Question

Prove √3 as irrational

Answer 1

HIO

FIND YOUR ANSWER IN THE ATTACHMENT I GAVE

HOPE THIS HELPS YOU

Answer 2

$\huge{\textbf{SOLUTION-}}$

Let's assume, √3 is rational,

This implies that there are co-prime numbers , a and b such that,

√3 = $\frac{a}{b}$

Squaring both sides,

(√3)² = $\frac{a^{2}}{b^{2}}$

3 = $\frac{a^{2}}{b^{2}}$

3b² = a²

⇒ 3 is a factor of a²

⇒ 3 is a factor of a             (1)

Let 3c= a

Squaring both sides, we get

9c² = a²

9c² = 3b²      (a²=3b²)

3c² = b²

⇒ 3 is a factor of b²

⇒ 3 is a factor of b        (2)

(1),(2) implies that 3 is a common factor of a and b

But a and b are co prime

⇒ a contradiction

A contradiction has arisen due to a false assumption

⇒ √3 is irrational

NOTE - Co prime numbers are numbers which do not have any common factor other than 1