证:假设\(\displaystyle \sqrt{2}\)为有理数
则\(\displaystyle \sqrt{2}=\frac{q}{p}\)(p,q互质)
∴ \(\displaystyle 2=\frac{{{{q}^{2}}}}{{{{p}^{2}}}}\)
∴ 2p2=q2
∴ \(\displaystyle \begin{array}{l}2\left| {{{q}^{2}}} \right.\\2\left| q \right.\\4\left| {{{q}^{2}}} \right.\\4\left| {2{{p}^{2}}} \right.\\2\left| {{{p}^{2}}} \right.\\2\left| p \right.\end{array}\)
如果\(\displaystyle 2\left| q \right.,2\left| p \right.\)则p,q不互质,与假设矛盾
∴ 假设不成立
∴ 得证
