mm-152
>Since what is true but unprovable depends on the
coding scheme (in>Godel's proof), it might be a question of
choosing a system where we>avoid including our intended
theorems (Number Theory) among the>undecidable sentences. But
the coding scheme is quite arbitrary (with>simple
requirements), pretty independent of the axiomatic
system>itself.This is not right. The speci'c example of an
unprovable (but true inthe intended model) statement found by
Goedel's method depends on thecoding, but the set of true
statements does not depend on the coding(only on the intended
model), and the set of provable statements onlydepends on the
formal language and the axioms (and the underlying
logic).Forty years after Goedel some non-contrived statements
of number theorywere found that are unprovable in elementary
(Peano) number theory, butare provable by trans'nite
induction
up to ordinals larger than can bedescribed using elementary
number theory (a crude characterisation ofthe situation,
sorry). The one I'm familiar with is
Goodstein's Theorem;I
think the Parris-Harrington Theorem of Ramsey Theory was
earlier.Michel.
I've never been able to decide whether
that should count.> The thing is, I'm not sure whether
I'm
badly confused about> second-order logic here - maybe someone
who knows can> say:> On the one hand it seems clear that the
second-order PA> axioms characterize N. On the other hand it
seems clear> that we can do second-order arithmetic in
'rst-order> set theory... (my guess is that the point is that
we're talking> about two different notions of second-order
arithmetic> here, but when I assume that then I get all
confused> over exactly what notion the 'rst sentence refers
to.)I don't see that you've asked any explicit
question here,
so Ihave to kind of guess, but I suppose what you're asking
issomething like ... so how come ZFC doesn't decide thetruth
value of all sentences of arithmetic?.One way of answering
that is to note that not all 'rst-ordermodels of ZFC
correctly
interpret the second-order semanticsof the natural numbers,
because certain subsets of the naturalnumbers of the model may
not be elements of the model. (Hereperhaps elements of should
be in quotes -- what I really mean is:Suppose N^M is the set
of all things that M believes to be natural numbers,and
suppose S is a subset of N^M, then there may be no object
S^Min M such that, for each element e of S, M thinks e is in
S^M).In fact, if N^M is not isomorphic to the real N, then N^M
mustbe illfounded, and its wellfounded part is a subset of
N^Mthat is not in M.
One way of answering that is to note
that not all 'rst-order> models of ZFC correctly interpret
the
second-order semantics> of the natural numbers, because
certain subsets of the natural> numbers of the model may not
be elements of the model. (Here> perhaps elements of should be
in quotes -- what I really mean is:> Suppose N^M is the set of
all things that M believes to be natural numbers,> and suppose
S is a subset of N^M, then there may be no object S^M> in M
such that, for each element e of S, M thinks e is in
S^M).Whoops, that's not quite right. ...there may be no
object
S^M inM such that, for each element e of M, e is in S if and
only ifM thinks e is in S^M.
How to prove that the
function> x -> inf_{t >= 0} { t in R, such that f(x,t)>0 }> is
continuous> if one knows that f is continuous with respect to x
and t ?> use an epsilon-delta proof
How to prove that
the function> x -> inf_{t >= 0} { t in R, such that f(x,t)>0
}> is continuous> if one knows that f is continuous with
respect to x and t ?> NicolasAre you sure that this the
claim you are to prove? It seemsto me that the claim is false
as stated. Think about thecollection U of pairs (x,t) such
that f(x,t). If f is a continuousfunction of the pair (x,t),
then U is open. But the boundary of U,may have parts that are
line segments parallel to the coordinate axes.This will result
in jumps for the value of your function.Mind you, a more
typical case giving similar behavior is that ofU consisting of
several components.Jyrki Lahtonen, Turku, Finland
>How
to prove that the function>x -> inf_{t >= 0} { t in R, such
that f(x,t)>0 }>is continuous>if one knows that f is
continuous with respect to x and t ?>use an
epsilon-delta proof> What about: f(x,t) = t^2-t+x,
investigate continuity at x=0.> I should miss something since
:inf{t^2-t+x} = inf{t^2-t} + x = -0.25 + xwhich is obviously
continuous wrt x ...However, i have to specify that- f(x,t) is
in R- for all t in R and x in R^n, f(x,t) exists- for all x in
R^n, lim_{t -> infty} f(x,t)=+inftyin other words, the inf is
a real number that always exists
=Jyrki Lahtonen a .8ecrit:>How to prove that the function>>x -> inf_{t >= 0} { t in
R, such that f(x,t)>0 }>>is continuous>>if one knows that f is
continuous with respect to x and t ?>>Nicolas> Are you
sure that this the claim you are to prove? It seems> to me
that the claim is false as stated. Think about the> collection
U of pairs (x,t) such that f(x,t). If f is a continuous>
function of the pair (x,t), then U is open. But the boundary
of U,> may have parts that are line segments parallel to the
coordinate axes.> This will result in jumps for the value of
your function.> Mind you, a more typical case giving similar
behavior is that of> U consisting of several components.>
Jyrki Lahtonen, Turku, Finland
>How to prove that the
function>x -> inf_{t >= 0} { t in R, such that f(x,t)>0
}>is continuous>if one knows that f is continuous with
respect to x and t ?>use an epsilon-delta proof>
What about: f(x,t) = t^2-t+x, investigate continuity at x=0. I should miss something since :> inf{t^2-t+x} = inf{t^2-t} +
x = -0.25 + x> which is obviously continuous wrt x ......which
is not at all what the original poster asked...I interpreted
the question like this:g(x) = inf_{t>=0} {t in R, such that
t^2-t+x > 0},g(0) = inf_{t>=0} {t in R, such that t^2-t > 0} =
inf {t such that t > 1} = 1,but if x>0 theng(x) = inf_{t>=0} {t
in R, such that t^2-t +x> 0} = 0 since at least 0^2-0+x > 0.--
G. A. Edgar http://www.math.ohio-state.edu/~edgar/
=Could
any reader please recommend a good introduction level book on
FourierTransformations. My maths is a reasonable level but
rather rusty.TIAEd Walsh
=I'd be interested to see an
elementary proof of the identity:2_F_0 (n,a;c) = (c-a)_(-n) /
c_(-n)I've got a proof of Chu-Vandermonde's
identity, which
relies on the binomialidentity: 1/(1-x)^a = sum((a)_n*x^n/n!)
but I can't see how to generalize inorder to prove
Gauss's
hypergeo. th. (is the method totally different?)thx for any
idea,--Julien Santini,Universit.8e de Provence,France.
=I'd
be interested to see an elementary proof of the identity:2_F_1
(n,a;c) = (c-a)_(-n) / c_(-n)I've got a proof of
Chu-Vandermonde's identity, which relies on the
binomialidentity: 1/(1-x)^a = sum((a)_n*x^n/n!) but I can't
see how to generalize inorder to prove Gauss's hypergeo. th.
(is the method totally different?)thx for any idea,--Julien
Santini,Universit.8e de Provence,France.
=This problem,
posted a week or so ago is still unsolved. I've had
somehelpful responses from the sci.math group but nothing
leading me to asolution so far.I should point out that I am
not a mathematician myself although I haveseveral close
friends who are maths graduates from years ago. They
haven'tcome up with the answer but at least they can help me
to understand yours!The problem involves 'nding the position
of an archery arrow as it strikesthe target using three
sensors around the circumference.My description of the problem
is at www.telco.demon.co.uk/targ/target.jpgand it is for a
help received,Paul
=[..]> All right, I am probably going
over already-covered ground here, but> I'll leave it to
you-all to decide if this is a useful or useless>
revisitation. Let's say I want to make a mathematical model
of
plant> growth which would permit me to predict, from the exact
geometric> properties of a seed, the exact shape and maximum
size of the mature> plant that the seed will yield. I spend
years in observation and> formulate a progression of such
models, trashing or revising them> whenever my observations
disagree with them. Finally, I have a model> that neither my
observations, nor the observations of anybody else so> far,
has been at odds with. Now: I would say that this has indeed
yielded> some new information. Not only has it yielded some
new predictive> capabilities, but more importatntly, to put it
crudely, it has possibly> yielded an insight into the way
nature works in a certain area. And it> has certainly yielded,
through all my discarded models, insight as to some> ways in
which nature _doesn't_ work.>> I am curious to understand
how
one would argue, instead, that the only> thing my model of
plant growth does is serve to make explicit the> conventions
surrounding the mathematical terms in scienti'c
statement.Your
model is just that - model and not what its modelling - and in
order tobe a perfect model it should be impossible to tell the
difference- but amodel must by convenience leave something out.
The study of twins offersnear perfect models - but they are not
*identical*. If your model wascorrect you would arrive at a
determined universe- i think this is no longerthought
possible- your model would have to model quantum
indeterminateeffects. The reason you can be con'dent of your
model lies in thedeterminism of mathematics? If mathematics
isn't determined then why botherto use it as a model?
Whatever
- i think you make another mistake - ofbelieving your model has
a *relationship* with nature- explains what natureis. This is
surprising as scientists used to go to great pains to
avoidothers making this mistake. I'll upset Ted and quote
Sir
James Jeans...science cannot hope ever to discover the true
nature of the ingredientsof the material universe; from the
nature of things, this lies beyond ourNot only this but there
are some underlying philosophical problems with thekinds of
knowledge you are offering.But to answer your question your
model uses the nature -the substance ofmathematics so as you
need not worry about it too much- (its guaranteedapriori)
unlike modelling it in cream cheese or the i ching, gut
feelingsetc. The problem will be in putting the non formal
system into the formalone - and this is insurmountable. BTW
this is i think the best picture- ifmathematics is not a
prirori you'll have something else to worry about.
=Is A.
Grothendieck still alive? Does he have any contacts with
individual mathematicians?Someone should write up his life
story. It would be just as or more interesting than John
Nash's (The Beautiful Mind), and his contributions to the
20th
century mathematics are much more important, even if he
doesn't
have a Nobel prize to show for it.Nemo
Is A. Grothendieck
still alive? Does he have any contacts with > individual
mathematicians? I was reading about this recently. Apparently,
he lives in a cabin ofsorts in the Pyrenees, claiming that the
devil is out there and conspiringto change things - like the
speed of light, from a round value of 300,000km/s to the ugly
299,795 km/s. I don't know how current or accurate this info
is though.
Is A. Grothendieck still alive? Does he have
any contacts with > individual mathematicians?> Someone
should write up his life story. It would be just as or more >
interesting than John Nash's (The Beautiful Mind), and his
contributions > to the 20th century mathematics are much more
important, even if he > doesn't have a Nobel prize to show
for
it.There is no Nobel prize for mathematics. The closest is the
Field's Medal and the awarding is age limited.Bob Kolker
Is A. Grothendieck still alive? Does he have any contacts with
> individual mathematicians?> Someone should write up his
life story. It would be just as or more > interesting than
John Nash's (The Beautiful Mind), and his contributions > to
the 20th century mathematics are much more important, even if
he > doesn't have a Nobel prize to show for it.in the
Pyrranes. He received the Fields Medal in 1966 so, in a sense,
he got his Nobel PrizeI do not see how Grothedieck's work in
agebraic varieties, algebraic geometry and category theory
could be as useful as Nash's work in game theory which has
applications to economics.Bob Kolker
> Is A. Grothendieck
still alive? Does he have any contacts with >> individual
mathematicians? Someone should write up his life story.
It would be just as or more >> interesting than John Nash's
(The Beautiful Mind), and his contributions >> to the 20th
century mathematics are much more important, even if he >>
doesn't have a Nobel prize to show for it.>> in the
Pyrranes.
He received the Fields Medal in 1966 so, in a sense, he > got
his Nobel Prize>> I do not see how Grothedieck's work in
agebraic varieties, algebraic > geometry and category theory
could be as useful as Nash's work in game > theory which has
applications to economics.Maybe it could be as useful because
all of those things[1] haveapplications in theoretical
computer science, including algorithmveri'cation? Moreover,
these applications are considerably lesscontroversial than the
application of game theory in economics.Footnotes: [1] Well,
I'm not so sure about how directly algebraic
geometryapplies.
-- Jesse Hughes [I]f gravel cannot make itself into an animal
in a year, how could itdo it in a million years? The animal
would be dead before it gotalive. --The Creation Evolution
Encyclopedia
Is A. Grothendieck still alive? Does he have
any contacts with > individual mathematicians?> You might check
out:http://www.fermentmagazine.org/home5.htmlwhich used to have
some snippets from G.'s autobiography, but now
it's for paying
folk only, it seems. All the same, some of the links from this
page have more info.Also, the life in the Pyrenees and
obsession with the devil noted by it's on the archive.>
Someone should write up his life story. It would be just as or
more > interesting than John Nash's (The Beautiful Mind),
and
his contributions > to the 20th century mathematics are much
more important, even if he > doesn't have a Nobel prize to
show for it.> pursuit of stacks would make even most
mathematicians bored. Speaking as an algebraic geometer whose
left arm is currently resting on an open copy of G.'s Tohoku
paper, I wouldn't say that his mathematics has the sort of
universal appeal that Nash's work tied to econ does. Of
course, I'm more interested in Nash's work on
real algebraic
structures on manifolds and the embedding theorem for
Riemannian manifolds, but... Maybe if EGA had more
pictures...> Nemo>
C-113opstall@math.washington.eduhttp://www.math.washington.edu
/~opstall/
I do not see how Grothedieck's work in
agebraic varieties, algebraic> geometry and category theory
could be as useful as Nash's work in game> theory which has
applications to economics.I very much doubt if Grothendieck
would want to be useful in your sense.He is/was the greatest
mathematician of the 20th century,transforming geometry and
algebra.-- Timothy Murphy tel: +353-86-233 6090
= >>
Is A. Grothendieck still alive? Does he have any contacts with
>> individual mathematicians?  Someone should write up his
life story. It would be just as or more >> interesting than
John Nash's (The Beautiful Mind), and his >> contributions
to
the 20th century mathematics are much more >> important, even
if he doesn't have a Nobel prize to show for it. > the in
the Pyrranes. He received the Fields Medal in 1966 so, in a >
sense, he got his Nobel Prize > I do not see how
Grothedieck's work in agebraic varieties, algebraic >
geometry
and category theory could be as useful as Nash's work in >
game
theory which has applications to economics. > Bob Kolker >Now
of course it's always debatable whether one
mathematician'scontributions are more important than
another's. So if you don't agreewith me then I
am not going to
convince you.Consider this: Even if you skip over AG's
contributions to abstractanalysis (theory of locally convex
spaces), his work in algebraicgeometry revolutionalized the
'eld, led to the solutions o§ong-standing open problems,
and
opened new areas that hundreds ofmathematicians are still
exploring today.The shape of modern (i.e. last 30 years)
mathematics research would beconsiderably different without
the concepts introduced by AG. You cansay that someone else
would invent those concepts if AG didn't,
butthat's another
discussion.N
>I do not see how Grothedieck's work in
agebraic varieties, algebraic>>geometry and category theory
could be as useful as Nash's work in game>>theory which has
applications to economics.> I very much doubt if
Grothendieck would want to be useful in your sense.> He is/was
the greatest mathematician of the 20th century,> transforming
geometry and algebra.There are multiple metrics of greatness.
For example Wiles will have to be rated highly, in some sense,
for slaying Fermats Last Theorem which has been a thorn in the
side of mathematics for 400 years.Where would Erdos 't in?For
pure breadth of vision of practice you would have to rate von
Neuman logic are considerable and he and Goldstein made
notable contributions to the modern stored program computer
which is called in the trade a von Neuman machine ( a generic
description as is Turing machine).Bob Kolker
all,Please help me on this integral:Integrate(tan(t)^(1/2), t
from 0 to pi/2)I doubt it is integrable... :=(-Walala
Integrate(tan(t)^(1/2), t from 0 to pi/2)> I doubt it is
integrable... :=(Make a change of variable by letting tan(t) =
x. Residue theory will then come to the rescue. Watch out for
fractional powers.
=Walala,> Please help me on this
integral:> Integrate(tan(t)^(1/2), t from 0 to
pi/2)Travis
integral:> Integrate(tan(t)^(1/2), t from 0 to pi/2)> I
doubt it is integrable... :=(> -Walala> Of course it is
integrable. The integrand is continuous on [0,pi/2],and any
continuous function is integrable.-- G. A. Edgar
http://www.math.ohio-state.edu/~edgar/
The integrand is
continuous on [0,pi/2]????
on this integral: Integrate(tan(t)^(1/2), t from 0 to
pi/2) I doubt it is integrable... :=(> -Walala>Of course it is integrable. The integrand is continuous on
[0,pi/2],>and any continuous function is integrable.Actually,
it's integrable on [0,pi/2] since it is continuous on
[0,pi/2),and it is asymptotic to (pi/2 - t)^(-1/2) as t
approaches pi/2 frombelow.David
McAnally--------------
=There is also Shakespeare & Co. ,
but I am not sure if they have mathtextbooks. It is across the
Seine from Notre Dame .I think it is mostlyliterature.Lurch>> I
might have a friend traveling to France (Paris) soon and I
wanted to> get him to buy me W. Rudin's Real and Complex
Analysis (the> international small red and cheap paperback
edition, not the ugly very> costly green one), but I don't
know what are the bookstores in Paris> where he might 'nd it.
I searched the net, but couldn't 'nd any>
concrete
information.>> Any suggestions/addresses/etc ?>
nojb.
g(x, y):g_x == g_yi.e., g has the same partial derivative with
respect to x and with respectto y...-walala
What's the name of this kind of function g(x, y):> g_x ==
g_y> i.e., g has the same partial derivative with respect to
x and with respect> to y...> -walalaAre there any such
functions other that of form g(x,y) = f(x+y)?
=I also
suspect that... but it seems that I need support with a
rigorousproof...I spent like several hours on this problem...I
now clear that I want not only g_x == g_y, but also all the
second orderpartial derivative equals 0... for instance, g_xx
== 0, g_xy == 0, g_yy ==0...I need to 'nd all these kind of
this kind of function g(x, y):>> g_x == g_y>> i.e., g has the
same partial derivative with respect to x and withrespect> to
y...> -walala>> Are there any such functions other that of
form g(x,y) = f(x+y)?
Are there any such functions other
that of form g(x,y) = f(x+y)?Suppose g is differentiable in
the two variable sense and g_x = g_y identically. Let c be a
constant and consider x -> g(x,c-x). Taking its derivative wrt
x gives g_x(x,c-x)*1 + g_y(x,c-x)*(-1) = 0. So g is constant on
each line x + y = c, and that gives your result.
>I also
suspect that... but it seems that I need support with a
rigorous>proof...>>I spent like several hours on this
problem...>>[SNIP]You spent like several hours... Like, wow.
like, dude. you arelike really working like really hard.You
know, I really hope you get run over by a big truck.That way
you will like stop like posting like on
sci.like.mathbiaaaatch.
the name of this kind of function g(x, y):>>g_x == g_y>>i.e.,
g has the same partial derivative with respect to x and with
respect>to y...I don't think there is a special name for it.
I
believe that it is onlypossible if g(x,y) is of the form G(x+y)
for some function G of onevariable.--Daryl McCulloughIthaca,
NY
OK, I guess what I meant is that if we were to write
Sb(x v|y) in the> formal language we would be writing out a
long string of symbols and> not just a shortened version like
Sb(x v|y). Or do we merely write out> Sb as a variable of type
2 and x, v and y as variables of class 1 so> the formula would
be something like:> 17^2(19, 23, 29)> where Sb = 17 x = 19 v =
23 y = 29 > which in formal language would be> sssss[17^2
times]0 ([19 s]0, [23 s]0, [29s 0])> and then replace the s, 0
and brackets with there corresponding> numbers and put them as
the powers of the primes in the order in which> they occur so
that we calculate the Godel number. I am kind of> confused on
this point as you may have gathered.> Okay. First, remember
that there is a difference between a formula> and a name for a
formula.> A(x,y): x + ssss0 = y> Here's a good example. The
formula A(x,y) really is the sequence of> symbols: x + ssss0 =
y. Here the A(x,y) is the name and x +> ssss0 = y is the
formula.> The same thing happens with Sb. Sb(x v|y) is the
name; the actual> formula is a rather lengthy sequence of
symbols, which will include> the symbols x, v, and y
somewhere.> The Godel number of a formula has nothing to do
with any names you> might give the formula -- it only depends
on the sequence of symbols. > So the Godel number of the
formula named Sb(x v|y) would be the same> even if you decided
to call that formula Babe the big blue ox> instead.> You also
need to remember the difference between a variable in the>
formal language and a genuine number. x is the symbol for a>
variable. You use p as the Godel number of a formula, which
means> that p is a genuine number, not a symbol at all. z(p)
is the Godel> number of a formal expression: the sequence of
symbols consisting of p> ss followed by a 0.> Finally, keep
in mind what is the result of substituting an expression> for
a variable in a formula. Given that A(x,y) is the formula x +>
ssss0 = y, the result of substituting ss0 for y would be the>
formula x + ssss0 = ss0, which we could call A(x,ss0).> The
operation on formal strings that takes x + ssss0 = y, y, and>
ss0 as arguments and yields x + ssss0 = ss0 as a result, can
be> translated into an operation on Godel numbers. This
operation takes> the Godel number of x + ssss0 = y, the
variable number for y, and> the Godel number of ss0 as
arguments and yields the Godel number of> x + ssss0 = ss0 as a
result. This purely numerical operation can be> described by a
formula, and that formula is Godel's Sb. In other> words,
the
operation described by the formula Sb(x v|z) can be>
summarized by: 'nd the formula whose Godel number is x, in
that> formula replace each occurence of the variable numbered
v with the> expression whose Godel number is z, and then take
the Godel number of> the resulting formula.> This operation
does not involve performing any substitutions within> the
expression whose Godel number is z.> ok I'm pretty sure
I've
got all this so far.> i guess by the formal representation of
p i meant ssssss...0 with> the s repeated 19 times. Part of my
confusion stems from whether we> actually write out the
substituted formula y in the formula that we> are substituting
into or not.> This is getting confusing so I'll try to
demonstrate what I mean.> Without using the sssss0s and
stuff.> Say we have a formula> y != 47 with Godel Number of
879> and we have a formula > x = Sb(p y z(p)) with Godel
number 9657> Now we evaluate > Sb(9657 p z(879))> This
question doesn't make sense. You haven't said
what the value
of> p is, so> it's impossible to evaluate the expression.
Let
me change the> question a little: evaluate Sb(9657 q z(879)),
where q is the variable> number of p. I think that's what
you
had in mind.> Thats kind of why it confused me in the original
paper. Godelconstructs formulas such as Q(y): y = Sb(y 19
z(y)). Being that the yis a free variable it would be
impossible to evaluate the resultingGodel number of the
formula with G number y when its free variableshad been
replaced by z(y) because y is a free variable and
thusdoesn't
code for any real formula, with or without a free variable
of19. Given that anoter poster said that Sb(x y z) was a
numbertheoretic function, and thus every time we see
Sb(whatever) we are toreplace it with the Godel number that
results from following arecursive series of instructions, I
don't see how we can follow theinstructions if the formula
that we are to act on is a free variable.language. I am still
confused.> Do we actually write out> x = Sb(z(879) y z(z(879))
which is equal to> OK i'm not sure what that is equal to.
Perhaps it is the distinction> between the formal language and
what Godel has written his formulas> out as that is confusing
me. I'm real lost now.> You would write out the formula with
Godel number 9657, i.e., x  Sb(p y z(p)). Then you would
substitute the expression whose Godel> number is z(879) (i.e.,
ssss...[879 of em]0) for each occurrence of> the variable whose
number is q (i.e., for each occurrence of p). > The resulting
formula is> x = Sb(ssss...[879 of em]0 y z(sss...[879 of
em]0))> according to the other post by Pierre the Sb functions
are notobjects of theformal system P, so Godel does not assign
Godel numbers to functions.so we would then have to go on to
substitute z(ssssss[879 of em]0) forevery occurence of y in
the formula of Godel number ssssss[879 ofem]0. However the Sb
function and the z function are now taking asarguments not
numbers as they were originally but numerals asrepresented in
the formal system (ssss...0). This is what we end upwith when
Godel substitutes z(p) for every y variable in the formulaof
Godel number p of a formula which contains the function Sb(y
19z(y)).Is this allowable for a function to take numbers as
arguments andnumerals later.Am I completely off track
here.Appreciate your assistance.> which we could also write as
x = Sb(879 y z(879)). The 'nal value> is the Godel number of
this formula.> I hope this clears things up.> Alan
Stern
Don't mean to clutter up the place with another i
found a §aw in> Godels undecidable propositions proof post
because that is not what i> am claiming. This is more of an
this is confusing and bugging the> hell out of me please
explain post and I would appreciate if anyone> could take the
time to clarify a few things for me.> Welcome to the group. Anyways what interests me is what Godel calls Sb(x v|y)
which refers> to the formula that results when every occurence
of the free variable> v in the formula encoded by the numeral x
has been replaced by the> symbol y.> Or the Godel number of
the formula. You have to be very careful here.> These aren't
the same thing. I don't have the paper in front of
me,I'm
going by the one on the net
athttp://home.ddc.net/ygg/etext/godel/godel3.htmif that helps.
It seems more or less legit to my eyes.> but IIRC a FORMULA is
the Godel number of a formula. In general, when> Godel uses
small capitals, he is talking about Godel numbers.Actually
this was a subtlety that I was having dif'culty keepingtrack
of. You have made it a lot clearer.> Thus, VARIABLES, TERMS,
FORMULAS, SENTENCES are all numbers. These numbers> have
*numerals*, which are long strings of Os'
symbols followed by
a> O0' symbol. These numerals, being objects of
the formal
system,> have Godel numbers too and sometimes you need the
NUMERAL of a number.> With you so far.> Then there are
primitive-recursive functions (just recursive in> the paper).
These are number-theoretic functions, taking numbers as>
arguments and returning numbers. These functions are not
objects of the> formal system P, so Godel does not assign
Godel numbers to functions.> There are no FUNCTIONS, but there
are formulas (lower case) that represent> functions in the
formal system in a precise sense explained in the paper.> OK
this is good because I was never sure exactly how to
approachthese. Sometimes they seemed to be like a computer
function as youdescribed, taking numbers as arguments and
returning numbers but other times it seemed lessclear. Does
this include all those n Gl x type bits numbered 1 - 45that
Godel terms functions(relations).If that is so when we see
function 1: x/y as part of a formula oranother function do we
replace it with a 1 or 0 to indicate the truthor falsity of
the statement. I seem to recall this being doneelsewhere in
the paper. Or do we replace it with (Ez)(z <= x & x =y.z) and
then work out the Godel number from these symbols. That
seemsdifferent from the number in, number out concept but
would that beone of the formulas (lower case) that represent
functions in the formal system in a precise sense explained in
thepaper that you mentioned. Jesus this is getting complicated.
If thatis the case when do we replace the function with a
number in a formulaand when do we use the formula to represent
it in the formal system.The formula representation is i suppose
a way of getting a FUNCTIONversion of a function?> If I
remember correctly, Sb() is a function. It takes numbers x,>
v, y. More precisely, a FORMULA x, a VARIABLE v, and a TERM y.
Sb()> returns the FORMULA that obtains when the TERM y is
SUBSTITUTED for> every FREE INSTANCE of the VARIABLE v in the
FORMULA x. This is the> number-theoretic de'nition, using
only
number-theoretic concepts like> VARIABLE, SUBSTITUTION and the
like, the only de'nition that can be> formalized in system P,
which speaks only about numbers. Well, numbers> and sets of
numbers, but certainly not formulas.> Once again i am with
you.> Now presumably to be able to encode SB(x v|y) as a
Godel number it is> necessary to write down the formula x and
replace all vs with ys and> Sb(x v|y) he is writing shorthand
for what in the formal system would> be a long string of
symbols.> No, I think Sb(x v|y) is a large number. You and I
are not system P,> we can unwrap the Godel numbering and
understand the formulas, variables,> terms directly. My
reading of Sb(x,v|y) in these terms is> 1) Decode x (a
number) into the formula of which it is the> Godel number. If
x is not a FORMULA, this step fails and> Sb(x,v|y) is
unde'ned.Here is one of my main problems. Godel uses formulas
such asQ(x,y): ~{B[Sb(y 19|z(y))]}Godel later talks about a
relation sign q for this formula (by which Ipresume he means
what has become known as a Godel number) meaning thathe has
worked out the Godel number for Q. But in doing so, when we
getto the part of Q(x,y) that is Sb(y 19|z(y)) and we try to
follow theabove steps we 'nd that y is a free variable and
hence not a FORMULA,and therefore Sb(y 19|z(y)) is unde'ned
as
you point out and thismakes it impossible to 'nd the Godel
number q for Q(x,y) which isused later in the proof. Would
this be where we use the formularepresentation of the fuction
to represent Sb(y 19|z(y)) in the formalversion of Q(x,y). I
could see how this might get around the problemif it is
possible but it seems that Sb() is sometimes a FORMULA
andother times not quite the same and there is no explicit
instruction asto which to use, unless I've missed
saomething.>
2) Decode v (a number) into the variable symbol of which it is>
the Godel number. If v is not a VARIABLE, Sb(x,v|y) is
unde'ned.> 3) Find all the free instances of the variable (2)
in the> formula (1).> 4) Decode y into the term of which it is
the Godel number. If> y is not a TERM, this steps fails and
Sb(x v|y) is unde'ned.> 5) Substitute the term found in (4)
in
all the slots> found in (2). The result is a long formula.> 6)
Compute the Godel number of (5). This is the value> of
Sb(x,v|y).> That is pretty much how I see the process going
also if Sb() istreated as a number in number out formula.>
Now Godel goes on to describe a fromula Q(x,y) that involves a
term Sb> (y 19|z(y)) ie the substitution of the formal
representation of the> Godel number y (sssssss......sssss0,
where s is a successor symbol)> into a formula of Godel number
y.> Right. And z() must be the function that maps every
number to its> NUMERAL. z(4), the NUMERAL of 4, is the Godel
number of Ossss0', already> a big number.
