Question

root 2 is irrational ​

Answer 1

Let root 2 be rational.

$\sqrt{2} = \frac{a}{b} (a \: \: and \: \: b \: \: are \: \: coprime)$

$\sqrt{2} b = a$

Squaring both sides

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

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

$a^{2} is \: divisible \: \: by \: 2$

$a \: \: is \: divisible \: \: by \: 2$

Now,

$a = 2c \: (let)$

Squaring both sides...

$a ^{2} = 4c ^{2}$

$2b^{2} = 4c ^{2}$

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

$b ^{2} \: \: is \: \: divisible \: \: by \: \: 2$

$b \: \: is \: \: divisible \: \: by \: \: 2$

So,a and b are not coprime.

This contradicts the fact that root is rational.

So,root 2 is irrational.