Question

please answer this question 9​

Answer 1

Let √5 be rational

let its simplest form be √5= p/q, where p and q are integers, having no common factor, other than 1 and q not equals 0.

Then, √5= p/q

=> p^2= 5q^2...(i)

=> p^2 is a multiple of 5

=> p is also a multiple of 5

Let p=5m for some positive integer m. Then,

p= 5m => p^2= 25m^2

=> 5q^2=25m^2 (Using (i))

=> q^2= 25/5m^2

=> q^2= 5m^2

=> q^2 is a multiple of 5

=> q is a multiple of 5

Thus, p as well q is a multiple of 5.

This shows that 5 is a common factor of p and q.

This contradicts or opposes the hypothesis that p and q have no common factor, other than 1.

This proves that √5 is not a rational number and hence it is irrational.

Let (2+√5) be rational. Then,

(2+√5) is rational

=> (2+√5)^2 is rational

=> (2+√5) is rational

But, (2+√5) being the sum of a rational and an irrational is irrational. Thus, we arrive at contradiction.

This contradiction arises by assuming that (2+√5) is rational.

Thus, this proves that 2+√5 is irrational.