Part I :- We have to prove that √5 is an irrational number. 

Let us assume contrary that √5 is a rational number. 

∴√5 can be written as \(\frac{p}{q}\) form where q ≠ 0 and p & q has no common factor other than 1. 

∴ √5 = \(\frac{p}{q}\) 

⇒ p = √5q 

⇒ p² = 5q² … (1) 

(By squaring both sides) 

⇒ 5 divides p² 

⇒ 5 divides p 

(∵ if a prime number divides a² then that prime number must divides a) 

∴ p = 5m where m is an integer. 

Now,

Putting p= 5m in equation (1), we get 

25m² = 5q² 

⇒ q² = 5m² 

⇒ 5 divides q² 

⇒ 5 divides q 

i.e., 5 divides both p and q which implies that 5 is a common factor of both integers p and q which is a contradiction of the fact that p & q have no common factor other than 1. 

∴ Our assumption is wrong. 

∴ √5 is an irrational number. 

Part II :- 

Now,

We have to prove that (3 – √5) is an irrational number.

Let us assume contrary that 3 – √5 is a rational number 3 – √5 = \(\frac{p}{q}\) where q ≠ 0 and p & q ∈ I. 

(∵ Every rational number can be written in \(\frac{p}{q}\) form)

⇒ √5 = 3 − \(\frac{p}{q}\) = \(\frac{3q-p}{q}\) … (2) 

L.H.S = √5 = irrational number 

(According to result (1)) 

R.H.S = \(\frac{3q-p}{q}\), ≠ 0 & 3q − p ∈ I is a rational number. 

(Because every number which can be written in the form p q , q ≠ 0 & p, q ∈ is a rational number) 

But rational ≠ irrational which is contradiction of equation (2). 

Hence,

Our assumption is wrong. 

∴ (3 – √5) is an irrational number.