Past exam of the year 1 of the computer science course of the University of Oxford / 2016 / Discrete mathematics, continuous mathematics and probability / 3 / b

For a proof by contradiction, suppose $\sqrt{3}=a/b$ 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 $\sqrt{3}$ is an irrational number.