Question

1. Prove that √5 is irrational.​

Answer 1

Proof:-

If possible , let √5 be rational and lets its simplest from be a/b.

Then , a and b are integers having no common factor other than 1 , and b≠0.

Now, √5 = a/b ⇒ 5 =a²/b²                      [squaring on both side]

⇒5b²=a²                     ..(i)

⇒5 divides a²                 [∵5 divides 5b²]

⇒5 divides a     [∵5 is prime and 5 divides a²⇒5 divides a]

let a =5c for some integer c

putting a = 5c in (i), we get

5b²=25c²⇒b²=5c²

⇒5 divides b²   [∵5 divides 5c²]

⇒5 divides b    [∵5 is prime and 5 divides b²⇒5 divides b]

Thus , 5 is a common factor of a and b

but, this contradicts the fact that a and b have no common factor other than 1

the contradiction arise by assuming that √5 is rational

Hence,√5 is irrational