# An irrational number $x$ such that $x^x$ is rational [duplicate]

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 $x$?

• 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. – Vidyanshu Mishra Dec 3 '16 at 19:37
• Here is where I already asked a related question. – AlgorithmsX Dec 3 '16 at 19:38
• @AlgorithmsX This is a related but different question since it requires $a=b$. – dxiv Dec 3 '16 at 19:42
• @dxiv I see that now. – AlgorithmsX Dec 3 '16 at 19:42
• @AlgorithmsX hi, I think my question is different. – Sara.T Dec 3 '16 at 19:43

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

Proof:

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

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

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 $p=2k$.

$$(2k)^{2k}=q^p2^q$$

$$2^{2k}k^{2k}=q^{2k}2^q$$

$$2^{2k-q}k^{2k}=q^{2k}$$

Then $q$ must be even. Contradiction.

Note:

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

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

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

• I can not see that how does it relevant to my question ? – Sara.T Dec 3 '16 at 19:55
• As I said in my question I knew such a number do exists but I am looking for an example. – Sara.T Dec 3 '16 at 19:57
• @Sara.T: this IS an example. – Yves Daoust Dec 3 '16 at 20:01
• @Sara.T I gave the explicit expression for $x$ in my previous comment. – dxiv Dec 3 '16 at 20:04
• @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). – 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

$$x^x=2.$$

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

$$p^p=2^qq^p.$$

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

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

• Crossed with Ahmed S. Attaalla. – Yves Daoust Dec 3 '16 at 19:59
• The conclusion depends on $q$ being strictly less than $2r$. I think it would help to make this explicit. – Barry Cipra Dec 3 '16 at 20:24
• I wonder if it's possible to prove the more general: if $x$ is an algebraic irrational then $x^x$ is irrational. – dxiv Dec 3 '16 at 20:27
• @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. – Yves Daoust Dec 3 '16 at 22:19
• @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! – Barry Cipra Dec 3 '16 at 22:45