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Topic: Number theory question continued
Replies: 9   Last Post: Mar 27, 2006 5:13 PM

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Robert Israel

Posts: 11,902
Registered: 12/6/04
Re: Number theory question continued
Posted: Mar 27, 2006 5:13 PM
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In article <e09liv$3cf$1@nntp.itservices.ubc.ca>,
Robert Israel <israel@math.ubc.ca> wrote:
>In article <44283c39$1@news.ysu.edu>,
>Eric J. Wingler <wingler@math.ysu.edu> wrote:


>>"ManOfLight" <mladensavov@yahoo.com> wrote in message
>>news:1143276915.440583.51260@z34g2000cwc.googlegroups.com...

>>> Hello everybody,
>>> I post it again to rapair one mistake
>>> Could you give a clue how I can start the following problem or
>>> propose a solution.
>>>
>>>
>>> "Is it true that for every sufficiently large interval there will be a
>>> integer in it of the form
>>> 2^n-3^m where m,n are integers?"
>>>
>>>
>>> As far as I understand it we are supposed either to prove that there
>>> exists number H : every interval with length H contains such a number
>>> or disprove it.

>>
>>If x = 2^n - 3^m, then log(1 - x/2^n) = m*log(3) - n*log(2), so it's
>>possible that you may be able to make use of the density of the set of
>>numbers of the form m*log(3) - n*log(2).

>
>I doubt it.
>
>This implies there are infinitely many integers of the form x = 2^n - 3^m
>with c < x/2^n < d for any c and d with 0 < c < d < 1. But unfortunately
>the interval for x grows as n -> infty. You can do slightly better
>using results about Diophantine approximation, but still you won't get
>an interval of constant length.


On the other hand, even a rather weak lower bound on the approximation of
log(3)/log(2) by rationals would be useful in the other direction.
I believe there are such bounds, due to Gelfond (with later improvements
by Baker and others).

Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada






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