How easy is it to prove that if $2^x,3^x, 5^x, 7^x, 11^x ... $ are integers then $x$ is an integer as well? I have read the definition of the exponent functions as given in my calculus text, and the question propped up.
Asked
Active
Viewed 171 times
6
Jam
- 10,325
-
2Is $x$ a real number? A rational number? – Nishant Sep 22 '14 at 16:32
-
@Nishant Real, obviously. – Theluqfzhluswcpzflabucheecatne Sep 22 '14 at 16:33
-
@Nishant I assume it's real. – Akiva Weinberger Sep 22 '14 at 16:33
-
4I asked a similar question and the following page was shown. http://mathoverflow.net/questions/17560/if-2x-and-3x-are-integers-must-x-be-as-well – mathlove Sep 22 '14 at 16:36
-
Based on @mathlove's link, it's evidently enough to require that $2^x,3^x,5^x$ are integers; however, that version is evidently difficult to show. So a proof for the "bases are prime" case would indeed be neat. – Semiclassical Sep 22 '14 at 16:40
-
@Semiclassical The fun little problem ... :O – Theluqfzhluswcpzflabucheecatne Sep 22 '14 at 16:42
-
Yes, one which is strictly speaking easier than yours! (And, reading that thread, one which no one solved when it appeared on a certain year's Putnam exam.) Hence a proof for your case would be interesting, but not necessarily very easy. – Semiclassical Sep 22 '14 at 16:45
-
Here is a link to the question that @mathlove posted. – TonyK Sep 22 '14 at 16:50
-
4@mathlove Here is the Putnam solution, https://mks.mff.cuni.cz/kalva/putnam/psoln/psol716.html :) Hard really. – Theluqfzhluswcpzflabucheecatne Sep 22 '14 at 17:02
-
Curiously, if you only know that $2^x$ and $3^x$ are integers, then the question if $x$ must be an integer is open (no proof and no counter-example is known). – Vladimir Reshetnikov Oct 07 '14 at 00:10
1 Answers
-2
Given $ 2^x $, $ 3^x $, $ 5^x $,... are integers. Now suppose $ x $ is not an integer. Suppose it is rational of the form $ p/q $ where $ gcd(p,q) = 1 $. So $ 2^{p/q} $, $ 3^{p/q} $, ... can't be integer, since base of exponents are all prime.
MATHSMARATHON
- 140
-
2
-
2
-
If $ x $ is irrational, irrational power of prime can't be integer. – MATHSMARATHON Sep 22 '14 at 17:08
-
5$2^{\log(3)/\log(2)}=3$, and $\dfrac{\log 3}{\log 2}$ is definitely not rational. – Semiclassical Sep 22 '14 at 17:10
-
We need $ x $ for all primes, means $ 2^x $, $3^x$, $ 5^x$, .... – MATHSMARATHON Sep 22 '14 at 17:20