Question

prove that is an irrational number​

Answer 1

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Question:-

Prove that $\sqrt {2}$ is irrational number.

☆Solution:-

Let, us assume that [tex]\sqrt {2}[/tex]is rational number.

Such that, $\sqrt {2}=\frac{a}{b}$ where, a and b are co-primes.

Now,

$b\sqrt{2} = a$

Squaring on both the sides, we get :-

$(b { \sqrt{2} })^{2} = ( {a)}^{2}$

$2 {b}^{2} = {a}^{2}$

${b}^{2} = \frac{ {a}^{2} }{3}$

$a = 2c$

By substituting for a, we get :-

$a = 2c$

$3 {b}^{2} = (3 {c})^{2}$

$3 {b}^{2} = 9 {c}^{2}$

${b}^{2} = \frac{ {9c}^{2} }{3}$

${b}^{2} = 3 {c}^{2}$

Therefore a and b have atleast 2 as a common factor

But, this contradicts the fact that a and b are co-primes (having only 1 as a common factor).

This contradiction arised because of our wrong assumption that $\sqrt {2}$ is rational.

So, we conclude that $\sqrt {2}$ is irrational number.