,
> LIttle things like the MVT.
A single point is less than a drop in the bucket. Besides
doesn¹t the
countable model have a mean value for every function that can
be
proven to exist? So why do we need more points if their only
purpose
is to be the mean value for functions that we can¹t prove
exist
anyway?

>What is your operational deŽnition? Show me a power set in
the lab
>and say what is in it.
> The lab is a notebook. What is in it are marks on a piece
of paper,
> forming inferences from the axioms in accordance with the
rules of
> inference.
Then surely you can see how the class of sets you prove
theorems about
is not different in kind than the class of natural numbers.
There are
not MORE sets you prove theorems about than numbers you can
write
down.

>The problem is that WHICH elements are in a
>power set depend on OTHER axioms than ZF
> No. You can only talk about a set being an element of
another set in
> the context of a speciŽc theory.
In that case there, the existance of some sets, that if they
existed
would be elements of the power set depends on the OTHER
axioms.
>and you haven¹t CHOSEN those other axioms YET,
> If I chose other axioms then it wouldn¹t be ZF any more. If
you Žnd
> it more convenient to work with, e.g., inaccessible
cardinals, that
> doesn¹t change the theorems of ZF.
I don¹t need inaccessible cardinals, there are missing sets
in the
power set of the naturals. Why isn¹t this a problem to anyone
else?
I don¹t publish proofs of this because it seems obvious and
trivial.
It can¹t possibly be that no one else knows this except me, I
Žnd
that extremely hard to believe.

>so you just insist that there ARE more subsets than can be
proven
>to exist
> No. I note that it is a theorem in ZF that they exist; I
don¹t draw
> any metaphysical conclusions from that. If you want a
constructive
> theory, they¹re down the hall ---->
Show me this theorem. I have only seen theorem about lack of
bijections not about existances of many sets. The two
concepts are
different, as I HOPE the skolem paradox¹s resolution has
already
made clear to you.

>since IF another axiom was made to count the previously
existing ones
> Are you talking about inaccessible cardinals?
No.

>You toss around the word all in a very UNoperational way in
ZF.
> Not at all; you simply don¹t understand what the relevant
operations
> are. They¹re rules of inference.
There aren¹t proofs that all things that COULD consistently
be added
to the theory ARE already in the theory. And if I mention
that you
sometimes hide behind the tree that you can¹t add axioms, and
othertime behind the incompleteness tree. You can toss the
word all,
but you haven¹t proven all, and I can describe things that
should be
sets that you can¹t prove are.

>The THEORY is about things you can measure
> That hasn¹t been true for a century. The theories used in
Physics
> involve gauge Želds that are quite removed from things that
we can
> measure.
I¹m sorry that we will have to disagree on this (I would have
preferred for you to understand my arguement and either adopt
it
yourself or convince me of a better one, but you don¹t seem
to either
not want to or not be able to understand it). The theory is
about the
lab results and the calculation results, how an individual
chooses to
solve it is not part of the theory. I could take some
numerical
algorithms and translate these gauge theories into a program
that
quizes the lab reporters about the experiment in terms that
are
entirely observable and that gives a two printouts, one of
observable
results and second rules to compare the experimentally
observable
results to the Žrst printout. I could them printout the
source code
of the program and call THAT my theory. We don¹t do that all
the time
in practise because mathematics allows us to exploit
symmetries that
make the computations much more efŽcient than the brute force
machine
I¹d write. But there is nothing inherent about unobservable
parts
about a theory for it to work. Nothing at all.

>If I don¹t PUT IN absolute phases,
> What do you mean by that? Theories involving local gauge
invariance
> require absolute phases, but the dynamics are invariant
under gauge
> transformations.
Those are DESCRIPTIONS of the theory. There is more than one
way to
describe a thoery. See the helpful computer program mentioned
above
as an alternative.
>so approximate solutions are Žne if they are accurate enough.
> Then why do you care if the axiom systems aren¹t accurate
in some
> vague metaphysical sense if the results are accurate enough?
Each logical problem from math is serious because I¹d have to
trace it
through every simplication I used to see if the problem
infected my
results.

>For instance, I know I guy in England who is spending his
career
>looking for closed timelike curves that are solutions to
Einstein¹s
>Želd equations, and he can¹t seen to prove that they either
exist
>or can¹t exist.
> Huh? Wasn¹t that solved a long time ago? Or do you mean
that he is
> looking for solutions subject to a constraint that he is
interested
> in?
I already said Closed timelike curves, do you consider that
to be
merely a constraint that he is interested in, or should I
tell him
that it¹s already been done? I¹m sure he¹d eventually admit
that he
wanted to be told, because if it¹s true then he¹ll Žnd out
eventually
and it¹s better sooner than later.

>It would be nice to have a decidable model
> Kurt G.9adel demolished that hope a long time ago. And
you¹re confusing
> model with theory. Models aren¹t decidable or undecidable;
only
> theories are.
How about a decriptively complete model?

>If a small enough model now would avoid things like that
being ambiguis,
> A small *theory* would leave more statements undecidable
than a larger
> theory.
ZF seems to have it¹s undecidable statements buried in
inŽnite sets,
so it seems like a theory with smaller sets could remove the
undecidable sets. Just say no more often instead of reaching
for as
many yeses as possible that catches many maybes in the net.

>but some results are independant of teh ZF axioms,
> What are you trying to say? The results are inferences from
the
> axioms; they can¹t be independent of them. The best that
you can say
> is that certain results exist in multiple theories. The
results that
> appear to be relevant to Physics exist in both ZF and GBN.
There are things that should be sets, but that you can¹t
prove if they
are sets in ZF. ZF is descriptively incomplete. Therefore the
existance of these sets is independant of the axioms of ZF.

>I don¹t want an inŽnite regress based on a mathemtaically
undecidable
>proposition. It may be that no matter which axioms I choose
that will
>happen anyway, but I can¹t know unless I try the axioms out.
> You know that the theory will be incomplete. You can¹t tell
whether
> that incompleteness is relevant to a physical theory until
you have
> the physical theory. And if it is relevant, the solution is
to add
> axioms, not to take them away.
Axioms haven¹t been well ordered. Adding an axiom that makes
the
system inconsistent doesn¹t seem very helpful, and what if an
existing
axiom is inconsistent with the axiom we really should add,
how are we
supposed to know in advance. Isn¹t the prudent thing to do,
to start
out with as few as possible so that the new ones have the
freedom to
be what they need to be? If not, what¹s wrong with what I
said?

>You describe your inial conditions as a wavefunction, and
your
>observations as sections of a wavefunction.
> What do you mean by sections of a wavefunction? And is
isn¹t a wave
> function rather unphysical?
Observations are local, the wavefunction is not. Clearly only
a
portion is applicable to the verifyable parts of any
particular
experiment. The other parts can be Žlled in with anything
that makes
the computation easier and can be thrown away when you are
done. So
basically you only describe the parts of the wavefunction
that affect
your experiment and only those parts that affect the
experiments
affect the portions of the calculations you care about. I
don¹t
consider the wavefunction to be unphysical, but I consider
anything
else to be, luck for me there IS nothign else in my models.

>Huh? You do your calculations, no one walks in and forces
you to
>pretend things commute when they don¹t.
> The fact that they don¹t commute has some observable
properties. In
> fact, some physicists start with the commutation relations.
Some physicists might start by counting their toes and
drinking
coffee, it doesn¹t mean that affects me. What matters is the
initial
conditions and the potentials and the evolution equations.
Nothing
else matters, full stop.
>I¹m not an idiot,
> Perhaps, but when you write things like Who cares about
> commutations? and You need an inŽnite space because you
have an
> especially when you contradict yourself.
You ask such details questions sometimes that I forget you
may not
upperbound on energies, and the size granality of my
equipment such,
then I can get by with a Žnite number of equivalence classes
of
states that will look and act the same for the experiment
that is
complete in the sense that everything I could (and hence do)
observe
is modelled. The simplest mathematical models that the
members of
these equivalence classes live in that is probably familiar
to you
have uncountable cardinality and inŽnite dimension. But
that¹s more
about how I talk to YOU, than about how *I* do physics. You
keep
assuming that I do my physics badly just because I don¹t do
it like
you. That¹s more than annoying. I¹m trying to get your
perspective
and I treat you with respect in that I assume that you have
good
reasons for having the opinions you do and for doing the
things you
do. You seem to assume that I¹m dishonest just because you
aren¹t
familiar with my work.

>What experiment are you performing,
> I¹m measuring the position and momentum of an electron.
I have trouble believing you. What equipment are you using?.
What
results are you capable of measuring and distguishing, and
what theory
are you using to interpret your results? Everyone I meet does
position measurements, there¹s this constraint called
locality and it
applies to all the experiments I can do with my funding.
Maybe you
have better funding than me. And to be honest if your funding
is THAT
much better, then maybe I shouldn¹t make models for you,
because
that¹s probably out of my league.

>what is the tolerances and the precisions.
> Arbitrary. Except that I can¹t, BECAUSE OF THOSE COMMUTATION
> RELATIONS.
I have trouble believing you about arbitrary precision. What
equipment are you using to get these results?. What theory
are you
using to interpret your results? Everyone I meet does Žnite
precision measurements, there¹s this constraint called Žnite
accessible energy and it applies to all the experiments I can
do with
my funding. Maybe you have better funding than me. And to be
honest
if your funding is THAT much better, then maybe I shouldn¹t
make
models for you, because that¹s probably out of my league.

>Why are you computing something like that?
> Maybe I want to know the physical limits on what I can
measure. There
> are observable consequences.
There are observable consequences to how much funding you
have? Well,
I agree with you, that usually the biggest case made for
asking for
more funding. But what does this have to do with science
speciŽcally.
>The results you need/want are the shape (location) of the
wavefunction.

> How do I measure the wave funtion?
You subject your Žnger to a hamiltonian from your brain that
correlates the brain state with the position of your Žnger
such that
the position of you Žnger is correlated with the parts of the
experimental apparratus that you decide to observe. This
causes a
correlation between marks on your lab book and the
wavefunction of the
experimental apparratus. If you repeat the process many times
with
similar wavefunction but move you pen to a different page of
the lab
book each time, then after a while the correlations between
pages on
the lab books will approach the autocorrelation of the
wavefunction of
the apparatus. The whole process assumes that you have a
reliable
process to produce things to observe and to make things to
observe
them.

