Solution
= Solution
For a <proof by contradiction>, suppose $\sqrt3=a/b$ for positive <integers> $a,b$ in lowest terms, so $\gcd(a,b)=1$. Squaring gives $a^2=3b^2$, and part (a) implies $3\mid a$. Write $a=3c$; then $9c^2=3b^2$, hence $b^2=3c^2$, and part (a) again implies $3\mid b$. This contradicts the coprimality of $a,b$. Therefore \b[$\sqrt3$ is an <irrational number>].