Is a rational to the power of itself irrational?

Question

Take two coprime natural numbers $a>0$ and $b>1$

Let $x = \frac ab$.

Is it guaranteed, that $y = x^x$ is irrational? If not, which properties do $a$ and $b$ or $x$ need?

Edit 1:

Thank you @mathworker21 and @fleablood. Now its easy to prove that $x^y$ (the same as $x^{x^x}$) is transcendental according to gelfond-schneider.

if $(a/b)^{a/b}$ were rational, then by Bezout's lemma, $(a/b)^{1/b}$ is rational, but the standard "$\sqrt{2}$ is irrational" proof will show that $(a/b)^{1/b}$ is irrational. — mathworker21, Jul 15 '19 at 21:12
This boils down to proving that if $m$ and $n$ are coprime and $\sqrt[k]{m}$ and $\sqrt[k]{n}$ are both irrational that one is not a rational multiple of the other. — fleablood, Jul 15 '19 at 21:48
"Now its easy to prove that $x^y$ is transcendental". Whoa! Irrational and trancendental is not the same. And for rational $r$, $r^r$ is very easily shown to be algebraic and not transcendental. (Let $r=\frac ab$ then $(r^r)^b\cdot b^a = a^a \in \mathbb Z$ and $r^r$ is a solution to $b^ax^b - a^a = 0$.) — fleablood, Jul 15 '19 at 22:39
Oh, never mind. $rational^{irrational}$ is trancendental. (D'oh!) — fleablood, Jul 15 '19 at 22:42
And since $x$ and $y$ are both algebraically, $x^y$ is transcendental — Kinheadpump, Jul 16 '19 at 07:40

fleablood · Accepted Answer · 2019-07-15T22:29:14.847

We know a real number $\sqrt[k]{n}$ (for integer $k >1$ and integer $n$) is irrational unless $n$ is a perfect $k$ power.

We can conclude then if $\frac nm$ is fraction of coprime integers. that $\sqrt[k]{\frac nm}$ is irrational unless both $n$ and $m$ are both perfect $k$ powers.

(Pf: $\sqrt[k]{\frac nm} = \frac ab$ with $a,b$ coprime then

($a^km = b^kn$ and so assume $p$ is a prime factor of $n$. Then $p|a^km$ but $p\not \mid m$ so $pa^k$ so $p|a$ so $p\not \mid b$. The power to with $p$ divides $a^k$ is a multiple of $k$ so the power to which $p|n$ is a multiple of $k$. That is true of all prime factors so $n$ is a perfect $k$ power. [Or if $n$ has no prime factors which can only occur if $n=\pm 1$ which is a trivial perfect $k$ power. {$k$ must be odd if $\frac nm < 0$}]. Identical argument shows that $m$ is a perfect $k$ power.)

Okay, so if $r = \frac ab$ and $a,b$ are coprime integers with $b$ positive then

$r^r = \frac {\sqrt[b]{a^a}}{\sqrt[b]{b^a}}$. This is only rational if both $a^a$ and $b^a$ are perfect $b$ powers. As $\gcd(a,b) =1$, the only way any $k^a$ can be a perfect $b$ power is if $k$ is a perfect $b$ power.

So for this to be rational there must exist $j,k$ so that $b = j^b$ and $a=k^b$.

But $b = j^b$ is ... fishy.

Claim: If $j\ge 2$ then for any natural $n$, $j^n > n$.

Pf: simple by induction. ($j^1 =j> 1;$ and if $j^n> n$ then $j^{n+1} > j*n \ge 2*n = n+ n \ge n+1$.)

So $j=1$ and $b=1$.

Thus the only way for $r^r$ to be rational, for a rational $r$, is for $r= \frac a1 = a\in \mathbb Z$.

Obviously for integer $a$ we have $a^a$ is also an integer. But if $r$ is a non-integer rational then $r^r$ is irrational.

Is a rational to the power of itself irrational?

1 Answers1

Linked