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It can be proved that there are two irrational numbers $a$ and $b$ such that $a^b$ is rational (see Can an irrational number raised to an irrational power be rational?) and that for each irrational number $c$ there exists another irrational number $d$ such that $c^d$ is rational (see For each irrational number b, does there exist an irrational number a such that a^b is rational?).

My question is: Is there an irrational number $a$ such that $a^a$ is rational (and how could you prove that)?

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    $\begingroup$ $a^a=x$ has at least one real solution $a$ for each positive rational $x$, so all we need is to prove that there exists an $x$ such that $a$ must be irrational... (I suggest $x=2$ as a worthy candidate...) $\endgroup$ – abiessu Feb 5 '16 at 15:23
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Consider the unique (positive) solution $a$ to $x^x = 2$. If $a$ were rational, say, $a = \frac{p}{q}$, $p$ and $q$ are positive integers such that $\gcd(p, q) = 1$, we would have $$\left(\frac{p}{q}\right)^{p / q} = 2 ,$$ and rearranging gives $$p^p = q^p 2^q .$$ Since there is no integer $n$ such that $n^n = 2$, we must have $q > 1$ and hence $2 \mid p^p$. Because $2$ is prime, we have $2 \mid p$. So, $2$ occurs an even number of times the prime factorization of $p^p$ and likewise of $q^p$. Since $p^p = q^p 2^q$, we must have $2 \mid q$, but now $2 \mid p, q$, and this contradicts $\gcd(p, q) = 1$. Thus, $a$ is irrational but $a^a$ is rational (in fact, an integer).

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By continuity of the function $x^x$, there is an $a$ such that

$$a^a=2.$$

This number is irrational. Otherwise, let $a$ be the irreducible fraction $p/q$ and

$$\left(\frac pq\right)^{p/q}=2,$$ is equivalent to $$p^p=2^qq^p,$$

which implies that $p$ is even and $q$ is even, a contradiction.


By the way, as

$$\ln(a^a)=a\ln(a)=\ln(a)e^{\ln(a)},$$

we have

$$\color{green}{a=e^{W(\ln(2))}}$$

where $W$ is the Lambert function.

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  • $\begingroup$ (Independently of @abiessu) $\endgroup$ – Yves Daoust Feb 5 '16 at 15:28
  • $\begingroup$ I wholly approve :-) $\endgroup$ – abiessu Feb 5 '16 at 15:28
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If $a^a=2$, then $a$ is irrational: If $a=p/q$, then $(p/q)^p=2^q$ is an integer, so $p/q$ is an integer.

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