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!
**Theorem** $\ \color{blue}{\gcd(a,b)=1}\ \Rightarrow\ \log\_{\,b}(a)\not\in \Bbb Q,\,\ $ for $\ \,1<a,b\in \Bbb N$

**Proof** $\ $ Else $\, \log\_{\,b} a = c/d\,\Rightarrow\, a = b^{c/d}\Rightarrow a^\color{#c00}d = b^\color{#c00}c\,$ contra $\,\color{blue}{\gcd(a,b)=1},\,$ since $\,\color{#c00}{d,c} \ge 1.$