Question

prove under root of 3 is irrational
by contradiction method please ​

Answer 1

Assume, √3 is rational

and in the form of p/q where p and q are co-primes

so,√3=a/b

squaring both sides

(√3)^2= (a/b)^2

3=a^2/b^2

b^2= a^2/3 (eq1)

3 divides a^2 so it will also divide a

let, a/3 =c

a= 3c

put value of a in eq1

b^2= (3c)^2/3

b^2= 9c^2/3

b^2= 3c^2

b^2/3 = c^2

3 divides b^2 so it will also divide b

Therefore, a and b have common factor 3

It contradicts that a and b are co primes

This contradiction occurred due to our wrong assumption of √3 as rational

Therefore √3 is irrational.