Question

Prove that √3 is an irrational number.​

Answer 1

Hope this was helpful .

Answer 2

Let us assume that $\sqrt{3}$ is rational .

That is , we can find integers a & b $(≠0)$ such that $\sqrt {3} = \frac {a}{b}$

where a & b are co - primes

So , $b\sqrt{3} = a$

Squaring on both sides, we get

$3b² = a²$

$\therefore a²$ is divisible by 3 , & it follows that a is also divisible by 3 .

So , we can write $a = 3c$ for some integer c .

Substituting for a , we get

$\fbox {3b² = 9c² = b² = 3c² }$

This means that b² is also divisible by 3 , & also b is also divisible by 3

( using p = 3 ) .

$\therefore$ a & b have at least 3 as a common factor .

But it contradicts the fact that a & b are co prime .

This contradiction has arisen because of our wrong assumption that $\sqrt{3}$ is rational .

So , we can conclude that $\sqrt{3}$ is irrational .