If x is a positive rational number, show that ,

\({ x }^{ x }\)

is irrational unless x is an integer.

If x is a positive rational number, show that ,

\({ x }^{ x }\)

is irrational unless x is an integer.

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TopNewestLet x = a/b, and x^x = c/d, where a, b, c, d, are integers, such that a & b and c & d are co-primes. Then we have after some re-arranging:

(a^a) (d^a) = (b^b) (c^b)

Since both a & b and c & d are co-primes, a^a = c^b and d^a = b^b, neither which is possible because a & b are co-primes, unless b = 1 and d = 1. – Michael Mendrin · 3 years, 2 months ago

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– Damiann Mangan · 3 years, 2 months ago

Ow, nice! The only thing that I think your proof lack is LaTeX. Anyway, thank you.Log in to reply

– Michael Mendrin · 3 years, 2 months ago

Yeah, I should learn how to us LaTeX, you're right about that.Log in to reply

– Rishabh Jain · 3 years ago

hey michael you can use \(Daum equation editor\)Log in to reply

– Sandeep Kv · 2 years, 9 months ago

wen I used this Daum eqn editor , I wasn't able to copy the equation and paste it in the Field.. It just copies the code only !! Any solution ??Log in to reply

– Paul Ryan Longhas · 2 years, 7 months ago

It's right to assume that a,b,c,d are relative prime?Log in to reply

Here you go @Michael Mendrin:

Let \(x=\frac{a}{b}\) and \(x^x=\frac{c}{d}\), where \(a\), \(b\), \(c\), and \(d\) are integers such that \(a\) and \(b\) are coprime as well as \(c\) and \(d\). After some rearranging, we attain

\[(a^a)(d^a)=(b^b)(c^b)\]

Since both \(a\) and \(b\) are coprime, \(a^a=c^b\), and \(d^a=b^b\). Neither of these statements are possible, because \(a\) and \(b\) are coprime, unless the case where \(b=1\) and \(d=1\). – Finn Hulse · 3 years ago

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