Let us assume that √5 is a rational number.
∴ we can write it as √5 = q
Now we can reduce p and q and we get reduced value of p and q as a and b which are co-prime have no common factor other than 1.
So we have √5 = b
⇒ √5b = a
⇒ 5b2 = a2 [squaring both sides] ............ (i)
⇒ b2 = 5
5 divides a2 so a is also divisible by 5 (theorem 1.3) ..... (ii)
so 5 is a factor of a.
Now taking a = 5c for some integer c
⇒ 5b2 = (5c)2
⇒ 5b2 = 25 c2
⇒ b2 = 5 c2
⇒ c2 = 5
5 divides b2 so b is also divisible by 5 (theorem 1.3) ...... (iii)
from (ii) and (iii) we get that
a and b are divisible by 5 this implies that 5 is common factor of a and b, while we assume that a and b are co-prime having no common factor other than 1.
This contradiction has risen due to our wrong assumption so √5 cann't be
expressed as q, Hence √5 is an irrational number.
-Answered by Master Purushottam On 29 June 2021:11:10:35 AM