Problem
NT-B1-M06-P006 Irrationality of \(\sqrt{3}\)
#6
★★☆☆☆ Level 2 of 5
Prove that \(\sqrt{3}\) is irrational.
Repeat the proof for \(\sqrt{2}\), replacing parity by divisibility by \(3\).
Let \(\sqrt{3}=\frac{a}{b}\), \(\gcd(a,b)=1\). Then \(a^2=3b^2\), so \(3\mid a\), \(a=3c\). We get \(9c^2=3b^2\), hence \(b^2=3c^2\), and \(3\mid b\). This contradicts \(\gcd(a,b)=1\).
Good independent practice with a prime divisor of a square.