Prove that for any prime positive integer `p ,sqrt(p)` is an irrational number.

Explanation

Let us assume on the contrary that `\sqrt{{p}}` is rational. Then, there exist positive coprimes a and b such that


`\sqrt{{p}}=\frac{{a}}{{b}} `


`\therefore {p}=\frac{{a}^{2}}{{~b}^{2}} `


`\therefore {b}^{2} {p}={a}^{2} `


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