Prove the number $\log_2 3$ is irrational
My attempt:
Assume that $\log_2 3 $ is rational
By definition $\log_2 3 =y$
Also by definition $y=\frac{a}{b}$ such that $b \neq 0$
Now you have $ 2^{\frac{a}{b}}=3$
Thus raising both sides by $b$ you arrive at $2^{a}=3^{b}$
Therefore $b=a\frac{\log2}{\log3}$ and $a=b\frac{\log3}{\log2}$
Now I'm not sure what to do with this at this point. Any hints or mistakes I'm making will be helpful this is a new technique for me.
Since $2^a = 3^b$, $3^b$ is an even number unless $a = 0$. If $a = 0$, $3^b = 1$. This is impossible since $b\neq 0$.