Prove that is irrational.

Foundation

Mathematics Foundation

Properties of Real Number

Irrational Number

Rationalisation

Question

Prove that $\sqrt{3}$ is irrational.

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Let us assume, to the contrary, that $\sqrt{3}$ is rational.

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

\sqrt{3} = \frac{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.

So,

b \sqrt{3} = ab \sqrt{3} = a

Squaring on both sides, and rearranging, we get 3b² = a²

Therefore, a² is divisible by 3, and by Theorem 1.3, it follows

that a is also divisible by 3.

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

Substituting for a, we get 3b² = 9c², that is,b² = 3c².

This means that b² is divisible by 3, and so b is also divisible by 3(using Theorem 1.3 with p = 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

\sqrt{3}

is rational. So, we conclude that

\sqrt{3}

is irrational.