NetNews Usenet Archive 1992 #19/NN_1992_19.iso/spool/sci/math/10526

Internet Message Format | 1992-08-23 | 1.8 KB

Path: sparky!uunet!cis.ohio-state.edu!rutgers!igor.rutgers.edu!remus.rutgers.edu!clong
From: [e-mail] (Chris Long)
Newsgroups: sci.math
Subject: Re: Transcendence question (QUASI-SPOILER)
Message-ID: <[e-mail]>
Date: 24 Aug 92 03:08:28 GMT
References: <[e-mail]> <[e-mail]>
Organization: Rutgers Univ., New Brunswick, N.J.
Lines: 28

In article <[e-mail]>, Ray Steiner writes:

> Is there an irrational number t such that 2^t and 3^t are
> both rational? Is this question still open? The answer is
> yes if we just consider 2^t (2^{log_2 3} being an example)
> but no if we look at 2^t, 3^t and 5^t.(This is a case
> of the famous "6 exponentials theorem.") The general case
> of a^t and b^t, a and b integers, is still open.

If ln(a) and ln(b) are Q-linearly independent, this follows from
the unproven Schanuel conjecture.  Let a^t = q and consider
degree( Q(ln(a), ln(b), ln(q), ln(q)*ln(b)/ln(a), a, b, q, b^t) )
= degree( Q(ln(a), ln(b), ln(q), ln(q)*ln(b)/ln(a)) ) if b^t is
rational, and this must be >= 4 by Schanuel, which is clearly
false.  It remains to be proven that ln(a), ln(b), ln(q), and
ln(q)*ln(b)/ln(a) are Q-linearly independent, but this would
follow if ln(a), ln(b), and ln(q) were algebraically independent,
which would follow from Schanuel if they were Q-linearly independent.
But assuming not, we obtain a contradiction.

This isn't a proof, but at least it shows that such t probably
do not exist.
--
Chris Long, 265 Old York Rd., Bridgewater, NJ  08807-2618

"In a study of schoolboys, an educator discovered a correlation between size
 of feet and quality of handwriting.  The boys with the larger feet were,
 on the average, older."        Wallis & Roberts, _The Nature of Statistics_