Prove 5 irrational

1.	Prove 5 irrational
Answer» 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.

Answer»

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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