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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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