I'm guessing that
Q(x,y) is the formula (not FORMULA)> that represents the
primitive-recursive relation x is not a PROOF of> Sb(y,
19|z(y)).Exactly right.> Anyway there is a formula with
Godel number p that is the Q(x, y)> formula generalised for
all X. This is equivalent to the formula /x> Q(x, y) and let
us call it P(y).> So P(y) is the formula that represents
Sb(y, 19|z(y)) is not PROVABLE.> Yep.> Godels undecidable
proposition is Sb(p 19|z(p)) or the substitution of> the
formal representation of the Godel number p, into the formula>
encoded by the number p.> Right.> In order to 'nd the Godel
number for this> proposition which all such propositions in
this formal system must> have we have to write out the formula
encoded by p and then substitute> sssssssss....0 (s repeated p
times) into the places occupied by 19> the variable whose
code is 19> and then encode it via the Godel numbering
strategy.> Right. Although we will want to pause just before
the 'nal Godel> numbering to look at the formula.> Sounds
good
so far.> In writing out p we> get a formula that uses the
term Sb(y 19|z(y))> Ok, I don't think the formula P(y)
contains Sb(y 19|z(y)) because that> is a number-theoretic
function of one variable and system P does not> speak about
functions. P(y) must be something like> Exists t
<Big_Mess>(y,t) and <Other_Big_Mess>(x,t)> where
<Big_Mess>(y,t) is the formula that represents the binary
relation> t= Sb(y, 19|z(y)) and <Other_Big_Mess>(x,t) is the
formula that represents> x is not a PROOF of t. Represents in
the sense given in the paper.> OK so we can write out Exists t
= Sb(y 19|z(y)) to get around theproblem? And presumably we
expand the recursive formula de'niton ofSb(y 19|z(y)) in
terms
of the other functions it is de'ned as and weget a loooooong
string of formal symbols (Big Mess) that we canGodelize into a
number. and then when we substitute in the numeral ofp to this
formula we are replacing the occurences of y in the bigmess. since y is symbolised> in the formal language as 19 by Godel
in his paper we must replace the> variables 19 in the formula
encoded by p with the formal> representation of p> the
numeral of p.> yes. I haven't always been to clear on that
distinction.> and so we write out p and we get a formula
using> the term Sb(y 19|z(y)) etc ad in'nitum.> No, we write
out the numeral of p, a string of p symbols Os'
and one> O0'
symbol.> In fact how can you have the formal
representation of the terms Sb(y> 19|z(y)) with the symbols
de'ned by Godel as the formal language when> y itself is a
free variable. What would that look like?> Like a
<Big_Mess>, see above.> I mean if 437 was the Godel number
for y = ssss0 (ie y = 4) and Godel> had written Sb(437 y|z(4))
then we could express Sb(437 y|z(4)) as> ssss0 = ssss0 in the
formal language.> Godel number of the (false) formula
ssssssss(437 of Oem)ss0 = ssss0 .> yes. But if 3895 is the
Godel number ofSb(x y|z(x))and the Sb is represented as the
Big mess formula representation of afunction. ie something
likeEt ( t= y <x.z & x >................................) or
whateverand we then were to calculate the Godel nmber of
substituting 437 intothis formula ie Sb (3895 y |z(437)) then
wouldn't this give us adiffernet Godel number to the FORMULA
we would obtain if we actuallycalculated Sb(437 y|z(437)) by
the algorithm you gave above.In other words, by the algorithm,
Sb(437 y|z(437)) would give us ssssssss(437 of Oem)ss0 = ssss0
as you point outwhereas substituting 437 into the big mess
version of 3895 would giveEt (t = ssssss(437 of Oem)0 <x.z & x
>.........................)which would have to have a different
Godel number even though theyboth seem to represent the same
FORMULA Sb(3895 y |z(437))? whichseems weird. So there is a
difference between the FORMULA obtainedwhen substituting into
the formula representation of a function andbetween
calculating the function itself. Which would mean that it
isnot like a computer function where you can make recursive
calls to thefunction and the function will be called again
with diffferent valuesetc. When there is a free variable y in
Sb(y x |z(y)) we must use aformula representation and when
there is not we follow the algorithmto obtain the Godel
number. am I on the right track?> But when we have Sb (y
19|z(y)) and y is a variable as in the formula> denoted by
Q(x,y) above how do we express this in the formal language> as
given by Godel. We do not know what y codes for, as y is a>
placeholder so we cannot go through with the process of
substitution> as we do above. It seems like an unexpressible
statement in the formal> language.> By representing in
system P the number-theoretical function> y --> Sb(y,19|z(y))
through a formula. <Big_Mess> above.> OK so where have I
messed up? Can someone explain both the in'nite> expansion
problem and the expression of a substitution into a>
non-exitant formula in the formal language to me.> Ok, you
just need to read the paper slowly and carefully, like all of
us.> Pay attention to the SMALL CAPS because they matter. But
especially,> pay *very, very close* attention to the theorem
that explains what it> means for a formula to represent a
function. It's a very weak> notion of representation, but
just
strong enough for the proof to> carry through. That's where
your main gap is.a lot of things in this paper a lot clearer
though it seems i'm stilla little behind.
Say we have a
formula> y != 47 with Godel Number of 879> and we have a
formula > x = Sb(p y z(p)) with Godel number 9657> Now we
evaluate > Sb(9657 p z(879))> This question doesn't make
sense. You haven't said what the value of> p is, so>
it's
impossible to evaluate the expression. Let me change the>
question a little: evaluate Sb(9657 q z(879)), where q is the
variable> number of p. I think that's what you had in mind.>
Thats kind of why it confused me in the original paper. Godel>
constructs formulas such as Q(y): y = Sb(y 19 z(y)). Being
that the y> is a free variable it would be impossible to
evaluate the resulting> Godel number of the formula with G
number y when its free variables> had been replaced by z(y)
because y is a free variable and thus> doesn't code for any
real formula, with or without a free variable of> 19.Yes. If I
were to give you a value for y, then you could go ahead
andevaluate it. The same is true for the formula y + s0; you
can'tevaluate that until I tell you what the value of y
is.But
you don't have to evaluate anything to understand what Q(y)
is;it's a formula in which y occurs as a free variable.
It's
just asequence of symbols.> Given that anoter poster said that
Sb(x y z) was a number> theoretic function, and thus every time
we see Sb(whatever) we are to> replace it with the Godel number
that results from following a> recursive series of
instructions, I don't see how we can follow the>
instructions
if the formula that we are to act on is a free variable.I'm
sure the other poster didn't say _that_ exactly.
There's
aconfusion here between the function and the formula that
expresses thefunction. For example, addition is a function and
it can berepresented by the formula x+y. So while you may feel
perfectlyfree to replace 2 added to 2 by 4, you shouldn't
feel
free toreplace ss0 + ss0 with ssss0 -- they are distinct
formulas.To help keep things straight, let's use separate
words for thefunction and the formula. So let's say that
Sub(x
y z) is anumber-theoretic function and Sb(x y z) is the
corresponding formula. Whenever you see Sub(x y z), feel free
to replace it with the Godelnumber resulting from following
the instructions. Don't feel free todo something similar
with
Sb(x y z).> ie Sb(897 x z(4))> where 897 is G number of x =
ssss0 > and the formula itself:> ssss0 = ssss0> I also got the
impression that the Sb(x y z) was more or less> analogous to a
computer function, where the Sb() is always replaced by> a
number.No -- Sub is analogous to a function (it _is_ a
function). Sb() ismuch more like a computer _program_. I
don't
know if this analogywill help, but let's try it. The
programs
PRINT 3+5and PRINT 8are equivalent as functions, since they
always produce the sameoutput. But they aren't the same
program -- they contain differentnumbers of characters, for
one thing.> But if there is a free variable in the function
then we> cannot evaluate the function, hence determine the
number and then> substitute it into the formula to make a
proposition in the formal> language. I am still confused.A
free variable is a lot like an argument to a program. Consider
program will do when you run it, because youdon't have any
value for x. That's like saying you can't
evaluate aformula in
which x occurs as a free variable. But it doesn't stopyou
from
doing things like counting the number of characters in
theprogram. Or from 'guring out what program would result if
youchanged each occurrence of the O+' sign to a
O-' sign.> Do
we actually write out> x = Sb(z(879) y z(z(879)) which is
equal to> OK i'm not sure what that is equal to. Perhaps it
is
the distinction> between the formal language and what Godel has
written his formulas> out as that is confusing me. I'm real
lost now.> You would write out the formula with Godel number
9657, i.e., x  Sb(p y z(p)). Then you would substitute the
expression whose Godel> number is z(879) (i.e., ssss...[879 of
em]0) for each occurrence of> the variable whose number is q
(i.e., for each occurrence of p). > The resulting formula is x = Sb(ssss...[879 of em]0 y z(sss...[879 of em]0))>
according to the other post by Pierre the Sb functions are
not> objects of the> formal system P, so Godel does not assign
Godel numbers to functions.> so we would then have to go on to
substitute z(ssssss[879 of em]0) for> every occurence of y in
the formula of Godel number ssssss[879 of> em]0.No, you
don't
have to go on and substitute anything. Look, supposeyou use a
utility like the Unix Otr' program to replace
eachwould be
program and then run into theproblem of not knowing what x is.
The Sub function behaves in thesame sort of way. It just
manipulates strings of symbols (or rather,their Godel
numbers). It doesn't need to evaluate anything.> However the
Sb function and the z function are now taking as> arguments
not numbers as they were originally but numerals as>
represented in the formal system (ssss...0). This is what we
end up> with when Godel substitutes z(p) for every y variable
in the formula> of Godel number p of a formula which contains
the function Sb(y 19> z(y)).> Is this allowable for a function
to take numbers as arguments and> numerals later.Here you need
to think of Sb as a formula, not a function. It's justa
sequence of symbols. There's no problem with replacing some
ofthose symbols with strings of other symbols. And the
originalarguments weren't numbers, BTW, they were also
symbols.> Am I completely off track here.> Appreciate your
assistance.Yes, you are a bit off track. Let's take the
example you mentionedearlier: Q(y): y = Sb(y 19 z(y)).Or less
concisely but more accurately: Q(y): y = Sb(y
sssssssssssssssssss0 z(y)).This formula Q(y) has a Godel
number; let's call that number p. Let'salso
assume that the
variable number for y is 19. Now if we want,we can get a
closed formula (no free variables) by substituting thenumeral
of p for y. The resulting formula is Q(p): ssss...[p of em]0 =
Sb(ssss...[p of em]0 19 z(ssss...[p ofem]0)).Or in shorter
notation this is Q(p): p = Sb(p 19 z(p)).Since there are no
free variables, we can evaluate this formula to seewhether it
is true or false. The value of Sb(p 19 z(p)) is theGodel
number of the formula that results from substituting
theexpression with Godel number z(p) for variable 19 in the
formula withGodel number p. In other words, it is the Godel
number of the formulathat results from substituting the
expression ssss...[p of em]0 forthe variable y in the formula
Q(y). If you go through the exercise,you'll see that this is
the Godel number of the formula Q(p).So the value of Q(p) is
true iff p = the Godel number of the formulaQ(p). Well it
isn't -- p is the Godel number of the formula Q(y),which is
a
different formula -- so the value of Q(p) is false.Maybe that
example will help clear things up.Alan Stern
x and v are
independent random variables and they has the pdf f_X(x)
and>F_V(v) respectively. m is a constant.Please be consistent
with upper and lower case. I like to use upper caseX and V for
the random variables, and then x and v can be used for valuesof
the random variables. f_X and f_V can be the pdf's for X and
V
(whichare functions of the values x and v), and then F_X and
F_V will be the respective cdf's.>Suppose T=x/v, and
T<=m.>How
to calculate the random variable T's pdf or its expect
value?>If there is not the condition T<=m, it is easy to get
the pdf of T. But how>about the case with this condition?In my
notation, T = X/V. Now I don't know how to interpret the
conditionT <= m. If you say X and V are independent with
certain distributionsand T = X/V, then you might have T <= m
or you might not. It will betrue with probability 1 if, say, 0
< X < ma and V > a with probability 1 for some positive
constant a. Or do you mean you want conditionaldistributions,
conditioning on T<=m?The simplest way to deal with these
questions is usually to use the cdfrather than dealing
directly with pdf's. Assuming for simplicitythat X and V are
always positive,F_T(t) = Prob{0 < X/V <= t} = Prob{0 < X <= t
V} = int_{0}^in'nity f_V(v) F_X(tv) dvf_T(t) = d/dt F_T(t) =
int_0^in'nity f_V(v) f_X(tv) v dvThe conditional pdf given
T<=m is f_T(t)/F_T(m) for t < m, 0 for t > m.Robert Israel
israel@math.ubc.caDepartment of Mathematics
http://www.math.ubc.ca/~israel University of British Columbia
Vancouver, BC, Canada V6T 1Z2
Really? It has no residue?
So the residue = 0 ?> No, residue is UNDEFINED at a branch
point. When you learn about> Riemann surfaces, you may begin
to understand why.Well he can understand why by just reading
the de'nition of residue.
Really? It has no residue? So
the residue = 0 ?Unde'ned, not 0.>Because I want to evaluate
I=Integrate(z^(-3/4)*(1-z)^(-1/4), z from 0 to 1)>by making a
big circle excluding the two small holes at 0 and 1... The
usual way to do this uses a keyhole contour that goes around
mostof a little circle around 0, then along the bottom edge of
the branch cut from 0 to 1, then around most of a little circle
around 1, then back alongthe top edge of the branch cut from 1
to 0. Robert Israel israel@math.ubc.caDepartment of
Mathematics http://www.math.ubc.ca/~israel University of
British Columbia Vancouver, BC, Canada V6T 1Z2
=walala
says...>I want to evaluate I=Integrate(z^(-3/4)*(1-z)^(-1/4),
z from 0 to 1)>by making a big circle excluding the two small
holes at 0 and 1... by>walking z from 0 to 1 and then going
back from 1 to 0... I should be able to>get
2*i*I=LoopIntegrate of that big circle... then I need to
have>2*pi*i*{Res[z^(-3/4)*(1-z)^(-1/4),
z=0]+Res[z^(-3/4)*(1-z)^(-1/4), z=1]},>so I need the
residue... am I wrong?The problem is that the function
z^(-3/4)*(1-z)^(-1/4)is multivalued, so its integral is
ill-de'ned. Butwhat you can do is make the function
single-valuedeverywhere except for a small number of cuts
whichare line segments on which the function is unde'ned.What
I think we can do is this: we can integrate arounda huge
circle of radius R to get an answer. Call thatanswer I. Then
we can distort the circle, squashing ittowards the Real axis,
which won't change the value, aslong as we
don't cross a
singularity or a cut. Once we'vesquashed it, we can break
the
integral up into 4 pieces: I1 = integral along the straight
line from z = -ir to z = 1 - ir I2 = integral along the
semicircle z = 1 + r exp(it) from t= -pi/2 to t= +pi/2 I3 =
integral along the straight line from z = +ir to z = 1 + ir I4
= integral along the semicircle z = 1 + r exp(it) from t=+pi/2
to t=3pi/2.We can evaluate I (the integral around the big
circle) byrewriting z^(-3/4)*(1-z)^(-1/4) as
z^{-1}*(-1)^{-1/4}*(1-1/z)^{-1/4}.When |z| is very large, this
can be approximated by simply (-1)^{-1/4}/zIntegrating that
around a big circle to get I is easy.So we know that I = I1 +
I2 + I3 + I4. I2 and I4 caneasily be calculated explicitly, at
least in the limitas r --> 0. So we have I1 + I3 = I - I2 -
I4Now, the only thing left to do is to 'gure out the
relationshipbetween I1 and I3. They are almost identical,
except for threedifferences: (1) the direction of integration
is opposite forthe two of them, and (2) they are along
different branchesof the 4th root function. Anyway, they are
going to be constantmultiples of each other: I3 = c I4, where
c is a constant ofmagnitude 1 (the phase introduced by going
around the singularityat z=0 or z=1).So I1 + I3 = (1+c) I1.
The answer will then be I1 = (I - I2 - I4)/(1+c)I believe that
c will be equal to exp(i pi/4), but I'm not sureabout
that.--Daryl McCulloughIthaca, NY
problem, which make me think several days. Who
can---------Statement:---------For a standard second
model:X_{k} = Phi*X_{k-1} + Sigmma*W_{k},Where: x_{1} x_{2}
w_{x}X_{k}=( x_{3} )_{k}; W_{k}=( w_{y} )_{k}; x_{4} 1 1 0 0
0.5 0 Phi = ( 0 1 0 0 ); Sigmma=( 1 0 ); 0 0 1 1 0 0.5 0 0 0 1
0 1W_{k} is a zero mean Gaussian white noise process with
covariance Q:E[w_{k} w_{j}^{T}] = Q*Dirac_{jk}, Q= ( q 0 ); 0
q---------Question:---------How to get the pdf of
f(x_{1}_{k}|x_{1}_{k-1}, x_{1}_{k-2})
andf(x_{2}_{k}|x_{1}_{k}, x_{1}_{k-1})?THanks again!!
=The
problem is as follows:Suppose you have a number a s.t. 7
doesn't divideit. Proof that (a^3 + 1) or (a^3 - 1) is
divided
by 7.I realise that:1.a = 1  a^3 - 1 = 0a = 2  a^3 - 1 = 7a
= 3  a^3 + 1 = 28 = 4 x 7a = 4  a^3 - 1 = 63 = 6 x 7a = 5
a^3 - 1 = 126 = 18 x 7a = 6  a^3 + 1 = 217 = 31 x 72.
Every other (prime?) number i.e. 17 can be regardedas a
remainder of division by 7 and a number of 7's(here: 3 + 2 x
7) so that the operations can be carriedout regarding the
remainding part (here 3).Now, it's so nice that
is's got to be
something right here.The part i'd need help with is the
formalisation of theproof. That is: how do i prove that by
only regarding theremainder i get the same result as if i
regarded the wholenumber?.If anybody is clever and helpfull -
please give me a hint.--
V.8anligenKonrad----------------------------------------------
-----phone #1: (+46/0) 708 - 70 73 92phone #2: (+46/0) 704 -
79 96 95url:
http://konrads.webbsida.com-----------------------------------
Sleep - thing used by ineffective people as a substitute for
coffeeAmbition - a poor excuse for not having enough sence to
be
lazy---------------------------------------------------
=Are
you familiar with groups and order?Look at the multiplicative
group Z (modulo 7).Consider the order of all elements,
especially the order of all elementsraised to the 3rd
power.GREG> The problem is as follows:> Suppose you have a
number a s.t. 7 doesn't divide> it. Proof that (a^3 + 1) or
(a^3 - 1) is divided by 7.>> I realise that:>> 1.> a = 1
a^3 - 1 = 0> a = 2  a^3 - 1 = 7> a = 3  a^3 + 1 = 28 = 4 x
7> a = 4  a^3 - 1 = 63 = 6 x 7> a = 5  a^3 - 1 = 126 = 18 x
7> a = 6  a^3 + 1 = 217 = 31 x 7>> 2. Every other (prime?)
number i.e. 17 can be regarded> as a remainder of division by
7 and a number of 7's> (here: 3 + 2 x 7) so that the
operations can be carried> out regarding the remainding part
(here 3).>> Now, it's so nice that is's got to
be something
right here.> The part i'd need help with is the
formalisation
of the> proof. That is: how do i prove that by only regarding
the> remainder i get the same result as if i regarded the
whole> number?.>> If anybody is clever and helpfull - please
give me a hint.> -->> V.8anligen> Konrad>
---------------------------------------------------> phone #1:
(+46/0) 708 - 70 73 92> phone #2: (+46/0) 704 - 79 96 95> url:
http://konrads.webbsida.com>
----------------------------------->> Sleep - thing used by
ineffective people> as a substitute for coffee>> Ambition - a
poor excuse for not having> enough sence to be lazy>
--------------------------------------------------->
Now, it's so nice that is's got to be
something right
here.>The part i'd need help with is the formalisation of
the>proof. That is: how do i prove that by only regarding
the>remainder i get the same result as if i regarded the
whole>number?.Yes, you've solved it. Suppose a is (7n+k),
where 1 <= k <= 6. Thena^3 is (7^3.n^3 + 3.7^2.n^2.k +
3.7.n.k^2 + k^3), which is 7z + k^3(where z = 7^2.n^3 +
3.7.n^2.k + 3.n.k^2). So if k^3 +/- 1 is amultiple of of 7, so
is a^3 +/- 1.-- Richard-- FreeBSD rules!
Are you familiar
with groups and order?No, i'm not. Hence, i'd
prefere a hint
not involvingthose. However, i'm fully aware that the very
coreof the problem stated might be about looking intothe
thingies you just mentioned (and those only).I'll give it a
V.8anligenKonrad----------------------------------------------
-----phone #1: (+46/0) 708 - 70 73 92phone #2: (+46/0) 704 -
79 96 95url:
http://konrads.webbsida.com-----------------------------------
Sleep - thing used by ineffective people as a substitute for
coffeeAmbition - a poor excuse for not having enough sence to
be
lazy---------------------------------------------------
Yes, you've solved it.I guess so. I came up with exactly the
same 'nishas you right before checking the news. Ah, what
anice feeling when one wrestle down a proof like that.--
V.8anligenKonrad----------------------------------------------
-----phone #1: (+46/0) 708 - 70 73 92phone #2: (+46/0) 704 -
79 96 95url:
http://konrads.webbsida.com-----------------------------------
Sleep - thing used by ineffective people as a substitute for
coffeeAmbition - a poor excuse for not having enough sence to
be
lazy---------------------------------------------------
The problem is as follows:> Suppose you have a number a s.t. 7
doesn't divide> it. Proof that (a^3 + 1) or (a^3 - 1) is
divided by 7.a^(p-1) = 1 mod p for any prime p and any a not
divisible by p.a^(p-1) - 1 = (a^((p-1)/2) - 1) (a^((p-1)/2) +
1) when p is odd.If a prime p divides xy, then p|x or p|y.-- |
Jim Ferry | Center for Simulation
|+------------------------------------+ of Advanced Rockets ||
http://www.uiuc.edu/ph/www/jferry/ +------------------------+|
jferry@[delete_this]uiuc.edu | University of Illinois |
=The
thesis:8 x 2^(2^n) + 1 is not prime for positive integers.I
checked out in Mathematica and, as expected,it was true for
the 'rst couple of n's. I suspectthe solution
would be hidden
somewhere withinthe rings, and congruences modulo something
buthere, i'm VERY stuck.Any hint (a really good one) would
be
of help.And i mean not something to get me startedbut
something that gets me -way. :)--
V.8anligenKonrad----------------------------------------------
-----phone #1: (+46/0) 708 - 70 73 92phone #2: (+46/0) 704 -
79 96 95url:
http://konrads.webbsida.com-----------------------------------
Sleep - thing used by ineffective people as a substitute for
coffeeAmbition - a poor excuse for not having enough sence to
be
lazy---------------------------------------------------
The thesis:> 8 x 2^(2^n) + 1 is not prime for positive
integers.> Any hint (a really good one) would be of help.> And
i mean not something to get me started> but something that gets
me -way. :)2^{something odd} mod 3 -- Timothy Murphy tel:
+353-86-233 6090
The thesis:>> 8 x 2^(2^n) + 1 is not
prime for positive integers.> Any hint (a really good one)
would be of help.>> And i mean not something to get me
started>> but something that gets me -way. :)>> 2^{something
odd} mod 3Or beter, note that a^(2k+1)+1 is divisible by a+1,
as X^(2p+1)+1 has (-1)as a root
> The thesis:>> 8 x
2^(2^n) + 1 is not prime for positive integers.It is
well-known that: 2^n+1 is a prime  n = 2^k for some positive
integerk. (in other words 2^n+1 is a Fermat's number).
The
thesis:> 8 x 2^(2^n) + 1 is not prime for positive integers.>
n=0, 8*2+1=17, prime.Larger n, divisible by 3, so not prime.>
I checked out in Mathematica and, as expected,> it was true
for the 'rst couple of n's. I suspect> the
solution would be
hidden somewhere within> the rings, and congruences modulo
something but> here, i'm VERY stuck.> Any hint (a really
good one) would be of help.> And i mean not something to get
me started> but something that gets me -way. :)
I claim
that if a and b are prime numbers divisible by 4 then a+b is
also a> prime divisible by 4.> THE ONLY WAY THAT YOU CAN
a and b, that are divisible by 4 BUT a + b is not a prime>
number divisible by 4! Since you can't, it follows that the
statement if a> and b are prime numbers divisible by 4 then
a+b is not neccessarily a prime> divisible by 4 is FALSE.
Hence the statement if a and b are prime numbers> divisible by
4 then a + b is also a prime divisible by 4 must be true.No, it
might be undecidable.Charlie Volkstorf
I claim that if a
and b are prime numbers divisible by 4 then a+b is also a>
prime divisible by 4.Since a and b are prime numbers divisible
by 4 is always false, any implication of form if a and b are
prime numbers divisible by 4 then Osomething' is
always true,
regardless of the Osomething' says. Thus you may
also claim if
a and b are prime numbers divisible by 4 then a+b is neither a
prime nor divisible by 4.Or if a and b are prime numbers
divisible by 4 then Steven is the pope.
> I claim that if
a and b are prime numbers divisible by 4 then a+b is also a>>
prime divisible by 4.>> THE ONLY WAY THAT YOU CAN CONVINCE ME
that are divisible by 4 BUT a + b is not a prime>> number
divisible by 4! Since you can't, it follows that the
statement
if a>> and b are prime numbers divisible by 4 then a+b is not
neccessarily a prime>> divisible by 4 is FALSE. Hence the
statement if a and b are prime numbers>> divisible by 4 then a
+ b is also a prime divisible by 4 must be true.>>No, it might
be undecidable.No, you're confused about what undecidable
means.>Charlie Volkstorf************************David C.
Ullrich
> I claim that if a and b are prime numbers
divisible by 4 then a+b is also a>> prime divisible by 4.>>
show me two>> prime numbers, a and b, that are divisible by 4
BUT a + b is not a prime>> number divisible by 4![cut]>>No, it
might be undecidable.> No, you're confused about what
undecidable means.I thought Charlie-Boo was correct.If I have
an undecibable statement S and you claim S is true,then I
can't convince you that you are wrong by showinga particular
instance where the statement is false. But, thisdoes not make
the statement S therefore true.-- Bill Hale
> I claim
that if a and b are prime numbers divisible by 4 then a+b
isalso a>> prime divisible by 4.>> THE ONLY WAY THAT YOU CAN
a and b, that are divisible by 4 BUT a + b is not aprime>>
number divisible by 4!> [cut]>>No, it might be undecidable.>>
No, you're confused about what undecidable means.>> I
thought
Charlie-Boo was correct.>> If I have an undecibable statement
S and you claim S is true,> then I can't convince you that
you
are wrong by showing> a particular instance where the statement
is false. But, this> does not make the statement S therefore
true.>> -- Bill HaleBut the case above is very special: there
are no prime numbers a,b that aredivisible by 4.Therefore, not
only can Charlie-Boo not 'nd a counter example, it
is*provable*
that there ain't one.Therefore, the statement is not
undecidable at all.Or perhaps i'm totally missing your
point,
here?Herman Jurjus
hale@tulane.edu (William Hale)> I claim that if a and b
are prime numbers divisible by 4 then a+b is also a> prime
divisible by 4.> THE ONLY WAY THAT YOU CAN CONVINCE ME THAT
are divisible by 4 BUT a + b is not a prime> number
divisible by 4!>[cut]>>No, it might be undecidable. No,
you're confused about what undecidable means.>>I thought
Charlie-Boo was correct.I'd typed a few paragraphs of
de'nitions before I went back andrealized I was puzzled about
what you meant. Since we're talkingabout subtle things, it
would probably be more ef'cient if youclari'ed a
few things
before I tried to reply, cuz if I really don'tget what
you're
saying then I can't really answer it:>If I have an
undecibable
statement S and you claim S is true,>then I can't convince
you
that you are wrong by showing>a particular instance where the
statement is false. This sentence has me very puzzled. I mean
I don't get it at_all_ - of _course_ you could convince me
that way? Thisseems so clear that I can't help but wondering
whether therewas a word missing or something?>But, this>does
not make the statement S therefore true.My 'rst impulse is to
say I didn't say it did make S thereforetrue. But
I'm going to
wait on that, since the thisevidently refers to the previous
sentence and I'm assumingthat I don't know
what you meant to
write there...>-- Bill Hale************************David C.