>I¹m REALLY upset with people pretending that there is a
complete
>basis all the time.
> It may be a crooked game, but it¹s the only game in town.
QFT is a
> mess, but we don¹t yet know how to clean it up, and
meanwhile we get
> some very accurate predictions.
If it¹s gonna be approximate anyway, why not Žnd a better
basis then?

>If we only wanted to calculate the ones we make in
>the lab, then it¹s OK, but then we don¹t need an inŽnite
model,
>like you claim.
> What do you think the wave functions are? There is no Žnite
basis for
> them.
For bounded energy bound states there is.

>Especially because you don¹t like models,
> There you go again. If I tell you that I don¹t want
raspberries in my
> chili, that doesn¹t mean that I dislike raspberries. It
just means
> that they don¹t belong in chili, possibly because it¹s a
waste of
> perfectly Žne raspberries. What I don¹t like is confusing
models with
> things that aren¹t models.
Models tell you things about your axioms. In this case they
pointed
out the fact that lack of bijections and size are not
intrinsically
related.

>I understand that isomorphism, you are the one claiming we
have to
>use ONE particular model with an inŽnite algebra even though
it
>makes distinctions between events and states that we can¹t
>distinguish between.
> No, I am the one that realizes that we¹ve known for decades
that the
> commutation relations can not be satisŽed with a Žnite
dimensional
> representation. If you¹re not aware of it, it¹s not my job
to remedy
> the lack. That has nothing to do with the fact that we use
> approximations in our calculations.
But these lack of satisifactions only occur under the rug
where we
can¹t observe them anyway with our Žnite precision. As long
as the
basis is big enough for your current experiment, then it¹s
good
enough. And if you need a bigger one for another one, that¹s
OK too.
The fact that we use approximations in our calculations
forgives us
from using an incomplete basis. If two states are differnt in
your
BIG model, but can¹t be experimentally distinguished in a
particular
experiment, then you can put them in an equivalence class. For
instance if there is much lasers being superimposed, but all
you are
measuring is the color you observe with your eye, then
instead of an
basis as large as the number of lasers you had, you only need
a basis
of three elements.
===
Subject: Re: Skolem¹s Paradox and why is math the way it is?
> [...]
> |> I don¹t remember how one would say such things as there
> |> exists a countable model of ZFC. Is there a way that you
> |> know of, or do you base such statements on something
> |> else besides IF logic?
> |I based it on the Skolem Paradox.
> That¹s not a means of _deŽning_ what it means for there to
> exist a countable model of ZFC.
> Usually, the way one says that a model is countable depends
> on a little bit of set theory. But you don¹t accept set
theory as
> a foundation.
You are correct to say that we have to deŽne model and
countable
and such to be clear about what phrases like there exists a
countable
model of ZFC mean. I was imaging that if someone believed set
theory, then they can use set theory to devise a winning
strategy, and
if they believed a model then they could use that too when
devising a
winning strategy. If I didn¹t believe in the same things as
that
person did then they might have trouble proving that their
strategy
was a winning one, or maybe even explaining it at all. There
might be
IF-logic statements such that the existance of winning
strategies is
hard or impossible to prove, I don¹t know for sure. But the
countable
model of ZF made it hard for me to imagine how to even begin
to come
up with an IF-logic sentace that describes an uncountable
universe.
It¹d have to be stronger than the ZF axioms to describe it,
since I
don¹t anymore believe that the ZF axioms describe an
uncountable
universe.
> |If I put the right kind of negation
> |on the intuitively inŽnite universe claim, I think I can
get a
> |statement about an uncountable universe that is neither
true nor
> |false, but I can¹t get a statement that is true that says
the universe
> |is uncountable. Doesn¹t mean that someone else can¹t, but
I haven¹t
> |been able to.
> I thought that your statement that the universe is inŽnite
was a
> statement that it is inŽnite, not that it is countable.
Wouldn¹t
> negating it give you something more like, the universe is
Žnite
> rather than the universe is uncountable?
> Keith Ramsay
One can make an IF-sentance that the universe is Žnite by
putting the
negation right at the beginning of the inŽnite statement,
yes. I was
thinking you might be able to combine the sentance with it¹s
negation
in an independant way to get something that was as losing as
winning,
but I was trying to say that that wouldn¹t mean anything
since we¹d
need a true statement that says something is uncountable to
mean
anything. And since it woulnd¹t mean anything I never tried
to make
it, perhaps I shouldn¹t have mentioned it. What I wanted to
mention
was how I¹ve failed to describe an uncountable universe, and
that
example was just an indication that I have tried, but that it
wasn¹t
successful. And that¹s because I just don¹t know any way to
describe
anything as intuitively uncountable, the closest is mismash
sentances
that don¹t say anything at all. And when I went to set theory
for
inspiration on how to describe it I just found that there
wasn¹t
really any evidence in ZF that sets were large, just evidence
that
bijections were missing.
I thought believing set theory would help me make nice
sentances in
IF-logic, but instead studying IF-logic made be doubt the
reality of
uncountable sets in set theory.
===
Subject: Re: Skolem¹s Paradox and why is math the way it is?
> |This is much better to me, than the concept of
countability which is
> |only true IN A MODEL based on whether a set called a
bijection (IN A
> |MODEL) exists with certain properties. Countability then
is a
> |property of a set that depends, not on the quantity of
elements in the
> |set, but instead on whether the MODEL in which the set
lives contains
> |some other set. Countability is an extrinsic, contextual
deŽnition.
> I hope you¹ll excuse my going back and answering this two
> weeks later. It still seems topical.
> I don¹t think your arguments against realism are any good,
> although some variation on them is popular, because I
> think they rely on a petitio principii; you assume from
> the outset something much too much like the conclusion
> you wish to establish. For realism about set theory to work,
> it has to be the case that when we refer to the cumulative
> hierarchy of sets (consisting of the pure, well-founded
> sets), the reference succeeds.
> There are plenty of people who have qualms about whether
> there¹s a single, well-deŽned notion of pure, well-
> founded set. (These are the kind of sets that the axioms
> of ZF as supposed to be talking about. Well-founded means
> that if you follow a chain of members X, X2, X3, X4, ...
> where X2 is a member of X, X3 is a member of X2 and so on,
> you eventually have to stop. Pure means that when you
> have to stop, it¹s because you¹ve reached the empty set,
> as opposed to reaching some non-set object. Sets like
> {{1, 2}, {George W. Bush, John Kerry}} are not pure because
> you can get to a non-set object after two steps.)
> My opinion is that these qualms are better expressed another
> way than you¹re doing. I would say that simply being unsure
> whether the concept is well-deŽned is better. In your case,
> apparently you feel comfortable with the semantics of
> Hintikka¹s IF logic. I still don¹t know how one would use
> that as a foundation for mathematics, but maybe it can be.
> So much of what you¹ve been saying has relied, however, on
> proceeding as if the references to set and so on are in
> a sense phony. But this doesn¹t give you an argument against
> realism; it amounts to starting by assuming realism is
> incorrect and then revelling in it. You say countability is
> only true IN A MODEL. If you let me take the cumulative
> hierarchy as my model, that¹s Žne, but there isn¹t any
> only a model about it-- it¹s viewed as including all sets
> of a certain type. It leaves no room for you to turn around
> and treat it as if it were phony in some way.
> If you don¹t think I can take the cumulative hierarchy as
> my model, then it¹s essentially because you have assumed
> that I can¹t.
I don¹t mind going back, but I¹m not following what you¹re
saying
about realism, I didn¹t think I brought that up and don¹t
even know
what it means. If you want to consider all sets, then you need
something other than ZF(C) to do it with. ZF(C) just isn¹t
powerful
enough. The countable faithful models make it clear to me,
and I
don¹t know what I see that others don¹t or what mistake I
make that
makes me think it is clear that others don¹t see or make. I
don¹t
know what you mean by a cumulative hierarchy, but you¹re free
to pick
any model you want. Since we are talking about ZF(C) and not
models
of it, my arguements about my model should carry over to
arguements
about your model. My arguements show that one can
consistently add
AXIOMS to introduce sets that have accessible cardinality,
are pure,
and are well-founded, and that could not previously have been
proven
to exist. Maybe your model will even have an element that
ALREADY
corresponds to my new set. Great for you, it means you can
use the
same model for ZFC+ while I have to make a new one with my
poor
countable model. But it STILL shows that ZFC was deŽcient,
which
relates to the question I was trying to ask. Why do we like
ZF(C),
and is there a way to have a model that is the intended one
that has
all the sets that should have been in all the ZF(C) models,
but
unfortunately were lacking in some models? Did that make any
sense at
all to you?
> |> When we consider all bijections, it¹s like considering
all orange
> |> objects. It makes no more sense to ask whether we might
be
accidentally
> |> considering only a subset of the bijections (when we say
we¹re talking
> |> about all of them) than it would to say, But maybe when
