Question change √5 is irrational.
Assume contrary that √5 is a rational number.
Then √5 = a/b, where a and b are co-prime and b ≠ 0.
∴ HCF (a, b) = 1
⇒ a = √5 b
Now, a2 = 5b2 (By squaring both sides)
⇒ 5 divides a2
⇒ 5 divides a (∵ If any prime number divides a2 then that prime number must divide a)
⇒ a = 5m
⇒ a2 = 25 m2
⇒ 25 m2 = 5b2 (∵ a2 = 5b2)
⇒ b2 = 5 m2
⇒ 5 divides b2
⇒ 5 divides b
Since, 5 divides both a and b.
Therefore, HCF (a, b) ≠ 1, it is a multiple of 5.
Which is contradicts the fact that a and b are co-prime numbers.
Therefore, our assumption is wrong.
i.e., √5 is an irrational number.