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I want to find a proof of the fact that $a^b$ is irrational if $a$ is a prime and $b$ is not an integer. Motivation behind this question: I was posed a question , of finding what is the probability of picking a rational number from $(-\infty,\infty)$ . so i said any number $n$ can be written as $$n=\prod_{i=1}^{i=f(n)}p_{i}^{\alpha_i}$$ where $f(n)$ denotes number of primes less than equal to $n$ and $\alpha$ denotes the corresponding power of the primes. now the probability of picking a rational number will be zero as if any of the $\alpha$ is not an integer the number is not rational , and probability of finding an integer in $(-\infty,\infty)$ is $0$. Thus if i can prove the question i posed i can say probability of finding a rational numberis $0$, but i dont even know where to start

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    $\begingroup$ Counterexample: $a=2,b=\log_25 \implies a^b=5$. $\endgroup$ – barak manos Apr 24 '16 at 14:16
  • $\begingroup$ oh , lol, then is the probability of finding a rational number non zero? @barakmanos $\endgroup$ – avz2611 Apr 24 '16 at 14:18
  • $\begingroup$ Sorry, didn't bother to read the rest of the question... BTW, perhaps you meant $b$ is a non-integer rational? $\endgroup$ – barak manos Apr 24 '16 at 14:20
  • $\begingroup$ yes that makes more sense @barakmanos $\endgroup$ – avz2611 Apr 24 '16 at 14:20
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Your assertion is incorrect. Breaking down the cases, let $a^b = x$, $a$ is a prime, $b$ is not an integer.

$x$ is irrational and algebraic if $b$ is rational (but non-integral as stipulated).

$x$ is irrational and transcendental if $b$ is irrational but algebraic (this is a result of the Gelfond-Schneider theorem).

If $b$ is irrational and transcendental, not much can be said about $x$. It can even be an integer, as in the example @barakmanos gave.

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