Ullrich
hale@tulane.edu (William Hale)>>If I have an undecibable
statement S and you claim S is true,>>then I can't convince
you that you are wrong by showing>>a particular instance where
the statement is false. >>This sentence has me very puzzled. I
mean I don't get it at>_all_ - of _course_ you could
convince
me that way?Presumably what he means is that S is undecidable
implies that onecannot explicitly exhibit counterexamples to
S, since if explicitcounterexamples were available, they would
decide the truth of S.So if the counterexamples could in fact
be displayed, then they wouldconvince you, but unfortunately
there aren't any available for ostentation.-- Tim Chow
tchow-at-alum-dot-mit-dot-eduThe range of our
projectiles---even ... the artillery---however great,
willnever exceed four of those miles of which as many thousand
separate us fromthe center of the earth. ---Galileo, Dialogues
Concerning Two New Sciences
-0500, hale@tulane.edu (William Hale)> I claim that if
a and b are prime numbers divisible by 4 then a+bis also a>
prime divisible by 4.> THE ONLY WAY THAT YOU CAN CONVINCE ME
that are divisible by 4 BUT a + b is nota prime> number
divisible by 4!>[cut]>>No, it might be undecidable. No,
you're confused about what undecidable means.>>I thought
Charlie-Boo was correct.> I'd typed a few paragraphs of
de'nitions before I went back and> realized I was puzzled
about what you meant. Since we're talking> about subtle
things, it would probably be more ef'cient if you>
clari'ed a
few things before I tried to reply, cuz if I really don't>
get
what you're saying then I can't really answer
it:>If I have
an undecibable statement S and you claim S is true,>then I
can't convince you that you are wrong by showing>a
particular
instance where the statement is false. > This sentence has
me very puzzled. I mean I don't get it at> _all_ - of
_course_
you could convince me that way?I don't understand what you
are
questioning.Steven says S is true. He gives his reason that is
truebecause I can't show him an instance where S is false.I
agree that I can't show him an instance where S is
false.Steven wants then to say that therefore S is
true.Charlie-Boo and I are saying that this does not prove
that S is true.> This> seems so clear that I can't help but
wondering whether there> was a word missing or something?I
don't think I am missing any words. Remember that I am
notdiscussing whether I can determine S is decidable or
not.Charlie-Boo and I are discussing whether failure to be
ableto exhibit the falsity of a statement is proof that
thestatement is true.>But, this>does not make the statement S
therefore true.> My 'rst impulse is to say I
didn't say it
did make S therefore> true.But, we are talking about
Steven's
principle stated in his post.> But I'm going to wait on
that,
since the this> evidently refers to the previous sentence and
I'm assuming> that I don't know what you meant
to write
there...The statement under contention is THE ONLY WAY THAT
me an instancewhere S is false. If I can't show an instance
where S is false,then S has to be true (and thus proved).Since
S is undecidable, I can't show an instance where S is
false.But, I can't show an instance where not S is false
either.That is, S is neiter true or false: S is undecidable.
Thus,Steven's principle of proof does not always work.P.S.
Of
course, Steven may not have meant this. I think the
manyreplies in this thread itself give the impression that
there isa principle that p implies q is true whenever p is
false.Of course that is true, but I don't like to think of
it
that way.I gave a direct proof of the original poster's
request. But,maybe I show have given the proof by
contradiction. I thinkmost (all?) are taking it is obvious
that p is false. I thinkthat should be proved and it comes out
automatically whenyou work with the original statement that was
to be proved.If you throw away all the red herrings, you come
to thefollowing statement to be proved: if 4 divides a then a
is not prime.-- Bill Hale
-0500, hale@tulane.edu (William Hale)>> I claim that if a
and b are prime numbers divisible by 4 then a+b> is also a>>
prime divisible by 4.>> THE ONLY WAY THAT YOU CAN CONVINCE ME
that are divisible by 4 BUT a + b is not> a prime>> number
divisible by 4!>>[cut]>No, it might be undecidable.> No,
you're confused about what undecidable means.>>I thought
Charlie-Boo was correct.> Steven says S is true. He gives his
reason that is true> because I can't show him an instance
where S is false.> I agree that I can't show him an instance
where S is false.> Steven wants then to say that therefore S
is true.> Charlie-Boo and I are saying that this does not
prove that S is true.That's correct. The reason S is true is
not that you haven't found acounterexample yet; the reason S
is true is that there is a simple*proof* that no
counterexample can possibly exist.Let's consider an analogy.
Is it possible that Fermat's Last Theoremcould be
from your answer in1992, before Wiles announced his proof?
FLT is known, just as was the case in1992.Do you agree that
the presence of a proof makes a difference?-- Dave SeamanJudge
Yohn's mistakes revealed in Mumia Abu-Jamal
ruling.<http://www.commoncouragepress.com/index.cfm?action=
book&bookid=228>
=This may be somewhat vague; but I hope
someone can help.Some time ago, I saw a throw-away example
purporting to be a good modelof the hyperbolic plane. At least
a very intuitive/comprehensible model.It was a 'nite open
disk,
with the (hyperbolic) straight lines appearing assemicircles,
all perpendicular to the boundary. This including diameters
OC.Now this has the correct local and global topology for the
hyperbolic plane,and seems to obey the incidence and
parallelism axioms very well.But it makes no mention of metric
ideas; neither of what is congruent to what, in particular
regarding SAS triangles (the homogeneity axiom); nor,
alternatively & equivalently but better, what is a suitable
metricto ful'l its hyperbolic nature and make these
semicircles geodesics.This could be either an in'nitesimal
metric, or simply a distancefuntion on any two points of a
given semi-circle.So, can anyone help me out with the
appropriate metric please?I'm not sure if it can even be
done,
but it seems plausible.
TIA-----------------------------------------------------------
------------------- Bill Taylor
W.Taylor@math.canterbury.ac.nz--------------------------------
---------------------------------------------- There is no
force, however great, Can stretch a thread - however 'ne,
Into
an absolutely straight And level geometric
line.---------------------------------------------------------
---------------------
This may be somewhat vague; but I
hope someone can help.>Some time ago, I saw a throw-away
example purporting to be a good model>of the hyperbolic plane.
At least a very intuitive/comprehensible model.>It was a
'nite
open disk, with the (hyperbolic) straight lines appearing
as>semicircles, all perpendicular to the boundary. This
including diameters OC.Arcs of circles, rather than
semicircles.>Now this has the correct local and global
topology for the hyperbolic plane,>and seems to obey the
incidence and parallelism axioms very well.>But it makes no
mention of metric ideas; neither of what is congruent >to
what, in particular regarding SAS triangles (the homogeneity
axiom); >nor, alternatively & equivalently but better, what is
a suitable metric>to ful'l its hyperbolic nature and make
these
semicircles geodesics.>This could be either an in'nitesimal
metric, or simply a distance>funtion on any two points of a
given semi-circle.>So, can anyone help me out with the
appropriate metric please?>I'm not sure if it can even be
done, but it seems plausible. TIAFor the unit disk, try ds^2 =
(dx^2+dy^2)/(1-x^2-y^2).As an additional observation, circles
in the hyperbolic plane are represented by cirles within the
open disk, and conversely, any circlecompletely within the
open disk represents a circle in the
hyperbolicplane.>---------------------------------------------
---------------------------------> Bill Taylor
W.Taylor@math.canterbury.ac.nz>-------------------------------
-----------------------------------------------> There is no
force, however great,> Can stretch a thread - however 'ne,>
Into an absolutely straight> And level geometric
line.>--------------------------------------------------------
----------------------David McAnally--------------
This
may be somewhat vague; but I hope someone can help.> Some
time ago, I saw a throw-away example purporting to be a good
model> of the hyperbolic plane. At least a very
intuitive/comprehensible model.> It was a 'nite open disk,
with the (hyperbolic) straight lines appearing as>
semicircles, all perpendicular to the boundary. This including
diameters OC.> Now this has the correct local and global
topology for the hyperbolic plane,> and seems to obey the
incidence and parallelism axioms very well.> But it makes no
mention of metric ideas; neither of what is congruent > to
what, in particular regarding SAS triangles (the homogeneity
axiom); > nor, alternatively & equivalently but better, what
is a suitable metric> to ful'l its hyperbolic nature and make
these semicircles geodesics.> This could be either an
in'nitesimal metric, or simply a distance> funtion on any two
points of a given semi-circle.> So, can anyone help me out
with the appropriate metric please?> I'm not sure if it can
even be done, but it seems plausible. TIA> This is known as
the Poincare model for the hyperbolic plane.
<http://mathworld.wolfram.com/PoincareHyperbolicDisk.html>--
G. A. Edgar http://www.math.ohio-state.edu/~edgar/
=On 27
(Bill>(...)>Some time ago, I saw a throw-away example
purporting to be a good model>of the hyperbolic plane. At
least a very intuitive/comprehensible model.>It was a 'nite
open disk, with the (hyperbolic) straight lines appearing
as>semicircles, all perpendicular to the boundary. This
including diameters OC.>(...)>There must be also a simple
formula for the distance between 2 points.Similar to that of
the Klein model - which is the part of a projectiveplane
inside a conic section - inside being BTW that side of
thecurve that is simply connected, if I am right: it IS
actually simplyconnected (homeomorphic to an open euclidian
disk) but I imply herethat the other side is not ;-) of which
I am not quite sure. In theKlein model, straight lines are the
non-empty intersections ofstraight lines of the proj. plane
with the model region and thedistance of 2 points A, B is
(proportional to) the absolute value ofthe logarithm of
Anh(A,B,C,D) where C,D are the points where theprojective
straight line AB cuts the conic and by Anh I mean
theanharmonic ratio if that is called like that in Englisch
(French:rapport anharmonique). This ratio will be > 0 so that
the log is real.Now consider the plane of Poincar.8e's model
as C, the 'eld of complexnumbers. The disk can be {z in C
such
that |z|<1} and the same thingsholds, the ratio must be
determined considering C+a point at in'nityas 1-dim. complex
proj. space (=line) and C,D are where the non-eucl.straight
line (circle orthogonal to the unit circle |z|=1) through
A,Bcrosses the unit circle. May-be a factor of +/- sqrt(-1) is
needed toget a real
value.
=PRAJa.2566827$YZ.391040@news.easynews.com...>> Prove
that R:reals is a in'nite dimensional vector space
overQ:rationals.>R is uncountable ...--Julien
Santini,Universit.8e de Provence,France.
Moreover :>R is a
K-vector space with 'nite dimension (K C R) K=R.>> I give
up. How do you prove that?>The following proof was proposed by
you'll
be able to guess a translation, (most words arecorps = 'eld;
sous-corps = sub'eld; impair = odd, pair = even; puissance
=power):Oui : K = R...Et c'est m.90me le seul.Posons
K' =
K(i).C/K' est une extension galoisienne, notons G son groupe
de Galois.Si G n'est pas trivial, il existe un nombre
premier
p divisant l'ordrede G, et soit H un sous-groupe de G
d'ordre
p. Notons L sous-corps de C'xe par H.On a donc les extensions
: K c K' c L c C.C/L est galoisienne, et m.90me cyclique
d'ordre p. De plus, L contient lesracines p-i.8fmes de
l'unit.8e : soit w une racine primitive p-i.8fme
del'unit.8e,
on a 1+w+...+w^(p-1) = 0, donc L(w)/L est une extension
dedegr.8e au plus p-1, et divisant p, donc L(w) =
L.D'apr.8fs
le th.8eor.8fme de Kummer, C est donc le corps de rupture
d'unpolyn.99me irr.8eductible X^p - a, o.9d a est dans
L.Soit
b et c complexes tels que b^p = a, et c^p = b.Comme X^p - a =
(X-b)(X-b_2)...(X-b_p), on a : -a = (-1)^p b b_2 ... b_p, i.e.
-a = (-1)^p N(b)o.9d N est la norme associ.8ee .88
l'extension
C/L.Donc N(c)^p = N(b) = (-1)^(p+1) a.Si p est impair, a =
N(c)^p, et donc a est une puissance p-i.8fme dans L,ce qui
contredit l'irr.8eductibilit.8e de X^p - a dans L[X].Si p =
2,
on a que -a = N(c)^2 est un carr.8e dans L, et doncC = L(a) =
L(a/N(c)) = L(i) ; c'est absurde car par construction i
estdans K' et donc dans L.Ainsi le groupe de Galois G est
trivial, i.e. K' = C, donc C/K est uneextension de degr.8e
2,
et comme K c R, on a 'nalement K = R.
16:02:50 +0200, Julien Santini>>Moreover :>>R is a K-vector
space with 'nite dimension (K C R) K=R.>> I give up. How
do you prove that?>>The following proof was proposed by Yann
able to guess a translation, (most words are>corps = 'eld;
sous-corps = sub'eld; impair = odd, pair = even; puissance
power):bother trying to puzzle out the French, wouldn't
make
much senseto me in English, I'm afraid. Remind me to learn
some
math some day...(I like that phrase corps de rupture, though.
Sounds like somethingassociated with a fairly serious medical
condition...)>Oui : K = R...>>Et c'est m.90me le
seul.>>Posons
K' = K(i).>C/K' est une extension galoisienne,
notons G son
groupe de Galois.>>Si G n'est pas trivial, il existe un
nombre
premier p divisant l'ordre>de G, et soit H un sous-groupe de
G
d'ordre p. Notons L sous-corps de C>'xe par
H.>>On a donc les
extensions : K c K' c L c C.>>C/L est galoisienne, et m.90me
cyclique d'ordre p. De plus, L contient les>racines
p-i.8fmes
de l'unit.8e : soit w une racine primitive p-i.8fme
de>l'unit.8e, on a 1+w+...+w^(p-1) = 0, donc L(w)/L est une
extension de>degr.8e au plus p-1, et divisant p, donc L(w) =
L.>>D'apr.8fs le th.8eor.8fme de Kummer, C est donc le corps
de rupture d'un>polyn.99me irr.8eductible X^p - a, o.9d a
est
dans L.>>Soit b et c complexes tels que b^p = a, et c^p =
b.>>Comme X^p - a = (X-b)(X-b_2)...(X-b_p), on a :> -a =
(-1)^p b b_2 ... b_p, i.e.> -a = (-1)^p N(b)>o.9d N est la
norme associ.8ee .88 l'extension C/L.>>Donc N(c)^p = N(b) =
(-1)^(p+1) a.>>Si p est impair, a = N(c)^p, et donc a est une
puissance p-i.8fme dans L,>ce qui contredit
l'irr.8eductibilit.8e de X^p - a dans L[X].>>Si p = 2, on a
que -a = N(c)^2 est un carr.8e dans L, et donc>C = L(a) =
L(a/N(c)) = L(i) ; c'est absurde car par construction i
est>dans K' et donc dans L.>Ainsi le groupe de Galois G est
trivial, i.e. K' = C, donc C/K est une>extension de degr.8e
2,
et comme K c R, on a 'nalement K =
R.>>************************David C. Ullrich
The
following proof was proposed by Yann Villessuzanne on
translation, (most words are> corps = 'eld; sous-corps =
sub'eld; impair = odd, pair = even; puissance  power):>
Oui : K = R...> Et c'est m.90me le seul.> Posons
K' =
K(i).> C/K' est une extension galoisienne, notons G son
groupe
de Galois.> Si G n'est pas trivial, il existe un nombre
premier p divisant l'ordre> de G, et soit H un sous-groupe
de
G d'ordre p. Notons L sous-corps de C> 'xe par
H.> On a donc
les extensions : K c K' c L c C.> C/L est galoisienne, et
m.90me cyclique d'ordre p. De plus, L contient les> racines
p-i.8fmes de l'unit.8e : soit w une racine primitive
p-i.8fme
de> l'unit.8e, on a 1+w+...+w^(p-1) = 0, donc L(w)/L est une
extension de> degr.8e au plus p-1, et divisant p, donc L(w) =
L.> D'apr.8fs le th.8eor.8fme de Kummer, C est donc le corps
de rupture d'un> polyn.99me irr.8eductible X^p - a, o.9d a
est
dans L.> Soit b et c complexes tels que b^p = a, et c^p =
b.> Comme X^p - a = (X-b)(X-b_2)...(X-b_p), on a :> -a =
(-1)^p b b_2 ... b_p, i.e.> -a = (-1)^p N(b)> o.9d N est la
norme associ.8ee .88 l'extension C/L.> Donc N(c)^p = N(b) =
(-1)^(p+1) a.> Si p est impair, a = N(c)^p, et donc a est
une puissance p-i.8fme dans L,> ce qui contredit
l'irr.8eductibilit.8e de X^p - a dans L[X].> Si p = 2, on a
que -a = N(c)^2 est un carr.8e dans L, et donc> C = L(a) =
L(a/N(c)) = L(i) ; c'est absurde car par construction i est>
dans K' et donc dans L.> Ainsi le groupe de Galois G est
trivial, i.e. K' = C, donc C/K est une> extension de degr.8e
2, et comme K c R, on a 'nalement K = R.Below is my attempt
to
rewrite the proof. I know next to zero French,so the phrasing
has been changed. Please feel free to correct any errors.Note:
K < R = real numbers, i = sqrt(-1), C = R(i).Let K' = K(i),
then C/K' is a Galois extension with Galois gp G. If Gis not
trivial, then there is a prime number p which divides |G| and
a subgroup H < G of order p. Let L be the sub'eld of C
corresponding toH.We get 'eld extensions K < K'
< L < C.C/L is
a Galois and cyclic extension of degree p. Furthermore, L
mustcontain the p-th roots of unity: otherwise, let w be a
p-th root ofunity, whence L(w)/L is cyclic of order p-1 which
is not a mult of p.So by a theorem of Kummer, C is the
splitting 'eld over L of anirreducible polynomial of the form
X^p - a = 0 (a in L).Let b,c be complex such that b^p = a, c^p
= b. Then we can factor(X^p - a) = (X - b)(X - bw) .... (X -
bw^{p-1})and by taking the norm of the extension C/L:-a =
(-1)^p N(b) N(c)^p = (-1)^{p+1} a.If p is odd, a = N(c)^p,
then N(c) is a root of the equation X^p - a = 0contradicting
the assumption of its irreducibility.If p=2, -a = N(c)^2, and
we have C = L(a) = L(a/N(c)) = L(i). This isabsurd as i is in
K' < L by construction.Hence G is trivial and K=R.
=Given A
and B, each an n x n matrix with non-zero determinant,
andwhich I have already calculated the inverses A^-1 and
B^-1,is there a simple formula for the inverse of (A+B)? Is
therea computational method of calculating (A+B)^-1 that is
lesscomputational effort because A^-1 and B^-1 are known?I'm
thinking not for the general case, though maybe for
somespecial cases. For example, if A and B is each a multiple
ofthe identity matrix, there clearly is a simple formula,
butit's pretty trivial to directly do it in that case
anyway.And surely if A and B are in the L-U form, there is
simpli'cation.Socks
Given A and B, each an n x n matrix
with non-zero determinant, and> which I have already
calculated the inverses A^-1 and B^-1,> is there a simple
formula for the inverse of (A+B)? Is there> a computational
method of calculating (A+B)^-1 that is less> computational
effort because A^-1 and B^-1 are known?> I'm thinking not
for the general case, though maybe for some> special cases.
For example, if A and B is each a multiple of> the identity
matrix, there clearly is a simple formula, but> it's pretty
trivial to directly do it in that case anyway.> And surely if
A and B are in the L-U form, there is simpli'cation.> SocksA
search in the sci.math archives with the keywords
Sherman-Morrisonwill yield many interesting answers to your
question.HTH,Felix.
=This was a press release from Australia
skeptics: If, however, we are to regard the ACA rules as
applying to everything we do, we had better take this
opportunity to confess that we have indeed backed down on the
following issues: Peace in the Middle-East Landing a man on
Mars by 2020 Solving Fermat's Last Theorem Overcoming the
salinity problem in rural Australia Tracking down that man
behind the Grassy
KnollHerc--www.winternet.com/~mikelr/§ame76.html __ / / / / /
/ / / / / / /__/__ /________ ___________/Proof of numerology
at www.adamskingdom.comname one billionaire?what did Lady Di
do?Tiger :: golf :: ?star wars program was introduced by which
president ?Nic Cage stars in what kind of movies ?Who is the
smartest man, Haw.... ?
This was a press release from
Australia skeptics:> If, however, we are to regard the ACA
rules as applying to everything we do, we had better> take
this opportunity to confess that we have indeed backed down on
the following issues:> Peace in the Middle-East> Landing a
man on Mars by 2020> Solving Fermat's Last TheoremMaybe
they've been reading JSH's press releases.>
Overcoming the
salinity problem in rural Australia> Tracking down that man
behind the Grassy KnollWhat man? There was no man on the
Grassy Knoll.- Oswald was a patsy.- The ri§e and shell casings
were planted on the sixth §oor of thebook depository.against it
while his body lay in the morgue.- Now to make this setup work,
the real shooter _had_ to have been_behind_ Kennedy. The
conspirators would not commit an oh, what agive-away by
shooting from the front!Logic can also debunk the so-called
Magic Bullet. Part of theconspiracist dogma is that the bullet
found on Conally's stretcher wasplanted there. This is
obviously false because- The conspirators planted three empty
shell casings.- The conspirators would have no way of knowing
that the three actualbullets 'red would not be recovered.
They
would have assumed that allof them would be. To plant a bullet,
and thus making a fourth, wouldagain be an oh, what a
give-away.The minimum requirement for a conspiracy theory is
that it'sconsistent! But since to dismiss the Grassy Knoll
and
the Magic Bulletis heresy, the conspricy buffs continue to
embrace them. They are muchlike JSH in that regard.> Herc --> www.winternet.com/~mikelr/§ame76.html> __> / /> / / > /
/ / > / / / > / /__/__ > /________ > ___________/> Proof of
numerology at www.adamskingdom.com> name one billionaire?>
what did Lady Di do?> Tiger :: golf :: ?> star wars program
was introduced by which president ?> Nic Cage stars in what
kind of movies ?> Who is the smartest man, Haw.... ?
all,By this equation u_x * g_x + u_y * g_y =0, where u and g
are harmonicfunctions... is it possible to prove that the only
harmonic functions u andg satisfying this equation must satisfy
Cauchy-Rieman relationship?I know C-R can  the above
equation, but can the above equation  C-Rgiven both g and u
are entire harmonic functions?Is there any such
theorem?Walala
u_y * g_y =0, where u and g are harmonic> functions... is it
possible to prove that the only harmonic functions u and> g
satisfying this equation must satisfy Cauchy-Rieman
relationship?> I know C-R can  the above equation, but
can the above equation  C-R> given both g and u are entire
harmonic functions?> Is there any such theorem?>
WalalaNo. In general, it turns out that u and g are the real
and imaginaryparts of an analytic function. That is, there
exists an analyticfunction f such that f = u + ig.
name
one billionaire?J. Paul Getty>what did Lady Di do?Die>Tiger ::
golf :: ?James>star wars program was introduced by which
president ?Reagan>Nic Cage stars in what kind of movies
?Hollywood movies>Who is the smartest man, Haw.... ?The one
with the highest IQ.---john@thecodezone.com
http://www.shatnerology.com
When you have a Seifert
surface presented as a 2-disk (a 2-dimensional> 0-handle) with
bands (2-dimensional 1-handles) attached to its > boundary, the
loops determined by the bands (one loop for each band,>
consisting of the core arc of the band together with an arc on
the> 2-disk) can be (cyclically) ordered in a conventional
manner, say> by the (cyclic) order of their left attaching
regions on the boundary> of the (oriented) 2-disk; and,
presumably, in the book you refer to> some such conventional
order is used to order the rows (and columns)> of the Seifert
matrix. During the course of your loop operations>
(handle-slides), this order may change (by an odd
permutation),> and that is where the sign of the determinant
can change.What I don't understand is:If the rows are
changed
by an odd permutation, aren't the correspondingcolumns
the determinant change)?Marc
> When you have a Seifert
surface presented as a 2-disk (a 2-dimensional>> 0-handle)
with bands (2-dimensional 1-handles) attached to its >>
boundary, the loops determined by the bands (one loop for each
band,>> consisting of the core arc of the band together with an
arc on the>> 2-disk) can be (cyclically) ordered in a
conventional manner, say>> by the (cyclic) order of their left
attaching regions on the boundary>> of the (oriented) 2-disk;
and, presumably, in the book you refer to>> some such
conventional order is used to order the rows (and columns)>>
of the Seifert matrix. During the course of your loop
operations>> (handle-slides), this order may change (by an odd
permutation),>> and that is where the sign of the determinant
can change.to which marcdohm@web.de (Marc Dohm) responds:>What
I don't understand is:>If the rows are changed by an odd
permutation, aren't the corresponding>columns permutated in
determinant change)?answer your question because it didn't
go
far enough (and inparticular the last line was wrong). What I
should have addedis that the loops determined by the bands are
not merelyordered ... by the (cyclic) order of their left
attachingregions, they are *oriented* by that order. So when a
slideexchanges (as it can) the positions in that order (made
non-cyclicby the choice of a 'rst loop) of the left and right
attaching regions of some loop, the orientation of the loop is
reversed;and this is where the sign of the determinant changes.
I think.Try sliding the lower end of the lower loop in the
followingdiagram up over the upper loop and see what happens.
____ _____| | | _ || | | | | || |________| || __________|| | |
|| | | | _____| | | | | _ || | | | | | | || | | | | | | ||
|____________| || ______________|| | | | | || |____| | | ||
______| | || |________| ||_______________|Lee Rudolph
=It
seems to me that the lambda calculus is the formulation that
makes theChurch-Turing thesis most plausible. How come I've
never seen a book that usesit to prove the unsolveability of
the halting problem? Does anybody know of asource like
that?Thx
Is there a well-known generalization of the
Lambert W-function where:>> W(n,y) is such that:> y =
W(n,y)*exp(exp(exp(...exp(W(n,y))...)))>> for all y, where the
number of Oexp's is n?> No. In fact, there is
not even a
well-known function solving> y=u*exp(exp(u)) for u. The above
is incorrect. Such functions have been partially considered
byCorless and other authors that 'rst considered the
LambertW.One can easily construct the function that Leroy asks
for. Thedif'culty lies in 'guring out how they
behave. In
effect, theirbehavior may be more complex than the problems
they solve, that's whythe quest of introducing them has been
temporarily paused.That, of course, doesn't mean that they
Ioannishttp://users.forthnet.gr/ath/jgal/_____________________
______________________Eventually, _everything_ is
understandable.
assumption. Would it be equivalent to
state that set A is> well-approximated by G iff there exists a
sequence {B_n} in G such> that lim (n-->oo) B_n = A? This
appears to make it easier to show> that u and q agree on any
such set A.> B_n and A are sets, not functions. What would
you mean by > lim (n-->oo) B_n = A? > If you mean the
indicator functions converge almost everywhere for the >
measures u and q, that would be equivalent [in the case where
the measures > are 'nite]; notice that if u((B_n A) union (A
B_n)) -> 0 it doesn't > mean B_n -> A u-almost-everywhere,
but
it does imply that some subsequence > does.> Robert Israel
israel@math.ubc.ca> Department of Mathematics
http://www.math.ubc.ca/~israel> University of British
Columbia> Vancouver, BC, Canada V6T 1Z2OK Prof. Israel. Many
The Bible
has a 'ne one.> If they aren't against me,
they're with me> If
they aren't with me, they're against meThese
statements are
logically equivalent.> What about those who are neither with
you nor against?He's stating that there aren't
any. This may
or may not be true,depending on what is included in they, but
it certainly isn't a logical fallacy.Robert Israel
israel@math.ubc.caDepartment of Mathematics
http://www.math.ubc.ca/~israel University of British Columbia
Vancouver, BC, Canada V6T 1Z2
In real life people have
thermostats on hot tubs, to stop them from > boiling.And in
some cases the thermostats have failed. There are a few cases
of genuine hyperthermia due to faulty hot tubs.Bob Kolker
Good old Snopes:>
http://www.snopes.com/critters/wild/frogboil.htmDid good old
snopes put a lid on the pot? If you put a frog into cold
water, put a lid on the pot and bring it to a boil, the frog
will absolutely not jump out. He may try to, but he is doomed
to failure.Bob Kolker
= . . .>Why is it O! = 1 instead O! =
0?>>Some decisions are arbitrary; the person setting the
precedent makes the . . .Why is 0! = 1? To agree with the
gamma <URL: http://mathworld.wolfram.com/GammaFunction.html
>.Well, even that deserves a bit of explana-tion, more than
I'm willing to undertakenow. The point is, though, that,
yes,there are indeed many examples of conven-ention and
arbitrary choice, among whichare better ones than 0!.--
Cameron Laird <Cameron@Lairds.com>Business:
http://www.Phaseit.netPersonal:
http://phaseit.net/claird/home.html
=# .# .# .# >Why is it
O! = 1 instead O! = 0?# ># >Some decisions are arbitrary; the
person setting the precedent makes the# .# .# .# Why is# 0! =
1# ? To agree with the gamma <URL: http://#
mathworld.wolfram.com/GammaFunction.html >.Actually Gamma was
derived to match factorial, which had already decided 0! =
1.--Derk Gwen http://derkgwen.250free.com/html/index.htmlYou
hate people.But I love gatherings. Isn't it ironic.
|-|erc says...>>IOW, Turing Machines *can* calculate the
*values* of non computable functions,>many are just integers,
we just don't know what non computable function they>are the
value for. As I've tried to put it 5 times now lol.>>
That's
right---every noncomputable function can be approximately
by..> there will be *some* decimal place at which R and r
disagree.>No, calculating the exact values of NCF's is
trivial
if the function maps to integers.Once again I'm not
approximating any function, I'm de'ning the
range for its
result.? The result to a computable problem may bethe same
number as the solution to a non computable problem.1/
functions are from numbers (domain) to numbers (range)2/ there
is more than 1 function3/ if a function result is not X,
another function may have result X4/ if a funciton result is
X, another function may have result X5/ a NCF has an unknown
result, another function may have that result6/ eg. Halt(555)
= [1 if TM(555) terminates, 0 if TM(555) does not terminate]
0+1 = 1 0+0 = 0? The result to a computable problem may bethe
same number as the solution to a non computable
problem.Cantors proof assumes a countable but INFINITE
list,obviously to make reals countable the assumptions of
constructability of in'nitelists have to be
modi'ed,
alternatives to Cantor are what I'm demonstrating
here,diagonalisation is elementary afterall.Cantor assumes the
constructability of an in'nite list, because it has a
beginningand no end it is countable. He then differentiates
between rationals and irrationalssuch that rationals are
countable and irrationals are not countable.I am showing
irrationals are countable, hence the construction of the
in'nite listand the diagonal number are shown as irrelevant,
as the distinction it drawsbetween rationals and irrationals
is wrong. The diagonal is a self referencingformula at its own
digit and amenible to logical interpretation, we are
questioningthe validity of the diagonal technique.And if
anyone can see my argument that computing a value to a non
computablefunction is trivially possible then say so. I'm
going against all text books here andits hardly worth it if a
dozen people just keep quoting them and get stuck on sucha
simple notion that a single function doesn't dictate the
existence of the numberthat happens to be its answer.>I'm
using the fact that H(n) is noncomputable to prove that r
cannot>be equal to pi/4 *and* that r cannot be equal to pi/5.
