In the contradiction proof that $\sqrt{2}$ is irrational, after assuming $\sqrt{2}=\frac{a}{b}$ in lowest terms, which step leads directly to the contradiction?

From $\sqrt{2}=\frac{a}{b}$, we get $2b^2=a^2$, so 2 divides $a^2$, hence 2 divides $a$. Writing $a=2c$ and substituting back gives $b^2=2c^2$, so 2 divides $b$ as well. Thus both $a$ and $b$ are divisible by 2, contradicting the assumption that $\frac{a}{b}$ is in lowest terms.