Thanks! Good Stuff!
Joth
PS: There is no Nobel Prize in math. This is perhaps unfortunate, as the
winner of a Nobel Prize
seems to get over $500,000 (US) in cash and generally a new building at the
university of choice.
The closest math counterpart is the Fields Medal which does include a prize:
$15,000 Canadian
(or about $10,000 US) and sometimes a corner office.
Ah, the peaceful uses of dynamite.
----- Original Message -----
From: Alan Simpson <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED]>; <[EMAIL PROTECTED]>
Sent: Wednesday, October 27, 1999 2:07 AM
Subject: RE: Mersenne: More Schlag
>
> hi,
>
> recently Joth Tupper posted a message with the following
>
> "I seem to recall a non-intuititive theorem about rational approximations
to
> numbers (this is from c. 1968). If you can approximate a number too
closely,
> then it is transcendental. S.Lang wrote a book on trancendental numbers
and
> degrees around 1973 and a precise statement might be there.
>
> Does anyone recall this?"
>
> Yes!
>
> This is the result that Liouville proved to show that Liouville's number
> must be transcendental. The result is known, funnily enough as Liouville's
> theorem.
>
> It says that if \alpha is an algebraic number of degree n \geq 2, then
there
> exists a constant c(\alpha)>0 such that
>
> | \alpha - p/q | > c(\alpha)/|q|^{n}
>
> always holds.
>
> Or if you prefer, for any m>n,
> | \alpha - p/q | < |q|^{-m} ---- call this (*)
>
> has only finitely many solutions.
>
> At the root of the proof is the extremely "profound" fact that there are
no
> integers between 0 and 1 (or rather that the integers are discrete):
>
> Let f(x)=x^n+a_{n-1}x^{n-1}+...+a_{1}x+a_{0} be the "minimal polynomial"
for
> \alpha over the rationals. Let p/q be a rational number.
> Then
> q^{n}f(p/q) is an integer, it cannot be zero because f(x) is the minimal
> polynomial for \alpha, therefore
> |q^{n}f(p/q)| \geq 1.
>
> f(x) =(x-\alpha)g(x) where g(x) is some other polynomial, so
> |q^{n}(p/q-\alpha)g(p/q)| \geq 1 or
> |p/q-\alpha| \geq |1/(g(p/q)q^{n})|.
> One can estimate g(p/q) quite easily and use this estimate to get the
value
> for c(\alpha) mentioned above.
>
> You prove that the Liouville number (or any similar type number) is
> transcendental from this result is by noting that the rational
> approximations are getting more and more accurate at each step (you are
> basically adding lots more zeros each time). But Liouville's theorem
forbids
> this, if the number is algebraic. Thus the number is transcendental.
>
> It can be formulated somewhat different yet again (in terms of "heights"
of
> numbers) and it is an extremely important result that is used almost
> everywhere in diophantine analysis and transcendental number theory (so
> ultimately most proofs in these areas depend upon cleverly showing that
some
> "deep" result depends on the fact that there are no integers between 0 and
1
> :-)).
>
> Liouville proved his theorem around 1840. It has since been improved.
>
> Thue (1909) showed that you can weaken the condition on m in (*) above to
> m>n/2
> Siegel (1922) shows that m>2\sqrt{n} suffices
> Dyson and Gelfond independently showed in the 40's that m>\sqrt{2n}
suffices
> and finally (sort of) Roth in 1954 showed that m>2 suffices.
>
> Roth won a Fields medal for this (and some other) work.
>
> Does that answer your questions?
>
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