Question

prove that
$\sqrt{3} is \: irrational \: no$
​

Answer 1

Let us assume that √3 is a rational number.

That is, we can find integers a and b (≠ 0) such that √3 = (a/b)

Suppose a and b have a common factor other than 1, then we can divide by the common factor, and assume that a and b are co-prime.

√3b = a

Squaring both sides;

⇒ 3b²=a²  ___ (1)

Therefore, a² is divisible by 3

Hence a is also divisible by 3.

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

Equation (1) becomes,

3b² =(3c)²

⇒ 3b² = 9c²

∴ b² = 3c²

This means that b² is divisible by 3, and so b is also divisible by 3.

Therefore, a and b have at least 3 as a common factor.

But this contradicts the fact that a and b are coprime.

This contradiction has arisen because of our incorrect assumption that √3 is rational.

So, we conclude that √3 is irrational.