Concept : 1 Mark
Application :1 Mark
Method : 2 Mark
Let us assume, to the contrary, that √5 is rational.
So, we can find co-prime integers a and b (≠0) such that √5=ab
⇒√5b=a Squaring on both sides , we get
5b2=a2 Therefore, 5 divides a2
Therefore, 5 divides a
So, we can write a=5c for some integer c.
Substituting for a, we get 5b2=25c2 ⇒b2=5c2
This means that 5 divides b2, and so 5 divides b.
Therefore, a and b have at least 5 as a common factor.
But this is contradiction to the assumption that √5 is rational.
So, we conclude that √5 is irrational.