you talk
about
> |> Œall orange objects¹ you are accidentally leaving some
orange objects
> |> out.
> |You ARE leaving things OUT, because more functions from
the naturals
> |TO the reals COULD be added without breaking consistency.
> This is treating the model as if it were a box, which things
> could be added to or removed from. But this is a misleading
> image. To point to a model is to state a criterion for which
> things are to belong to it. Once I¹ve stated my criterion,
> whether something belongs can¹t be manipulated like this.
> Again, it makes no more sense than to argue that I can¹t
> refer to all oranges, because you could add oranges to
> reality in such a way as to make them exist but lie outside
> of the scope of the term all oranges.
> This is also why I say all the countable models are more
> complex than the cumulative hierarchy is. I have been
accepting
> your referring to countable models, since I agree they
exist.
> But if you want to use them in a non-circular argument for
> your case, you have to show that it¹s possible to build them
> up from IF logic (or something else that you accept), which
> you haven¹t done. Also, if you want to consider their
complexity,
> you can¹t just take for granted how they¹re deŽned by other
> people. Try actually writing out the deŽnition of one of
these
> countable models! By logicians¹ standards, it¹s not
terribly hard
> to specify a particular countable model, but on the other
hand,
> the deŽnition will NOT be terribly simple.
> If one were to write out the deŽnition of a particular
countable
> model in full, it would probably become obvious that nobody
doing
> mathematics is actually thinking of that particular model
as being
> the universe.
I HAVE written out deŽnitions of countable models of ZF(C),
and they
are well-founded and pure, AND obviously incomplete in that
there are
things that should be sets that aren¹t in it. They are not
complicated, the only problem with using them in practise is
the
OBVIOUSNESS of the INcompleteness of ZF(C). But since the
missing
sets aren¹t in a faithful model, it means that the existance
of those
sets is logically INDEPENDANT of the ZF(C) axioms, just as the
consistency of ZF (ZFC) is independant of the ZF (ZFC resp.)
axioms.
And *I* used to think of that model as the universe because I
naively
assumed that when we said all subsets, that we meant it. I was
wrong about that, but to me that means ZF(C) is deductively
incomplete
in the sense that sets that should exist can¹t be proven to
exist.
And I¹m tired of people saying that all subsets of naturals
exist,
when they can¹t prove it and when it doesn¹t follow FROM THE
AXIOMS of
ZF(C). And when people use second order logic they start out
by
assuming that all functions exist and I can¹t tell how that
DESCRIBES
any more sets than in the Žrst order theory did. And I have my
doubts about whether it really does because if I tried reading
Bourbaki, they¹d start out deŽning mathematics as pushing
around
formal symbols, and there aren¹t very many arrangements of
formal
symbols. But basically since these other sets that aren¹t
described
by ZF(C) don¹t seem to appear except when one makes a
countable model,
I wonder why cardinal arthimetic makes such a big deal about
pretending that weird subsets exist instead of being about a
lack of
bijections.
> |SpeŽcally
> |you only consider bijections that are ALSO sets. You aren¹t
> |considered because you called your universe the class of
all
sets,
> |but that doesn¹t MAKE it so. If you look in the BASE
MODEL, the idea
> |of being a set is about whether it is in the domain of the
model,
> |which is arbitrary.
> Note that the fact that a bijection is a set (or perhaps a
> class, if the bijection is one between two classes) is a
> matter of deŽnition. How is it that it can be plausible to
> you that something that¹s true by deŽnition might be wrong?
> Because you apply your habit of treating some of my terms as
> being phony in a selective way. You assume you¹re entitled
> to proceed as if by set I mean something other than set.
> And then since not all bijections are necessarily
phony-sets,
> you conclude I¹m wrong to consider all bijections (between
> the natural numbers and other sets) to be sets.
I¹m saying that the ZF axioms don¹t actually imply the
existance of an
uncountable number of bijections. It implies the existance of
a set
of bijections (like P(w)) that lacks a bijection to another
set (w),
but that doesn¹t DISTINGUISH between on the one hand having
few
bijections and on the other hand P(w) being large. I agree
that you
could DEFINE P(w) to be large by saying the lack of a
bijection
MEANS that it is large. The point is that largeness isn¹t IN
the
ZF(C) axioms, only lack of bijections, which is very
different since
it doesn¹t distinguish between two interpretations of its
lack.
Apparantly one interpretation makes many people happy, and
makes them
make fun of other people who don¹t share it. I agree that if
you
DEFINE bijections to be sets, then they are sets, and if you
want to
USE that deŽnition then you can, but then there are things
that COULD
HAVE been deŽned as bijections that now lack a name. Imagine
the
class, C, of subsets, S, of naturals, w, such that there
exists a
proof in ZF(C) that S exists. There is NO set in ZF that
corresponds
to that class. I think that režects badly on the concept of
treating
your intended model seriously, whatever model that is. The
point is
that ZF(C) doesn¹t accurately describe what people say it
does, namely
sets that are bigger than the naturals because they contain
lots of
elements.
> |I had Žgured that the models of ZF had a domain in ZF,
because
> |mathematicians don¹t seem to like to use anything
different than ZF,
> |this is VERY bad for this discussion, because the other
posters keep
> |jumping around and using a word deŽned in one ZF (the base
model) in
> |another ZF (the model).
> I think it is you who have been jumping around. We say
> something, and you take it as meaning a corresponding thing
> about a model, which makes it sound wrong to you.
Different models points out what is independant of an axiom
and what
isn¹t. People are saying that ZF(C) has all subsets of an
inŽnite
set. There is no proof of that, and choosing a model where the
missing sets are OBVIOUS should make it clear. But I was
wrong to
think it was clear to other people. It¹s missing anyway
because the
existance of these sets doesn¹t follow from the axioms. But a
poor
model, or ignoring models obscures this to some people. The
ZF(C)
axioms don¹t DESCRIBE uncountably many sets because they only
prove
that some sets exist, and while there is no set in ZF(C) that
corresponds to a bijection from the naturals to the the sets
ZF(C)
describes, one CAN consistently add a new axiom and then the
axioms
ZF(C)+ would contain a bijection from those two sets. Of
course, NOW
we lack a bijection from the naturals to the sets ZF(C)+
describes.
This incompleteness is unstatisfying to me, and I don¹t know
where
this faith about there actually being more numbers out there
comes
from so I can¹t share with people that have it.
===
Subject: Re: Skolem¹s Paradox and why is math the way it is?
> what I say wrong, independant of my spelling, that¹s why I
was using
> caps because people seemed to think I meant set when I said
> class and I wanted people to be able to easily go back and
read
> clearly what I tried to say before they responded to things
I
> didn¹t say. I¹m very frustrated that either some people here
> aren¹t trying to understand me or that I¹m not being clear
because
> people keep argueing against straw men and not against me.
Which
> is sad because I don¹t care about these straw men, but I¹d
love
> very much to have some succeed at explaining how I¹m wrong,
or just
> telling me why the ZF axioms are good.
>>
>> Ah yes, I suddenly realized you use the code ZF and don¹t
know at
>> all what it means. For instance, surprise, there are no
mention of R
>> (or its non-denumerability) in them... As for the reason
they are
>> good : 1) they are simple. 2) in your sense, they are not
Žxable :
>> *any* (consistent) axiom set has denumerable models (Skolem
>> theorem). And (surprise) ; those axioms *don¹t* speak of
models
>> either.
> Is there not a theorem that says that the set generated by
feeding the
> output of the axiom of inŽnity to the input of the axiom of
powers
> has no bijection to the set generated by the axiom of
powers?
> No. But it could be a typo (you mean axiom of inŽnity at
the end of the
> sentence). And, of course, there is no such thing as
generated by axiom
X
> . Anyway, as usual, your answer is beside the point :
1)those two axioms
> are simple. 2) they dont speak of models
You are quite right that it was a typo. What I am saying is
that the
Žrst order ZF set theory axioms don¹t prove the existence of
uncountably many subsets of the naturals. They prove the
existence of
ONE particular set of subsets of naturals (namely the one
asserted to
exist by feeding the output of the inŽnity axiom to the input
of the
powers axiom, let¹s call it P(w)), and the Žrst order ZF
axioms prove
the existence of countably many proper subsets of that set.
But one
either needs more Žrst order axiom to add to ZF to get more
subsets
P(w), or one needs an entire 2nd order theory from scratch.
There are
a small number of sets that can be proven to exist with Žxed
Žrst
order ZF axioms, and they could be put on a list (and that
list
corresponds to no set in ZF). The fact that I can describe
things
that should be in ZF that aren¹t is just a sign that Žrst
order ZF is
incomplete. But it appears to be incomplete in that it has
models
that are too small, so Skolem tells us that we either need
more than a
Žnite number of axioms, or we need other than 1st order
axioms.
> there could be a countable model with all the sets that are
> provable for just those axioms,
> A set is not provable . You probably mean its existence.
I¹m sorry, I thought that understanding was clear. For
instance there
is a set whose proof that it exists has just one call to one
axiom,
the axiom of inŽnity, which has no inputs.