First you>prove that r cannot be equal to pi/4. Then you prove
that it can't>be equal to pi/5. Together, these proofs show
that it is false that>r=pi/4 OR r=pi/5.Not if this is the step
in your proof, relying on the set solution for r:I should have
written r=[pi/4, pi/5]>Well, assume that r is equal to pi/4.
In that case, since pi/4 is>computable, we can easily compute
H(n) for any n---just computeYou're using the fact that a
function doesn't have an answer to exhaustthe range of
possible values. Halt(n) = 0 or 1. According to your
proofHalt(n) != 0 and Halt(n) != 1.Herc
[...]> Actually,
Canotr prved it for every list, irrespective of any>
allegations about completenss. In Cantor's version of the
diagonal> proof, there was no a priori assumption about any
list being> complete, Cantor showed in a direct proof that
EVERY list was> incomplete.Yes, I've actually seen you
explain
this before, and I have noreason to doubt you. It doesn't
seem
like a very important pointto me, though. Do some people have
more trouble following proofsby contradiction? Perhaps that
was true when we had a lessevolved mathematical notation to
use? I am thinking of the originalstatement of Fermat's Last
Theorem:: Cubem autem in duos cubos, aut quadratoquadratum in
duos : quadratoquadratos, et generaliter nullam in in'nitum
ultra : quadratum potestatem in duos ejusdem nominis fas est
dividere(Pierre de Fermat, c.
1630)http://www.bath.ac.uk/~ns1cc/fermats.htmlI could imagine
that a proof even slightly unclear could becomeunusable, given
that notation. But I don't see why a proofby contradiction
would be any less clear than a direct proof.> Since you've
given an explicit list of reals, it's possible > to give an
explicit real not on the list. Let>> T_n = the real output by
the Turing machine de'ned by n> T_nm = the m-th digit in the
decimal expansion of T_n> D_n = some digit other than T_nn,
eg, (T_nn + 1) mod 10> D = SUM_n D_n/10^n> Care should be
taken that the replacement digit used is never > zero and
never one less that the base of the basal representation >
(e.g., not 0 or 9 in base 10), to avoid the problem of dual
basal> representations of the same number. One must therefore
avoid base> two. This can easily be done by converting base 2
to base 4 by > using pairs of binary digits instead of single
digits, and using > only 01 or [10] as replacements.Ouch! You
are, of course, right. I wonder, did Cantor includethat point
in his original proof? I don't know my math historywell
enough
to say when the idea of limits became wellenough developed that
it was accepted that1.00000... = 0.99999...Jim Burns
=|-|erc
says...> |-|erc says...>IOW, Turing Machines *can*
calculate the *values* of non computable functions,>>many are
just integers, we just don't know what non computable
function
they>>are the value for. As I've tried to put it 5 times now
lol.>> That's right---every noncomputable function can be
approximately by>..>> there will be *some* decimal place at
which R and r disagree.>>No, calculating the exact values of
NCF's is trivial if the>function maps to integers.Sorry, I
don't know what you mean. The precise claim abouta
noncomputable real such as r = sum over all n of H(n) 10^{-n}
where foreach n, H(n) = 0 if Turing machine n halts on input
n, and H(n) = 1 otherwise.is this: there is no computable
function f such that forall n,f(n) = the nth decimal place of
r. Or to rephrase: given anyfunction f from N to N, there
exists a number n such that f(n)is not equal to the nth
decimal place of r.So the set of computable reals does not
include r.>Once again I'm not approximating any function,
I'm
de'ning the>range for its result.>>? The result to a
computable problem may be>the same number as the solution to a
non computable problem.>>1/ functions are from numbers (domain)
to numbers (range)>2/ there is more than 1 function>3/ if a
function result is not X, another function may have result
XAgain, I'm not sure what you mean, but
Turing's proof of
theunsolvability of the halting problem can be stated as
follows: For every total computable function f, there exists a
natural number n such that H(n) is unequal to f(n).So as a
function from N into N, H is unequal to any computable
function.>Cantors proof assumes a countable but INFINITE
list,A function from N into N is >I am showing irrationals are
countable,No, you're not. You're making the
claim that every
irrationalnumber is produced by some Turing machine. You're
not showingit. And you can't show it (since it is provably
false).>And if anyone can see my argument that computing a
value to a non computable>function is trivially possible then
say so.Of course it is possible to compute a value to a non
computable function.The claim is that if H is noncomputable,
then there is no computablefunction f such that for every
input n, f(n) = H(n).>I'm going against all text books here
and>its hardly worth it if a dozen people just>keep quoting
them and get stuck on such>a simple notion that a single
function>doesn't dictate the existence of the number>that
happens to be its answer.I haven't mentioned any
textbooks.>>I'm using the fact that H(n) is noncomputable to
prove that r cannot>>be equal to pi/4 *and* that r cannot be
equal to pi/5. First you>>prove that r cannot be equal to
pi/4. Then you prove that it can't>>be equal to pi/5.
Together, these proofs show that it is false that>>r=pi/4 OR
r=pi/5.>>Not if this is the step in your proof, relying on the
set solution for r:>I should have written r=[pi/4, pi/5]I
don't
know what you are talking about. I can prove that r isnot equal
to pi/4. I can prove that r is not equal to pi/5. Together,this
proves that it is false that r = pi/4 OR r = pi/5>>Well, assume
that r is equal to pi/4. In that case, since pi/4
is>>computable, we can easily compute H(n) for any n---just
compute>>You're using the fact that a function
doesn't have an
answer to exhaust>the range of possible values. Halt(n) = 0 or
1. According to your proof>Halt(n) != 0 and Halt(n) != 1.No,
I'm not saying that. I'm saying that there
exists some n
suchHalt(n) != the nth decimal place of pi/4. Since r is
de'nedin terms of Halt, we can say that there exists some n
in
whichthe nth decimal place of r is unequal to the nth decimal
placeof pi/4.--Daryl McCulloughIthaca, NY
|-|erc
says...> |-|erc says...>IOW, Turing Machines *can*
calculate the *values* of non computable functions,>>many are
just integers, we just don't know what non computable
function
they>>are the value for. As I've tried to put it 5 times now
lol.>> That's right---every noncomputable function can be
approximately by>..>> there will be *some* decimal place at
which R and r disagree.>>No, calculating the exact values of
NCF's is trivial if the>function maps to integers.>> Sorry,
I
don't know what you mean. The precise claim about> a
noncomputable real such as>> r = sum over all n of H(n)
10^{-n}> where foreach n, H(n) = 0 if Turing machine n halts
on input n,> and H(n) = 1 otherwise.>> is this: there is no
computable function f such that forall n,> f(n) = the nth
decimal place of r. Or to rephrase: given any> function f from
N to N, there exists a number n such that f(n)> is not equal to
the nth decimal place of r.That is unjusti'ed. It should
read:
there is no computable functionthat we can identify f such
that..>> So the set of computable reals does not include
r.invalid conclusion>>Once again I'm not approximating any
function, I'm de'ning the>range for its
result.>>? The result
to a computable problem may be>the same number as the solution
to a non computable problem.>>1/ functions are from numbers
(domain) to numbers (range)>2/ there is more than 1
function>3/ if a function result is not X, another function
may have result X>> Again, I'm not sure what you mean, but
Turing's proof of the> unsolvability of the halting problem
can be stated as follows:>> For every total computable
function f, there exists a natural number> n such that H(n) is
unequal to f(n).>> So as a function from N into N, H is unequal
to any computable function.>This is irrelavent to the claim
that reals are computable as long as *some* function,any
function, some uncategorised but TM indexed function covers
the outputof any H(n). Because a 1 state TM can output 1, and
another 1 state TM can output 0,and this is all any H(n) will
ever output, H being uncomputable is not an issuebecause the
numbers it theoretically and practically outputs are already
computed.>Cantors proof assumes a countable but INFINITE
list,>> A function from N into N is>>I am showing irrationals
are countable,>> No, you're not. You're making
the claim that
every irrational> number is produced by some Turing machine.
You're not showing> it. And you can't show it
(since it is
provably false).Its not provably false because you
misinterpret non computability andassume it puts gaps on the
number line which is wrong. I've shownit can catlog every
rational in a count and it shows every computablereal, that
leaves random numbers and non computable functions, andthere
is NO thoerem that the outputs of NCF are uniquely used fora
that single function. Integer output NCFs are examples of
this.>>And if anyone can see my argument that computing a
value to a non computable>function is trivially possible then
say so.>> Of course it is possible to compute a value to a non
computable function.> The claim is that if H is noncomputable,
then there is no computable> function f such that for every
input n, f(n) = H(n).>>I'm going against all text books here
and>its hardly worth it if a dozen people just>keep quoting
them and get stuck on such>a simple notion that a single
function>doesn't dictate the existence of the number>that
happens to be its answer.>> I haven't mentioned any
textbooks.>I'm using the fact that H(n) is noncomputable to
prove that r cannot>>be equal to pi/4 *and* that r cannot be
equal to pi/5. First you>>prove that r cannot be equal to
pi/4. Then you prove that it can't>>be equal to pi/5.
Together, these proofs show that it is false that>>r=pi/4 OR
r=pi/5.>>Not if this is the step in your proof, relying on the
set solution for r:>I should have written r=[pi/4, pi/5]>> I
don't know what you are talking about. I can prove that r
is>
not equal to pi/4. I can prove that r is not equal to pi/5.
Together,> this proves that it is false that r = pi/4 OR r =
pi/5>Well, assume that r is equal to pi/4. In that case,
since pi/4 is>>computable, we can easily compute H(n) for any
n---just compute>>You're using the fact that a function
doesn't have an answer to exhaust>the range of possible
values. Halt(n) = 0 or 1. According to your proof>Halt(n) != 0
and Halt(n) != 1.>> No, I'm not saying that.
I'm saying that
there exists some n such> Halt(n) != the nth decimal place of
pi/4. Since r is de'ned> in terms of Halt, we can say that
there exists some n in which> the nth decimal place of r is
unequal to the nth decimal place> of pi/4.>Ok, assume n is a
large number for a complex TM and it seems to runinde'nately
to no set pattern, it may halt or not. assume n =
555.Halt(555) = 1 if the program terminatesHalt(555) = 0 if
the program runs foreverMy claim is the result of Halt is 0 or
1.What is your claim?Herc
=|Yes, I've actually seen you
explain this before, and I have no|reason to doubt you. It
doesn't seem like a very important point|to me, though. Do
some people have more trouble following proofs|by
contradiction?People have preferred proofs without
contradiction for variousreasons. Once it was supposed that if
one wanted to give aproof that gave the cause of a mathematical
result beingtrue, it would have to be a proof not using
contradiction. Someforms of contradiction are nonconstructive
(but that's not anissue in this case, and I doubt Cantor
would
have minded).I think Cantor wanted to avoid any potential
philosophicaltaint that might increase skepticism toward his
controversialnew ideas.Keith Ramsay
=|-|erc says...>>
Sorry, I don't know what you mean. The precise claim about>>
a
noncomputable real such as>> r = sum over all n of H(n)
10^{-n}>> where foreach n, H(n) = 0 if Turing machine n halts
on input n,>> and H(n) = 1 otherwise.>> is this: there is no
computable function f such that forall n,>> f(n) = the nth
decimal place of r. Or to rephrase: given any>> [computable]
function f from N to N, there exists a number n>> such that
f(n) is not equal to the nth decimal place of r.>>That is
unjusti'ed. It should read: there is no computable
function>that we can identify f such that..No, it doesn't
have
anything to do with whether we can identify f.*Every*
computable function f is unequal to H for some input n.So if f
is any computable function, at least one of the following
isfalse: f(0) = H(0) f(1) = H(1) f(2) = H(2) f(3) = H(3) . . .
etc.>> Again, I'm not sure what you mean, but
Turing's proof of
the>> unsolvability of the halting problem can be stated as
follows:>> For every total computable function f, there
exists a natural number>> n such that H(n) is unequal to
f(n).>> So as a function from N into N, H is unequal to any
computable function.>>This is irrelavent to the claim that
reals are computable as long as>*some* function, any function,
some uncategorised but TM indexed>function covers the output of
any H(n).I don't know what you mean by covers the output of
any
H(n). Obviously,you can 'nd one Turing machine that will
output
H(0), and a differentTuring machine that will output H(1), and
yet another Turing machine thatwill output H(2). But there is
no *single* Turing machine that willinput any natural number n
and output H(n).But let me repeat what it means to say that a
real number is computable.If r is computable, then there is at
least one Turing machine program fsuch that for all n, f(n) =
the nth decimal place of r.For example, pi is computable.
There is a single program f such thatif you input 0, f will
output 3. If you input 1, f will output 1. Ifyou input 2, f
will output 4. Etc.There is no such function that will compute
the number r whose nthdecimal place is 0 if the nth Turing
machine halts on input n, and 1otherwise.>Because a 1 state TM
can output 1, and another 1 state TM can output 0,>and this is
all any H(n) will ever output, H being uncomputable is not>an
issue because the numbers it theoretically and practically
outputs>are already computed.I really don't understand what
you are getting at. You seem to realizethat H is
noncomputable. But H is noncomputable if and only if r
isnoncomputable. By de'nition, a real number is noncomputable
if and onlyif it's decimal expansion is noncomputable, and
the
decimal expansionof r is H.>> No, you're not.
You're making the
claim that every irrational>> number is produced by some Turing
machine. You're not showing>> it. And you
can't show it (since
it is provably false).>>Its not provably false because you
misinterpret non computability and>assume it puts gaps on the
number line which is wrong.I never said anything about number
lines or gaps, so I'm de'nitelynot assuming
anything about
them. All I've said is that there is nocomputable function f
such that f(1) = 'rst decimal place of r f(2) = second
decimal
place of r f(3) = third decimal place of r etc.>> No, I'm
not
saying that. I'm saying that there exists some n such>>
Halt(n) != the nth decimal place of pi/4. Since r is de'ned>>
in terms of Halt, we can say that there exists some n in
which>> the nth decimal place of r is unequal to the nth
decimal place>> of pi/4.>>Ok, assume n is a large number for
a complex TM and it seems to run>inde'nately to no set
pattern,
it may halt or not. assume n = 555.>>Halt(555) = 1 if the
program terminates>Halt(555) = 0 if the program runs
forever>>My claim is the result of Halt is 0 or 1.Of course
that's true, but it is irrelevant.The question
isn't whether
Halt(555) is computable.The question is whether there exists a
single program that will computeHalt(0), Halt(1), Halt(2),
Halt(3), etc.Think about computing pi. If I asked for a
program to compute pi, itwouldn't be good enough for you to
give me one program that computes 3,a second program that
computes 1, a third program that computes 4, afourth program
that computes 1, a 'fth program that computes 5, etc.I want a
*single* program that takes n as input and outputs the
nthdecimal place of pi.>What is your claim?That there is no
program f that takes n as an input and outputs thenth decimal
place of r. Yes, there is a program that can output the'rst
decimal place of r. Yes, there is a program that can output
thesecond decimal place of r. Yes, there is a program that can
output thethird decimal place of r. But there is no *single*
program that cantake an arbitrary input n and output the nth
decimal place of r.For comparison, here is a (very
inef'cient)
program for computingsquare-root(2): int sqrt-two(int n) { int
x = 2; int i = 1; while (i <= n) { x = x*10; i = i+1; } // at
this point, x = 2 * 10^n int y = 1; while ((y+1)*(y+1) <= x) {
y = y+1; } // at this point, y = the largest integer less than
or // equal to sqrt(x) int z = y/10; // Because of roundoff,
this throws away the // 'rst decimal place of y. return y -
10*z; // This is the 'rst decimal place of y. }This program
takes an *arbitrary* input n and outputs the nthdecimal place
of square-root(2).My claim is that there is no similar program
that computesthe real number r whose nth decimal place is
Halt(n).--Daryl McCulloughIthaca, NY
> Sorry, I don't know
what you mean. The precise claim about>> a noncomputable real
such as>> r = sum over all n of H(n) 10^{-n}>> where foreach
n, H(n) = 0 if Turing machine n halts on input n,>> and H(n) =
1 otherwise.>> is this: there is no computable function f
such that forall n,>> f(n) = the nth decimal place of r. Or to
rephrase: given any>> [computable] function f from N to N,
there exists a number n>> such that f(n) is not equal to the
nth decimal place of r.>>That is unjusti'ed. It should read:
there is no computable function>that we can identify f such
that..>> No, it doesn't have anything to do with whether we
can identify f.> *Every* computable function f is unequal to H
for some input n.> So if f is any computable function, at least
one of the following is> false:>> f(0) = H(0)> f(1) = H(1)>
f(2) = H(2)> f(3) = H(3)>H is not computable, you can't
perform such an identi'cation.I'm not saying a
computable
function exists that equals H, as a matterof fact I doubt it,
I'm saying there could be but we would never beable to show
it.there is no computable function f such that forall n,... is
not to betaken literally, its not a phrase designed for the
context of total mappingsof functions. computable function in
this context means known computablefunction. if you can show
otherwise do so but I doubt you can.obviously you would
interpret there IS a computable function.. as we wouldbe able
to use the function wouldn't you? The opposite meaning of
this
is thereis no function we can use, not there is no
function.>>Ok, assume n is a large number for a complex TM and
it seems to run>inde'nately to no set pattern, it may halt or
not. assume n = 555.>>Halt(555) = 1 if the program
terminates>Halt(555) = 0 if the program runs forever>>My claim
is the result of Halt is 0 or 1.>> Of course that's true,
but
it is irrelevant.Well, assume that H(555) is equal to 1. In
that case, since 1 iscomputable, we can easily compute H(555),
then H(555) = 1, thenit follows that H(n) is computable.I'm
using the fact that H(n) is noncomputable to prove that H(555)
cannotbe equal to 1 *and* that H(555) cannot be equal to 0.
First youprove that H(555) cannot be equal to 1. Then you
prove that it can'tbe equal to 0. Together, these proofs
show
that it is false thatH(555)=1 OR H(555)=0.Herc
=|-|erc
says...> So if f is any computable function, at least one of
the following is>> false:>> f(0) = H(0)>> f(1) = H(1)>> f(2)
= H(2)>> f(3) = H(3)>>H is not computable, you can't perform
such an identi'cation.>I'm not saying a
computable function
exists that equals H, as a matter>of fact I doubt it, I'm
saying there could be but we would never be>able to show it.No
there couldn't be. It doesn't have anything to
do with
whetherwe are able to show it or not.>there is no computable
function f such that forall n,... is not to be>taken
literally,Yes, it certainly is. We have an enumeration of all
partialrecursive functions (a function is partial if it fails
tohalt on some inputs). So let f_n be the nth partial
recursivefunction. The claim is literally forall n, exists m,
f_n(m) doesn't halt, or f_n(m) halts and is unequal to
H(m)>its not a phrase designed for the context of total
mappings>of functions. computable function in this context
means known computable>function.No, it doesn't. We can
enumerate all *possible* partial recursivefunctions. We can
prove that *none* of them computes the same functionas H. So
it's not just that no *known* computable function computes
H,no *possible* computable function computes H.>if you can
show otherwise do so but I doubt you can.>obviously you would
interpret there IS a computable function..>as we would be able
to use the function wouldn't you? The opposite>meaning of
this
is there is no function we can use, not there is>no
function.But the proof works for *all* computable functions,
not just theones that we know. More speci'cally, for every
index n, we canprove that f_n does not compute H. Forevery n,
there exists an nsuch that f_n(m) fails to halt or f_n(m) !=
H(m).>I'm using the fact that H(n) is noncomputable to prove
that H(555) cannot>be equal to 1 *and* that H(555) cannot be
equal to 0.Well, that doesn't make any sense. The fact that
H
is a noncomputablefunction doesn't imply that H(555) is
noncomputable.We've already been through this. There is
certainly a computablefunction that outputs H(0). There is a
*different* computablefunction that outputs H(1). There is yet
another computable functionthat outputs H(2). There is yet
another computable function thatoutputs H(555). All that is
irrelevant to the question of whetherH is computable. For H to
be computable, there must be a singlecomputable function f such
that f(0) = H(0) AND f(1) = H(1) ANDf(2) = H(2), etc.>First you
prove that H(555) cannot be equal to 1.How do you prove
that?>Then you prove that it can't>be equal to 0.How do you
prove that?>Together, these proofs show that it is false
that>H(555)=1 OR H(555)=0.Yes, if you gave a proof that H(555)
is unequal to 0,and you also gave a proof that H(555) is
unequal to 1, thenyou would have shown that it is false
H(555)=1 or H(555)=0.That's completely correct.
Unfortunately,
you don't havea proof that H(555) can't be
equal to 0, and you
don't havea proof that H(555) can't be equal
to 1. So you
haven't shownanything.This is in contrast to my
noncomputable
real r. We *can* provethat r is not equal to pi/4. And we can
also prove that r isnot equal to pi/5. So we can show that it
is false thatr=pi/4 or r=pi/5.Here's the proof. Let f be the
function that computes thedecimal expansion for pi/4. (So for
all n, f(n) = nth decimalplace of pi/4). De'ne a new function
g as follows: int g(int n) { int i = 1; if (f(n) == 0) { while
(i > 0) { i = i+1; } return 1; } else { return 0; } }So g is
de'ned so that if f(n) = 0, then g(n) never halts,and if f(n)
> 0, then g(n) = 0. We can compute the Turingnumber of g. Call
that k. Now consider the computation g(k).By de'nition of H,
if
g(k) never halts, then H(k) = 1. Ifg(k) halts, then H(k) = 0.
So we have two cases: 1. g(k) never halts, so H(k) = 1. But by
the explicit de'nition of g, the only way for g(k) to fail to
halt is if f(k) = 0. So we have H(k) = 1 and f(k) = 0. So f(k)
is unequal to H(k). 2. g(k) halts, so H(k) = 0. But by
de'nition of g, if g(k) halts, it must be because f(k) is
unequal to 0. So we have H(k) = 0 and f(k) is nonzero. So once
again, f(k) is unequal to H(k).So it is impossible for f(k) to
be equal to H(k). Butf(k) = the kth decimal place of pi/4, and
H(k) = thenth decimal place of r. So r and pi/4 can't be
thesame real numbers, since they differ in position k.The same
proof goes through for every computable real.No computable real
can be equal to r. --Daryl McCulloughIthaca, NY
>And if
anyone can see my argument that computing a value to a non
computable>function is trivially possible then say so. I'm
going against all text books here and>its hardly worth it if a
dozen people just keep quoting them and get stuck on such>a
simple notion that a single function doesn't dictate the
existence of the number>that happens to be its answer.For a
function from N to N, yes, this is trivial: Every value (being
aninteger) is computable by a constant function, and constant
functions toN are computable. But so what?For a function from
N into R, this not true, however. The difference isthat a
single value in N can be 'nitely speci'ed, but
some values in
RcanNOT be 'nitely speci'ed. So the value of a
function from N
into Rmay be non-computable. Even constant functions (from 1
into R) need notbe computable -- unlike functions from 1 into
N.Michel.
>I'm using the fact that H(n) is noncomputable
...That *fact* is wrong. H(n) *is* computable (for every n).
H(n) is simplyan integer; in fact, quite restricted: a member
of {0, 1}. But H (thefunction) is NOT computable.Daryl said it
well: you need a *different* computable function (a
constantfunction will do) for each n. (In fact, you only need
two such computablefunctions). Computability of a *function*
(as opposed to a value in thatfunction's range) is a
different
matter.All you're saying is that every real number is
computable, because we haveten constant functions into
{0,1,2,3,4,5,6,7,8,9} that together permit anydigit of any
real number to be computed -- and those ten functions are
allcomputable! Wowee.Michel.
To sum up, though, I
wouldn't say that all these failures to adopt> alternatives
to
the formalist position add up to being a formalist even> in an
armchair sort of way, and I have my doubts whether enough>
mathematicians are actually opinionated enough to count as
de'nitely> belonging to one camp or the other.
I'd be
interested to know if I'm> mistaken about that.> Keith
RamsayI read an interesting paper in a journal on logic.
(Logic, perhaps? It's only a few months old, at any rate.)
The
authors design formallanguages which they demonstrate describe
the ontologies ofWittgenstein's factualism, platonism, and
realism. They then provethat all three formal languages are
equivalent. That is, anythingwhich can be described by one
language can be described by the othertwo. So there is no a
priori preferred language. So my thinking isthat a convention
should be established which allows mathematicians
tocommunicate clearly. And I think that's already happened
with ZFC andits platonist leanings.I can 'nd a reference next
time I go to campus.Alex SollaJuniorReed College
=I'm
looking for everything I can 'nd about
Moessner's algorithm
forfactorials.Any help appreciatedSandra
>What is your
de'nition of an even number?>> Two is an even number, and if
x is an even number, then x+2 is an even>number.>A slightly odd
de'nition of an even number. It means that zero is not
an>even
number.You're right, of course - I was trying to be as brief
as possible while stillcoming up with the simplest possible
example of an induction proof.Doug
Can someone please
provide an example of a VERY simple theorem --
something>trivial NOT involving sums, divisibility, modular
arithmetic, etc. -- that>n < 2^n for nonnnegative n.2n <= 2^n
for nonnegative n, with strict inequality when n > 2.The sum
of the angle measures of an n-sided convex polygon is 180(n-2)
degrees.Frankly, I don't see what is nontrival about the
veri'cation of formula for the sum of the 'rst n
positive
integers. I remember 'nding it extremely neat (i.e., cool)
the
'rst time I saw it.-- Stephen J. Herschkorn
herschko@rutcor.rutgers.edu
What is your de'nition of an
even number?>> Two is an even number, and if x is an even
number, then x+2 is an even number.>> Then why don't you use
this for the base step?>> He did, didn't he?>>Consider the
'rst even number, 2. 2 is clearly divisible by 2,>since 2/2 =
1, an integer>> The base step must establish that the 'rst
even number is divisible> by 2. By de'nition, 2 is the
'rst
even number. We show that it is> divisible by 2. The base step
is complete.Yep, I was thinking he stated it (since 2 is
divisible by 2, then 2 is even),but he correctly did it the
other way around, already knowing 2 was even.More an
association than a proof, based on the fact even is usually
de'ned such :Defn : even numbers are divisible by 2Theorem :
if x is even x + 2 is evenProof:consider the value 2, 2 is
divisible by 2, therefore its evenNow, assume that the theorem
is true for a general even number x=2n. Anothereven number
would therefore be 2n+2, since 2n+2 = 2(n+1), it followsthat
(2n+2)/2 = n+1, and therefore, 2n+2 is divisible by
2.Therefore 2n+2 is even. if 2n is even 2n+2 is
even.Herc
Can someone please provide an example of a
VERY simple theorem -- something>trivial NOT involving sums,
divisibility, modular arithmetic, etc. -- that>One more:A
'nite set with n elements has 2^n distinct subsets. (|P(X)| =
2^|X| for any 'nite set X.)-- Stephen J. Herschkorn
herschko@rutcor.rutgers.edu
Can someone please provide
an example of a VERY simple theorem -- something>trivial NOT
involving sums, divisibility, modular arithmetic, etc. --
that> THEOREM: All even numbers are divisible by two.>
PROOF: Consider the 'rst even number, 2. 2 is clearly
divisible by 2,There is no 'rst even number because if n is
even so is n - 2 .GC> since 2/2 = 1, an integer.> Now,
assume that the theorem is true for a general even number 2n.
