Answer :

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.

So, √3b = a

⇒ 3b^{2} = a^{2} (Squaring on both sides)

Therefore, a^{2} is divisible by 3.

∴ ‘a’ is also divisible by 3.

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

Equation (1) becomes,

3b^{2} = (3c)^{2}

3b^{2} = 9c^{2}

b^{2} = 3c^{2}

This means that b^{2} 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 co - prime.

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

So, we conclude that √3 is irrational.

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Prove that followKC Sinha - Mathematics

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