×

Prove This If You Can

If x is a positive rational number, show that ,

$${ x }^{ x }$$

is irrational unless x is an integer.

Note by Harsh Depal
3 years ago

Sort by:

Let 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. · 3 years ago

Ow, nice! The only thing that I think your proof lack is LaTeX. Anyway, thank you. · 3 years ago

Yeah, I should learn how to us LaTeX, you're right about that. · 3 years ago

hey michael you can use $$Daum equation editor$$ · 2 years, 10 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 ?? · 2 years, 7 months ago

It's right to assume that a,b,c,d are relative prime? · 2 years, 5 months ago

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 $$d=1$$. · 2 years, 10 months ago