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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.

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