Question

prove that √3 is a irrational number. ​

Answer 1

Let us assume to the contrary that √3 is a rational number.

It can be expressed in the form of p/q

where p and q are co-primes and q≠ 0.

⇒ √3 = p/q

⇒ 3 = p2/q2 (Squaring on both the sides)

⇒ 3q2 = p2………………………………..(1)

It means that 3 divides p2 and also 3 divides p because each factor should appear two times for the square to exist.

So we have p = 3r

where r is some integer.

⇒ p2 = 9r2………………………………..(2)

from equation (1) and (2)

⇒ 3q2 = 9r2

⇒ q2 = 3r2

Where q2 is multiply of 3 and also q is multiple of 3.

Then p, q have a common factor of 3. This runs contrary to their being co-primes. Consequently, p / q is not a rational number. This demonstrates that √3 is an irrational number.

Answer 2

Answer :

let's assume that √3 is rational

thus, √3 =p/q (where p and q are coprime)

√3q = p

squaring

3q² = p² (.... 1)

3 is a factor of p², 3 is a factor of p

let 3m = p

thus, 9m² = p²

9m² = 3q² (from 1)

thus 3m² = q²

thus 3 is a factor of q², 3 is a factor of q

thus, 3 is a factor of both p and q

thus p and q are not coprime

thus our assumption was wrong

hence √3 is irrational

hence proved