Question

prove that root 3 is an irrational​

Answer 1

Contradiction Method

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.

Hence the Proof.

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