The number log_{2} 3 is irrational

Proof by contradiction: Assume log_{2} 3 is rational. Choose a, b ∈ N^{+}, log_{2} 3 = \frac{a}{b}, in lowest terms.

log_{2} 3 = \frac{a}{b}

3 = 2^{\frac{a}{b}}

3^{b} = 2^{a}

Since 3^{b} is always an odd number, and 2^{a} is always an even number, we have that an odd number is equal to an even number. This is a contradiction. Thus, the hypothesis is true: log_{2} 3 is irrational.