The next> even number would therefore be 2n+2, and since 2n+2
= 2(n+1), it follows> that (2n+2)/2 = n+1, and therefore, 2n+2
is divisible by 2.> Doug
Can someone please
provide an example of a VERY simple theorem --
something>>trivial NOT involving sums, divisibility, modular
arithmetic, etc. -- that THEOREM: All even numbers are
divisible by two. PROOF: Consider the 'rst even number,
2. 2 is clearly divisible by 2,>There is no 'rst even number
because if n is even so is n - 2 .Depends on your de'nition
of
even number, now doesn't it?If I might, you're
being a bit
pedantic, since the OP was only asking fora very simple
induction proof, which I provided. If you're going to bean
asshole about it, then 'ne: all POSITIVE even numbers are
divisibleby two. Happy?Doug
Can someone please provide an
example of a VERY simple theorem -- something> trivial NOT
involving sums, divisibility, modular arithmetic, etc. --
that> Dan> Show that any amount of postage greater than or
equal to 2 cents can bemade up with only 2 cent and 3 cent
stamps.The Tower of Hanoi. (Google Tower of Hanoi if you
aren't familiarwith the problem.)-- Paul SperryColumbia, SC
(USA)
Can someone please provide an example of a
VERY simple theorem -- something>>trivial NOT involving sums,
divisibility, modular arithmetic, etc. -- that>> THEOREM:
All even numbers are divisible by two.>> PROOF: Consider the
'rst even number, 2. 2 is clearly divisible by 2,>>There is
no
'rst even number because if n is even so is n - 2 .>> Depends
on your de'nition of even number, now doesn't
it?>> If I
might, you're being a bit pedantic, since the OP was only
asking for> a very simple induction proof, which I provided.
If you're going to be> an asshole about it, then
'ne: all
POSITIVE even numbers are divisible> by two. Happy?>Just
clarifying that part about pedantic asshole, 0 is not negative
would thatmake it positive? If so 0 would be the 'rst
even.Herc
Can someone please provide an example of a VERY
simple theorem -- something> trivial NOT involving sums,
divisibility, modular arithmetic, etc. -- that>RTP 2x < 3x x E
NBASE STEP:x=12x = 23x = 3THERFORE 2x < 3xINDUCTIVE STEP:Assume
2x < 3x2(x+1) = 2x + 23(x+1) = 3x + 32x + 2 < 3x + 3, (as 2x <
3x, and 2 < 3)2x < 3x is true for x=1, and when true for x it
is true for x+1THEREFORE 2x < 3xHerc
Can someone please
provide an example of a VERY simple theorem -- something>
trivial NOT involving sums, divisibility, modular arithmetic,
etc. -- that> Dan> Theorem. All sets of n cars are the same
color.Proof: The case n=1 is easy. Let us assume the case n,
and prove the case n+1. Consider a set A = {x1,x2,...,x(n+1)}
of n+1 cars. By the induction hypothesis, the cars B =
{x1,x2,...,xn} are the same color, as are the cars C =
{x2,...,x(n+1)}. Since xn and x(n+1) are both in C, they both
have the same color. Hence all the cars in set A = B union
{x(n+1)} have the same color.Corollary. All cars are
black.There exists at least one black car. Now apply the above
theorem, where n is the cardinality of all cars.-- Stephen
Montgomery-Smithstephen@math.missouri.eduhttp://
www.math.missouri.edu/~stephen
There exists at least one
black car. Now apply the above theorem, where n is >the
cardinality of all cars.For those trying to 'nd the §aw
in
this proof, consider what the base caseis (and should
be).Doug
> Can someone please provide an example of a VERY
simple theorem -- something> trivial NOT involving sums,
divisibility, modular arithmetic, etc. -- that> can be proven
by mathematical induction in about 've lines or so?>>
Doug's
example certainly seems sound, once you've adopted his>
inductive de'nition of even numbers.>> Other possiblities are
to prove various properties shared by all> natural numbers
(which are normally inductively de'ned in the
'rst> place):
all natural numbers are positive, divisible by 1, 'nite, and>
so on.>> Why are you asking? Who is your audience?>> If
you're
trying to come up with an induction for kids example, then> the
things we've suggested so far won't be very
interesting for
them.>I am writing an introduction to the axioms for the
natural numbers (based onPeano's axioms) for a fairly
general
audience of advanced high schoolstudents and non-specialist
undergrads. If possible, I want to illustratethe induction
principle without introducing the other topics -- sums,etc. --
which are usually the basis for most introductions to induction
thatI can recall.more of them.Dan
Can someone please
provide an example of a VERY simple theorem --
something>trivial NOT involving sums, divisibility, modular
arithmetic, etc. -- thatTheorem: Beards don't exist.Proof:
Let
N be the number of hairs in a putative beard.Base case: Clearly
a chin bearing N=1 hair does not sport a beard.Inductive step:
If N hairs don't make a beard, then surely the addition of
one
more hair isn't going to make it a beard, is it?By induction
on
N, we see that for every natural number N, a set ofN hairs is
not a beard.QED
Can someone please provide an example of a
VERY simple theorem -- something>trivial NOT involving sums,
divisibility, modular arithmetic, etc. -- that>> Theorem:
Beards don't exist.>> Proof: Let N be the number of hairs in
a
putative beard.>> Base case: Clearly a chin bearing N=1 hair
does not sport a beard.> Inductive step: If N hairs don't
make
a beard, then surely the addition> of one more hair isn't
going
to make it a beard, is it?>> By induction on N, we see that for
every natural number N, a set of> N hairs is not a
beard.>Theorem: Camels have strong backs....Herc
Can
someone please provide an example of a VERY simple theorem --
> something> trivial NOT involving sums, divisibility, modular
arithmetic, etc. -- > that> can be proven by mathematical
induction in about 've lines or so?>> Doug's
example certainly
seems sound, once you've adopted his> inductive
de'nition of
even numbers.>> Other possiblities are to prove various
properties shared by all> natural numbers (which are normally
inductively de'ned in the 'rst> place): all
natural numbers
are positive, divisible by 1, 'nite, and> so on.>> Why are
you
asking? Who is your audience?>> If you're trying to come up
with an induction for kids example, then> the things we've
suggested so far won't be very interesting for them.> I am
writing an introduction to the axioms for the natural numbers
(basedon> Peano's axioms) for a fairly general audience of
advanced high school> students and non-specialist undergrads.
If possible, I want to illustrate> the induction principle
without introducing the other topics -- sums,> etc. -- which
are usually the basis for most introductions to inductionthat>
I can recall.>> more of them.>The simplest theorem provable by
induction (so far) seems to be theThe successor of any number
k cannot be k itself.I like this one because it makes use of
nothing beyond Peano's axioms -- noteven addition!Dan
I
am writing an introduction to the axioms for the natural
numbers (based on> Peano's axioms) for a fairly general
audience of advanced high school> students and non-specialist
undergrads. If possible, I want to illustrate> the induction
principle without introducing the other topics -- sums,> etc.
-- which are usually the basis for most introductions to
induction that> I can recall.In that case your audience might
appreciate the well-known analogy between induction and those
fancy arrangements of blocks of dominos:1) The induction step
is the law that an appropriately placeddomino block will
induce the next block to fall, when it itself isfalling.2) The
need for an initial case requiring separate checking is
clear.3) Behold! In the end all the blocks are down (Some
studentsfeel cheated upon after the 'rst lecture on
induction:
theirargument being that how can we assume the thing we are
supposedto prove:)Good luck,Jyrki
Can someone please
provide an example of a VERY simple theorem -- something>
trivial NOT involving sums, divisibility, modular arithmetic,
etc. -- that> DanThe best you can do is to look over the
book :The method ofmathematical induction by I.S. SOMINSKII .
Blaisdell Publishing(1961)L Rodriguez
> Can someone please
provide an example of a VERY simple theorem -- something>>
trivial NOT involving sums, divisibility, modular arithmetic,
etc. -- that DanTheorem. All sets of n cars are the
same color.>>Proof: The case n=1 is easy. Let us assume the
case n, and prove the case n+1. > Consider a set A =
{x1,x2,...,x(n+1)} of n+1 cars. By the induction >hypothesis,
the cars B = {x1,x2,...,xn} are the same color, as are the
cars C = >{x2,...,x(n+1)}. Since xn and x(n+1) are both in C,
they both have the same >color. Hence all the cars in set A =
B union {x(n+1)} have the same color.>>Corollary. All cars are
black.>>There exists at least one black car. Now apply the
above theorem, where n is >the cardinality of all cars.Well, a
topology professor of mine (with a somewhat whimsical sense
ofhumor, we once challanged him to make a pun on every
statement made bya bunch sitting around a table, and he went
on for about a half hour)presented that theorem in its older
form - all horses are the samefollows : Asuume the conclusion
is false. But that would be a horseof another color. That
doesn't work with cars.Larry(this space unintentially left
blank .....
=For a set of vectors S = {v1, v2, . . . , vn} ,
prove that span (S) is theintersection of all subspaces that
contain S.I am stuck on this one. I think i dont really get
the question. I mean, ifwe have two subspaces A and B who both
contain S. Then the interesection ofA and B would just be S
wouldnt it ??Any help apreciated!
For a set of vectors S
= {v1, v2, . . . , vn} , prove that span (S)> is the
intersection of all subspaces that contain S.> I am stuck on
this one. I think i dont really get the question. I> mean, if
we have two subspaces A and B who both contain S. Then the>
interesection of A and B would just be S wouldnt it ??Is that
the way it would work for sets rather than
subspaces?Bart
=Balcha Abanefso <fake@none.com> scribbled
the following:> For a set of vectors S = {v1, v2, . . . , vn}
, prove that span (S) is the> intersection of all subspaces
that contain S.> I am stuck on this one. I think i dont really
get the question. I mean, if> we have two subspaces A and B who
both contain S. Then the interesection of> A and B would just
be S wouldnt it ??> Any help apreciated!I read the question
as: S is the intersection of all those sets T, whereS is a
subset of T.Here's my idea of a proof: S itself is one such
a
set T. Theintersection of any collection of sets cannot be
larger than thesmallest set in that collection. Go from
there.-- /-- Joona Palaste (palaste@cc.helsinki.')
---------------------------| Kingpriest of The Flying Lemon
Tree G++ FR FW+ M- #108 D+ ADA N+++||
http://www.helsinki.'/~palaste W++ B OP+
|----------------------------------------- Finland rules!
------------/
For a set of vectors S = {v1, v2, . . . ,
vn} , prove that span (S)> is the intersection of all
subspaces that contain S.Here are some hints.Span(S) is the
set of all linear combinations c1 v1 + c2 v2 + ... + cn
vnwhere each ci is a scalar. Span(S) is a subspace because (c1
v1 + c2 v2 + ... + cn vn) + (d1 v1 + d2 v2 + ... + dn vn) = (c1
+ d1) v1 + ... + (cn + dn) vn is again in span(S)and C (c1 v1 +
c2 v2 + ... + cn vn) = (C c1) v1 + ... + (C cn) vn is in
span(S).Let U be a subspace containing all the vectors in
S.Then span(S) is contained in U, because the subspace Umust
contain all linear combinations of the vectors in S.Thus
span(S) is the smallest subspace containing S. Verify that the
intersection W of all subspaces containingS is a subspace that
contains S. Then span(S) is containedin W because span(S) is
smaller than W. On the other hand,if v is in W, then v is in
every subspace that contains S,and in particular v is in
span(S), which means that W iscontained in span(S). So span(S)
= W.> I am stuck on this one. I think i dont really get the
question. I> mean, if we have two subspaces A and B who both
contain S. Then the> interesection of A and B would just be S
wouldnt it ??The answer to this should be clear from the
remarks above. WhenA and B are subspaces containing S, each
mustcontain all linear combinations of the vectors in S.--
______________________________________________________________
___________Levent Kitis
lk3a@esmeralda.mech.virginia.edu______________________________
___________________________________________
For a set of
vectors S = {v1, v2, . . . , vn} , prove that span (S) isthe>
intersection of all subspaces that contain S.>> I am stuck on
this one. I think i dont really get the question. I mean,if>
we have two subspaces A and B who both contain S. Then the
interesectionof> A and B would just be S wouldnt it ??>> Any
help apreciated!> I'm kind of rusty on this, but
isn't the SPAN
de'ned as the set of allvector sums of the elements of S.I
might start out this way.. just to form an example, and then
generalizeit.Let v1 = <1,2>and v2 = < 2,3>then span (v1,v2) =
the whole plane.The problem here is to be more speci'c about
the de'nition of a vectorsubSPACE. because of the requirement
that a subspace itself must have theessential de'ning
properties of the space ( otherwise it is merely a subSET).)So
the vectors <0,0> and <1,1> automatically get included.and
consequently the whole plane.The intersection with itself is
still the whole plane.Is this coo-coo or am I on the right
track?Bob Pease
Balcha Abanefso <fake@none.com> scribbled
the following:> For a set of vectors S = {v1, v2, . . . , vn} ,
prove that span (S) isthe> intersection of all subspaces that
contain S.>> I am stuck on this one. I think i dont really get
the question. Imean, if> we have two subspaces A and B who both
contain S. Then the interesectionof> A and B would just be S
wouldnt it ??>> Any help apreciated!>> I read the question as:
S is the intersection of all those sets T, where> S is a subset
of T.> Here's my idea of a proof: S itself is one such a set
T.
The> intersection of any collection of sets cannot be larger
than the> smallest set in that collection. Go from there.>>
--> /-- Joona Palaste (palaste@cc.helsinki.')
---------------------------> | Kingpriest of The Flying Lemon
Tree G++ FR FW+ M- #108 D+ ADA N+++|> |
http://www.helsinki.'/~palaste W++ B OP+ |>
----------------------------------------- Finland rules!
------------/ I think you are confusing S with span (S). Span
(S) (or <S>) is the setof ALL linear combinations of set S
together with the binary operations ofaddition and scalar
multiplication inherited from parent vector space (callit V) >
S. It turns out that span(S) is a vector subspace of V. Now
simply show equality of span (S) and the intersection of
allsubspaces that contain S (also a subspace of V!). This is
simply a case ofshowing that each is a subset of other...Hope
this helps...Ari
=A.G. <ag@notrue.net> scribbled the
following:>> Balcha Abanefso <fake@none.com> scribbled the
following:>> For a set of vectors S = {v1, v2, . . . , vn} ,
prove that span (S) is> the>> intersection of all subspaces
that contain S.>> I am stuck on this one. I think i dont
really get the question. I> mean, if>> we have two subspaces A
and B who both contain S. Then the interesection> of>> A and B
would just be S wouldnt it ??>> Any help apreciated!>> I
read the question as: S is the intersection of all those sets
T, where>> S is a subset of T.>> Here's my idea of a proof:
S
itself is one such a set T. The>> intersection of any
collection of sets cannot be larger than the>> smallest set in
that collection. Go from there.> I think you are confusing S
with span (S). Span (S) (or <S>) is the set> of ALL linear
combinations of set S together with the binary operations of>
addition and scalar multiplication inherited from parent
vector space (call> it V) > S. It turns out that span(S) is a
vector subspace of V.It appears I was indeed confusing the
two. My proof is for showing thatS is the intersection of all
supersets of S, which is pretty muchtrivial.For span(S) it is
less trivial. I haven't studied enough linear algebrato
suggest a proof for that.-- /-- Joona Palaste
(palaste@cc.helsinki.') ---------------------------|
Kingpriest of The Flying Lemon Tree G++ FR FW+ M- #108 D+ ADA
N+++|| http://www.helsinki.'/~palaste W++ B OP+
|----------------------------------------- Finland rules!
------------/Bad things only happen to scoundrels. -
Moominmamma
=Then span(S) is contained in W because span(S)
is smaller than WLets say that we have a number of subspaces
ONLY containing S. Theintersection of those subspaces, W would
then only contain S. W would notcontain all the elements of
Span(s) and thus Span(s) would not be a subsetof W , which is
the same as saying thats span(S) is not contained in W. ?I
know you can prove me wrong here, which i hope you do. Because
this is theonly part i dont get.Count Dracula
<lk3a@esmeralda.mech.Virginia.EDU> skrev i meddelandet>> For a
set of vectors S = {v1, v2, . . . , vn} , prove that span (S)>
is the intersection of all subspaces that contain S.>> Here
are some hints.>> Span(S) is the set of all linear
combinations> c1 v1 + c2 v2 + ... + cn vn> where each ci is a
scalar. Span(S) is a subspace because> (c1 v1 + c2 v2 + ... +
cn vn) + (d1 v1 + d2 v2 + ... + dn vn)> = (c1 + d1) v1 + ... +
(cn + dn) vn is again in span(S)> and> C (c1 v1 + c2 v2 + ... +
cn vn) = (C c1) v1 + ... + (C cn) vn> is in span(S).>> Let U be
a subspace containing all the vectors in S.> Then span(S) is
contained in U, because the subspace U> must contain all
linear combinations of the vectors in S.> Thus span(S) is the
smallest subspace containing S.>> Verify that the intersection
W of all subspaces containing> S is a subspace that contains S.
Then span(S) is contained> in W because span(S) is smaller than
W. On the other hand,> if v is in W, then v is in every
subspace that contains S,> and in particular v is in span(S),
which means that W is> contained in span(S). So span(S) = W.>>
I am stuck on this one. I think i dont really get the question.
I> mean, if we have two subspaces A and B who both contain S.
Then the> interesection of A and B would just be S wouldnt it
??>> The answer to this should be clear from the remarks
above. When> A and B are subspaces containing S, each must>
contain all linear combinations of the vectors in S.>> -->
______________________________________________________________
___________> Levent Kitis lk3a@esmeralda.mech.virginia.edu>
______________________________________________________________
___________
=A recent report shows high-pitched rapidly
repeated subliminalmessages in a large number of television
broadcasts. For the mostpart the messages are of a political
nature, insisting such things asBush is good and The war in
Iraq is going well. But what ispuzzling some political
scientists is the occasional subliminalmessage regarding the
credibility of a certain mathematician. Thefollowing message
was deciphered repeatedly by audio analysts inbroadcasts from
around the nation:Cantor knows all. Cantor cannot be
disproved. Bush's economicplatform is based on
Cantor's
theorems and is therefor invincible. Cantor is omniscient.
Bush must be reelected to save the nation fromTerrorismAri
Fleisher said that the messages were sheer coincidence,
claimingthat with the appropriate 'lters it would be possible
to discoversuch messages in any broadcast whatsoever The
president was notavailable for comments.
[... something
with irony maybe (?) ... ]I didn't 'nd a
6-prime multi-set
yet.Rainer Rosenthalr.rosenthal@web.de
> [... something
with irony maybe (?) ... ]>>I didn't 'nd a
6-prime multi-set
yet.Okay, I made a slight mistake in my earlier statement. An
(n-1)-AP of primesyields an n-prime (multi)set only if the
difference is less than the initialterm. So while
5,11,17,23,29 is an arithmetic progression, we don't onlyget
a
6-prime set unless you allow {-1,6,6,6,6,6}. To get a valid
6-primeset from this method we have to go a little higher, to
{7,30,30,30,30,30}:30+7, 30+30+7, 30+30+30+7, 30+30+30+30+7,
and 30+30+30+30+30+7 are prime.There's probably a smaller
(in
terms of sum?) 6-prime set than this.Just don't ask me for a
24-prime set, I don't have one yet :). -- Erick
=Erick Bryce
Wong > To get a valid 6-prime set from this method we have to >
go a little higher, to {7,30,30,30,30,30}:> 30+7, 30+30+7,
30+30+30+7, 30+30+30+30+7, > and 30+30+30+30+30+7 are prime. There's probably a smaller (in terms of sum?) 6-prime >
set
than this.Here is a smaller one, but may be still not the
smallestone (in terms of the maximum): { 1, 1, 2, 3, 4, 8
}with the prime-adding
families
elements: 1+1+2+3+4+8=195 elements: 1+1+2+3+4=11 and
1+1+3+4+8=174 elements: 1+1+2+3=7 and 2+3+4+8=173 elements:
1+2+4=7 and 2+3+8=132 elements: 1+2=3 and 2+3=5 and 1+4=5 and
3+8=11
that every element of 1, 2, 3, 4, 8 is memberof some family
with any given number of elements.For example (for the
interested reader): Element 4 isa member of {1,4}, {1,2,4},
{2,3,4,8}, {1,1,2,3,4} and{1,1,2,3,4,8}, i.e. member of a
prime-adding family with2, 3, 4, 5 and n=6 elements.I like
this puzzle. I always like things once I've understood :-)Is
an OEIS-sequence starting to grow? a(6) = 8? Or a(6) <
8?Hopefully we get an idea where the OP David found thissort
of question.Rainer Rosenthalr.rosenthal@web.de> Just don't
ask
me for a 24-prime set, I don't have one yet :).> -- Erick>
Erick Bryce Wong >> To get a valid 6-prime set from this
method we have to >> go a little higher, to
{7,30,30,30,30,30}: There's probably a smaller (in terms
of sum?) 6-prime >> set than this.>>Here is a smaller one, but
may be still not the smallest>one (in terms of the maximum): {
1, 1, 2, 3, 4, 8 }>with the prime-adding
families>
elements: 1+1+2+3+4+8=19>5 elements: 1+1+2+3+4=11 and
1+1+3+4+8=17>4 elements: 1+1+2+3=7 and 2+3+4+8=17>3 elements:
1+2+4=7 and 2+3+8=13>2 elements: 1+2=3 and 2+3=5 and 1+4=5 and
3+8=11>
like this puzzle. >I always like things once I've understood
:-)>Is an OEIS-sequence starting to grow? >a(6) = 8? Or a(6) <
8?I did a computer search, and the 'rst solution in
lexicographicright-to-left order (which minimises a(6)) is:
{1,1,2,2,3,4}:6 elements: 1+1+2+2+3+4 = 135 elements:
1+1+2+3+4 = 114 elements: 1+1+2+3 = 7 and 2+2+3+4 = 113
elements: 1+1+3 = 5 and 1+2+4 = 72 elements: 1+2 = 3 and 3+4 =
7.The 'rst 7-prime set has an even smaller maximum:
{1,1,1,2,2,3,3}.Then the 8- and 9-prime sets just consist of
1's and 2's. Then for10-prime, it goes back up
to 3:
{1,1,1,2,2,2,2,2,3,3}, and for 11,2's and 1's
again.I'm
starting to think it should be easy to prove that for
suf'cientlylarge n, there is an n-prime set consisting only
of
1's and 2's. Somaybe we should make the
problem more
interesting by requiring distinctvalues? Even that harder
version may be tractable, by using someappropriate
perturbation of {1,2,...,n}. -- Erick
> [... something
with irony maybe (?) ... ]>>I didn't 'nd a
6-prime multi-set
yet.> Okay, I made a slight mistake in my earlier statement.
An (n-1)-AP of primes> yields an n-prime (multi)set only if the
difference is less than the initial> term. So while
5,11,17,23,29 is an arithmetic progression, we don't only>
get
a 6-prime set unless you allow {-1,6,6,6,6,6}. To get a valid
6-prime> set from this method we have to go a little higher,
to {7,30,30,30,30,30}:> 30+7, 30+30+7, 30+30+30+7,
30+30+30+30+7, and 30+30+30+30+30+7 are prime.> There's
probably a smaller (in terms of sum?) 6-prime set than this.>>
Just don't ask me for a 24-prime set, I don't
have one yet
:).Nor I, but I've found the following n-prime sets:n = 5:
two
2's and three 1'sn = 7: two 2's
and 've 3'sn = 8: 've
2's and
three 1's n = 9: four 2's and
've 1'sn = 11: four 6's and
seven 5'sn = 12: seven 2's and
've 1'sn = 13: six 2's and
seven 1'sn = 16: seven 5's and nine
6'sn = 17: eight 2's and
nine 3'sn = 28: 15 2's and 13
1'sn = 29: 14 2's and 15 1'sn =
43: 20 2's and 23 3'sn = 44: 23
2's and 21 3'sn = 49: 24 2's
and 25 1'sRobert Israel israel@math.ubc.caDepartment of
Mathematics http://www.math.ubc.ca/~israel University of
British Columbia Vancouver, BC, Canada V6T 1Z2
I did a
computer search, and the 'rst solution > in lexicographic
right-to-left order (which > minimises a(6)) is:
{1,1,2,2,3,4}:> 6 elements: 1+1+2+2+3+4 = 13> 5 elements:
1+1+2+3+4 = 11> 4 elements: 1+1+2+3 = 7 and 2+2+3+4 = 11> 3
elements: 1+1+3 = 5 and 1+2+4 = 7> 2 elements: 1+2 = 3 and 3+4
= 7.> Grumpfff... I re-read my non-computer search andfound my
mistake, when I studied this multi-set:I missed the 2+2+3+4
multi-set :-(I wonder, why Robert Israel didn't
'nd ist. I
justread his posting with the list of multi-sets.>... more
interesting by requiring distinct values? > Even that harder
version may be tractable, by using > some appropriate
perturbation of {1,2,...,n}.Do you have examples already? It
is an interesting question too, I agree.But the original
question ist still funny enough,because I don't see a
generating rule in thesemultimultimulti-sets, which you found
and whichRobert's posting shows.Rainer
Rosenthalr.rosenthal@web.de
Grumpfff... I re-read my
non-computer search and>found my mistake, when I studied this
multi-set:>I missed the 2+2+3+4 multi-set :-(>I wonder, why
Robert Israel didn't 'nd ist. I just>read his
posting with the
list of multi-sets.I was looking for examples with just two
values.Robert Israel israel@math.ubc.caDepartment of
Mathematics http://www.math.ubc.ca/~israel University of
British Columbia Vancouver, BC, Canada V6T 1Z2
sci.math'ers,I've been contemplating on the
following question
and I'm sure there'sa trivial answer, but I
cannot 'nd it.Given
a composite number N, what kind of restriction can be set on
thenumber of digits in N's divisors (call them A and B,
A,B,N
are allintegers.)For instance, if N has 5 digits, what
possible digit lengths can A andB have? For instance N=15000,
two possible digit lengths for A and Bis 2 and 4 (15 x 1000).
1 and 4 is another solution (3x5000).How do you solve this in
general when N is known, but not itsdivisors? How do you
get/enumerate all possibilities?Perhaps the carry-propagation
problem comes into play here, but I'mnot sure.Any ideas
would
be appreciated!
Given a composite number N, what kind of
restriction can be set on the>number of digits in N's
divisors
(call them A and B, A,B,N are all>integers.)>This sounds like
junior high school math concepts.Generally, (assuming if you
say composite you also mean whole number orinteger) if N=A*D,
the number of digits in N is the sum of the numbers ofdigits
in A and D. ...But not exactly so.consider most two-digit
numbers. The factors are either a one digit and a onedigit, or
a one digit and a two digit. 85 (two digit) = 17 (two digit) *
5 (one digit)20 (two digit) = 2 (one digit) * 10 (two digit)20
(two digit) = 4 (one digit) * 5 (one digit)A better statement
than the 'rst Generally,... should be more like:the number of
digits expected for an integer product may be as large as the
sumof the number of digits of its factors; or this sum minus
1. division and multiplication calculation answers by using
this idea. If forexample, one divides a 5 digit number by a 2
digit number, one could expect aresult of a 5 - 2 = 3 digit
number (as far as the WHOLE number part of theanswer). The
student Might get a two digit result, or a three digit
result,but a one digit result or a four, 've, or 6 digit
result would not bereasonable. (Yes, sometimes students in
junior high school are still learning how toperform divisions
with whole number dividends and divisors).. G C
sci.math'ers,>> I've been contemplating on the
following
question and I'm sure there's> a trivial
answer, but I cannot
'nd it.>> Given a composite number N, what kind of
restriction
can be set on the> number of digits in N's divisors (call
them
A and B, A,B,N are all> integers.)The number of (decimal)
digits in an integer N is d(N) = §oor(log_10(N)) + 1.Now, if N
= AB we have: d(N) = d(AB) = §oor(log_10(AB)) + 1 =
§oor(log_10(A) + log_10(B)) + 1 = §oor(log_10(A)) + 1 +
§oor(log_10(B)) + 1 - E = d(A) + d(B) - Ewhere E is either 0
or 1.In other words, either d(N) = d(A) + d(B) ord(N) = d(A) +
d(B) - 1.Obviously, not all combinations of d(A) and
d(B)satisfying one of these equations are possible.If for
instance N is prime, then the only possibility isd(A) = 1 and
d(B) = d(N) (or the other way around).Asger.
=good starting
point.Out of sheer curiousity Purush., Whats your major at
Powai.> to split n in r parts some may be 0> (n+r-1 choose r)>
for positive solns, split n-r in r parts.> Given a number
n is there any formula that can tell us in how many> different
ways it can be split.I understand ramanujan had some formula>
for it.> To be more speci'c I am concerned with the question
of given an> arbitrary n how many ways can I split it as a sum
of three positive> integers(namely n=x+y+z)>
=This is
obvious, perhaps, but it is still interesting to me, for
itcombines number-theory and calculus (as do MANY other
well-knowntheorems in other ways).Let f(x) be any real -> real
analytic (about x = 0) function, wheref(0) = 0, and all of
f's
derivatives at x = 0 are integers.(..and where f's and
exp(f(x))'s Taylor-series expansions {about x =0}
converge.)For simplicity, let a[j] = the j_th derivative of
f(x) at x = 0,and letb[j] = the j_th derivative of exp(f(x))
at x = 0.Now, not only are all a[j]'s integers, but so are
the
b[j]'s.And also...For p = any prime,p divides (b[p+1] -
b[p]*a[1] - a[p+1]) -Examples:If g = exp(sin(x)) or
exp(arctan(x)),and a[j] = the j_th deivative of g at x =
0,then:p divides (a[p+1] - a[p]).If g(x) =
e^(x*e^(x*e^(x*e^...))) = exp(x*g(x)),then:p divides (a[p+1] -
2*a[p]).(perhaps)All of the above seems to be just a
meaningless curiosity. Leroy Quet
=Has anyone noticed this:q
= (P_n - P_n-1) - 1q is *almost* always 1 or prime. Where P_n-1
and P_n are consecutive primes.Of course it is not true for 2
and 3, for which q = 0. But for *almost* allthe primes (P_n-1
>= 3), this is true---at least in my experiment with the'rst
1000 primes.Has this been observed before? or has it been seen
in primes > 1000?Navin Kadambi---Complaints to
news@netfront.net
Has anyone noticed this:> q = (P_n -
P_n-1) - 1> q is *almost* always 1 or prime. Where P_n-1 and
P_n are consecutive primes.> Of course it is not true for 2
and 3, for which q = 0. But for *almost* all> the primes
(P_n-1 >= 3), this is true---at least in my experiment with
the> 'rst 1000 primes.> Has this been observed before? or
has it been seen in primes > 1000?I tested this for the 'rst
10^7 primes and didn't observe apreference for primes per
se.
