×

# 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
4 years ago

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in $$...$$ or $...$ to ensure proper formatting.
2 \times 3 $$2 \times 3$$
2^{34} $$2^{34}$$
a_{i-1} $$a_{i-1}$$
\frac{2}{3} $$\frac{2}{3}$$
\sqrt{2} $$\sqrt{2}$$
\sum_{i=1}^3 $$\sum_{i=1}^3$$
\sin \theta $$\sin \theta$$
\boxed{123} $$\boxed{123}$$

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.

- 4 years ago

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

- 4 years ago

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

- 4 years ago

hey michael you can use $$Daum equation editor$$

- 3 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 ??

- 3 years, 7 months ago

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

- 3 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 $$d=1$$.

- 3 years, 10 months ago