Question: > 
> Prove that $\log\_2 3$ is irrational
> 
> 
> 

Seemingly simple homework assignment. Was never the best with logarithms, how would I go about proving?

Answer: 
Let $\log\_23 = p/q$ where $p\in\mathbb{Z}$ and $q\in\mathbb{N}$ (since surely $\log\_23 > 0$ you may directly assume that $p\in\mathbb{N}$ as well.)

Now it must hold

$$ 2^p = 3^q$$

But note that one side is even and the other one is odd!

Hence $\log\_23$ is **not** rational!
Assume, $x := log\_2(3) = \frac{a}{b}$ 

As x > 0 , we can assume that a and b are positive integers.

Now $$2^x=3$$

leads to $$2^\frac{a}{b}=3$$

Therefore $$2^a=3^b$$

If a and b are positive, the left side is even, the right side odd.
 A clear contradiction.