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This question already has an answer here:

The following fact is known:

If $a$ and $b$ are both irrational numbers, then $a^b$ can be a rational number.

Proof. Suppose $a^b$ is always irrational. Then $\sqrt{2}^\sqrt{2}$ is an irrational number, which in turn implies that $\left(\sqrt{2}^\sqrt{2}\right)^\sqrt{2}$ must also be irrational. However: $$ \left(\sqrt{2}^\sqrt{2}\right)^\sqrt{2} = \sqrt{2}^{\sqrt{2}\times\sqrt{2}} = 2 $$


Now assume $a$ is a rational number and $b$ is irrational. I wonder if $a^b$ can be a rational number. Any way to prove, or a reference maybe?

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marked as duplicate by Xander Henderson, user21820, B. Mehta, user223391, Namaste May 25 '18 at 21:59

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ $1^\sqrt{2}$ is rational. Any combination is possible. $\endgroup$ – max_zorn May 3 '18 at 8:28
  • $\begingroup$ Indeed, assume $a$ to be positive and different from 1 $\endgroup$ – cyanide May 3 '18 at 8:36
  • $\begingroup$ Alternatively, this answer addresses the point. $\endgroup$ – Xander Henderson May 25 '18 at 2:35
  • $\begingroup$ The observation "above the fold" is a classic illustration of the principle of excluded middle. One can in fact identify $\sqrt{2}^\sqrt{2}$ as irrational and even transcendental (I think due to Gelfond-Schneider), I think constructively, but if all that one wants are irrational $a, b$ with $a^b$ rational, it's much easier and constructive to take $a = \sqrt{2}$ and $b = 2\log_2 3$, as an answer below indicates. No law of excluded middle needed. In view of this, the popularity of the example above slightly baffles me. $\endgroup$ – user43208 Aug 4 '18 at 16:51
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Let $\log_a$ be the logarithm to the base $a$. $$a^{\log_a c}=c$$ If $$a^\frac{q}{p}=c$$ then $$a^q=c^p$$ and each prime that divides $a$ must divide $c$ and each prime that divides $c$ must divide $a$. So if this is not the case then $\log_ac$ must be irrational.

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  • $\begingroup$ What is $b$ in your last paragraph? $\endgroup$ – IanF1 May 3 '18 at 9:32
  • $\begingroup$ @IanF1. two typos, thx $\endgroup$ – miracle173 May 3 '18 at 11:01
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$2^b = 3$ where $b = \log_2(3)$ is irrational. In fact, if $2^b = c$ where $b$ and $c$ are rational, $b$ must be an integer.

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You can use the Gelfond-Schneider theorem to prove that $a^b$ is not only irrational, but transcendental. The theorem states that:

If $a$ and $b$ are algebraic numbers with $a ≠ 0,1$ and $b$ non-rational, then any value of $a^b$ is a transcendental number.

This not only gives rise to trivial results such that $2^{\sqrt 2}$ is irrational, but to some more interesting ones as well. For instance, by taking $a=-1$ and $b=-i$, where $i=\sqrt {-1}$, we can obtain the following result:

$$ (-1)^{-i} = \left( e^{i \pi} \right)^{-i} = e^\pi, $$

showing that $e^\pi$ is transcendental.

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    $\begingroup$ While interesting, this doesn't actually answer the question as asked. The question is whether or not a^b could be rational, and you only ruled out that being the case for algebraic numbers. $\endgroup$ – Triclops200 May 3 '18 at 17:55

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