The strongest tendency I observedis for the prime gap P_n -
P_(n-1) to be a multiple of 6, whichaccounts for the
relatively high frequencies of q = 5, 11, 17, 23,and 29 for
the 'rst 1000 primes.The tendency for the prime gap to be a
multiple of 6 makes sensebecause multiples of 3 are composite.
For example, we can estimatethe relative freqency of prime gaps
of size 2, 4, 6, and more than6 by supposing that primes occur
with frequency x among the numbers6n+1, 6n+5, 6n+7, 6n+11,
etc:6n+1 6n+3 6n+5 6n+7 6n+9 6n+11 6n+13 6n+15 ...MAYBE NO
MAYBE MAYBE NO MAYBE MAYBE NO ...If 6n+1 is prime, then these
are the probabilities for various sizesof the ensuing prime
gap: 2: 0 4: x 6: (1-x) x > 6: (1-x)^2If 6n+5 is prime, then
these are the probabilities: 2: x 4: 0 6: (1-x) x > 6:
(1-x)^2Assuming 6n+1 and 6n+5 primes are equally common, then
the relativefreqency of gaps of size 2, 4, 6, and more than 6
are x/2, x/2, (1-x) x, and (1-x)^2,where x is about 3/log k
for numbers of size k.This heuristic analysis is borne out by
the entries from thefrequency table below. Gaps of 2 and 4 are
equally common, buta gap of 6 is almost twice as common.Here is
the complete frequency table for the 'rst 10^7 primes:1 12
7385974 7387176 12975408 56615110 72980812 92066114 50352416
37167718 66773420 35426722 30723024 45321526 21120328 22917730
39871332 12312334 12904336 20672238 9468240 11154642 15995644
6486646 5493148 9369350 5218352 3880054 6415756 3222458
2798560 5530562 1676364 1737466 3096068 1236870 1747572
1725574 854076 725378 1375880 676082 479184 981886 341188
345490 705692 225994 205896 354498 1831100 1923102 2374104
1168106 933108 1634110 941112 711114 1125116 439118 433120
948122 287124 318126 533128 183130 211132 301134 128136 100138
210140 140142 90144 123146 46148 67150 94152 52154 43156 57158
19160 27162 27164 20166 9168 25170 18172 4174 10176 11178
12180 10182 5184 4186 3188 1190 1192 3194 1196 1198 6202 2204
3210 4220 1222 1-- | Jim Ferry | Center for Simulation
|+------------------------------------+ of Advanced Rockets ||
http://www.uiuc.edu/ph/www/jferry/ +------------------------+|
jferry@[delete_this]uiuc.edu | University of Illinois
|
=Start with a regular m-gon.If we are to draw a path
consisting of connected straightline-segments within the
polygon, so that each segment starts/ends onthe vertexes of
the m-gon, and where each vertex is visited exactlyonce (1st
and last segment need not be connected),then the total path
could cross itself several times.As an example, lame-looking
ascii-art for m = 4 (4 is an m too low tobe interesting):1
2
4The path starts at vertex 3, goes to 2, then 1, and 'nishes
at 4.There is one crossing, at the X.So, a couple of
questions:What is the maximum number of crossings for an
m-gon?(The minimum # is 0, the path being along the m-gon
perimeter.)If we give a positive integer <= the maximum number
of crossings foran m-gon, can we determine the number of paths
with this number ofcrossings??What if we de'ne crossing as
being each time a segment crosses aparticular segment, which
would usually double the number ofcrossings, but might count
some crossings (by old de'nition) 3 ormore times, if 3 or
more
line-segments intersect at the same point?What if we restrict
the path as to keep it from visiting adjacentvertexes (along
the perimeter) consecutively?What if the path must be closed,
ie the 'rst and last segment must beconnected??-Do not
confuse
this post's topic with the game I posted the other
day:http://mathforum.org/discuss/sci.math/t/524658In this
post's situation, the line-segments must be
connectedend-to-end, for one thing.But you can still make a
puzzle/game out of this.If one player(puzzle proposer) knows a
possible number of crossingscan occur, then the other
player/solver can try to 'nd a path (thereare, of course,
several which are at least due to rotation andre§ection) that
has that particular number of crossings.Leroy Quet
=Has
anybody done PET scans to 'nd up what areas of the brain
light
Has anybody done PET scans to 'nd up what areas of the brain
light up > when people try to prove/understand higher
math?Based on other abstract tasks, this would be the
prefrontal lobe, almost for certain.Bob Kolker
> Has
anybody done PET scans to 'nd up what areas of the brain
light
up >> when people try to prove/understand higher math?>
Based on other abstract tasks, this would be the prefrontal
lobe, almost > for certain.> Bob Kolker> But is it the area
for language processing, or the area for spatial reasoning? Or
both?
But is it the area for language processing, or
the area for spatial > reasoning? Or both?For spatial thinking
the temporal lobe lights up in a pet scan, but for pure
abstract thinking it is most often the prefrontal lobe that
lights up.Bob Kolker
=Please recommend some entertainment
math books. My undergraduate major was math, but, no graduate
training, I hadgraduate training in electrical engineering.I
am looking for some entertainment math books for my
background. Thebooks should cover some new and depth results.
Most entertainment mathbooks are for general public, too
simple and too shallow. I am lookingfor some easy and fun to
read books with depth, and suitable for mathundergraduate
major. (Not trivial subject of chaos or fractal, orthings like
those).Ideally the book can cover some recent active research
results, but,in someway, people without graduate training can
understand and enjoy.
Please recommend some entertainment
math books. >>My undergraduate major was math, but, no
graduate training, I had>graduate training in electrical
engineering.>>I am looking for some entertainment math books
for my background. The>books should cover some new and depth
results. Most entertainment math>books are for general public,
too simple and too shallow. I am looking>for some easy and fun
to read books with depth, and suitable for math>undergraduate
major. (Not trivial subject of chaos or fractal, or>things
like those).>>Ideally the book can cover some recent active
research results, but,>in someway, people without graduate
training can understand and enjoy.It would be hard to cover
recent active research results in a mannerwhich satis'es all
of your criteria. Most recent research results aretoo far
removed for anyone to make the exceedingly hard effort itwould
take to accomplish a presentation which is entertaining, not
toosimple, not too shallow, suitable for an undergraduate math
major withlittle or no graduate training. Not to mention that
the market forsuch a book would be pretty small...In any case,
try to 'nd the reprints of Martin Gardner's
columns inScienti'c
American. Those would be the grand-daddy of all
popularpresentations of mathematics both interesting,
accessible, anddeep. They are certainly not recent, but I
think it is about the levelyou are looking for. There are many
of these books, among them:Hexa§exagons and other mathematical
diversions, The SecondScienti'c American Book of Mathematical
Puzzles and Diversions, TheUnexpected Hanging, Wheels, Life,
and other MathematicalAmusements, New Mathematical Diversions,
Penrose Tiles to TrapdoorCiphers, Fractal Music, Hypercards,
and more..., Time travel andother mathematical bewilderments,
and Knotted Doughnuts and othermathematical entertainments.Ian
Stewart is pretty good, as is Ivars Peterson. I'm sure you
willget other
suggestions.
very selective
about what I accept as reality. --- Calvin (Calvin and
Hobbes)
Please recommend some entertainment math books.>> My
undergraduate major was math, but, no graduate training, I
had> graduate training in electrical engineering.>> I am
looking for some entertainment math books for my background.
The> books should cover some new and depth results. Most
entertainment math> books are for general public, too simple
and too shallow. I am looking> for some easy and fun to read
books with depth, and suitable for math> undergraduate major.
(Not trivial subject of chaos or fractal, or> things like
those).>> Ideally the book can cover some recent active
research results, but,> in someway, people without graduate
training can understand and enjoy.A lot here depends on what
subjects you are interested in, and what you mean
byentertaining and not too simple.Looking at my own shelves
and applying your criteria, I would suggest Needham'sVisual
Complex Analysis, K.9arner's Fourier Analysis,
Stewart's
GaloisTheory, and Reid's Undergraduate Commutative Algebra.
These are de'nitelynot general-public pop math books, and it
is probably a stretch to consider themeasy in the sense of
bedtime reading, but anyone with an undergraduatebackground
should not 'nd them hard. I think they are entertaining, and
theyare certainly not too shallow.
For the sequence
a_{n}de'ned by a_{1}=k, a_{i+1}=k^a_{i}, k is an integer >>
0,
prove or disprove that a_{n} for suf'ciently large n has a
constant> least residue (mod j), j > 0.>j > 0 allowing (mod 1)
??Would you clarify the role of j? All j, some j, or what?Let p
be any factor > 1 of k. Then for all n a_n = 0 (mod p)except if
k = 1 when a_n = 1 (mod any.integer>1)> IF the statement is
false, for what k does the statement hold true?>
For the
sequence a_{n}de'ned by a_{1}=k, a_{i+1}=k^a_{i}, k is an
integer >> 0, prove or disprove that a_{n} for suf'ciently
large n has a constant> least residue (mod j), j > 0.>> IF the
statement is false, for what k does the statement hold true?[I
assume that you mean: For all j>0, a_n will be constant (mod
j)for suf'ciently large n.]For k=1 the statement is trivially
true, so let's assume that k>1.I will prove the statement by
induction on j.For j=1 the statement is obviously true, so
assume j>1 and assumethe statement proven for all smaller
values of j.First of all, notice that if j is divisible by a
primepower p^m andk is divisible by p, then a_n will be
divisible by p^m for suf'cientlylarge n. By the chinese
remainder theorem, we can therefore assumethat gcd(k, j)=1.Now
let r be the smallest positive integer such that k^r = 1 (mod
j).Since r divides phi(j) [phi being the Euler phi-function]
and j>1,r must be less than j. By the induction hypothesis we
know thata_n will be constant (mod r) for suf'ciently large
n,
and it followsthat a_(n+1) = k^(a_n) will be constant (mod j)
for (the same)suf'ciently large values of n.Asger
=I have to
proof some theorem. Part of that theorem I have proofed but
withone thing I can't.Please help me.Theorem: Lef f be a
function from a space X into a space Y where X islocally
compact and Y is regular and dense in itself. Then f is open
andcontinuous if and only if f is pre-semi-open and
irresolute.I have done everything without openess of function
f.that:Let us consider a compact neighborhood F of a ponit x
in X. Thus f(F) iscompact, and since Y is Hausdorff, f(F) is
closed. And that's all. I don'thave idea what
have to be
following.-- michaelSerwis Usenet w portalu Gazeta.pl ->
http://www.gazeta.pl/usenet/
=There were mistake in subject.
My fault.michael-- Serwis Usenet w portalu Gazeta.pl ->
http://www.gazeta.pl/usenet/
Of course there are several
ways to *construct* the exponential function, but is> there
also a pure existence proof? I am very interested in such a
theorem if it> exists.> Peace,>
EJ
1 . For each a>0 there exists an unique continuous function
E:R-->(0,infty) satisfying (1) E(x)E(y)=E(x+y) for all real
x,y , and(2) E(1)=a . De'nition. Let E be the solution of (1)
with E(1)=a .Then E(x) is the exponential function a^x .
THEOREM 2 . There exists only one continuous solution
F:R--->(0,infty) offunctional equation (1) with F(1)=a where
a^x >= x^a for all positive x .De'nition. Let F:R-->(0,infty)
satisfying F(x)F(y)=F(x+y) for all real x,y a^x >= x^a for all
positive x , with a:=F(1).Then F(x) is the exponential e^x .
ln x + arctan x, 'ndg'(pi/4)Let y=f(x)then
dy/dx=(x^2+x+1)/(x[x^2+1]) sodx/dy = x(x^2+1)/(x^2+x+1)
.........(1)So, to 'nd g'(pi/4) I need the
value of x
whichsatis'es pi/4 = ln x + arctan x to substitiute into eqn
1.The answer is the nice number 2/3. How do I get x?Any help
much much appreciated.Brad
If g is the inverse function
of f(x) = ln x + arctan x, 'nd> g'(pi/4)> Let
y=f(x)> then
dy/dx=(x^2+x+1)/(x[x^2+1]) so> dx/dy = x(x^2+1)/(x^2+x+1)
.........(1)> So, to 'nd g'(pi/4) I need the
value of x
which> satis'es pi/4 = ln x + arctan x to substitiute into
eqn
1.> The answer is the nice number 2/3. How do I get x?How did
you know that?In the back of the book, no doubt :-)So to solve
pi/4 = log x + arctan x ...(i) you might draw a graph and
guess the answer,(ii) there's not many numbers whose log and
arctan you knowexactly .... so try one of them.-- Robin
Chapman, www.maths.ex.ac.uk/~rjc/rjc.html The League of
Gentlemen
If g is the inverse function of f(x) = ln x +
arctan x, 'nd> g'(pi/4)> Let y=f(x)> then
dy/dx=(x^2+x+1)/(x[x^2+1]) so> dx/dy = x(x^2+1)/(x^2+x+1)
.........(1)> So, to 'nd g'(pi/4) I need the
value of x
which> satis'es pi/4 = ln x + arctan x to substitiute into
eqn
1.> The answer is the nice number 2/3. How do I get x?> Any
help much much appreciated.> Brad> Note that ln(1) = 0 and
arctan(1) = pi/4.
Suppose that instead of adding 2
angular momenta, you need to add 3> So, instead of <J M|j1
j2 m1 m2> you have> <J M|j1 j2 j3 m1 m2 m3>> Is there a
formula to calculate the latter?> I know how to compute <J
M|j1 j2 j3 m1 m2 m3> using raising and> lowering operators,but
I need something which can> be implemented in a computer
program (I allready have a utility which> calculates <J M|j1
j2 m1 m2>, so if there is a formula linking the> 2-momenta
coef'cients with 3-momenta coef'cients,
it'd do).> New
complications arise.When you combine three angular momenta, in
general the result will haveseveral angular momenta with
multiplicity greater than one. Forexample, if you combine two
objects with spin 1, you will have anobject with spin 0,
another with spin 1, and another with spin 2. Nowcombine all
these with a third object of spin 1; the results are:1 object
with spin 32 objects with spin 23 objects with spin 31 object
with spin 1.Call the initial objects A, B, and C; you will get
a different set ofcombinations depending on whether you 'rst
couple (AB) and then couplethe results to C, or couple (BC) to
A, etc. The 6-j symbols are thematrix elements of
transformations that relate one of these couplingsto the
others. And there's more: I think I have seen 12-j
symbols.There's lots of written material available. The most
thoroughtreatment may be a two-volume set by Biedenharn and
Louck.-- Chris HenrichPaddy Solemn likes to use
precise-sounding terminology in a vague way, andderives from
this a bracing sense of intellectual rigour. -- Conor Cruise
O'Brien
=I was perusing the sci.math FAQ the other day and
ran into a fascinating topic I hadn't seen before.
It's called
Freiling's Axiom of Symmetry. It appears about 3/4 down the
page at
http://www.faqs.org/faqs/sci-math-faq/AC/ContinuumHyp/.There'
s
a point of the discussion I didn't understand. Perhaps
someone
can shed some light. In what follows I'm paraphrasing the
FAQ
to clarify my own understanding.Let A be the collection of
functions from the reals to countable sets of reals. In other
words if f is an element of A, then f is a function from the
reals R to the power set of the reals P(R) such that for each
x in R, f(x) is a countable set.One example of such an f would
be a constant function, for example f(x) = Q where Q is the
rationals. A slightly more interesting f would be given by
f(x) = Q+x, the set of rationals translated by x.Given f in A
and any two reals x and y, f(y) is a countable set, which has
Lebesgue measure 0. Therefore x is an element of f(y) with
probability 0, and x is not an element of f(y) with
probability 1. Likewise, y is not an element of f(x) with
probability 1.This motivates the Axiom of Symmetry (AX), which
says that for any f in A, there exist two reals x and y such
that x is not in f(y) and y is not in f(x). AX is plausible
because, as we've seen, we can pick any old x and y at
random
and satisfy x not in f(y) and y not in f(x) with probability
1, so we certainly must be able to 'nd speci'c x
and y that
work.AX seems harmless, but there's a surprise. AX is
logically equivalent to the negation of the Continuum
Hypothesis.The proof in one direction goes like this. If CH is
true, then we can well-order the reals (working in ZCF) as r_0,
r_1, r_2, r_3, . . . r_x, . . . where the subscripts x are
ordinals with x < aleph_1. I'm a little shaky at this point,
but my understanding is that this is analagous to saying that
we could well-order a countable set using 'nite ordinals 0,
1,
2, 3, . . . with each index ordinal strictly less than w, the
'rst in'nite ordinal. Likewise we can enumerate
an uncountable
set of reals using countable ordinals strictly less than
aleph_1, as long as we know that the cardinality of the reals
is aleph_1.Now de'ne f by f(r_x) = {r_y : y >= x}. The claim
is that this f falsi'es AX. I can see that for any reals r_a
and r_b, we may assume a <= b. By de'nition f(r_a) is {r_y :
y
>= a}, therefore r_b is an element of f(r_a). Since r_a and r_b
are arbitrary, AX can not possibly be satis'ed for this f.My
problem is that I don't see why f must be in A. For example
why is f(r_0) countable? There are uncountably many ordinals
less than aleph_1 and greater than or equal to 0, otherwise
there wouldn't be enough ordinals to index the reals. So it
seems to me that f(r_0) must be uncountable, and therefore I
am missing something crucial.
='shfry says...>>I was
perusing the sci.math FAQ the other day and ran into a
fascinating >topic I hadn't seen before. It's
called
Freiling's Axiom of Symmetry.Yes, that's an
interesting
topic.>The proof in one direction goes like this. If CH is
true, then we can >well-order the reals (working in ZCF) as
r_0, r_1, r_2, r_3, . . . r_x, >. . . where the subscripts x
are ordinals with x < aleph_1. >>I'm a little shaky at this
point, but my understanding is that this is >analagous to
saying that we could well-order a countable set using >'nite
ordinals 0, 1, 2, 3, . . . with each index ordinal strictly
less >than w, the 'rst in'nite ordinal. Likewise
we can
enumerate an >uncountable set of reals using countable
ordinals strictly less than >aleph_1, as long as we know that
the cardinality of the reals is aleph_1.>>Now de'ne f by
f(r_x) = {r_y : y >= x}. The claim is that this f >falsi'es
AX.I think you've got the de'nition of f
backwards. If you
wantf(r) to always be a countable set, then de'ne it as
follows f(r_x) = { r_y : y < x }Then for every r, f(r) is a
countable set (because each index x isa countable ordinal, so
there are only countably many smaller ordinals).With f de'ned
as { r_y : y >= x }, f(r) is always an uncountable set.--Daryl
McCulloughIthaca, NY
'shfry says...>>I was perusing the
sci.math FAQ the other day and ran into a fascinating >topic I
hadn't seen before. It's called
Freiling's Axiom of Symmetry. Yes, that's an interesting topic.>The proof in one
direction goes like this. If CH is true, then we can
>well-order the reals (working in ZCF) as r_0, r_1, r_2, r_3,
. . . r_x, >. . . where the subscripts x are ordinals with x <
aleph_1. >>I'm a little shaky at this point, but my
understanding is that this is >analagous to saying that we
could well-order a countable set using >'nite ordinals 0, 1,
2, 3, . . . with each index ordinal strictly less >than w, the
'rst in'nite ordinal. Likewise we can enumerate
an >uncountable
set of reals using countable ordinals strictly less than
>aleph_1, as long as we know that the cardinality of the reals
is aleph_1.>>Now de'ne f by f(r_x) = {r_y : y >= x}. The
claim
is that this f >falsi'es AX.> I think you've
got the
de'nition of f backwards. If you want> f(r) to always be a
countable set, then de'ne it as follows> f(r_x) = { r_y : y
< x }> Then for every r, f(r) is a countable set (because
each index x is> a countable ordinal, so there are only
countably many smaller ordinals).> With f de'ned as { r_y : y
>= x }, f(r) is always an uncountable set.> That's what I
thought too. But not only the sci.math FAQ. but about half a
dozen other sites on the Net have the other de'nition, the
one
that seems wrong. Could all these sites have copied from each
other without anyone noting a typo?
This is the correct
original equation, just to avoid any confusion.> |2X-3| <
|5X+4|X satis'es this if and only if (2X-3)^2 < (5X+4)^2,
which simpli'es to0 < (3X+7)(7X+1).-- | Jim Ferry | Center
for
Simulation |+------------------------------------+ of Advanced
Rockets || http://www.uiuc.edu/ph/www/jferry/
+------------------------+| jferry@[delete_this]uiuc.edu |
University of Illinois |
=
>on the interval [0,2Pi] with respect to the inner product >
<f,g> = int [0, 2Pi] f(t)*conj(g(t)) dt.Whoops, I had to
asked.> But really, the best proof of FTA is the one that
shows> If p is a polynomial with no zeros in C,> then |p|
cannot achieve a minimum unless p is constant. >They know the
analysts have the right proof.Mean value theorem quickly
proves odd degree p in R[x] has real root.What algebraist can
prove that with even twice or thrice the labors? >> Consider
f(x,y) = |p| as a function of the two variables >> in z = x +
iy; use theorem if f is a bounded below real >> continuous
function of R^2 then f attains minimum m, since >> f -> oo as
(x,y) -> oo.Let m = inf f. Find some large closed, hence
compact square withf(x,y) > m out side the square. Thus within
the square f attains m. >Right, suppose p is a nonconstant
polynomial that is nowhere 0. Then >from the above, |p|
attains an absolute minimum, and this minimum is >positive. By
translating and multiplying by a constant, we can assume >this
minimum is 1 and that p(0) = 1. Write p(z) = 1 + az^m + q(z),
>where m > 0, a is nonzero and q(z) is the sum of the higher
order >terms (if any).For some c in C, p(z+c) is nonconstant
polynomial with no roots and minimun > 0 at z = 0. >Now if we
were lucky enough to have p(z) = 1 + az^m, with no higher
>order terms, then it's easy to see we have a contradiction.
Why? >Because for any r > 0, the map 1 + az^m takes D(0,r)
onto D(1,r^m), >and every D(1,r^m) contains points of absolute
value < 1.What's D(0,r) ? An open disk or ball centered at 0
with radius r?Seems to me the map takes D(0,r) into
D(1,|a|r^m), so an adjustment?For some small r in R, let z^m =
-|a|r/a, |p(z)| = | 1 - r|a| | = 1 - r|a| < 1 >Unfortunately
life is not that simple, and we have to deal with q(z). >But
note the following: If |z| is small, then |q(z)| will be much
>smaller than |az^m|. So the strategy is this: Choose t such
that >exp(imt) = -|a|/a. Then you'll see that for small r >
0,
>|p(rexp(it))| < 1 - (|a|r^m)/2, and that contradiction proves
FTA.Let k > 0 be the smallest with ak /= 0, c = max{ 0, |aj| :
k < j <= n }for |z| < 1 |p(z)| <= |a0 + ak z^k| + sum(j=k+1..n)
|aj| |z|^j < |a0 + ak z^k| + c(n-k)|z|----
Now if we were
lucky enough to have p(z) = 1 + az^m, with no higher>order
terms, then it's easy to see we have a contradiction. Why?Because for any r > 0, the map 1 + az^m takes D(0,r) onto
D(1,r^m),>and every D(1,r^m) contains points of absolute
value < 1.> What's D(0,r) ? An open disk or ball centered at
0
with radius r?Yes.> Seems to me the map takes D(0,r) into
life is not that simple, and we have to deal with q(z).>But
note the following: If |z| is small, then |q(z)| will be muchsmaller than |az^m|. So the strategy is this: Choose t such
that>exp(imt) = -|a|/a. Then you'll see that for small r >
0,>|p(rexp(it))| < 1 - (|a|r^m)/2, and that contradiction
proves FTA.> Let k > 0 be the smallest with ak /= 0,> c =
max{ 0, |aj| : k < j <= n }> for |z| < 1> |p(z)| <= |a0 + ak
z^k| + sum(j=k+1..n) |aj| |z|^j> < |a0 + ak z^k| + c(n-k)|z|>
That last line of your looks questionable. You have |p(z)| <=
|1 + az^m| + |q(z)|, and |q(z)|/|z|^m -> 0 as z -> 0.
Therefore |q(z)| <= (|a||z|^m)/2 for small z. So |p(rexp(it))|
<= 1 - |a|r^m + (|a|r^m)/2 < 1 for small r > 0.What you want to
remember though is the basic intuition: Nonconstant polynomials
take open sets to open sets. We don't need to prove all of
that, but that's the basic idea. If you know that, then FTA
is
clear. And this is obvious for 1 + az^m; you're going from
discs to discs. Now you add on q(z), which is small compared
to z^m. How could that ruin things? It can't.PS: When you
respond without skipping a line, like below, it's hard to
read. >They know the analysts have the right proof.Mean value
theorem quickly proves odd degree p in R[x] has real root.What
algebraist can prove that with even twice or thrice the
labors?
Mean value theorem quickly proves odd degree p in
R[x] has real root.> What algebraist can prove that with even
twice or thrice the labors?The algebraist's proof uses every
odd degree real polynomial has aroot and every complex number
has a squareroot. Then, with no moreanalysis, proceeds to
prove every nonconstant complex polynomial has aroot.-- G. A.
Edgar http://www.math.ohio-state.edu/~edgar/
So the use
of the Riemann Hypothesis in> in counting primes is just about
the> rate of convergence, not about the> validity of the
formula?> What is the ratio of the rate without> RH and the
rate with RH?I'm not sure I understand your question, but
without RH one can prove that pi(x) = li(x) + A(x), where A(x)
= O(x exp( -c (log x)^(3/5) / (log log x)^(1/5) ) ) with c =
0.009. With RH one can take A(x) = O( sqrt(x) log x ). Here
li(x) is integral from 0 to x of (1 / log x).--
= >> An
element u of F is an algebraic integer over K when >> the
minimal polynomial of u is in D[x] >> equivalently >> some
monic polynomial p in D[x] with p(u) = 0 >There is indeed a
general notion of integral over a ring. >If R is a subring of
S, then an element x of S is said to >be integral over R if x
is a root of a monic polynomial >with coef'cients in R, i.e.
>
n n-1 > x + r x + ... + r = 0, with coefs r in R > 1 n i >Many
(but not all) of the properties of algebraic integers >and
number rings generalize to arbitrary integral extensions.The
'rst I notice is when S is an integral domain, then the
quotient'eld of S may be used to show the integers of S over
R
are a subring of Sby much the same proof as used to show the
ring of algebraic integers is asubring.When S isn't an
integral domain, I doubt the linear algebra proofmentioned
above will be valid. Am I right, that when S isn't an
integraldomain, the integers of S over R may not be a subring
of S and when S isan integral domain, the integers are a
subring?proof of William's Metatheorem: Whatever math I
dream
up is already old hat.----
=Is there a schauder basis for R
over Q? I know that there exists a Hamel basis, and this
allows you to provesome interesting results, but I don't
know
if there is a schauderbasis.
Is there a schauder basis
for R over Q? > I know that there exists a Hamel basis, and
this allows you to prove> some interesting results, but I
don't know if there is a schauder> basis.I guess you will
have
to de'ne what you mean.The usual de'nition is
for a topological
vector space over the reals.But since every real number is the
value of an in'nite serieswith rational terms, do you mean
that {1} is your Schauder basis?Or the constant sequence, {1,
1, 1, 1, ...} ??That doesn't seem useful. So what is your
de'nition?-- G. A. Edgar
http://www.math.ohio-state.edu/~edgar/
=I have a table
containing
numbers:a1,a2,a3,a4,a5,a6,a7,...b1,b2,b3,b4,b5,b6,b7,...c1,c2,
c3,c4,c5,c6,c6,......Typically, these tables have 1000 columns
and 1500 rows.The data is organized like this: a1 < a2 < a3 <
a4 ... < anb1 < b2 < b3 < b5 ....< bn...that is, row cells are
in increasing order.Also, the column cells are in increasing
order, so thata1 < b1 < c1 ...a2 < b2 < b2 ......How can I
ef'ciently search for a given number?Any ideas would be
appreciated.
How can I ef'ciently search for a given
number?between the cells having values less than x and those
having values greater than x look like?-- David Eppstein
http://www.ics.uci.edu/~eppstein/Univ. of California, Irvine,
School of Information & Computer Science
=This seems like a
fairly standard-looking question, though I don't
recallseeing
it before.How big can be a set of natural numbers, so that one
plus the product ofANY TWO of them is a prime?A very crude
estimation of densities suggests that the cardinalities of
suchsets must be bounded.Here is such a set of cardinality 5:-
{1, 2, 6, 18, 30}(The pair-products are 2 6 18 30 12 36 60 108
180 540; each + 1 is a prime)How much bigger can they be?
