Problem
NT-B1-M06-P007 Irrationality of \(\sqrt{5}\)
#7
★★☆☆☆ Level 2 of 5
Prove that \(\sqrt{5}\) is irrational.
If \(5\mid a^2\), then \(5\mid a\).
Let \(\sqrt{5}=\frac{a}{b}\) in lowest terms. Then \(a^2=5b^2\), so \(5\mid a\). Let \(a=5c\). Then \(25c^2=5b^2\), hence \(b^2=5c^2\), and \(5\mid b\). This contradicts lowest terms.
Can be generalised to any prime \(p\).