1.3 Revisiting Irrational Numbers: Proofs

Assume $\sqrt{3}$ is rational: $\sqrt{3} = \frac{a}{b}$ where $a, b$ are coprime[cite: 213].

Square both sides: $3 = \frac{a^2}{b^2} \implies 3b^2 = a^2$[cite: 216].

$a^2$ is divisible by 3, so $a$ is divisible by 3 (Theorem 1.2). Let $a = 3c$[cite: 218].

Substitute back: $3b^2 = (3c)^2 = 9c^2 \implies b^2 = 3c^2$[cite: 219].

$b^2$ is divisible by 3, so $b$ is divisible by 3[cite: 220].

Both $a$ and $b$ share a factor of 3, contradicting coprimality[cite: 225].

Therefore, $\sqrt{3}$ is irrational.