Theory or data both
welcome.------------------------------------------------------
------------------------ Bill Taylor
W.Taylor@math.canterbury.ac.nz--------------------------------
---------------------------------------------- Chebychev said
it, I'll say it again; There is always a prime between n and
2n.
-Erdos--------------------------------------------------------
----------------------
=|How big can be a set of natural
numbers, so that one plus the product of|ANY TWO of them is a
prime?Based on a standard conjecture, in'nite. If a1 < a2 <
a3
< ... < an arepositive integers, then from the conjecture,
there should exist in'nitely manyintegers k for which a1*k+1,
a2*k+1, ..., an*k+1 are all prime, just in casethere doesn't
exist a prime q such that for every k, at least one of them
isdivisible by q. But for these particular polynomials, if k
is divisible by qthennone of a1*k+1,...,an*k+1 is divisible by
q, so it satis'es the condition.Thus, as long as the set you
have satis'es the condition, you can keepextending it
inde'nitely (conjecturally). In fact, by the same
reasoningthere's an in'nite set of natural
numbers such that 1
plus the product ofany subset of them is prime.Keith
Ramsay
This seems like a fairly standard-looking
question, though I don't recall> seeing it before.> How
big can be a set of natural numbers, so that one plus the
product of> ANY TWO of them is a prime?> A very crude
estimation of densities suggests that the cardinalities of
such> sets must be bounded.> Here is such a set of
cardinality 5:- {1, 2, 6, 18, 30}> (The pair-products are 2
6 18 30 12 36 60 108 180 540; each + 1 is a prime)> How much
bigger can they be? Theory or data both welcome.combinatorial
explosion makes them hard to 'nd as the set size expands.
I'm
pretty sure any size is possible though.Cardinalty 6 from
{1,2,6,18,a,b} wherea=30 : 270 606 2790 3180 6036 21396 22110
22446 23856 24096 44646 6831077490 92346 102396 114660 116190
123120 127596 129126 134850 136500 160650174990 176040 176316
186876 188316 211740 234066 235440 247716 274710 299670299856
341826 382620 398916 420966 432660 433626 472836 475596 490200
493896a=270 : 576 606 1866 6036 9396 11886 25410 31266 37506
43626 44646 4695664236 65550 69246 69696 80406 86856 88650
102396 110436 113796 117306 119616123120 135510 157290 168480
172410 173670 174330 176040 205200 214176 226200229266 240006
247380 274710 288246 299856 345576 350280 355590 420966
448056etc. etc.
[GP:forprime(p=19,1000,if(p%6==1,a=p-1;if(ispseudoprime(2*a+1)
&&ispseudoprime(6*a+1)&&isspeudoprime(18*a+1),print1(a=a);
forprime(q=p+1,100000,if(q%6==1,b=q-1;if(ispseudoprime(2*b+1)&
&ispseudoprime(6*b+1)&&isspeudoprime(18*b+1),print1(
b)))))))]Cardinality 7 from {1,2,6,18,30,a,b} wherea=270 : 606
6036 44646 102396 123120 176040 274710 299856 420966a=606 :
22110 77490 123120 129126 174990 186876 188316 235440 299856
398916 433626etc. etc.
[GP:test7(n)=forprime(p=n+2,100000,if(p%6==1,a=p-1;if(
ispseudoprime(2*a+1)&&ispseudoprime(6*a+1)&&ispseudoprime(18*a
+1)&&ispseudoprime(n*a+1),print1(a=a :
);forprime(q=p+1,500000,if(q%6==1,b=q-1;if(ispseudoprime(2*b+1
)&&ispseudoprime(6*b+1)&&ispseudoprime(18*b+1)&&ispseudoprime(
n*b+1)&&ispseudoprime(a*b+1),print1(
b))));print())))test7(30)]Cardinality 8 from
{1,2,6,18,30,270,a,b} wherea=606 : 123120 299856 888456
1343430 2072100 2530110a=6036 : 102396 123120 176040 420966
2025216 2372370 2433900 2530110 3044400etc. etc. (No GP script
this time, they came from my Otuplet detector'
program
GenSv)Cardinality 9 from {1,2,6,18,30,270,606,123120,a} where
a=<<<_tup8: 888456_tup8: 1343430_tup8: 2072100_tup8:
2530110>from {1,2,6,18,30,270,606,299856,a} where
a=<<<_tup8: 888456>from {1,2,6,18,30,270,6036,102396,a}
where a=<<<_tup8: 64229760_tup8: 81256770_tup8: 123120_tup8:
37535460_tup8: 37464126_tup8: 57792336_tup8: 71233806_tup8:
4798476>from {1,2,6,18,30,270,6036,123120,a} where
a=<<<_tup8: 58034130_tup8: 2530110_tup8: 3044400_tup8:
80365770_tup8: 18273630_tup8: 26400420_tup8: 84862740_tup8:
6648516_tup8: 57792336_tup8: 54485916_tup8: 23180616_tup8:
28703826_tup8: 44589696>So cardinality 9's a piece of cake
still.Cardinality 10:from
{1,2,6,18,30,270,606,123120,888456,a} where a=<<<_tup9:
100195860_tup9: 23070450>from
{1,2,6,18,30,270,606,123120,1343430,a} where a=<<<_tup9:
100195860>from {1,2,6,18,30,270,606,299856,888456,a} where
a=<<<_tup9: 64269216>from
{1,2,6,18,30,270,6036,102396,123120,a} where a=<<<_tup9:
57792336>etc. etc. Cardinality 11:from
{1,2,6,18,30,270,606,123120,888456,23070450,a} where
a=<<<_tup10: 331609770_tup10: 238550160>from
{1,2,6,18,30,270,6036,102396,123120,57792336,a} where
a=<<<_tup10: 1547586180>etc. etc.Cardinality 12:from
{1,2,6,18,30,270,606,123120,888456,23070450,238550160,a} where
a=<<<_tup11: 8282903640>from
{1,2,6,18,30,270,606,123120,888456,23070450,331609770,a} where
a=<<<_tup11: 8282903640>from
{1,2,6,18,30,270,6036,102396,123120,57792336,1547586180,a}
where a=<<<_tup11: 14795106960_tup11: 193473104286_tup11:
291271054410>Cardinality 13:from
{
1,2,6,18,30,270,606,123120,888456,23070450,238550160,8282903640
,a} where a=<<<_tup12: 299666515500_tup12: 317804196756>from
{1,2,6,18,30,270,606,123120,888456,23070450,331609770,8282903640
,a} where a=<<<_tup12: 143176650306_tup12:
275449471320>Hardy and Littlewood tell me that I can go on
for ever, and I trust them onthat.Note - none of the above
have been checked for pseudoprimality, or for bugs!Phil
In fact, by the same reasoning> there's an
in'nite set of
natural numbers such that 1 plus the product of> any subset of
them is prime.Except the empty subset ;-)Phil
This seems
like a fairly standard-looking question, though I don't
recall> seeing it before.> How big can be a set of natural
numbers, so that one plus the product of> ANY TWO of them is a
prime?> A very crude estimation of densities suggests that
the cardinalities of such> sets must be bounded.Hmmm, really?>
Here is such a set of cardinality 5:- {1, 2, 6, 18, 30}> (The
pair-products are 2 6 18 30 12 36 60 108 180 540; each + 1 is a
prime)I don't see why there should not exist some n such
that
n+1, 2n+1, 6n+1,18n+1 and 30n+1 are all prime, although such
an n may be hard to 'nd.Heuristically, the density should be
something like k / (ln x)^5and the integral converges.> How
much bigger can they be? Theory or data both welcome.---J K
Hauglandhttp://www.neutreeko.com
> In fact, by the same
reasoning>> there's an in'nite set of natural
numbers such
that 1 plus the product>> of any subset of them is prime.>
Except the empty subset ;-)Including the empty subset :-)--
Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html The League of
Gentlemen
In fact, by the same reasoning> there's
an in'nite set of natural numbers such that 1 plus the
product> of any subset of them is prime. Except the
empty subset ;-)> Including the empty subset :-)That's twice
in as many days. Bollocks.Phil
>> In fact, by the
same reasoning>> there's an in'nite set of
natural numbers
such that 1 plus the product>> of any subset of them is
prime.> Except the empty subset ;-) Including the
empty subset :-)> That's twice in as many days.Twice what? >
Bollocks.So, Phil, what do you get for the empty product?--
Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html The League of
Gentlemen
> In fact, by the same
reasoning> there's an in'nite set of natural
numbers such
that 1 plus the product> of any subset of them is
prime. Except the empty subset ;-)> Including
the empty subset :-) That's twice in as many days.>
Twice what?I've completely ruined what I had hoped would be
an
intelligent contribution to a thread with something that was
complete crap.I blame the hot weather.>> Bollocks.> So,
Phil, what do you get for the empty product?Yeah, yeah, no
need for one to rub it in.Phil
>> Twice what?> I've
completely ruined what I had hoped would be an intelligent>
contribution to a thread with something that was complete
crap.> I blame the hot weather.Hot weather! where?-- Robin
Chapman, www.maths.ex.ac.uk/~rjc/rjc.html The League of
Gentlemen
Chapman>> Twice what? I've completely ruined what I
had hoped would be an intelligent>> contribution to a thread
with something that was complete crap. I blame the hot
weather.>>Hot weather! where?Here in Oklahoma it's been
above
or just below 100 F for a while;that's what
I've been blaming
for _my_ idiocies.************************David C.
Ullrich
I have to sovle AX= B for matrix X,> where A is
nXd matrix, X = dXd matrix and B = nXd matrix and n>>d> The
constraint is X should be symmetric and postive de'nite.>
navneetUse Lagrange multipliers to adjoin the symmetric
constraint:J = (AX-B)'(AX-B)+ L(X-X'), and
make J stationary
wrt X,L.Adjoining the SPD constraint is much tougher since
that leadsto inequality constraints, and a NLP.
= I have a
few discrete math questions that need to be answered. Theyare
questions for a practice exam. I need them to be answered
andexplained well so I understand them and can use the answers
to preparefor the actual exam. The practice exam as well as the
notes requiredfor question eight are in Adobe Acrobat format,
so I decided to createa web site so they can be easily viewed.
I also included the answersthat I have so far in a Microsoft
Word document. Some of thequestions are answered, but are
missing an explanation. I need all ofthe questions (except
number seven) to be answered and explained.Also, please notice
that some questions require drawing a diagram. Here is the URL
to the site: http://members.cox.net/520/ I am studying for
'nals right now. Having the test completed for mewill allow
me
to make the most ef'cient use of my study time. If youare
willing to do this, I will pay you via PayPal or a similarweb
site. * Remember: Answering question number seven is not
required.
=the First and Second Real Auxiliary Exponential
Equations.It turns out we have a bonus and most of the work
has been done in thoseEquations, which, as it turns out,
behave very much like the 'rst
two.<http://users.forthnet.gr/ath/jgal/math/
AuxiliarynExponential.html>In my Math
Section:<http://users.forthnet.gr/ath/jgal/math/>Enjoy the
read.--
Ioannishttp://users.forthnet.gr/ath/jgal/_____________________
_____________JC <baga14$rc1$1@newsg4.svr.pol.co.uk>
<bcepod$esh$1@news8.svr.pol.co.uk>
<LEOGa.6764$Jw6.3726116@news1.news.adelphia.net>
<QNOGa.6769$Jw6.3728911@news1.news.adelphia.net>
<871xxouhdh.fsf@phiwumbda.localnet>
<874r2i5wn4.fsf@phiwumbda.localnet>
<vatJa.15237$Jw6.6456830@news1.news.adelphia.net>
<AetJa.15239$Jw6.6457907@news1.news.adelphia.net>
Logical
expressions have a truth table (even Wittgenstein knew> this).
No, no, no, no, no. This is woefully inadequate for making any
senseout of Hahn or mathematics generally.Formulas in a
propositional logic have truth tables. First orderlogic has no
such thing. Propositional logic is completely inadequatefor
representing mathematical arguments. One must have quanti'ers
todo mathematics.Whether or not the denizens of alt.postmodern
need a tutorial in logicand mathematics in order to understand
Hahn, you are apparently verypoorly equipped to provide the
tutorial. The issues here haveabsolutely naught to do with
truth tables. No matter what you may have learned in
electrical engineering courses.-- Jesse HughesYes, I'm one
of
those arrogant people who tries to be quotable.There is
actually at least one person who quotes me often. -- James
Harris
=OK, I'll take you're word for it.>> Logical
expressions have a truth table (even Wittgenstein knew>
this).>> No, no, no, no, no. This is woefully inadequate for
making any sense> out of Hahn or mathematics generally.>>
Formulas in a propositional logic have truth tables. First
order> logic has no such thing. Propositional logic is
completely inadequate> for representing mathematical
arguments. One must have quanti'ers to> do mathematics.>>
Whether or not the denizens of alt.postmodern need a tutorial
in logic> and mathematics in order to understand Hahn, you are
apparently very> poorly equipped to provide the tutorial. The
issues here have> absolutely naught to do with truth
tables.> No matter what you may have learned in electrical
engineering courses.>> -- > Jesse Hughes> Yes, I'm one of
those arrogant people who tries to be quotable.> There is
actually at least one person who quotes me often.> -- James
Harris
=Given a binary relation G(x,y) and T1 de'ned
as:T1(x,y) = I / G(x,y) / G(x,y) / T1(x,y)and T2 de'ned
as:T2(x,y) = I / G(x,y) / T2(x,y) / T2(x,y)where I is identity
relation, prove that T1 = T2.
=I've seen the sum over all
unordered pairs of numbers being written as: sum_{i<j} ...The
formula has i<j to prevent pairs from occurring twice, for
example (1,2) and (2, 1), and to prevent pairs of a number
with itself, for example(1, 1).But what if instead of numbers,
i and j are objects with no order?In that case, I cannot write
i < j. But perhaps I could write somethinglike: sum_{{i, j} in
{{a, b} | a in O and b in O and a ne b }} ...where O is the set
of all objects.Is this too messy? If so, what would you
recommend?With index variables would be the standard way of
writing sums andproducts, but introducing dummy ordering of
the objects seems a bitinelegant to me.-olli
I've seen
the sum over all unordered pairs of numbers being written as:>
sum_{i<j} ...>> The formula has i<j to prevent pairs from
occurring twice, for example (1,> 2) and (2, 1), and to
prevent pairs of a number with itself, for example> (1,
1).>And it assures i < j. Consider sum_i<j i+j^2 /= sum_j<i
i+j^2.> But what if instead of numbers, i and j are objects
with no order?> In that case, I cannot write i < j. But
perhaps I could write something> like:>> sum_{{i, j} in {{a,
b} | a in O and b in O and a ne b }} ...> where O is the set
of all objects.>> Is this too messy? If so, what would you
recommend?Yes, remove all thoseuseless Os.When O
isn't
numbers, then sum has no meaningand when O is numbers, what
does your expression say?When O has more than one number,
it's
equivalent to i+j in O.I suggest a collection of unordered
pairs { {x,y} | x,y in O; x /= y }Which skips {y,y} = {y} and
doesn't list {x,y} = {y,x} twice.> With index variables
would
be the standard way of writing sums and> products, but
introducing dummy ordering of the objects seems a bit>
inelegant to me.>Indeed, for ordered pairs, an arbitrary
ordering would give random meaningto (x,y) as contrasted with
(y,x) and as pointed out in example at top,can give different
results for different orders.
I suggest a
collection of unordered pairs> { {x,y} | x,y in O; x /= y }>
Which skips {y,y} = {y} and doesn't list {x,y} = {y,x}
twice.wasn't clear.But how to write a sum over such a
collection?I'm correcting parts of what I said below so it
perhaps becomes morein monospace ASCII as well...> I've seen
the sum over all unordered pairs of numbers being written as:
sum f(i, j) i<jwhere f(i, j) = f(j, i)> The formula has i<j to
prevent pairs from occurring twice, for example (1,> 2) and (2,
1), and to prevent pairs of a number with itself, for example>
(1, 1).>> But what if instead of numbers, i and j are objects
with no> [natural] order? In that case, I cannot write i < j.
But perhaps I> could write something like: sum f(i, j) {i, j}
in {{x, y} | x, y in O; x /= y}> where O is the set of all
objects.>> With index variables would be the standard way of
writing sums and> products, but introducing dummy ordering of
the objects seems a bit> inelegant to me.
>I've seen the
sum over all unordered pairs of numbers being written as:>>
sum_{i<j} ...>>The formula has i<j to prevent pairs from
occurring twice, for example (1,>2) and (2, 1), and to prevent
pairs of a number with itself, for example>(1, 1).>>But what if
instead of numbers, i and j are objects with no order?>In that
case, I cannot write i < j. But perhaps I could write
something>like:>> sum_{{i, j} in {{a, b} | a in O and b in O
and a ne b }} ...>>where O is the set of all objects.>>Is this
too messy? If so, what would you recommend?>>With index
variables would be the standard way of writing sums
and>products, but introducing dummy ordering of the objects
seems a bit>inelegant to me.>Given f: I -> R with either I
f(i)In your case, I would write sum_{ (i,j) in I x J} f(i,j).
Did you really mean to exclude all terms where i = j? If so,
add the second line i != j under the capital sigma.-- Stephen
J. Herschkorn
herschko@rutcor.rutgers.edu
=http://maxima.sourceforge.net/=
= (-1)^3 = (-1)^(6/2) = ((-1)^6)^(1/2) = 1^(1/2) = 1 ...I lie.
(Is it true or false?)The second sentence is true. The 'rst
is
false.(Which is true and which is false?)Zenon paradox (today
not paradox)Russel paradoxI am very curious about other
paradoxes!Benjamin
-1 = (-1)^3 = (-1)^(6/2) =
((-1)^6)^(1/2) = 1^(1/2) = 1 ...The last step, where you take
the sqr(1) is false. It is = +/- 1. Thisisn't really a
paradox
at all...>I lie. (Is it true or false?)True. You do lie
sometimes. Not really a paradox as stated>The second sentence
is true. The 'rst is false.>(Which is true and which is
false?)Neither>Zenon paradox (today not paradox)I hope you
mean Zeno's Paradox... (an arrow that travels 1/2, 1/4,
1/8etc. of the way to the target...>Russel paradoxI am
unfamilliar with this (possibly just the name) What is it?>I
am very curious about other paradoxes!>Benjamin
>Russel
paradox>>I am unfamilliar with this (possibly just the name)
What is it?>I am very curious about other
paradoxes!>>BenjaminRussel's paradox is about Russel sets
and
non-Russel sets.A russel set is a set that is an element of
itself. For example,the set of all abstract ideas is itself an
abstract idea.A non-russel set is a set that is not an element
of itself. Forexample, the set of all pencils is not a
pencil.esSo, the paradox goes like this... Consider the set of
all non-russelsets, lets call it OS'. Is S
russel or
non-russel?If S is russel then it is a member of itself and
thus non-russel.If S is non-russel then it must be included in
the set of all non-russel sets (S), and thus it is
russel.That's the paradox.
-1 = (-1)^3 = (-1)^(6/2) =
((-1)^6)^(1/2) = 1^(1/2) = 1 ...>> The last step, where you
take the sqr(1) is false. It is = +/- 1. This> isn't really
a
paradox at all...He's right, you're wrong. The
square root of
one is 1. Just 1.>I lie. (Is it true or false?)>> True. You do
lie sometimes. Not really a paradox as stated>>The second
sentence is true. The 'rst is false.>(Which is true and which
is false?)>> Neither>>Zenon paradox (today not paradox)>> I
hope you mean Zeno's Paradox... (an arrow that travels 1/2,
1/4, 1/8> etc. of the way to the target...You are unkind.
Have you got any paradoxes to share?Yes, but it's a paradox
whose existence cannotbe revealed to you in fewer than twenty
words.
> Have you got any paradoxes to share?>> Yes, but
it's a paradox whose existence cannot> be revealed to you in
fewer than twenty words.>>Er...? What do you mean by
that?
> Have you got any paradoxes to share?>> Yes,
but it's a paradox whose existence cannot>> be revealed to
you
in fewer than twenty words.>Er...? What do you mean by that?He
just used 17 words.Robert Israel israel@math.ubc.caDepartment
of Mathematics http://www.math.ubc.ca/~israel University of
British Columbia Vancouver, BC, Canada V6T
1Z2
=-------------------------------------------------------
-------------->Have you got any paradoxes to share? Write
them here!>Mathematics and logic are subsets are of each
other, but they are not the same 'eld.When you know a
function
almost everywhere, you don't know it anywhere-- Stephen J.
Herschkorn herschko@rutcor.rutgers.edu
=Suppose we have a
large composite integer N whose prime factors arenot known. Is
there any practical way to test if a given integer is adivisor
of Euler's Totient function Phi(N), short of factoring
N?Would
there be any use for such a test?
While I have no
particular objection to your new>terminology, I don't think
you are contributing anything>useful to this thread.> Do you
have a notion of a useful contribution other than> David,
you're absolutely right? I don't think you
are> right---I
think that you are being inconsistent. I don't see where you
are disagreeing with me. You aremerely changing the
terminology so that it sounds likeyou are disagreeing with
me.
=Parting comments:One of the de'ning characteristics of
religionis a belief in a realm of existence beyond whatwe can
observe. Cantor introduced such a realminto mathematics, and
there is no doubt he hadreligious motivations.This current
debate is very similar to the debatebetween the creation
scientists and the evolutionists.It is a science vs. religion
debate. Both sidesknow they're right.The claim of the
evolutionists is that we don'tneed to invoke a realm of
existence beyond whatwe observe to explain the observable
universe, and I am merely claiming that that idea can
beapplied to mathematics.My only reason for getting involved
in this thread was to offer Mike Deeth advice on howto 'ght
the evil Cantorian religionists. It'ssomething he wants to
do.
I have little desireto do so myself.
<vhsihi3vtgo5f6@corp.supernews.com>
<8765ltbnw8.fsf@phiwumbda.localnet>
<87znj4l6p9.fsf@phiwumbda.localnet>
<87n0f38zl9.fsf@phiwumbda.localnet>
<87fzku5vx8.fsf@phiwumbda.localnet>
Parting
comments:>> One of the de'ning characteristics of religion>
is
a belief in a realm of existence beyond what> we can observe.
Cantor introduced such a realm> into mathematics, and there is
no doubt he had> religious motivations.How many times must it
be said? Read your history. Cantor did notintroduce any such
realm when he proved |N| < |R|. He proved atheorem in an
existing mathematical theory. He didn't introduce a
newmathematical setting. He made explicit what was implicit in
theexisting setting.You repeat the error over and over and
over. I have explicitlycorrected this error on many occasions,
and yet you persist in statingthese 'ctions. Your view of
mathematics is simple enough: Pre-Cantorwas about the
observable, post-Cantor was not. Unfortunately (evenignoring
the extraordinarily facile distinction of
theobservable/calculational/computational/whatever), it is
also clearthat this view is wrong.While Cantor perhaps
contributed to the axioms of set theory, hisproof that |N| <
|R| was not foundational in this sense. He workedwith the
existing mathematical structures of the day.> This current
debate is very similar to the debate> between the creation
scientists and the evolutionists.> It is a science vs.
religion debate. Both sides> know they're right.No, I have
never said that mathematics as it is currently done isright.
I've only said that your revolution requires a real,coherent
thought, a basic understanding of the philosophy and historyof
mathematics involved which you haven't provided.
I'm not
arguing against your alternative philosophy, because I
haven'tseen any coherence there. I've seen
vague terms
involvingcalculation, computation, existence and observe.
There's nosubstance behind these terms, as far as I can see,
and so I won'tdispute a thesis which is half-formed at
best.I
am instead correcting repeated factual errors. Your
understandingof how Cantor's results should be seen in the
history of mathematicsis just wrong.-- Even if [...] a
communistic regime should come [to China], the oldtradition
[...] will break Communism and change it beyond
recognition,rather than Communism [...] break the old
tradition. It must be so. -- Lin Yutang on Socialism with
Chinese characteristics in
1935
=david_lawrence_petry@yahoo.com (David Petry)
says...>One of the de'ning characteristics of religion>is a
belief in a realm of existence beyond what>we can observe.The
more annoying traits of religious people are thetendency to
lecture those who don't follow their viewof morality.
Mathematicians aren't like that.>This current debate is very
similar to the debate>between the creation scientists and the
evolutionists.>It is a science vs. religion debate. Both
sides>know they're right.Not at all. Mathematicians study
in'nite sets becauseit's *interesting*. They
don't have a
dogmatic belie'n the existence of those objects.
Mathematicians (unlikereligious believers) are perfectly
willing to entertaindifferent possibilities. They will look
into constructivemathematics, or various alternative theories
besides ZFCin an attempt to 'nd interesting mathematics.For
some reason, a common attack against mainstream physicsand
mathematics is to call it a religion. That is an
absolutelyridiculous charge. It's a lot like creationists
attackingevolution as a religion.--Daryl McCulloughIthaca,
NY
But, the de'nition of R precedes Cantor. Take
Dedekind's de'nition,> say, or
Cauchy's, and it is easy to
derive that the usual canonical> decimal representation works.
Cantor did *not* offer a new or> different de'nition of R.
Nor
does the de'nition of R come from> Cantor's
work. Rather,
Cantor proved an important theorem about the> existing
de'nition of R.> You need to check your history.One thing
(of many others) that's confusing me in following
thisdiscussion is that nobody involved seems to have a good
idea of whatCantor did and did not actually do. In other
words, nobody has theirhistory straight.First, Dedekind's
de'nition of the reals was formulated around the sametime as
Cantor's de'nition, and yes, Cantor did
formulate a de'nition
ofthe reals, which *was* new at the time: it's called
(unjustly) the Cauchycompletion of the rationals. Dedekind
was, in fact, corresponding withCantor heavily at the time,
and so was inspired by Cantor to some degree. The most
important point here, I think, is that there was no
existingde'nition of R before Cantor. Some mathematicians,
like Cauchy, hadmade some attempts, but their de'nitions were
circular.Second, the whole decimal representation thing is
partly what motivatedCantor to de'ne the reals in the
'rst
place, but decimal representationswere not a focus of his
work. In fact, not until almost two decades laterdid he give
his famous diagonal argument using decimal expansions, andthat
argument isn't even his, having been given a few years
earlier
byinvolved using nested intervals of the real line.A good
source for all this is Dauben's book on Cantor. I
don't have
theexact page numbers handy, but the book is clearly organized
and itshouldn't be hard to 'nd the relevant
sections (near the
beginning).
Cantor introduced a quasi-religious
mythology into> mathematics, and the acceptance of that
mythology> does have sociological implications which the >
mathematics community has been unwilling to face> up to. I am
accusing the mathematics community of> irresponsibility.> An
easy mistake to make is that because Cantor had quasi-religious
beliefsthat motivated his work, acceptance of his work by the
mathematicalcommunity indicates acceptance of his
beliefs.What's funny is that the situation is quite
different
from what you say. Acceptance of Cantor's work came about
*because* mathematicians realizedthat arguing about the
metaphysics of mathematical work was silly andunfruitful. As
long as it is logically consistent, it is validmathematics.
So, in a sense, Cantor's revolution is really about
theridding
of the quasi-religious mythology in mathematics.
Ironically,this revolution was started by a man who sought to
replace the dominantmetaphysics of the time (as exempli'ed by
Kronecker, et al) with his own.
=In all this, one of the
things I 'nd most peculiar is David
Petry'sapparent attitude
toward constructivism. It seems clear to me that it wouldshed
a lot of light on the issues he raises, and also serve a lot
of thepurposes he intends to pursue. Errett Bishop is famous
for having come outin favor of giving mathematics
computational meaning as much as possible.Why reinvent the
wheel at this point?|>But would you really call that
Cantor's
proof?||It is structurally the same proof, except that instead
of|assume that f(n) is an enumeration of all reals|you say
assume that f(n) is a computable enumeration|of all the
computable reals.The proof is not completely constructive the
way it's usually done, becauseof its reliance on each real
having a decimal expansion (which isn'tconstructive--
there's
no way to tell in general whether a number close to0.1 is
>=0.1 or <0.1). However, an adjusted version of the proof that
isconstructive is not hard to give. Take a nested sequence of
intervals wherethe n-th interval has length 1/3^n and does not
contain the 'rst n realsin the enumeration; the intersection
is
a real not in the enumeration.Then, we may note that although
one isn't required by constructivism toassume that each
function from the natural numbers to the natural numbersis
recursive (computable), it's consistent with constructive
mathematicsto assume so. This stuff about the computable real
numbers is just the sameresults in the context of making that
extra assumption.|In particular, there was one assumption that
was|always accepted implicitly, but never made explicit.|That
assumption was central to the very de'nition|of mathematics.
