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
⇒ 3b2 = a2 (Squaring on both sides)
Therefore, a2 is divisible by 3.
∴ ‘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 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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