Question

25. Prove that √3 is an irrational number​

Answer 1

Answer:

let us assume that √3 is rational

√3 = a/b where b ≠0 , they have co prime

squaring both side

(√5)² = a²/b²

5 = a²/b²

b²5 =a²

b²= a²/5

therefore a² is divisible by 5

so, a is also divisible by 5 - i

let a = 5m for any integer m

(√5)² = (a/b)²

5 = (5m/b)²

5 = 25m²/b²

b²5= 25m²

b² = 25m²/5

b² = 5m²

b²/5 = m²

therefore b² is divible by  5

so, b is also divisible by 5 -ii

from (i) and(ii)

a and b have atleast 2 comman factor

This contradicts the fact that a and b have no comman factor except 1.

this contradiction has arisen because of our incorrect assumption.

so, √5 is irrational

Step-by-step explanation: