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!
You can't because it isn't. Assume $\log\_2(3)=\frac pq$, then$ 3=2^{\frac pq}, 3^q=2^p$ so???