Question

prove that
$\sqrt{3}$
is irrational

Answer 1

Sol: 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 root 3 is irrational.
IF YOU LIKE THE ANSWER THEN MARK IT AS THE BRAINLIEST ANSWER.