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From: eklhad@ihnet.UUCP (K. A. Dahlke)
Newsgroups: net.math
Subject: Re: Need proof for density problem
Message-ID: <308@ihnet.UUCP>
Date: Fri, 27-Sep-85 10:18:47 EDT
Article-I.D.: ihnet.308
Posted: Fri Sep 27 10:18:47 1985
Date-Received: Sun, 29-Sep-85 05:02:41 EDT
References: <58@unc.unc.UUCP>
Organization: AT&T Bell Laboratories
Lines: 45

> ... The question
> is, can a number of the form 2^m * 3^n be found between every two positive
> rational numbers?
>    Scott Southard

This is an interesting problem, let me attempt a proof.
Restated: for each pair of positive reals R < S, there exists
integers M and N, such that R < 2^M * 3^N < S.
R and S need not be rational.
Since log base 3 is a continuous, monotonic, cheerful function,
we can take logs across the board, transforming the inequality:
log3(R) < log3(2)*M + N < log3(S).
In other words, it is sufficient to prove that numbers of the form
log3(2)*M + N are dense in the reals.
Next, we can shift the interval (log3(R), log3(S)) by an integer offset,
so that 0 <= log3(R) < 1.  The naked integer "N" in "log3(2)*M + N"
takes up the slack.
Thus, it is sufficient to prove that numbers of the form log3(2)*M + N are
dense in the interval (0, 1).
This problem was discussed here a few months ago (don't know if you are
a regular subscriber).  For any irrational X (e.g. log3(2)),
the set of values K*X mod 1 is dense in (0, 1).
Several people offered clever proofs, though the exact details elude me.
Let me try to recreate the reasoning.  Take a circle 1 unit in 
circumference, and take quantized steps around it, each step of length X.
Is there a section of the circle you never land in (say (R, S))?  No.
The first time you step over 0 (where you started), you will
land within X/2 of 0.  In the case of log3(2) (approximately 0.630929),
it only takes 2 steps, and you are at 0.261859.  Now group your steps
together, traveling 0.261859 each time, and adding 2 to M instead of 1.
After 3 more steps (now only 0.261859 in length),
you will again cross 0, landing at 0.047438, and you can again
modify your step size.  After another 20 such steps, you are at 0.996198.
Each iteration divides your step size in half (at least).
When it is finally less than S-R, you will surely land in the (R, S)
interval on your next trip around.
We must also show that log3(2) is irrational, but that is easy.
Assume log3(2) = P/Q (rational), and consider log3(2^Q) = (P/Q)*Q = P.
Thus 3^P = 2^Q, contradicting the fundamental theorem of arithmetic.
After all this, you have probably lost sight of the original problem.
The bottom line is, the values 2^M * 3^N are dense in the positive reals.
Hope this helps.
-- 
	This .signature file intentionally left blank.
		Karl Dahlke    ihnp4!ihnet!eklhad


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