Answer to Question #147044 in Discrete Mathematics for Kamyar

Question

Prove that log2 3 is irrational.

Answer 1

Suppose that $\log_23$ is a rational number. Since $\log_23>0,$ $\log_23=\frac{a}{b},$

where $a,b\in\mathbb N.$ Then $3=2^{\frac{a}{b}},$ and therefore, $3^b=2^a.$ But the last equality is impossible beacause the natural number $2^a$ is divisible by 2, but the number $3^b$ is not divisible by 2. This contradiction proves that the assumption was wrong, hence $\log_23$ is irrational.