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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 $x$?

user390026
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2 Answers2

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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:

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

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

Ahmed S. Attaalla
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    you got mine too – Vidyanshu Mishra Dec 03 '16 at 19:54
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    I can not see that how does it relevant to my question ? – user390026 Dec 03 '16 at 19:55
  • The irrational number $x$ produces a rational number $x^x=2$ @Sara.T – Ahmed S. Attaalla Dec 03 '16 at 19:56
  • @Sara.T You can write $x$ explicitly in terms of the Lambert $W$ (or productlog) function: $x = e^{W(\ln 2)}$. The answer above proves that $x$ is irrational and $x^x=2$ as can be verified on Wolfram Alpha. – dxiv Dec 03 '16 at 19:56
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    As I said in my question I knew such a number do exists but I am looking for an example. – user390026 Dec 03 '16 at 19:57
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    @Sara.T: this IS an example. –  Dec 03 '16 at 20:01
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    @YvesDaoust disagree, it only shows that an example EXISTS – user390026 Dec 03 '16 at 20:02
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    @Sara.T: this number exists thanks to the continuity of $x^x$. You confuse existence and closed-form formula. –  Dec 03 '16 at 20:04
  • @Sara.T I have edited maybe this will satisfy you. – Ahmed S. Attaalla Dec 03 '16 at 20:04
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    @Sara.T I gave the explicit expression for $x$ in my previous comment. – dxiv Dec 03 '16 at 20:04
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    @YvesDaoust, I agree with sara as OP clearly want an example which follow the condition mentioned. But the answers do not address that thing.(sorry if I m wrong) – Vidyanshu Mishra Dec 03 '16 at 20:05
  • @THELONEWOLF.: $x$ is irrational while $x^x$ is rational. What other condition are you referring to ? The solution is constructive as $x^x=2$ unambiguously defines a real number. –  Dec 03 '16 at 20:06
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    I m not saying anything else.I agree that question specifically mention that x is irrational while x^x is rational, but I think OP is demanding for an example in numbers. A number which satisfy the above mentioned condition. – Vidyanshu Mishra Dec 03 '16 at 20:08
  • It might be quicker to say $p^p=q^p2^q\implies q\mid p^p\implies q=1$ since $(p,q)=1$, and then note that $2$ is not of the form $p^p$ for any integer $p$. – Barry Cipra Dec 03 '16 at 20:09
  • @THELONEWOLF.: such a number cannot be written down as it is irrational. –  Dec 03 '16 at 20:09
  • @YvesDaoust I think the issue arises from using multiple representations of the variable (x), even though these representations are equivalent. –  Dec 03 '16 at 20:09
  • @YvesDaoust, please don't laugh at me if I m arguing about a wrong thing. But $\sqrt{2}$ is irrational but we can still write it. – Vidyanshu Mishra Dec 03 '16 at 20:11
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    @floorcat: I think the issue comes from the OP wanting a closed-form expression without stating it and believing that the root of an equation is not a valid definition. –  Dec 03 '16 at 20:12
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    @THELONEWOLF.: I can write $x^x=2$, can't I ? By the way, $\sqrt2$ is originally defined as the solution of $x^2=2$. –  Dec 03 '16 at 20:13
  • @YvesDaoust A closed expression is an elegant answer to this question, given the parameters. Elegance is relative though, but it seems you make it difficult to glance over this answer and immediately see the example she's asking for. –  Dec 03 '16 at 20:14
  • Maybe you can,But still I can't get you. Please elaborate a bit, I know you will say that "see my answer", but I can't get it from there too. – Vidyanshu Mishra Dec 03 '16 at 20:15
  • @floorcat: closed-form was not specified. The OP has to know and accept that there are other ways to define a number, for instance infinite series. Anyway the answerer gave an analytical expression too. –  Dec 03 '16 at 20:16
  • @YvesDaoust I'm not saying, and did not say that closed form was specific in OP. But it also did not specify open-form and yet we both answered in our own ways. I will not engage in speculations about what the OP should or should not know or accept. –  Dec 03 '16 at 20:23
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    @AhmedS.Attaalla Thanks, But I can not accept it as an answer to my question. – user390026 Dec 03 '16 at 21:06
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    @Sara.T Just curious, if you can not accept it as an answer then what answer were you actually looking for? You should edit into the question what exactly are your requirements for an answer to qualify. – dxiv Dec 03 '16 at 21:19
  • I find it very interesting that the most upvoted answer has been clearly rejected by the author of this question. –  Dec 03 '16 at 23:25
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    @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 04 '16 at 00:18
  • sure. She doesn't understand her own question, which you know better enough to give an answer she doesn't want. Makes sense. Good night. –  Dec 04 '16 at 00:30
  • Thank you! And happy new year. @gowrath – Ahmed S. Attaalla Jan 01 '17 at 02:01
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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. –  Dec 03 '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 03 '16 at 20:24
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    I wonder if it's possible to prove the more general: if $x$ is an algebraic irrational then $x^x$ is irrational. – dxiv Dec 03 '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. –  Dec 03 '16 at 22:19
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    @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 03 '16 at 22:45