> but that there can always be other
> sets in other models that are based on axioms that *could*
have been
> consistantly added to the ZF axioms.
> Sure. Inaccessible cardinals, say
I¹m not trying to talk about inaccesible cardinals, there are
small
sets that aren¹t in ZF. SpeciŽcally there are sets that new
axioms
could consistently assert to exist that would be subsets of
the
naturals. To me, that¹s a žaw with the Žrst order ZF axioms.
> Do you see how I¹m using model
> theory to interpret the theorem,
> So what? Your interpretation doesn¹t change anything to its
validity *in
all
> models* and *in all reasonable Žrst order set theories*
The point is that elements that should exist can¹t be proven
to exist.

> and that this interpretation is
> different than the usual interpretation that says that
there is
> actually lots and lost of subsets of the inŽnite set?
> Yes
Why don¹t we use second order axioms instead then? Does it
create new
problems, or is there some OTHER reason that that is the
standard
approach in the schools and textbook I see?

> one has to use second order things to actually force all
> models to be bigger than the inŽnite set.
> That¹s called Skolem thorem. I would have thought you would
have
understand
> it by now, since you are so found of citing it, and since I
said it, for
> instance , in the point 2) you mentioned at teh beginning...
It¹s possible that we¹ve all been talking past each other the
whole
time, I¹ve been doing my best to try to understand you and
everyone
else, and I apologize for any shortcomings on my part. I¹m
still
trying to Žgure out, if Žrst order ZF is incomplete (in the
sense I
mentioned before), then why does everyone like it? I still
have that
question, even after talking to you. Is it better than other
Žrst
order theories? Are Žrst order theories better than second
order
theories? What¹s the reason?
===
Subject: Re: Skolem¹s Paradox and why is math the way it is?
J.E. says...
>What I am saying is that the Žrst order ZF set theory axioms
don¹t
>prove the existence of uncountably many subsets of the
naturals.
Sure it does. In set theory, to say that a set S is
uncountable is to
say that there does not exist a bijection between that set
and the
naturals. ZF proves that there does not exist a bijection
between P(N)
and N.
There is a distinction between
A. There are uncountably many sets S such that ZF proves S
exists.
B. ZF proves that there are uncountably many sets S.
There are only countably many sets that ZF can name (via an
explicit
deŽnition). But ZF can prove that there are more sets than
can be named.
--
Daryl McCullough
Ithaca, NY
===
Subject: Re: Skolem¹s Paradox and why is math the way it is?
> J.E. says...
>What I am saying is that the Žrst order ZF set theory axioms
don¹t
>prove the existence of uncountably many subsets of the
naturals.
> Sure it does. In set theory, to say that a set S is
uncountable is to
> say that there does not exist a bijection between that set
and the
> naturals. ZF proves that there does not exist a bijection
between P(N)
> and N.
> There is a distinction between
> A. There are uncountably many sets S such that ZF proves S
exists.
> B. ZF proves that there are uncountably many sets S.
> There are only countably many sets that ZF can name (via an
explicit
> deŽnition). But ZF can prove that there are more sets than
can be named.
What is the cardinality of the smallest sets that cannot be
named? I
don¹t that there is REALLY an existance proof for sets that
cannot be
proved to exist. I think you are using language in a sloppy
manner.
Proving a lack of a bijection proves the class of bijections
is small,
NOT that the class of subsets is large. Look at the countable
model
to see this result. This REALLY isn¹t falling off a log kind
of
obvious to you?
Your proof about a bijection just says that bijections can
construct
elements that aren¹t in their image. But if the bijection
does NOT
exist, then you have NOT proven that that element exists. You
can¹t
ASSUME something incorrect and then get an intermediate
result and
then get a contradiction and THEN pull out the intermediate
result.
The PROOF of that additional element is FALLACIOUS. The proof
is that
the bijection doesn¹t exist, not that the range of the
non-existant
bijection is different than the power set. There is no proof
of that
one way or another, it is independant of the ZF axioms.
===
Subject: Re: Skolem¹s Paradox and why is math the way it is?
J.E. says...
>What is the cardinality of the smallest sets that cannot be
named?
There are countable sets that cannot be named (using ZF).
>I don¹t that there is REALLY an existance proof for sets
that cannot be
>proved to exist.
I don¹t know what that means. As I said,
ZF proves there exists a set that is uncountable.
>I think you are using language in a sloppy manner.
That¹s the reason people use mathematics. The claim that the
set of reals is uncountable is perfectly precisely rendered
into
mathematics, and is provably true.
>Proving a lack of a bijection proves the class of bijections
is small,
>NOT that the class of subsets is large.
Like I said, to say that a set is uncountable is to say that
there is no bijection between that set and N.
>Look at the countable model to see this result.
>This REALLY isn¹t falling off a log kind of
>obvious to you?
A countable model of ZF obviously doesn¹t contain all the
reals,
because the set of reals is uncountable.
--
Daryl McCullough
Ithaca, NY
===
Subject: Re: Skolem¹s Paradox and why is math the way it is?
> J.E. says...
>What I am saying is that the Žrst order ZF set theory axioms
don¹t
>prove the existence of uncountably many subsets of the
naturals.
> Sure it does. In set theory, to say that a set S is
uncountable is to
> say that there does not exist a bijection between that set
and the
> naturals. ZF proves that there does not exist a bijection
between P(N)
> and N.
> There is a distinction between
> A. There are uncountably many sets S such that ZF proves S
exists.
> B. ZF proves that there are uncountably many sets S.
> There are only countably many sets that ZF can name (via an
explicit
> deŽnition). But ZF can prove that there are more sets than
can be named.
What is the cardinality of the smallest sets that cannot be
named? I
doubt that there is REALLY an existance proof for sets that
cannot be
proved to exist. I think you are using language in a sloppy
manner.
Proving a lack of a bijection proves the class of bijections
is small,
NOT that the class of subsets is large. Look at the countable
model
to see this result. This REALLY isn¹t falling off a log kind
of
obvious to you?
Your proof about a bijection just says that bijections can
construct
elements that aren¹t in their image. But if the bijection
does NOT
exist, then you have NOT proven that that element exists. You
can¹t
ASSUME something incorrect and then get an intermediate
result and
then get a contradiction and THEN pull out the intermediate
result.
The PROOF of that additional element is FALLACIOUS. The proof
is that
the bijection doesn¹t exist, not that the range of the
non-existant
bijection is different than the power set. There is no proof
of that
one way or another, it is independant of the ZF axioms.
===
Subject: Re: Skolem¹s Paradox and why is math the way it is?
J.E. says...
>The proof is that the bijection doesn¹t exist, not that the
range
>of the non-existant bijection is different than the power
set.
I can¹t make any sense of that statement. The claim is that
there
is no bijection between the reals and the naturals. That¹s
what
it *means* to say that the reals are uncountable.
--
Daryl McCullough
Ithaca, NY
===
Subject: Re: Cardinality of a sigma-algebra
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id i98JYrO09673;
>Let A1, A2, ..., An be n subset of Omega.
>Then the smallest sigma-algebra generated by {A1, A2, ...,
An} has at
>most 2^2^n elements.
>How to prove it?
>And if A1, A2, ..., An make a Žnite partition of Omega, what
about the
>cardinality of the sigma-A. generated?
>P.S. Please also correct my English :-)
>Josh.
For the Žrst question, you might as well be asking for the
(smallest) Boolean algebra in Omega generated by A_1, ...,
A_n.
You can think of this Boolean algebra as a ring in which the
addition is symmetric difference and multiplication is
intersection; obviously every element A is idempotent (i.e.
satisŽes A^2 = A), and 2A = 0 (i.e. the additive group is a
vector space over Z_2). Notice that 1 + A is the complement of
A.
Using distributivity, idempotence, and 2A = 0, every element
in the Boolean algebra can be expressed in normal form as a
sum
of type
Sum_{subsets J of {A_1, ..., A_n}} (n_J)(Intersection(J))
where each integer n_J is either 0 or 1. The number of such
J is 2^n, and the number of combinations for such sums is
therefore
2^{2^n}.
As for the second question, the partition assumption means we
can restrict attention to those J with exactly one element.
(why?)
Your English is more or less Žne -- more easily followed than
that of many of the posts in this newsgroup.
Todd Trimble
===
Subject: Re: Prime numbers and the RSA algorithm
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id i98JYrL09677;
>I have been reading about this and I have a question. It is
probably
>naive or simplistic, but I would like to understand.
>The question is not about the RSA algorithm itself but the
nature of
>prime numbers and factorization.
>For RSA encryption, you choose two large prime numbers p and
q and the
>public key N is the product of p and q. The strength of the
encryption
>depends upon the impracticality of factoring N back into p
and q.
True. (AFAAK)
>The question is how to Žnd a large prime number p. Say you
randomly
>choose p. You either check if it is prime by factoring it or
by
>dividing it by every number between 2 and the square root of
p (with
>the understanding that there may be more efŽcient algorithms
for
>checking primes.)
Noone checks for primality by trial division. may be more
efŽcient
is an understatement......There are much much much faster
methods.
>If it is easy to check p for primeness, then it must be
relatively
>easy to check N for its factors. I.e., by factorization or
by dividing
>it by every number between 2 and the square root of N.
Non sequitur. Please explain how you reach the conclusion
that factoring is
easy if prime testing is easy. I am curious why
you think the former follows from the latter.
I suggest that you do a little arithmetic.. Suppose N is a
100 digit
number. How many divisions do you think you will need to
factor it
if it is the product of nearly equal primes?????
>As a mathematical layman, I must be missing something.
You need to look at the amount of arithmetic involved in
trying to factor by trial division. There are about 8.7 x
10^47
primes less than 10^50. How long do you think it will take to
test them all?
>Or is the answer that factoring large numbers is
impractical. But
>testing large numbers for primeness IS practical. This must
be the
>answer, but I haven¹t found any information on how this is
done.
Do a web search. There are many sources of information about
prime testing algorithms and factoring algorithms.
===
Subject: Re: Prime numbers and the RSA algorithm
days. My association with the Department is that of an
alumnus.
[.snip.]
>>If it is easy to check p for primeness, then it must be
relatively
>>easy to check N for its factors. I.e., by factorization or
by dividing
>>it by every number between 2 and the square root of N.
>Non sequitur. Please explain how you reach the conclusion
that
>factoring is easy if prime testing is easy. I am curious why
you
>think the former follows from the latter.
I suspect that the OP Žgured that if one checks p for
primeness by
trial division (and hence this is easy), then it would
naturally be
just as easy to check N for its factors by trial division, as
he
mentions. Of course, he forgot to include the possibility of
those
other methods for checking primality.
--
It¹s not denial. I¹m just very selective about
what I accept as reality.
--- Calvin (Calvin and Hobbes)
Arturo Magidin
magidin@math.berkeley.edu
===
Subject: WILL THREE COLORS DO THIS ONE?
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id i98KCBe13296;
If one begins with a unicursal network (one in which each
edge can be
traversed exactly once in a continuous fashion), at most two
of its
vertices can be of odd order. My question is whether three
colors
are always enough to color it.
Here is why I ask.
If the map has no vertices of odd order, any unicursal path
Žnishes
where it starts; and one can use a Jordan curve type inside is
white, outside is black argument to show that two colors
sufŽce.
If the map has two vertices of odd order, then any unicursal
path
starts at one of them and ends at the other. So, what if we
join
these ends by another sequence of edges inserted between
existing
vertices; creating a map of the Žrst type, for which two
colors are
sufŽcient. Those faces split by our new edges might require
the
third color.
However, this argument by itself does not hold water, as the
path
might split faces sharing a common boundary. If we can show a
path
may always be chosen to avoid this, then we may have an
argument. I
have tried many such networks, and have always been able to
do this.
Perhaps you have an idea to complete the proof (or, egad, a
===
Subject: Tropic Geometry
posting-account=2LgBuQ0AAAC2HZQp160dLwjzCEeNC7or
What is tropic geometry all about, elementary notions etc?
What are the
big (including answered) questions there? What kind of Želds
does it
include?
===
Subject: Re: Zenkin¹s paper on Cantor
> I try to be civilized, Keckman. And even when I am dealing
with an
> extremely rude person like Robin Chapman,
> Google groups records 17 hits for
> author:erayo@bilkent.edu.tr
> while only 5 for
> author:rjc@ivorynospamtower.freeserve.co.uk
> As far as rudeness is concerned, I am lagging well behind
> Ozkural. I must sharpen up my act!
No, you are just lucky enough to not have come across someone
like David
Longley.
I¹d say, let¹s just both try to be more polite.
--
Eray
===
Subject: Re: Zenkin¹s paper on Cantor
> In Cantor¹s second proof (diagonal proof), there are
> inŽnitely many steps, *after* which we take another step.
> Excellent! You have laid bare the duplicity of Cantor, in
sneakily
> taking another step after an inŽnity of steps, as though
this could
> be done in a reasonable and orderly fashion. It only
remains to pack
> the old scoundrel off to the Chamber of Eloquence, where he
will be
> roundly reminded of his delinquency by means of Q-tips
piercing his
> brain.
> Now we need to take on those other stupid old mathematical
codgers.
LOL
I must admit that I found the above piece quite cunning. :)
--
Eray Ozkural
===
Subject: Re: Zenkin¹s paper on Cantor
> Well, yes, as I said, when I can help it, but in the case
of Kent Paul
> Dolan, I don¹t think I could.
Your self-image seems to be out of whack.
You are such an idiot...
Now go away you idiot!
...he is a ing idiot just like you!
You are a ing idiot.
Raatikainen is an idiot...
===
Subject: Re: Zenkin¹s paper on Cantor
> Well, yes, as I said, when I can help it, but in the case
of Kent Paul
> Dolan, I don¹t think I could.
> Your self-image seems to be out of whack.
The quotes below belong to which context? Please don¹t take
quotes out
of context. I did not say I never swore.
> You are such an idiot...
Said in response to David Longley¹s drivel.
> Now go away you idiot!
This was in response to you. Why did I say such a thing?
Because you
kept on your pointless one-liners for no reason. Here is what
I said:
:Now go away you idiot! You have raised my temper! Go back to
dealing
:with those in your league, you redundant self-appointed
USENET
:expert!
I had actually thought that you were a self-appointed USENET
expert
like Mr. Dolan or Doktor Dynasoar, because you kept asserting
some
points which you believed in, without Žrst introducing
yourself, or
bothering to explain anything. (No, that¹s not a good
behavior) Nobody
is supposed to know your research interests, or why you
thought X is
the deŽnitive Godel resource or other things, or believe in
what you
said because you believe that you are an expert. In public
places, you
Žrst introduce yourself, listen to others and and then talk
politely.
If you don¹t listen to others and talk politely, don¹t expect
others
to behave differently.
I can say that your corrective one-liner posts were not
things to
like, especially because we were talking of quite vague
things. And,
AFAICT, nobody likes one-liners. It¹s called trolling.
> ...he is a ing idiot just like you!
> You are a ing idiot.
These two must be in response to Longley. He well deserves
far worse
insults.
> Raatikainen is an idiot...
This one does not belong to me. Here is the context, below
the line of
Œ-¹s. I have said that his criticism is trivial, I didn¹t say
he was
an idiot. He may well have many accomplishments beyond the
scope of
the disputed paper.
Well, at least you¹re not a good journalist. You knowingly
distorted
my USENET presence, that was not objective at all!
But you have a point. I shouldn¹t have sworn to you, just
because you
tried to insult me publicly in your own private language. I
should
have maintained civil discourse, like any amateur philosopher
must.
And I must not have assumed things about you. For these I must
sincerely apologize to you. (Though you yourself seem to make
several
bad assumptions about others. Many of your replies to me
contained ad
hominem arguments. Not all of them were funny.)
My only requests: do not give incorrect or inaccurate quotes
from me,
and please do not continue these off-topic attacks to me.
(change the
subject with a preŽx [OT] if you so wish to žame me)
-------------------------------------------------------------
---------------

