In the proof that \(\sqrt{3}\) is irrational (using \(\sqrt{3}=\frac{p}{q}\) and coprime integers), what is the correct “middle objective” after squaring?

Correct answer: Show that \(3\mid p\) and therefore set \(p=3k\) for some integer \(k\).

Starting with \(\sqrt{3}=\frac{p}{q}\) in lowest terms and squaring gives \(3q^2=p^2\). Since the left side is divisible by \(3\), \(p^2\) is divisible by \(3\). That implies \(3\mid p\). So the correct middle objective is to conclude \(3\mid p\) and write \(p=3k\) for some integer \(k\).