Question

Prove that \(\sqrt{6}\) is irrational.

Answer 1

Suppose that \(\sqrt{6}\) was rational. We show that this leads to a contradiction. We may write \(\sqrt{6}=x / y\) where \(x\) and \(y\) are integers with \(y \neq 0\) and \(\operatorname{gcd}(x, y)=1\). Squaring this equation and cross-multiplying we get that \(6 y^{2}=x^{2}\) or \(2 \cdot 3 \cdot y^{2}=x^{2}\). Therefore, 2 divides \(x^{2}=x \cdot x\). Since 2 is prime we must have that 2 divides \(x\). Similarly, 3 divides \(x^{2}=x \cdot x\).

And since 3 is prime we must have that 3 divides \(x\). Since \(2 \mid x\) and \(3 \mid x\) and \(\operatorname{gcd}(2,3)=1\), by the first part of this problem, we have that \(6=2 \cdot 3\) must divide \(x\). So \(x=6 u\) where \(u\) is a non-zero integer. Subbing this into \(6 y^{2}=x^{2}\) gives us that \(6 y^{2}=6^{2} u^{2}\). Thus \(y^{2}=6 u^{2}\). Following the same reasoning as above, this forces that 6 must divide \(y\). Therefore, 6 is a common divisor of \(x\) and \(y\) which contradicts the fact that \(\operatorname{gcd}(x, y)=1\).