Question

how can we prove the number ✓3 as irrational​

Answer 1

By contradiction method

To prove √3 is irrational

So , Let √3 be rational

$\sqrt{3} = \frac{a}{b}$

squaring on both sides

${ \sqrt{3} }^{2} = {( \frac{a}{b}) }^{2}$

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

${a}^{2} = 3 {b}^{2} ....(1)$

${a}^{2} \: is \: divisible \: by \: 3$

a is divisible by 3.....(2)

Now, Let a=3K

Squaring on both sides

${a}^{2} = (3 {k)}^{2}$

$3 {b}^{2} = 9 {k}^{2} ....from(eqn1)$

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

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

${b}^{2} \: is \: divisible \: by \: 3$

b is divisible by 3......(3)

from (2) and (3)

a and b are divisible by 3

a and b have common factor 3

But it is not possible

This contradicts our assumption that a/b is rational

(i.e) a and b do not have any common factor other than 1

So, a/b is not rational

√3 is not rational

√3 is an irrational number