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Re: Number theory question continued
Posted:
Mar 27, 2006 5:13 PM


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^n3^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