> No, no, sure, you are right. Raatikainen is an idiot and the
> criticisms he has worked on for some years and published
several
> technical papers on must be ultimately misguided, because
they
> contradict Chaitin¹s claims.
> Or perhaps you could respond to the criticisms. It¹s a
strategy that
> seems to have eluded you so far.
I don¹t think he¹s an idiot, but I¹m somehow suspecting that
he is
making a fuss about a trivial point, which has not been
challenged by
any mathematician... Of course making a fuss is not a
problem, the
problem is that he thinks he has *conclusively* shown that
Chaitin¹s
incompleteness result is wrong!
--
Eray
===
Subject: Re: Zenkin¹s paper on Cantor
>> Raatikainen is an idiot...
> This one does not belong to me.
True, you didn¹t describe Raatikainen as an idiot, but only
as an
intellectual vacuum.
===
Subject: Re: Zenkin¹s paper on Cantor
<4CP8d.254$MT4.137@reader1.news.jippii.net>
<ck1cfg$qug$1@news6.svr.pol.co.uk>
<opsfhiewo03uk9lu@cs81133.pp.htv.Ž>
<vcb1xga90ej.fsf@beta19.sm.ltu.se>
<vcbvfdlmt5q.fsf@beta19.sm.ltu.se>
<vcb3c0o3ib5.fsf@beta19.sm.ltu.se>
Discussion, linux)
> Raatikainen is an idiot...
>> This one does not belong to me.
> True, you didn¹t describe Raatikainen as an idiot, but only
as an
> intellectual vacuum.
This isn¹t quite true either.
What our esteemed Chaitin scholar said was, I believe that
Chaitin¹s
work must not be criticized from an intellectual vacuum, like
the way
Raatikainen did.
prepositions, shall we?
--
Jesse F. Hughes
Do not click any hyperlinks that you do not trust. Type them
in the
Address bar yourself. -- Microsoft gives security advice.
===
Subject: Re: Zenkin¹s paper on Cantor
> prepositions, shall we?
As in sworn to you?
===
Subject: Re: Zenkin¹s paper on Cantor
>> Eray are there integers with an inŽnite number of digits?
Ozkural
> I try to be civilized,
>> That¹s a tall tale!
> Keckman. And even when I am dealing with an
> extremely rude person like Robin Chapman,
>> You can¹t have met many rude people then.
> I try to keep such primitive
> behavior to myself.
>> behaviour.
> Any reason you insist on behaviour but not civilise?
> Too French, is it?
Certainment!
--
Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html
Lacan, Jacques, 79, 91-92; mistakes his penis for a square
root, 88-9
Francis Wheen, _How Mumbo-Jumbo Conquered the World_
===
Subject: Re: Zenkin¹s paper on Cantor
> I will have to disagree, Robin. Philosophy of mathematics
is directly
> relevant to foundational matters. Think of the environment
in which
> set theory was developed. I do not say that the role of
inŽnity
> within mathematics is pre-theoretic, but I do think that the
Thinking again :-(
> conception of actual inŽnity is based on pre-theoretic, or
> pre-mathematical conceptions of inŽnity, which is the stuff
Zeno had
> argued against (as well as motion, etc. to support
Parmenides). He was
> one of the Žrst mathematicians.
Zeno? a mathematician? You amaze me!

> At any rate, this modernist/careerist perspective that
mathematics has
> no dealing with philosophy is absurd, regardless of whether
Zeno had a
> point or not. Mathematicians should not be deaf and blind to
> philosophical foundations.
Philosophical considerations have generally acted as a brake
on mathematical progress. Recall the Kantian objections
to non-Euclidean geometry.
--
Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html
Lacan, Jacques, 79, 91-92; mistakes his penis for a square
root, 88-9
Francis Wheen, _How Mumbo-Jumbo Conquered the World_
===
Subject: Tropical Geometry
posting-account=2LgBuQ0AAAC2HZQp160dLwjzCEeNC7or
What is tropical geometry all about, what are the elementary
notions
etc? What are the big, including already answered, questions
there?
What kind of Želds does it include?
===
Subject: Re: Tropical Geometry
>What is tropical geometry all about, what are the elementary
notions
>etc? What are the big, including already answered, questions
there?
>What kind of Želds does it include?
on this subject which have showed up over the past few years
at http://arxiv.org/math/ . If you search there I¹m sure you
will Žnd answers to your questions. At the edges of my memory
there¹s something teasing me to the effect of tropical
geometry
is like real-algebraic geometry except that it¹s over the
reals
with min(x,y) as the product, but that can¹t be right, can it?
Anyway, among the authors¹ names have been several that I
recognized as perfectly respectable people.
Lee Rudolph
===
Subject: Re: Tropical Geometry
|
|>What is tropical geometry all about, what are the
elementary notions
|>etc? What are the big, including already answered,
questions there?
|>What kind of Želds does it include?
|
|on this subject which have showed up over the past few years
|at http://arxiv.org/math/ . If you search there I¹m sure you
|will Žnd answers to your questions. At the edges of my memory
|there¹s something teasing me to the effect of tropical
geometry
|is like real-algebraic geometry except that it¹s over the
reals
|with min(x,y) as the product, but that can¹t be right, can
it?
|Anyway, among the authors¹ names have been several that I
|recognized as perfectly respectable people.
i thought that it was with min(x,y) as the sum, and x+y as the
product, or something like that. (that at least keeps the
distributivity of product over sum.) i¹ve worked on similar
ideas,
but not as part of the subculture that calls the subject
tropical
geometry or tropical algebra or tropical algebraic geometry or
whatever. the basic idea is (at least if you read into things)
hundreds of years older than the use of the terminology
tropical to
describe it, as far as i know. there is even a joke argument
to the
effect that the terminology tropical algebra is as bad as
possible
because it is actually the algebra of statistical mechanics
at zero
temperature.
--
[e-mail address jdolan@math.ucr.edu]
===
Subject: re:Help Constructing the Umbilic Torus / Mobius Band
Awesome!
when searching earlier. Jim, I¹ll deŽnitely try out that
surface
plan.
===
Subject: Re: New paper, algebraic integers, Galois Theory
<deleted>
> Now divide f^2 from both sides, which gives
>
> P(m)/f^2 = (m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1 + mf^2)
xu^2 + u^3 f
>
> P(m)/f^2 = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf)/f^2
>
> and you note that P(0)/f^2 = u^2(3x + uf), which means that
now your
> constant terms are u for the Žrst, u for the second and 3x
+ uf for
> the third.
>
> True enough, but if what you want at the end is a
factorization
> with algebraic integer coefŽcients, this is the wrong way to
> divide f^2 out of the right side. It has been proved that
> there is a RIGHT way. By putting unjustiŽed stress on
preserving
> the constant terms AFTER division by f^2, you have lost
sight of
> the primary goal of ending up with a factorization with
algebraic
> integer coefŽcients. That is the root of some of your
subsequent
> problems with this paper. But see below also.
The terms constant with respect to m, are just that, constant
with
respect to m.
In this exposition I¹ve stepped through an argument quite
drily
without extra.
Now you¹re adding extra, but giving no mathematics.
> Now then, if m=1, what are the constant terms now?
>
> They are u for the Žrst, u for the second, and 3x + uf for
the third.
>
> If m = 2938479378, what are the constant terms now?
>
> They are u for the Žrst, u for the second, and 3x + uf for
the third.
>
> How can the constant terms of the Žrst two go from uf to u?
>
> They must be divided by f.
>
> Now, the constant term of a_1 x + uf, is uf, but when f^2
is divided
> from P(m), it is u; therefore, a_1 x + uf is divided by f,
and you
> have
>
> a_1 x/f + u
>
> and the constant term of a_2 x + uf is uf, but when f^2 is
divided
> from P(m), it is u; therefore, a_2 x + uf is divided by f,
and you
> have
>
> a_2 x/f + u
>
> while the constant term of a_3 x + uf is 3x + uf, and after
f^2 is
> divided off, it is 3x + uf, so you have
>
> a_3 x + uf
>
> so, dividing P(m) by f^2 gives
>
> P(m)/f^2 = (a_1 x/f + u)(a_2 x/f + u)(a_3 x + uf).
>
> There is no way to mathematically argue with the result.
>
> See above. You have lost sight of what you needed to do
> in factoring out f^2, and as a result have chosen the wrong
> factorization. The equation just above is correct, but when
> m > 0, the coefŽcients a_1/f and a_2/f are not algebraic
> integers, which (if you remember!) is what you wanted.
Huh?
Your statement is just bizarre, and I¹ll stop at this point
as I¹ve
shown your strategy which is to make VERY LONG POSTS and
ignore
pointed questions--basically bullshitting your way through
until I get
tired of arguing with you.
That long posting style is just freaking annoying, and I
think you
learned that a long time ago, which is why you keep at it.
Now the math here is quite DRY, and concise which it is to
leave out
room for b.s. and the reality is that the terms constant with
respect
to m are constant with respect to m.
You will lie about it, dance around it, and make elaborate and
overlong posts to try and dodge the basic mathematics, but it
doesn¹t
change it.
James Harris
===
Subject: Re: New paper, algebraic integers, Galois Theory
> Moved draft under top post to allow more deletions.
>
> Additional deletions would provide further improvement ...
Oh, I wouldn¹t be surprised if there are more deletions.
James Harris
===
Subject: Re: New paper, algebraic integers, Galois Theory
The problem is that the third section
is now rather thin. Here is a suggestion as
to what it could say.
III. Third section