Informally, the assumption is that|mathematics has something
to do with computation.This strikes me as being a bit of
historical mind-reading.I don't think that geometry was
understood as necessarily having to do withcomputation
throughout history. It possessed (not for ideological
reasons)such noncomputable features as the belief that a point
on a plane eitherlies on a given line or lies off it, and that
two lines are either parallelall the way out to in'nity or
that they intersect at some de'nite
point.Descartes' method of
coordinates made geometry much closer to arithmetic,and hence
to computation, but muddles over concepts like that of
anarbitrary curve persisted on into the 19th century. Around
1800 leadingmathematicians were considering whether an
arbitrary real function (whichis closely related to the idea
of an arbitrary curve) should be consideredto include
mechanically generated ones. The idea that it should be
givenby a formula (or a sequence of formulas on intervals) was
being consideredwhen Fourier complicated matters by claiming a
proof that quite a broadclass of functions could be given as
Fourier series.If this intuition about computation had really
been in the blood of allmathematicians before Cantor, all
these discussions would have been muchmore one-sided. A
function would have of course been given in such a wayas to
make its values computable, and if there wasn't a logical
guaranteethat a curve generated by a natural mechanical
process was always like that,so much the worse for nature,
eh?|It's my claim that Cantor exploited this defect|in the
mathematics of his time to come to the|startling conclusion
that there exists a super-|in'nite world having nothing to do
with computation.|That idea that this super-in'nite world
should be|accepted as part of mathematics is original
with|Cantor (I believe so, anyway).It's your own idea that
it
has nothing to do with computation.I think this is something
like the mistake Mach made, when he arguedagainst the
existence of atoms, because they weren't observable,
meaningin
fact unobservable in a direct enough way to be satisfactory to
him.|When Cantor 'rst made his results available,
many|mathematicians noticed there was something very|peculiar
about his ideas. But at that time, they|were never able to
identify the problem.They were more articulate than you give
them credit for. It's just thatpeople like Kronecker failed
to
persuade the community to go their way,Poincare didn't
persuade
the community to go his way, and so on.And now there's you,
of
course, writing as if you've found yet anotherpath out of
the
dilemma which in some unclear way is superior to thoselaid out
at the time, or their later re'nements.One of the turning
points was when Kronecker and others (Dedekind?) wereboth
considering so-called ideal divisors in algebraic number
theory.Kronecker made a concrete, relatively computational
de'nition which washowever harder to work with. The
de'nition
which won out was like theone we now use, which is an
arbitrary set I closed under both additionand multiplication
by elements of the ring in question. I think it's apity that
this particular episode didn't work out differently. In
myopinion, as one does constructive mathematics, one need not
be so carefulto avoid the concept of an arbitrary subset of a
given set; in hindsight,I think the usual de'nition of ideal
was more compatible with hisphilosophy than they would have
known at the time.|One of the de'ning characteristics of
religion|is a belief in a realm of existence beyond what|we
can observe.No, this may be common, but it's not part of the
de'nition. Therede'nitely are Buddhists who
don't believe in
disconnected realms outthere.|Cantor introduced such a
realm|into mathematics, and there is no doubt he had|religious
motivations.Cantor was concerned with opposition from the
religious authorities ofthe time, who were generally
dyed-in-the-wool 'nitists when it came tomathematics. Heaven
forbid anybody should step in and try to turn in'nityinto
just
another mundane topic to be analyzed rationally.The kind of
thing you're describing as utterly fantastic is such
thingsas
the idea of the set of all subsets of a given set, or perhaps
the ideaof {1, 2, 3, 4, ...} considered as being a set. This
is not one of theplaces where mathematicians have allowed
constructive content to bleedaway, though.|This current debate
is very similar to the debate|between the creation scientists
and the evolutionists.|It is a science vs. religion debate.
Both sides|know they're right.I would say it's
more like the
debate between Mach and people who acceptedatomic theory
before the evidence was entirely conclusive, or maybe likethe
disagreement between Popper and Lakatos. Popper like you had
an ideaof what kinds of restrictions science *ought* to follow
in order to countas a science. But then one has to deal with
the fact that in practice itdoesn't quite work that way, not
without making a lot of allowances forcontinuing to tinker
with theories after they've been falsi'ed, or
beforethey've
made successful new predictions, which makes it a lot harder
to saywhen it becomes actually unscienti'c to keep working on
them.As Lakatos pointed out, you can't really expect to have
a
methodology witha _fast-acting_ way of weeding out the junk. Of
course science has to beable to rid itself of junk, but
typically it requires that you be workingout a successful
alternative to the defective paradigms in question, andthat
time for the issue to be sorted out be allowed.|The claim of
the evolutionists is that we don't|need to invoke a realm of
existence beyond what|we observe to explain the observable
universe, |and I am merely claiming that that idea can
be|applied to mathematics.Certainly introducing such
intangibles as atoms into science was a bigmistake when it was
done. I mean, good grief, next thing they'll beclaiming that
the universe might have originated in a space-timesingularity
that we have no way of recreating, or that protons and
neutronsobserve them.|My only reason for getting involved in
this |thread was to offer Mike Deeth advice on how|to 'ght
the
evil Cantorian religionists. It's|something he wants to do.
I
have little desire|to do so myself.Did offering the
permanently 11-year-old lad this advice in public helphim
'ght
the good 'ght? Isn't he doing a better job of
parodying thegood
'ght than of 'ghting it?Keith Ramsay
<874r1dem8d.fsf@phiwumbda.localnet>
<vhsihi3vtgo5f6@corp.supernews.com>
<8765ltbnw8.fsf@phiwumbda.localnet>
<87znj4l6p9.fsf@phiwumbda.localnet>
<87n0f38zl9.fsf@phiwumbda.localnet>
<87fzku5vx8.fsf@phiwumbda.localnet>
But, the de'nition of R precedes Cantor. Take
Dedekind's
de'nition,>> say, or Cauchy's, and it is easy
to derive that
the usual canonical>> decimal representation works. Cantor did
*not* offer a new or>> different de'nition of R. Nor does the
de'nition of R come from>> Cantor's work.
Rather, Cantor
proved an important theorem about the>> existing de'nition of
R. You need to check your history.>> One thing (of many
others) that's confusing me in following this> discussion is
that nobody involved seems to have a good idea of what> Cantor
did and did not actually do. In other words, nobody has their>
history straight.>> First, Dedekind's de'nition
of the reals
was formulated around the same> time as Cantor's
de'nition,
and yes, Cantor did formulate a de'nition of> the reals,
which
*was* new at the time: it's called (unjustly) the Cauchy>
completion of the rationals. Dedekind was, in fact,
corresponding with> Cantor heavily at the time, and so was
inspired by Cantor to some degree. > The most important point
here, I think, is that there was no existing> de'nition of R
before Cantor. Some mathematicians, like Cauchy, had> made
some attempts, but their de'nitions were circular.I
appreciate
the corrections. You're right that I was fuzzy on
thehistory,
and didn't have the appropriate texts at hand. I see
thatDedekind published his work on irrationals and continuity
in 1872.When did Dedekind and Cantor begin correspondence? I
have a book here(by Boyer) that says they 'rst met in 1874,
but it is silent onwhether they had previously corresponded.
In any case, Boyer iswriting a broad book for a wide audience,
and perhaps should not betaken as a reliable source on these
matters.Let's see to what extent these corrections alter my
main criticism,namely, that it is wrong to credit Cantor with
changing the existingmathematical setting to one in which the
trans'nite hierarchy can befound. I've largely
ignored sets in
this latest discussion, but I do not cedethe point that the
axioms of modern set theory were primarily chosento formalize
Cantor, as David Petry indicates. On the contrary, theset
theory which is behind Cantor's proofs is essentially the
same
asthat of Frege[1] and Dedekind as well. Whether or not they
were incontact with one another and in§uenced one another to
some degree, itseems apparent that this set theory was a
natural means of makingexplicit the implicit use of sets in
mathematics of the day. Ofcourse, the postulation of in'nite
sets was controversial, but I takeit that David is not
disputing the axiom of in'nity.I don't mean to
belittle
Cantor's foundational work, but his aim wasnot to introduce
a
new mathematical setting, but to make explicit theimplicit
setting of the day in order to bring rigor to the subject.This
is in stark contrast to David's claim that Cantor introduced
notonly a new theoretical setting but also abandoned the
workingassumption that mathematics is
aboutcalculation/computation/observables/whatever word David
uses today.This is not at all how I see Cantor's intention
or
effect. > Second, the whole decimal representation thing is
partly what motivated> Cantor to de'ne the reals in the
'rst
place, but decimal representations> were not a focus of his
work. In fact, not until almost two decades later> did he give
his famous diagonal argument using decimal expansions, and>
that argument isn't even his, having been given a few years
earlier by> involved using nested intervals of the real line.I
didn't know that the diagonal argument was 'rst
given by
someoneelse, but in this case, I think the historical
development is not sogermane to my point.David claims Cantor
introduced the setting in which different sizes
o'n'nity
coexist. On the contrary, I claim that the proof |N| < |R|is
not due to any feature of analysis introduced by Cantor, at
leastnot any feature solely introduced by him. Instead, all
one needs forthe diagonal proof is the usual decimal
representation of R (and thatN and R are sets, of course). So,
it is misleading in the extreme toclaim that Cantor is to blame
for the mathematics in which |N| < |R|is provable. This claim
is partly historical (as far as whether the
decimalrepresentation was taken for granted prior to Cantor),
but largelylogical (as to what assumptions are suf'cient for
the proof that |N|< |R|). I am not so concerned with who
proved it and when, butwhether it *could have been* proven
prior to Cantor. I acknowledgethat this question is not so
well-de'ned, since the mathematics ofthe day was not explicit
in its foundations, but this is my aim.Again, I appreciate the
corrections on Cantor's contributions to realnumbers and his
correspondence with Dedekind. These two points arerelevant and
damaging to my argument as I had stated it, but not to mythesis
as a whole, I think.Footnotes: [1] I hesitate to include Frege,
because I would guess that he wasvery directly in§uenced by
Cantor's work.-- [N]ow for once I might actually have an
audience that realizes that[my proof of Fermat's Last
Theorem
is correct], because you see,they'll 'nally
know what's in it
for them--cold, hard cash. --James Harris embarks on a new
mathematical strategy.
In'nitary objects exist in the
sense that we> can construct 'nitary approximations to the
objects.> Is this or is this not a rejection that R exists?
It really does depend on what you mean by exists.> Is R
constructed by> 'nitary approximations in your view? The
notion of R as it is de'ned within the axiomsystems inspired
by Cantor's theory cannot beconstructed by
'nitary
approximations. In fact,that's what the diagonalization
argument proves.Nevertheless, we can construct something
veryclose to our intuitive notions of R by
'nitaryapproximations. I would claim that
Cantor's theory
leads tocounter-intuitive conclusions about R, thoughI
realized that those who have spent yearsstudying Cantor's
theory may have changed theirintuitions.
> In any case,
I have overemphasized the set-theoretic components of>>
Cantor's theorem the |X| < |P(X)|, to the neglect of his
theorem that>> |N| < |R|. For that proof[1], one needs only>> (1) The sets N and R> In'nitary objects exist in the
sense that we> can construct 'nitary approximations to the
objects.Any distinction between in'nitary objects
andin'nite
objects.> Cantor claims to have proven that there exist>
in'nitary objects for which we cannot construct>
'nitary
approximations.I can't imagine that Cantor would have made
any
such claim.> If you're going to mock the very idea of a
world>
of computation, which you seem to be doing, then> there's
little point for me to respond to you.Good. Then why did
you?-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html The
League of Gentlemen
So, it is misleading in the extreme
to> claim that Cantor is to blame for the mathematics in which
|N| < |R|> is provable. Cantor introduced that idea that the
sizes o'n'nite sets can be compared.Why
aren't you arguing
with Darryl? He hasintroduced a notion of uncountability
wherethe sizes of sets can be compared to N butnot to each
other in general.
You repeat the error over and over and
over. I have explicitly> corrected this error on many
occasions, and yet you persist in stating> these 'ctions.
Right.This is a religious battle. You repeat your errorsover
and over, and I repeat my errors over and over.If our goal was
to come to an agreement, then Iwould win. My whole point is to
reduce mathematicsto the study of things we all agree exist,
and toignore the rest.If the goal to prove ourselves right and
the otherwrong, then you win. You have more energy than Ido.
Congratulations.
=|||> So, it is misleading in the extreme
to|> claim that Cantor is to blame for the mathematics in
which |N| < |R||> is provable. ||Cantor introduced that idea
that the sizes of|in'nite sets can be compared.no, galileo
considered that idea long before cantor. but then ofcourse
galileo was one of those fantasyland types who never
exhibitedany interest in the real world.--
As Lakatos
pointed out, you can't really expect to have a methodology
with> a _fast-acting_ way of weeding out the junk.That's
actually a very good point.I'm claiming that with the advent
of computers, therehas been a huge shift in our thinking about
the natureof computation. Computation is no longer seen as
somethingthat exists merely in our minds, but rather, it is
seenas something having a virtual reality. Certainly
theyounger generation has this view. The mathematicscommunity
has proven to be very resistant to thisparadigm shift.This
shift in our way of thinking about computationdoes give us a
way of weeding out junk. I don't expectthe process of
weeding
out the junk to be rapid, butI do believe it to be
inevitable.> |My only reason for getting involved in this >
|thread was to offer Mike Deeth advice on how> |to 'ght the
evil Cantorian religionists. It's> |something he wants to
do.
I have little desire> |to do so myself.> Did offering the
permanently 11-year-old lad this advice in public help> him
'ght the good 'ght? Isn't he
doing a better job of parodying
the> good 'ght than of 'ghting it?Hmmm.I
don't think our
11-year-old friend is on the righttrack, but he is energetic
and enthusiastic, so I thought it might be worth while to get
him on theright track. <8765ltbnw8.fsf@phiwumbda.localnet>
<87znj4l6p9.fsf@phiwumbda.localnet>
<87n0f38zl9.fsf@phiwumbda.localnet>
<87fzku5vx8.fsf@phiwumbda.localnet>
<874r17donm.fsf@phiwumbda.localnet So,
it is misleading in the extreme to claim that Cantor is to>>
blame for the mathematics in which |N| < |R| is provable.>>
Cantor introduced that idea that the sizes of> in'nite sets
can be compared.He gave a de'nition of cardinality, one which
apparently seemednatural to Frege, Dedekind and other
contemporaries.Surely you're not suggesting that his
*de'nition* is the source ofall problems? If you
don't think
that cardinality is an appropriatesense of the naive or
intuitive notion of size, I reckon we havelittle to discuss.
But, you seem to be claiming something stronger.You seem to be
claiming that introducing (or extending) the relationthe
cardinality of X is less than the cardinality of Y for
in'nitesets was bad.I can't see how anything
bad comes from
giving a name to a logicalformula.> Why aren't you arguing
with Darryl? He has> introduced a notion of uncountability
where> the sizes of sets can be compared to N but> not to each
other in general.I haven't seen any such thing. Maybe a
message-id?-- Jesse F. HughesThat's what's
annoying about
Usenet as some loser will state a case,get their ass kicked,
but STILL keep coming back as if nothinghappened. -- James
Harris explains his strategy.
<87znj4l6p9.fsf@phiwumbda.localnet>
<87n0f38zl9.fsf@phiwumbda.localnet>
<87fzku5vx8.fsf@phiwumbda.localnet>
<87he57egcp.fsf@phiwumbda.localnet You
repeat the error over and over and over. I have explicitly>>
corrected this error on many occasions, and yet you persist in
stating>> these 'ctions. >> Right.>> This is a religious
battle. You repeat your errors> over and over, and I repeat my
errors over and over.I should point out that Chan Ho Suh
corrected some of my errors.These errors are evident in the
arrogant tone I used above. Thehistorical situation is subtler
than I stated.-- Jesse HughesWell, if I can get [my proof of
FLT accepted], then I hopefully get abook deal down the road,
and maybe I get to go on OOprah'. James Harris,
on the rewards
of mathematical endeavours.
If not, can you help me out
with an example of a function that is >> differentiable
everywhere but the derivative is unbounded? It will >>
probably be obvious as soon as you mention it, but I'm just
not >> seeing the distinction.>>f(x) = x^2.-- Stan Brown, Oak
Road Systems, Cortland County, New York, USA
http://OakRoadSystems.comFortunately, I live in the United
States of America, where we aregradually coming to understand
that nothing we do is ever ourfault, especially if it is
really stupid.
Let f be a differentiable, real valued
function de'ned on all of R.> Suppose f is also uniformly
continuous.> Prove or disprove: the derivative of f is
bounded.> Then multiply that by a smooth function that -> 0
at oo. The result might > just be the example you're looking
for.> If this is not enough to prove that f' is bounded,
then what extra> conditions does one need--like continuity of
the derivative, or f being> bounded.........?> Certainly
neither of those conditions will work; see above. Hmmm ...
what > if the extra condition is that f' is uniformly
continuous?THEOREM: IF f is differentiable on [0,in'nity),
lim_{x -> in'nity}f(x) exists, andf' is
uniformly continuous
on [0,in'nity), THEN lim_{x -> in'nity}
f'(x) = 0.(And, in
particular, f' is bounded.)Proof is LTRAE.-- A.
>> Let f
be a differentiable, real valued function de'ned on all of
R.>>
Suppose f is also uniformly continuous.>> Prove or disprove:
the derivative of f is bounded. >> Then multiply that by a
smooth function that -> 0 at oo. The result might >> just be
the example you're looking for. >> If this is not enough to
prove that f' is bounded, then what extra>> conditions does
one need--like continuity of the derivative, or f being>>
bounded.........? >> Certainly neither of those conditions
will work; see above. Hmmm ... what >> if the extra condition
is that f' is uniformly continuous?>THEOREM: >IF >f is
differentiable on [0,in'nity), lim_{x ->
in'nity}f(x) exists,
and>f' is uniformly continuous on [0,in'nity),
>THEN >lim_{x
-> in'nity} f'(x) = 0.But that lim_{x ->
in'nity} f(x) exists
is much too strong an assumption.All you need to conclude f'
is bounded isf differentiable and uniformly continuous on R,
and f' uniformly continuous on R.|f(x) - f(y)| < 1 and
|f'(x)-f'(y)| < 1. Suppose
f'(a) > N, and show that
f(a+epsilon/2) - f(a) > 1 if N istoo big.Robert Israel
israel@math.ubc.caDepartment of Mathematics
http://www.math.ubc.ca/~israel University of British Columbia
Vancouver, BC, Canada V6T 1Z2
Try Gain*gain, or
gain^eexponent where exponent is greater than 1That did the
trick. I formula I ended up with was...x =
-gain*((5-rating)^2)I wanted the rating to have more weight
than the gain.I'm still playing with the exponent.2 seems a
little low, 3 is too high.I may end up using a fractional
exponent....Krick
there exists a unique...> But this
should not be used in a published paper. The words should be>
written out. For quick writing on a chalkboard, where you
speak as you> write, OK. But for written work, no. Except
maybe for a paper in> symbolic logic, where it is part of the
symbolic system you are talking> about.I agree on your
opinion. For other symbols, I saw a lot of them in
thetextbook. As an engineering student, I haven't seen this
symbol in abook before. This symbol appears on an IEEE paper.
It has no context,confused me.Xiaoying
If this is a
backwards OE' and a
O!', then this might mean> there exists
exactly one .... (formally, there exists such an> object, and
if Oa' and Ob'
satisfy the condition, then a=b).> What is
MRF?> Best wishes,> MikeXiaoying
=Why are martingales
called martingales?TiaGC
Why are martingales called
martingales?> I have a theory about this, but I would love to
know for sure whether it is true. (I out together various
stories I heard here and there.) A martingale is a part of a
horse's bridle, which I believe is designed to stop the
horse
from somewhat like pulling oneself up using ones shoelaces.
Now someone comes along and designs this unbeatable gambling
strategy for a fair game, where one initially places a bet of
say $1, and then if one loses, keeps doubling the bet.
Eventually one wins, and then one is up by $1. (The §aw in
this scheme is that one becomes bankrupt at some point.) Since
this manner of making money, if it worked, would be like pullng
oneself up by ones shoelaces, it was called a martingale. Later
someone worked out the theory of fair games, and used the same
name, martingales, to describe the whole theory.-- Stephen
Montgomery-Smithstephen@math.missouri.eduhttp://
www.math.missouri.edu/~stephen
Why are martingales
called martingales?>> I have a theory about this, but I
would love to know for sure whether it is> true. (I out
together various stories I heard here and there.) A martingale
is> a part of a horse's bridle, which I believe is designed
to
stop the horse from> somewhat like pulling oneself up using
ones shoelaces. Now someone comes along> and designs this
unbeatable gambling strategy for a fair game, where one>
initially places a bet of say $1, and then if one loses, keeps
doubling the bet.> Eventually one wins, and then one is up by
$1. (The §aw in this scheme is> that one becomes bankrupt at
some point.) Since this manner of making money, if> it worked,
would be like pullng oneself up by ones shoelaces, it was
called a> martingale. Later someone worked out the theory of
fair games, and used the> same name, martingales, to describe
the whole theory.> --> Stephen Montgomery-Smith>
stephen@math.missouri.edu>
http://www.math.missouri.edu/~stephenI knew the horse's
bridle
& gambling strategy meanings but saw noconnection between them.
You may be right.GC
Why are martingales called
martingales? According to Karlin and Taylor, the name
Omartingale' derives form a French acronym for
the gambling
strategy of doubling ones [sic] bets until a win is
secured.According to
http://members.aol.com/jeff570/mathword.html, the OED de'nes
the word martingale to mean this strategy (no mention of
acronyms) and cites a usage of the word dating from 1815.--
Stephen J. Herschkorn
herschko@rutcor.rutgers.edu
According to Karlin and
Taylor, the name Omartingale' derives form a
>French acronym
for the gambling strategy of doubling ones [sic] bets >until a
win is secured.And I believe that this comes from a French
region where there was a style oftrousers that were backwards.
Hence, the Martingale style of betting isimplied to be
stupid.--irascible since 1957
Why are martingales called
martingales?>>Tia>GCMy dictionary says the word is from the
provencal martegalo whichmeans from Martigues, a city near
Marseille. (it's in the south onthe coast of the
mediterranean, where the casinos are). So maybethey're
called
like that because they started there, or became popularthere,
etc
Why are martingales called
martingales?>>Tia>>GC>My dictionary says the word is from
the provencal martegalo which>means from Martigues, a city
near Marseille. (it's in the south on>the coast of the
mediterranean, where the casinos are). So maybe>they're
called
like that because they started there, or became popular>there,
etc Jacqueline Pichoche's _Dictionnaire Etymologique du
francais_(Robert, 1979) clari'es this, and, I think, settles
wherethe betting system got its name. My translation:
Martingale, 16th century. Originally _chausses `a la
martingale_, put on backwards, that is, in the style of
Martigues: alteration of the Provencal _martegalo_, feminine
of _martegal_, name of the inhabitants of this little isolated
village on the edge of the marsh of Berre, often ridiculed by
others. 16th century, _`a la martingale_, in an absurd
fashion. 17th century, way of betting by always doubling the
preceding loss, Provencal _jouga a la martegalo_, to play in
the style of Martigues. It seems clear enough (to me) that
_jouga a la martegalo_ didn'trefer to any *actual* style of
gaming practiced in Martigues, butthat it was just a way of
ridiculing the gaming system by namingit for a generic butt of
ridicule.Lee Rudolph
Why are martingales called
martingales?>>Tia>>GC>My dictionary says the word is from
the provencal martegalo which>means from Martigues, a city
near Marseille. (it's in the south on>the coast of the
mediterranean, where the casinos are). So maybe>they're
called
like that because they started there, or became popular>there,
etc> Jacqueline Pichoche's _Dictionnaire Etymologique du
francais_> (Robert, 1979) clari'es this, and, I think,
settles
where> the betting system got its name. My translation:>
Martingale, 16th century. Originally _chausses `a la>
martingale_, put on backwards, that is, in the style of>
Martigues: alteration of the Provencal _martegalo_, feminine>
of _martegal_, name of the inhabitants of this little
isolated> village on the edge of the marsh of Berre, often
ridiculed> by others. 16th century, _`a la martingale_, in an
absurd> fashion. 17th century, way of betting by always
doubling the> preceding loss, Provencal _jouga a la
martegalo_, to play in> the style of Martigues.> It seems
clear enough (to me) that _jouga a la martegalo_ didn't>
refer
to any *actual* style of gaming practiced in Martigues, but>
that it was just a way of ridiculing the gaming system by
naming> it for a generic butt of ridicule.> Lee RudolphYes,
it does seem clear now (or, as you say, clear
enough).GC
>Why are martingales called martingales?>My
dictionary says the word is from the provencal martegalo
which>means from Martigues, a city near Marseille. (it's in
the south on>the coast of the mediterranean, where the casinos
are). So maybe>they're called like that because they started
there, or became popular>there, etc> Jacqueline Pichoche's
_Dictionnaire Etymologique du francais_> (Robert, 1979)
clari'es this, and, I think, settles where> the betting
system
got its name. My translation:> [snip def supporting above]Maybe
it settles it, maybe not. The previously-suggested bridle idea
is supported by Webster's 1913 Dictionary(according to
http://www.hyperdictionary.com anyhow):3. (Gambling) The act
of doubling, at each stake, that which has been lost on the
preceding stake; also, the sum so risked; -- metaphorically
derived from the bifurcation of the martingale of a harness.
[Cant] --Thackeray. That is, hyperdictionary'sentry makes no
mention of this usage of martingale deriving fromMartigues,
but instead says it derives from martingale as
harness.-jiw
>Why are martingales called martingales?>My
dictionary says the word is from the provencal martegalo
which>means from Martigues, a city near Marseille. (it's in
the south on>the coast of the mediterranean, where the casinos
are). So maybe>they're called like that because they started
there, or became popular>there, etc>> Jacqueline Pichoche's
_Dictionnaire Etymologique du francais_> (Robert, 1979)
clari'es this, and, I think, settles where> the betting
system
got its name. My translation:>> [snip def supporting above]>
Maybe it settles it, maybe not. The previously-suggested>
bridle idea is supported by Webster's 1913 Dictionary>
(according to http://www.hyperdictionary.com anyhow):> 3.
(Gambling) The act of doubling, at each stake, that which has>
been lost on the preceding stake; also, the sum so risked; -->
metaphorically derived from the bifurcation of the martingale>
of a harness. [Cant] --Thackeray. That is,
hyperdictionary's>
entry makes no mention of this usage of martingale deriving
from> Martigues, but instead says it derives from martingale
as harness.> -jiwOh! Previously I saw no connection between
harnesses and gamblingstrategy. Sorry if I've opened a can
of
worms <giggle>.GC
=...>> Jacqueline Pichoche's _Dictionnaire
Etymologique du francais_>> (Robert, 1979) clari'es this,
and,
I think, settles where>> the betting system got its name. My
translation:>>[snip def supporting above]>>Maybe it settles
it, maybe not. The previously-suggested >bridle idea is
supported by Webster's 1913 Dictionary>(according to
http://www.hyperdictionary.com anyhow):>3. (Gambling) The act
of doubling, at each stake, that which has >been lost on the
preceding stake; also, the sum so risked; -- >metaphorically
derived from the bifurcation of the martingale >of a harness.
[Cant] --Thackeray. That is, hyperdictionary's>entry makes
no
mention of this usage of martingale deriving from>Martigues,
but instead says it derives from martingale as harness.My 1924
printing of the 1909 edition of the Merriam-WebsterNew
International Dictionary (commonly called Webster's
FirstInternational) doesn't have that citation from
Thackeray.
Idon't know which Thackeray it was, but assume it to be
theEnglish novelist and literary man of that name. In that
casecertainly (and in any case, maybe with less certainty) I
wouldgreatly favor the scholarly opinion of Jacqueline Picoche
(a professional lexicographer writing in 1979) over the(very
likely less scholarly, and in any case many decadesearlier)
opinion of Thackeray. The metaphorical derivationgiven by
Thackeray has all the earmarks of folk-etymology(if not his
own fertile imagination). Also, though Picochedoesn't give a
citation for jouga a la martegalo, she doesdate it to the 18th
century, whereas the Oxford English Dictionary's earliest
citation for martingale in thegaming sense is from 1815 (ah,
and followed by Thackeray-the-novelist, in a novel, 1854) as a
noun, and from 1823 asa verb (whose etymology asks the reader
to compare to French martingaler without positively claiming
that that's its source;the OED etymology of the *noun*
martingale is even less help).Lee Rudolph
> Why are
martingales called martingales? >According to Karlin and
Taylor, the name Omartingale' derives form a
>French acronym
for the gambling strategy of doubling ones [sic] bets >until a
win is secured.>According to
http://members.aol.com/jeff570/mathword.html, the OED >de'nes
the word martingale to mean this strategy (no mention of
>acronyms) and cites a usage of the word dating from
1815.Well, I checked the OED.The OED derives the word
ultimately from the name of the town of Martiguesin
south-eastern France. In French the phrase chausses a la
martingale, hose that fasten at the back, is known from 1491.
There's meaning (1) for a strap on a horse bridle, (2) for a
rope on a ship (to keep the boom from rising), as well as (3)
the gambling system. Although meaning (1) is known before (2),
apparently there's some argumentthat (1) came from (2),
which
could come not from the reputation of thepeople of Martigues
for being eccentric and naive, but from the importanceof the
town as a port and ship-building centre. The use in French
fora gambling system goes back to 1754. I don't think anyone
really knows whether usage (3) comes from the reputation of
the people of Martigues or from meaning (1) or (2). I rather
suspect it's (1) or (2) because with this system, although
youcan go broke you won't end up winning more than the
original bet - thisis in some way analogous to the horse not
being able to raise its head or the boom not being able to
rise too far.Robert Israel israel@math.ubc.caDepartment of
Mathematics http://www.math.ubc.ca/~israel University of
British Columbia Vancouver, BC, Canada V6T 1Z2
> I need
an algorithm that can give me the number of people I will need
to> populate N daily shifts, Y people per shift (including
weekends) for a> business plan.>> I need to incorporate this
in an excel sheet so I am not looking for any> programs, just
a way to calculate the aproximate number of people I will>
need to hire.>> JCWith no other constraints, just multiply the
two 'gures together.Bob Pease