Question

prove all step that √3 is irrational

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 coprime.

√3b = a

⇒ 3b2=a2 (Squaring on both sides) → (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.

Answer 2

let √3 be a rational
then it's simplest form will be a/b
a/b=√3 =>a=b√3
b=a/√3
do it again and show that √3 has another common factor other than a so it contradiction our assumption so it is an irrational no.