So, even if a_1/f is not an algebraic integer, you can Žnd
w_1 an
algebraic integer.

But if a_1/f is an algebraic integer and w_1 is not, they
cannot be
equal.

But I have

P(m)/f^2 = (w_1 x + v_1)(w_2 x + v_2)(w_3 x + v_3)

and

P(m)/f^2 = (a_1 x/f + u)(a_2 x/f + u)(a_3 x + uf)

so how do you reconcile a case where a_1 x/f is not an
algebraic
integer?

There must exist z_1, z_2, and z_3 such that

w_1 = (a_1 x z_1)/f, w_2 = (a_2 x z_2)/f and w_3 = a_3 x z_3

and z_1 z_2 z_3 = 1,

Now set s_1=f/z_1, s_2 = f/z_2 and s_3 = 1/z_3
Clearly s_1 s_2 s_3 = f^2
and
P(m)/f^2 = 1/f^2 (a_1 x + uf)(a_2 x + uf)(a_3 x + uf)
= (a_1 x/s_1 + uf/s_1)(a_2 x/s_2 + uf/s_2)(a_3 x/s_3 + uf/s_3)
= (a_1 z_1 x/f + z_1 u)(a_2 z_2 x/f + z_1 u)(z_3 a_3 x + z_3
uf)
= (w_1 x + v_1)(w_2 x + v_2)(w_3 x + v_3)
So, noting that the a_i,w_i,v_i and s_i are all functions of
m,
There exist s_1(m), s_2(m), s_3(m) with s_1(m) s_2(m) s_3(m)
= f^2
such that
P(m)/f^2= (a_1(m) x/s_1(m) + uf/s_1(m))(a_2(m) x/s_2(m) +
uf/s_2(m))
(a_3(m) x/s_3(m) + uf/s_3(m))
and a1(m)/s_1(m), uf/s_1(m), a_2(m)/s_2(m), uf/s_2(m),
a_3(m)/s_3(m)
and uf/s_3(m) are all algebraic integers
Isn¹t that a more impressive conclusion than
> so algebraically the z¹s are units, but z_1, z_2 and z_3
are not units
> in the ring of algebraic integers, since a_1/f is not.
-William Hughes
===
Subject: What would it take?
Now I face a situation where I claim to have important
mathematics
which other people claim is false.
I¹m just one person, there are many of them, and they refuse
to just
talk out the math logically and objectively.
years over the same things, and they cheat.
And before you say I¹m being paranoid consider what happened
with a
paper of mine, which a group of sci.math posters successfully
managed
to censor out of publication in a peer reviewed math journal
by gang
emailing the editors:
See http://rattler.cameron.edu/swjpam/vol2-03.html
I mean, come on, I¹m just one guy, but if a group of
sci.math¹ers are
willing to use those kind of tactics, what can I do?
The chief editor yanked my paper basically immediately upon
getting
*claims* that it was wrong, without there being time enough
for him to
have actually bothered to check, when the journal had my
paper for
over NINE MONTHS before!
You can see this group operating on these newsgroups now as
they make
sure to try and reply to my posts when I post content,
mathematical or
otherwise.
If there¹s math, then usually it¹ll be Nora Baron or Dik
Winter
replying to me, so that Arturo Magidin can reply to them
since he says
that he won¹t reply to me now.
If there¹s a lot of commentary then it¹ll be Jesse Hughes,
David
Ullrich, or Gib Bogle, among others.
That¹s just 6 of the standard crew, and you can see some of
the others
that pop in and out.
I¹m ONE GUY. I need to know what it would take to convince a
sizeable
number of you that there¹s a chance I could be right and that
all of
this dedicated activity is actually--oddly and bizarrely
enough--meant
to hide the truth.
Now I¹ve demonstrated my willingness to talk about the
details of my
work from prime counting to advanced polynomial factorization
but the
people who reply to me are part of the cabal.
Are there any others willing to talk out the math?
Is there anyone who has a solution?
And, consider, I tried publication. The sci.math¹ers gang
emailed the
editors of the Southwest Journal of Pure and Applied
Mathematics.
Other editors are saying to me that I should go to other
journals!!!
It¹s a massive passing of the buck on a huge academic scale.
So what do I do? What would you do?
James Harris
===
Subject: Re: What would it take?
> Now I face a situation where I claim to have important
mathematics
> which other people claim is false.
Um, speaking as one of those other people, I have provided
simple arithmetic that refutes your conclusion in the paper
you¹re so hot under the collar about. I have shown it in
who so unceremoniously yanked your paper.
> I¹m just one person, there are many of them, and they
refuse to just
> talk out the math logically and objectively.
Oh, is that why you refuse to respond to my requests to show
me where my argument goes awry? After all, if you have a
correct argument that [mumble] is true, and someone shows
that [not mumble] is true instead, either the original
argument
or the refutation must be in error. You have kept this
nonsense
up for months, without responding to my argument (except to
make
some bogus claim about a weird assumption).
Do you really think I¹m about to slink away?
> years over the same things, and they cheat.
It¹s cheating to tell folks the truth?
> And before you say I¹m being paranoid consider what
happened with a
> paper of mine, which a group of sci.math posters
successfully managed
> to censor out of publication in a peer reviewed math
journal by gang
> emailing the editors:
> See http://rattler.cameron.edu/swjpam/vol2-03.html
> I mean, come on, I¹m just one guy, but if a group of
sci.math¹ers are
> willing to use those kind of tactics, what can I do?
If my rebuttal had been full of holes, do you really think the
actions, but you can¹t imagine that they acted solely out of
fear
of displeasing the Great and Powerful OZ, can you?
> The chief editor yanked my paper basically immediately upon
getting
> *claims* that it was wrong, without there being time enough
for him to
> have actually bothered to check, when the journal had my
paper for
> over NINE MONTHS before!
ooooh. *claims*.
You imagine that my argument takes over Žve minutes to verify?
Have you tried to test it?
I provided enough arithmetic for whichever author it was to
go to his
computer and construct all the products that I claimed were
true. The
argument is absolutely transparent to anyone who knows the
least bit
of algebra, and is unassailable, unlike the gauze that you
hope to
fashion into an airtight vessel.
> You can see this group operating on these newsgroups now as
they make
> sure to try and reply to my posts when I post content,
mathematical or
> otherwise.
> If there¹s math, then usually it¹ll be Nora Baron or Dik
Winter
> replying to me, so that Arturo Magidin can reply to them
since he says
> that he won¹t reply to me now.
Hmm... It¹s your notion that these folks step in *for the
purpose* of
getting Magidin to step in? It seems to me that he only
enters the
fray to pose corrections or simpliŽcations.
BTW, clever swipe at Arturo Magidin¹s integrity,
since he says that he won¹t reply to me now.
Everyone knows that you speciŽcally requested (yes, at his
offering)
him to stop interacting with you. You sure want it both ways,
don¹t
you?
> If there¹s a lot of commentary then it¹ll be Jesse Hughes,
David
> Ullrich, or Gib Bogle, among others.
Oh, Commentary.
That¹s a real hoot. JSH pontiŽcating about logic, about
Galois Theory,
about algebra.
Commentary
teeheehee.
> That¹s just 6 of the standard crew, and you can see some of
the others
> that pop in and out.
> I¹m ONE GUY. I need to know what it would take to convince
a sizeable
> number of you that there¹s a chance I could be right and
that all of
> this dedicated activity is actually--oddly and bizarrely
enough--meant
> to hide the truth.
(MUSIC SWELLS AS THE LIGHTS DIM,
EXCEPT FOR A SMALL SPOT ON JSH)
To dream ... the impossible dream ...
To Žght ... the unbeatable foe ...
To bear ... with unbearable sorrow ...
To run ... where the brave dare not go ...
To right ... the unrightable wrong ...
To love ... pure and chaste from afar ...
To try ... when your arms are too weary ...
To reach ... the unreachable star ...
This is my quest, to follow that star ...
No matter how hopeless, no matter how far ...
To Žght for the right, without question or pause ...
To be willing to march into Hell, for a Heavenly cause ...
> Now I¹ve demonstrated my willingness to talk about the
details of my
> work from prime counting to advanced polynomial
factorization but the
> people who reply to me are part of the cabal.
Sorry, only us caballeros here!
> Are there any others willing to talk out the math?
> Is there anyone who has a solution?
> And, consider, I tried publication. The sci.math¹ers gang
emailed the
> editors of the Southwest Journal of Pure and Applied
Mathematics.
> Other editors are saying to me that I should go to other
journals!!!
> It¹s a massive passing of the buck on a huge academic scale.
> So what do I do? What would you do?
