Question

Prove that √2 is an irrational number.

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Answer 1

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Answer 2

$\huge \boxed{ \underline{ \underline{ \bf{Answer}}}}$

Let √2 be a rational number which can be written in the form of p/q, where p and q are co-prime and q≠0.

$\implies \: \sqrt{2} = \frac{p}{q}$

$\implies \sqrt{2} q \: = \: p$

Squaring both sides we get,

$\implies \: 2 {q}^{2} \: = \: {p}^{2}$

Now, 2 is a factor of p^2

Therefore, 2 is a factor of p.

Now, Let p = 2m

$\implies \: 2 {q}^{2} \: = \: {(2m)}^{2}$

$\implies \: 2 {q}^{2} \: = \: 4 {m}^{2}$

$\implies \: {q}^{2} \: = \: 2 {m}^{2}$

So, 2 is a factor of q^2

Therefore, 2 is a factor of q.

Hence 2 is a factor of p and q.

But here p and q are co-prime.

So it contradicts our supposition.

$\boxed{ \sf{ \therefore \: \sqrt{2} \: is \: an \: irrational \: number}}$