Under root 3 is irrational.prove that

1.	Under root 3 is irrational.prove that
Answer» Let us assume that √3 is a rational number.That is, we can find integers\xa0a\xa0and\xa0b\xa0(≠ 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⇒\xa03b2=a2\xa0(Squaring on both sides) → (1)Therefore, a2\xa0is divisible by 3Hence ‘a’ is also divisible by 3.So, we can write a = 3c for some integer c.Equation (1) becomes,3b2\xa0=(3c)2⇒\xa03b2\xa0= 9c2∴ b2\xa0= 3c2This means that b2\xa0is 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.

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