Question

# If $$p$$ is a prime number, prove that $$\sqrt{p}$$ is irrational.

Solution

## Let $$\sqrt{p}$$ be a rational number and$$\sqrt{p}=\dfrac{a}{b}$$$$\Rightarrow p=\dfrac{{a}^{2}}{{b}^{2}}$$$$\Rightarrow {a}^{2}=p{b}^{2}$$$$\therefore p$$ divides $${a}^{2}$$But when a prime number divides the product of two numbers, it must divide atleast one of them.here $${a}^{2}=a\times a$$$$p$$ divides $$a$$Let $$a=pk$$        ......$$(1)$$$${\left(pk\right)}^{2}=p{b}^{2}$$$$\Rightarrow {p}^{2}{k}^{2}=p{b}^{2}$$$$\Rightarrow {b}^{2}=p{k}^{2}$$$$\therefore p$$ divides $${b}^{2}$$But $${b}^{2}=b\times b$$$$\therefore p$$ divides $$b$$Thus, $$a$$ and $$b$$ have atleast one common multiple $$p$$But it arises the contradiction to our assumption that $$a$$ and $$b$$ are coprime.Thus, our assumption is wrong and $$\sqrt{p}$$ is irrational number.Mathematics

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