3

I want to know, given a rational number $x=\frac ab$ where $a$ and $b$ are distinct positive integers with $b\ne1$ and $\gcd(a,b)=1$, whether $x^{-x}$ can be rational as well.

What I achieved so far is as follows.

$$ \begin{align*} x^{-x} &= \bigg(\frac{a}{b}\bigg)^{\big(-\frac{a}{b}\big)} \\ &= \bigg(\frac{b}{a}\bigg)^{\big(\frac{a}{b}\big)} \\ &= \frac{\sqrt[b]{b^a}}{\sqrt[b]{a^a}} \\ \end{align*} $$

For this to be rational, either (i) both the numerator and the denominator must be rationals or (ii) they must share an irrational factor $r$ which will be canceled out by the division.

I can see that the first condition is unsatisfiable: As $a$ and $b$ have no common prime factors, both $\sqrt[b]{a}$ and $\sqrt[b]{b}$ must be rationals but $\sqrt[b]{b}$ cannot be a rational.

I cannot, however, prove or disprove whether the second condition is also unsatisfiable. If you can provide me a simple proof or point me to a complicated one, I would be grateful.

Bill Dubuque
  • 272,048
Eser Aygün
  • 151
  • 1
    Why do you write $x^{-x}$ you could have asked $x^x$ only, it looks simple and is equivalent! – Arpan1729 May 11 '17 at 02:49
  • This would require b=k^b for some integer k. That is easily shown to be impossible – fleablood May 11 '17 at 02:59
  • If they have an irrational in common, that irrational is a root. If your rise it to a and to be it becomes rational. So it a root of a power that is a common factor of a and b. But a and b is relatively prime. So that's unsatisfactory. – fleablood May 11 '17 at 03:03
  • 1
    all these are wrong, both the denominator and the numerator can be irrational, in that case t's difficult to tell whether the number is rational or not. – Arpan1729 May 11 '17 at 03:06

3 Answers3

3

Assuming the contrary, let $x^{-x}$ be rational. Hence, it can be expressed as $\dfrac m n$, with $m$ being an integer and $n$ being a positive integer. Moreover, we can assume that $m$ and $n$ are co-prime.

$$\begin{array}{rcl} x^{-x} &=& \dfrac m n \\ \left(\dfrac b a\right)^{a/b} &=& \dfrac m n \\ \dfrac{b^a}{a^a} &=& \dfrac{m^b}{n^b} \\ b^a n^b &=& a^a m^b \\ \end{array}$$

Let $p^k$ be a prime power factor of $b$ with maximum $k$. Then, $p^{ka}$ is a prime power factor of $b^a$ with maximum power. Since $b$ and $a$ are co-prime, the factorization of $a^a$ cannot contain $p$ at any power. Therefore, $p^{ka}$ must be a prime power factor of $m^b$.

Let $p^h$ be a prime power factor of $m$ with maximum $h$. Then, $p^{hb}$ is a prime power factor of $m^b$ with maximum power. Since $m$ and $n$ are co-prime, by the same argument, $p^{hb}$ must be a prime power factor of $b^a$.

Hence, $ka \le hb$ and $hb \le ka$, so $ka=hb$. Since $a$ and $b$ are co-prime, $k$ is divisible by $b$.

Since $p$ is arbitrary, this holds for any prime factor of $b$. In other words, if for any prime $p$, $p^k$ is a maximum prime power factor of $b$, then $k$ is divisible by $b$. This means that $b$ can be expressed as $c^b$ where $c$ is a positive integer.

We know that $c$ cannot be $1$, or else $b$ would be $1$. If $c$ is any larger, then we must have $c^b > b$. Therefore, it is impossible.

Kenny Lau
  • 25,049
  • That was very easy to follow. Thank you very much. I'm a bit sad about the result though; I was hoping to find a rational solution. – Eser Aygün May 11 '17 at 12:19
  • @EserAygün It's simpler to reduce it to a basic result about mutual coprime powers - see my answer. – Bill Dubuque May 14 '22 at 11:47
1

Suppose $b>1$ and that $x^{-x}=\frac{\sqrt[b]{b^a}}{\sqrt[b]{a^a}}=\frac pq$ for some $p,q\in\Bbb Z$. Cross-multiplying and raising to the power of $b$ we get $$a^ap^b=q^bb^a$$ Now consider factors of $a$ and $b$ in $q^b$ and $p^b$:

  1. Because $\gcd(a,b)=1$, $a$ and $b$ occur at least $a$ times in $q^b$ and $p^b$ respectively.
  2. To keep $p^b$ and $q^b$ perfect powers we need at least $-a\bmod b$ more factors of $a$ or $b$ on both sides.
  3. To preserve the equality we need $-a\bmod b$ more factors each of $b$ in $q^b$ and $a$ in $p^b$ respectively.
  4. To keep $p^b$ and $q^b$ perfect powers we need at least $+a\bmod b$ more factors of $a$ or $b$ on both sides.
  5. To preserve the equality we need $+a\bmod b$ more factors each of $b$ in $p^b$ and $a$ in $q^b$ respectively.

Because $\pm a\not\equiv0\bmod b$, steps 2-5 are repeated infinitely, so $a^ap^b=q^bb^a$ is never satisfied if $b>1$. Since $b=1$ trivially yields a rational number, $x^{-x}$ is rational iff $x$ is an integer.

Parcly Taxel
  • 103,344
1

Hint: if $\,x\!=\!\frac{a}b,\ (a,b)\!=\!1\,$ & $\,x^{x}\! = \frac{m}n,\ (m,n)\!=\!1\,$ then $\,\large \frac{a^a}{b^a} \!=\! \frac{m^b}{n^b}\,$ so $\,\color{#0a0}b^a \!= n^{\color{#c00}b}$ (or $\,m^\color{#c00}{b}\,$ if $\,a<0),\,$ thus the coprime expts $\,(a,\color{#c00}b)\!=\!1\Rightarrow \color{#0a0}b\,$ too is a $\,\color{#c00}b$'th power, i.e. $\exists\:\! n\!\in\!\Bbb Z\!:\ b = n^b\, [\:\!\ge 2^b > b\,\Rightarrow\!\Leftarrow]$

Bill Dubuque
  • 272,048