I¹ve suggested learning some algebra, so you get at least a
sense
of the fact that this isn¹t all just made up to spite you.
Hell,
you could probably learn a little from cracking open a book
and
working through it. Getting in on the actual language would
buy
you a little bit of credit.
You would have a lot more if you had some knowledge of, say,
Galois Theory, before announcing that it was overinterpreted
(whatever on Earth that means).
By the way, if you refrained from announcing that the stuff
you
had just stumbled across was beyond the capabilities of
everyone
else on Earth, that wouldn¹t hurt.
Want an example? Try this:
It¹s a stunning and argument killing result as no
mathematician on the planet can deliver the factor,
and if challenged they may do odd things.
Like Kenneth Ribet or Andrew Wiles might at Žrst
pause, as they consider the question, and maybe
scribble a bit, but eventually, realize it can¹t
be done.
It¹s also the kind of demonstration that travels
quickly over the Internet.
Now you can Žnd an algebraic integer factor of 6,
like 2 or 3, or sqrt(2), or sqrt(3), and possibly
many have been lulled into naive complacency because
of the inŽnite decomposability of 6 in the ring
of algebraic integers.
The full result--more formally so that there¹s no
room for confusion--is that in the ring of algebraic
integers, given a non-unit irrational algebraic integer
A, that is a factor of a composite natural C it is
IMPOSSIBLE to Žnd a non-unit algebraic integer factor
of A that is coprime to any prime factor of C.
To refute, someone needs to just demonstrate the contrary,
by delivering a factor.
At least some one of you should apologize to me.
For the record, I¹ll note that it didn¹t take more than a day
or two
for you to see the need to retract the claim (but not the
boasting
that announced the claim, which you never retracted).
I¹ve also suggested responding to my bit of arithmetic, to
tell me
what you think is incorrect with it. Notice that people are
saying
that you reach incorrect conclusions via your arguments. This
means
that your arguments are invalid, and it makes no sense to
wade through
them until you agree that the conclusions you reach are
invalid.
As long as you¹re trying to defend as valid an argument that
everyone
else can plainly see as being deeply žawed, there¹s not going
to be any
real cooperation.
The knee-jerk responses of you¹re lying or the algebraic
integers
are žawed don¹t make your case (just in case you didn¹t know
this).
Your habit of trotting out the same old trivial examples that
work
the way you want them to work is foolish, since a single
example can
*never* prove a general result; you should pay much more
attention to
the counterexamples that you apparently have admitted exist.
Don¹t you
know, it only takes *one* counterexample to rebut an argument?
> James Harris
Just my two bits¹ worth.
Dale
===
Subject: Re: What would it take?
Discussion, linux)
> Now I face a situation where I claim to have important
mathematics
> which other people claim is false.
> I¹m just one person, there are many of them, and they
refuse to just
> talk out the math logically and objectively.
> years over the same things, and they cheat.
[...]
> You can see this group operating on these newsgroups now as
they make
> sure to try and reply to my posts when I post content,
mathematical or
> otherwise.
> If there¹s math, then usually it¹ll be Nora Baron or Dik
Winter
> replying to me, so that Arturo Magidin can reply to them
since he says
> that he won¹t reply to me now.
> If there¹s a lot of commentary then it¹ll be Jesse Hughes,
David
> Ullrich, or Gib Bogle, among others.
> That¹s just 6 of the standard crew, and you can see some of
the others
> that pop in and out.
> I¹m ONE GUY. I need to know what it would take to convince
a sizeable
> number of you that there¹s a chance I could be right and
that all of
> this dedicated activity is actually--oddly and bizarrely
enough--meant
> to hide the truth.
Why would I care to hide the truth? Why on earth would you
place your
biggest publicist on that list? I rarely speak about your
mathematics
claims. When your delusions on logic come up, then I feel
compelled to
react, but aside from that, I¹m mostly interested in your
self-effacing humorous commentary.
I¹ve always loved your ability to slyly poke fun at yourself
by making
ridiculously grandiose claims. I can¹t fathom how you can be
so
darned funny, posturing about the other side¹s alleged sins
while
giving a laundry list that reads like JSH¹s greatest hits.
You¹re a
funny guy, ain¹t no doubt.
Now which truth might I be trying to hide for pointing this
out and
why?
Say, you haven¹t Žgured out me and Uncle Andy¹s secret, have
you?
Why, the whole Wiles clan will be mighty mad if you have.
--
Jesse F. Hughes
And I¹m one of my own biggest skeptics as I had *YEARS* of
wrong
ideas, and attempts that failed. Worse, for some of them it
took
*MONTHS* before I Žgured out where I screwed up. -- James
Harris
===
Subject: Re: What would it take?
posting-account=UtgH7gwAAACpBhTelVPOXNP7RAfbtQrK
> Now I face a situation where I claim to have important
mathematics
> which other people claim is false.
> I¹m just one person, there are many of them, and they
refuse to just
> talk out the math logically and objectively.
> years over the same things, and they cheat.
> And before you say I¹m being paranoid consider what
happened with a
> paper of mine, which a group of sci.math posters
successfully managed
> to censor out of publication in a peer reviewed math
journal by gang
> emailing the editors:
> See http://rattler.cameron.edu/swjpam/vol2-03.html
> I mean, come on, I¹m just one guy, but if a group of
sci.math¹ers are
> willing to use those kind of tactics, what can I do?
Well, you could try joining a high-IQ club and publishing
stuff on
their web-site.
> The chief editor yanked my paper basically immediately upon
getting
> *claims* that it was wrong, without there being time enough
for him
to
> have actually bothered to check, when the journal had my
paper for
> over NINE MONTHS before!
> You can see this group operating on these newsgroups now as
they make
> sure to try and reply to my posts when I post content,
mathematical
or
> otherwise.
That¹s kind of what a newsgroup is about: dialogue. Usenet is
not
your private little journal.
> If there¹s math, then usually it¹ll be Nora Baron or Dik
Winter
> replying to me, so that Arturo Magidin can reply to them
since he
says
> that he won¹t reply to me now.
> If there¹s a lot of commentary then it¹ll be Jesse Hughes,
David
> Ullrich, or Gib Bogle, among others.
Shit, I don¹t rate a mention? What do I gotta do, crank up the
abuse intensity?
> That¹s just 6 of the standard crew, and you can see some of
the
others
> that pop in and out.
> I¹m ONE GUY. I need to know what it would take to convince a
sizeable
> number of you that there¹s a chance I could be right
Try actually _being_ right. Duh.
> and that all of
> this dedicated activity is actually--oddly and bizarrely
enough--meant
> to hide the truth.
> Now I¹ve demonstrated my willingness to talk about the
details of my
> work from prime counting to advanced polynomial
factorization but the
> people who reply to me are part of the cabal.
I don¹t get cabal where I live, gotta use a sattelite dish.
> Are there any others willing to talk out the math?
Nobody here but us chickens, boss.
> Is there anyone who has a solution?
Yes, but I don¹t think you¹re going to like it.
> And, consider, I tried publication. The sci.math¹ers gang
emailed
the
> editors of the Southwest Journal of Pure and Applied
Mathematics.
You tried to commit fraud and got caught. Boo-hoo.
> Other editors are saying to me that I should go to other
journals!!!
I thought you said you had _important_ mathematics?
> It¹s a massive passing of the buck on a huge academic scale.
> So what do I do?
Give up. Or why don¹t you at least work on some other
problems?
I¹d love to hear your analysis of the Collatz Conjecture.
> What would you do?
Kill myself.
> James Harris
===
Subject: Suggestions for grad school?
X-RFC2646: Format=Flowed; Original
I¹d like recommendations on suggested mathematics courses for
preperation of

mathematics grad school. I¹m not sure yet which school but
I¹d like a
general scenario.
I¹ve taken up to calculus II and Maxtrix Theory. That was
about seven years

