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: 
**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 $\,\color{#c00}{d,c} \ge 1.$
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.