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I can show that there exists an irational number $x$ such that $x^x$ is rational. But I have no example. Can you give a pricise example of such number $x$?


marked as duplicate by Watson, zhoraster, Namaste, Martin Sleziak, saz Dec 4 '16 at 19:56

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  • $\begingroup$ Hi @sara.T and Welcome to MSE. Take a tour to the site and read some questions/answers. These things will help you to understand the site in a better way. $\endgroup$ – I am Back Dec 3 '16 at 19:37
  • $\begingroup$ Here is where I already asked a related question. $\endgroup$ – AlgorithmsX Dec 3 '16 at 19:38
  • $\begingroup$ @AlgorithmsX This is a related but different question since it requires $a=b$. $\endgroup$ – dxiv Dec 3 '16 at 19:42
  • $\begingroup$ @dxiv I see that now. $\endgroup$ – AlgorithmsX Dec 3 '16 at 19:42
  • $\begingroup$ @AlgorithmsX hi, I think my question is different. $\endgroup$ – Sara.T Dec 3 '16 at 19:43

The positive real solution to $x^x=2$ is irrational.


Assume $x=\frac{p}{q}$ where $p$ and $q$ have no common factor but $1$ then:


We must have $p^p=q^p2^q$

Note $p>q$ because $x>1$ (this can be shown).

Then $p$ must be even. So we may set $p=2k$.




Then $q$ must be even. Contradiction.


We have $x^x=2$, then $x \ln x=\ln 2$ and $\ln xe^{\ln x}=2$ so using the lambert W function:

$$\ln x=W(\ln 2)$$

$$x=e^{W(\ln 2)}=\frac{\ln (2)}{W(\ln 2)}$$

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    $\begingroup$ I can not see that how does it relevant to my question ? $\endgroup$ – Sara.T Dec 3 '16 at 19:55
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    $\begingroup$ As I said in my question I knew such a number do exists but I am looking for an example. $\endgroup$ – Sara.T Dec 3 '16 at 19:57
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    $\begingroup$ @Sara.T: this IS an example. $\endgroup$ – Yves Daoust Dec 3 '16 at 20:01
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    $\begingroup$ @Sara.T I gave the explicit expression for $x$ in my previous comment. $\endgroup$ – dxiv Dec 3 '16 at 20:04
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    $\begingroup$ @floorcat That's simply because after receiving a correct and deservedly upvoted answer, the OP had a change of heart and decided that s/he meant to ask a presumably different question (which hasn't been articulated, yet). $\endgroup$ – dxiv Dec 4 '16 at 0:18

As $x^x$ is a continous function, $1^1=1$ and $2^2=4$, then there is an $x$ such that


As the function is monotonic in this range, the solution is unique.

This number is irrational, otherwise let $x$ be the irreducible fraction $p/q$:

$$\left(\frac pq\right)^{p/q}=2$$ implies


Then $p$ is even, $p=2r$ with $q$ odd, and

$$2^{2r}r^{2r}=2^qq^p,$$ so that $q$ is even !

  • $\begingroup$ Crossed with Ahmed S. Attaalla. $\endgroup$ – Yves Daoust Dec 3 '16 at 19:59
  • $\begingroup$ The conclusion depends on $q$ being strictly less than $2r$. I think it would help to make this explicit. $\endgroup$ – Barry Cipra Dec 3 '16 at 20:24
  • $\begingroup$ I wonder if it's possible to prove the more general: if $x$ is an algebraic irrational then $x^x$ is irrational. $\endgroup$ – dxiv Dec 3 '16 at 20:27
  • $\begingroup$ @BarryCipra: I don't see why. As $q^p$ is odd, $q$ is also the multiplicity of $2$ in the LHS, which is certainlly even. $\endgroup$ – Yves Daoust Dec 3 '16 at 22:19
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    $\begingroup$ @YvesDaoust, ah, that argument works as well. I was thinking of rewriting $2^{2r}r^{2r}=2^qq^p$ as $2^{2r-q}r^{2r}=q^p$. But I kind of like your way: If $q$ is odd, it must be even! $\endgroup$ – Barry Cipra Dec 3 '16 at 22:45

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