ago. What is the best way to review my previous courses
without taking time

to go through those courses again? I¹d like to use that time
for attending

higher level courses such as abstract algebra and real
analysis.
Brett
===
Subject: Component of vector perpendicular to other vector
posting-account=XVykTQ0AAAAllcfBRnVTR8EjmEbnoy-N
please consider the following vectors, A, B and P:
P B
| /
|/ (theta)
-------------A
/
So B intersects A. P is at right angles to A, but
in the direction of B. It¹s the component of B that
is at right angles to A. Theta is the angle between
A and B.
This is what I do currently, but is there a neater way
to get P?
theta = AngleOf(B) - AngleOf(A);
theta = (theta+360) mod 360; // make -270 into +90, etc
P = Rotate(A, 90 * Sign(theta));
P = P / Length(P); // Normalize P to unit length
P = P * (Length(B) * Sin(theta)); // Scale P accordingly
where
AngleOf(V) = Atan(V.y,V.x);
Dez
===
Subject: Simple disproof, eerily simple
There¹s been a lot of arguing back and forth for years, but I
think I
have a simple demonstration which will end it for those of
you who
actually care more for the mathematics than social crap.
First again I need
P(m) = f^2 ((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1 + mf^2) xu^2
+ u^3 f)
and
P(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf)
and to solve for the a¹s, using
a_1 a_2 a_3 = f^2 ((m^3 f^4 - 3m^2 f^2 + 3m)
a_1(a_2 + a_3) + a_2 a_3 = 0
and
a_1 + a_2 + a_3 = - 3(-1 + mf^2)
so the a¹s are the negatives of the three roots of the cubic
a^3 - 3(-1 + mf^2)a^2 + f^2 ((m^3 f^4 - 3m^2 f^2 + 3m) = 0
and at issue is my claim of proof that algebraically *two* of
the
roots have f as a factor.
Posters arguing with me have claimed that instead *all* of
the a¹s
would have factors of f based, supposedly on Galois Theory.
Now let f=sqrt(5), m=1, to get an example where several
posters have
claimed counterexamples to my argument, as then you have
a^3 - 12a^2 + 65 = 0
and the claim then is that Galois Theory proves that *all* of
the a¹s
have some factors in common with sqrt(5).
Well then, let a = 65/y, and substitute, which gives
65^3/y^3 - 12(65)^2/y^2 + 65 = 0, which is
y^3 - 12(65)y + 65^2 = 0.
Now let
a_1 = w_1 v_1, a_2= w_2 v_2, and a_3 = w_3 v_3 where
w_1 w_2 w_3 = 5, and v_1 v_2 v_3 = 13, then
a_1 = 65/y_1, w_1 v_1 = (w_1 w_2 w_3)(v_1 v_2 v_3)/y_1, so
where y_1, y_2 and y_2 are the three roots of
y^3 - 12(65)y + 65^2 = 0, and I have easily enough
y_1 = (w_2 w_3)(v_2 v_3), and similarly
y_2 = (w_1 w_3)(v_1 v_3) and
y_3 = (w_1 w_2)(v_1 v_2)
and from
y^3 - 12(65)y + 65^2 = 0.
the sum of the y¹s is 0, so
(w_2 w_3)(v_2 v_3) + (w_1 w_3)(v_1 v_3) + (w_1 w_2)(v_1 v_2)
= 0
proving the all of the w¹s CANNOT be coprime to each other!!!
Ok, did I screw up some algebra here? That¹s too easy.
I probably did. Oh well, I¹ll toss it out anyway.
James Harris
===
Subject: i dont like irrationals
Sorry, this is stupid, i agree. I just could not get sleep
and wondered how to get rid of endless decimal numbers,
which i don¹t like
---
We say that there are irrational numbers like sqrt(2)
because the ratio between square¹s diagonal and side is
sqrt(2), but the diagonal¹s length itself could
be some n in N, if we decide to make a standard
measure unit so.
We have no method to tell about some measure unit
if it is irrational or not. And the choose of
that measure is not in math, nature or god but
it is from human.
So. I accept only Natural numbers. Even i had
to deny rational numbers. Then there is not that
ratio between square and diagonal but only
natural numbers and some function out of them
which is not number but result.
--
amount and biggest
1+1+1+...= inŽnite and Žnite in math today
Petri Keckman
===
Subject: Re: i dont like irrationals
> Sorry, this is stupid, i agree. I just could not get sleep
> and wondered how to get rid of endless decimal numbers,
> which i don¹t like
> ---
> We say that there are irrational numbers like sqrt(2)
> because the ratio between square¹s diagonal and side is
> sqrt(2), but the diagonal¹s length itself could
> be some n in N, if we decide to make a standard
> measure unit so.
Then the length of the diagonal and the length of a side
could NOT be
measured with a common unit. The technical term is
incomensurable. There
is no common unit for the side of a square and its diagonal.
That is it.
Period. End of the matter.
> We have no method to tell about some measure unit
> if it is irrational or not. And the choose of
> that measure is not in math, nature or god but
> it is from human.
You have missed the point. No unit will work to commonly
measure the
side of a square and its diagonal. None. Zero. Zip. Nada.
Nichts.
Nichevo. Got it?
> So. I accept only Natural numbers. Even i had
> to deny rational numbers. Then there is not that
> ratio between square and diagonal but only
> natural numbers and some function out of them
> which is not number but result.
I take it you do not like solving quadratic equations
exactly? If you
restrict yourself to intengers and integer ratios you will be
unable to
solve some very simple equations. You wont have limits and
you wont have
calculus. That means you wont have physics. Is that really
what you want?
Bob Kolker
===
Subject: Re: i dont like irrationals
They don¹t like you either. In fact, they think you stink.
===
Subject: Triples correspond to sequences
For every triple of positive integers (a,b,c) there is
associated a sequence
{
(a^n + b^n) mod c }.
Is this a one-to-one correspondence?
With the condition a < b < c < (a+b), is this one-to-one?
With the addtional condition gcd(a,b,c), how about now?
Doug
===
Subject: Re: Triples correspond to sequences
> For every triple of positive integers (a,b,c) there is
associated a
sequence {
> (a^n + b^n) mod c }.
> Is this a one-to-one correspondence?
You expect to get all sequences this way? There are
uncountably
many sequences, but only countably many triples (a,b,c).
Or have I misunderstood your intention?
--
G. A. Edgar
http://www.math.ohio-state.edu/~edgar/
===
Subject: Re: Triples correspond to sequences
>With the addtional condition gcd(a,b,c), how about now?
With the additional condition gcd(a,b,c)=1, I meant to write.
>For every triple of positive integers (a,b,c) there is
associated a
sequence
>(a^n + b^n) mod c }.
For every triple of positive integers (a,b,c) there is
associated a
sequence
{ (a^n + b^n) mod c }
Darn this clunky news reader. I can¹t get Outlook and Outlook
Express to
work
together.
Yours,
Doug Goncz ( ftp://users.aol.com/DGoncz/incoming )
Student member SAE for one year.
I love: Dona, Jeff, Kim, Mom, Neelix, Tasha, and Teri,
alphabetically.
I drive: A double-step Thunderbolt with 657% range.
===
Subject: Re: Deep Thoughts # 17: Liar Paradox is a Formal
Metamathematical
Theorem
>the liar paradox is just the result of not considering time
>in the equation;
It sure doesn¹t look that way: there is no explicit reference
to time in
This statement is false. By what deeper analysis to do you
impute some

time reference?
>it is simply decategorized, as they say.
Eh? By it do you mean time, or the Liar sentence, or its
paradoxical
nature, or what? And whatever it is, how is it being
decategorized
at
all (which I take as being some po-mo jargon adapted from
bureaucratic
jargon, as in decategorized funds), much less simply? And who
are
they? Inquiring minds need to know.
> this is the same as with his Village Barber paradox.
Who¹s he? And again, where is the explicit or implicit
reference to
time
in the Barber paradox? And the Liar seems much more
intractable than the
Barber: the simple solution to the Barber paradox is that
there can not be
any such barber (just as the usual solution to the
Russell-set paradox is
that there can not be any such set). But applying that to the
Liar would be

saying that there can not be any such statement, but in
actual fact there it

is right before your eyes.
>what is a wff?
If you don¹t know that then you almost certainly don¹t know
much at all
about formal logic (beyond Aristotle), which IMO rather
disqualiŽes you
from having an informed opinion on the Liar and related
issues. But hey,
this is Usenet after all.
--
---------------------------
| BBB b Barbara at LivingHistory stop co stop uk
| B B aa rrr b |
| BBB a a r bbb | Quidquid latine dictum sit,
| B B a a r b b | altum viditur.
| BBB aa a r bbb |
-----------------------------
===
Subject: Re: Deep Thoughts # 17: Liar Paradox is a Formal
Metamathematical
Theorem
>> If, as it appears, you think provability is a property of
sentences, you
>> obviously don¹t have any idea what you are talking about.
>
> Of course it is. What¹s the problem?
>
> OK, is (x)(y)(fx = fy -> x=y) provable? Yes or no.
>
> It¹s a function of your axioms and rules (and deŽnitions.)
>
> C-B
> Which is exactly why provability isn¹t a property of
*sentences.*
> Œcid Œooh
for approval. He had only 2 or 3 questions. Then I noticed
that all
of his questions had to do with material from pages 1 and 2.
(I don¹t
know if its incompetence or laziness that makes people pass
judgment
without having really examined an issue.)
So your comment is that provability isn¹t just a function of a
sentence, but also of the axioms and rules used? Well, yeah,
I think
we all know that.
So how¹d you like my two Metamathematical theorems?
C-B
===
Subject: Re: Deep Thoughts # 17: Liar Paradox is a Formal
Metamathematical
Theorem
>> If, as it appears, you think provability is a property of
sentences, you
>> obviously don¹t have any idea what you are talking about.
>
> Of course it is. What¹s the problem?
>
> OK, is (x)(y)(fx = fy -> x=y) provable? Yes or no.
>
> It¹s a function of your axioms and rules (and deŽnitions.)
>
> C-B
> Which is exactly why provability isn¹t a property of
*sentences.*
> Œcid Œooh
for approval. He had only 2 or 3 questions. Then I noticed
that all
of his questions had to do with material from pages 1 and 2.
(I don¹t
know if its incompetence or laziness that makes people pass
judgment
without having really examined an issue.)
So your comment is that provability isn¹t just a function of a
sentence, but also of the axioms and rules used? Well, yeah,
I think
we all know that.
So how¹d you like my two Metamathematical theorems?
C-B
===
Subject: Re: Deep Thoughts # 17: Liar Paradox is a Formal
Metamathematical
Theorem
> If, as it appears, you think provability is a property of
sentences,
you
> obviously don¹t have any idea what you are talking about.
>>
>> Of course it is. What¹s the problem?
>>
>> OK, is (x)(y)(fx = fy -> x=y) provable? Yes or no.
>
> It¹s a function of your axioms and rules (and deŽnitions.)
> In other words, provability is not a property of sentences.
Congrats.
Typically the axioms and rules are Žxed.
C-B
===
Subject: Re: Deep Thoughts # 17: Liar Paradox is a Formal
Metamathematical
Theorem
>> If, as it appears, you think provability is a property of
sentences, you
>> obviously don¹t have any idea what you are talking about.
>
> Of course it is. What¹s the problem?
>
> OK, is (x)(y)(fx = fy -> x=y) provable? Yes or no.
>>
>> It¹s a function of your axioms and rules (and deŽnitions.)
>> In other words, provability is not a property of
sentences. Congrats.
> Typically the axioms and rules are Žxed.
Right, Žx axioms and rules and you get a notion of
provability. In
other words, provability is not a property of sentences.
Congrats.
Chris Menzel