Let \(\sqrt{3}\) be a rational number.
Then \(\sqrt{3}=\frac{q}{p}\)
\(\operatorname{HCF}(p, q)=1\)
Squaring both sides
\((\sqrt{3})^{2}=\left(\frac{q}{p}\right)^{2}\) \(3=\frac{p^{2}}{q^{2}}\) \(3 q^{2}=p^{2}\)
3 divides \(p^{2}\) » 3 divides \(p\)
3 is a factor of \(p\)
Take \(p=3 c\)
\(3 q^{2}=(3 c)^{2}\)
\(3 q^{2}=9 c^{2}\)
3 divides \(q^{2}\) » 3 divides \(q\)
3 is a factor of \(q\)
Therefore 3 is a common factor of \(p\) and \(q\)
It is a contradiction to our assumption that q/p is rational. Hence \(\sqrt{3}\) is an irrational number.
