Question

prove that √2 is irrational

Answer 1

Given that √2 is irrational number.

We assume that √2 is a rational number

Let √2 = a / b where a,b are integers b ≠ 0 we also suppose that a / b is written in the simplest form.

Now √2 = a / b ⇒ 2 = a2 / b2 ⇒ 2b2 = a2 ∴ 2b2 is divisible by 2 ⇒ a2 is divisible by 2 ⇒ a is divisible by 2 ∴ let a = 2c a2 = 4c2 ⇒ 2b2 = 4c2 ⇒ b2 = 2c2 ∴ 2c2 is divisible by 2 ∴ b2 is divisible by 2 ∴ b is divisible by 2 ∴a are b are divisible by 2 . this contradicts our supposition that a/b is written in the simplest form Hence our supposition is wrong ∴ √2 is irrational number.

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Answer 2

To Prove :-

• √2 is an irrational number.

SoluTion :-

Let's assume on the contrary that √2 is a rational number.

Then, there exists two rational numbers a and b

such that √2 = a/b where, a and b are co primes.

(√2)² = (a/b)²

→ 2 = a²/b²

→ 2b² = a²

2 divides a²

So, 2 divides a.

a = 2k , (for some integer)

a² = 4k²

2b² = 4k²

b² = 2k²

2 divides b²

2 divides b

Now, 2 divides both a and b but this contradicts that a and b are co primes.

It happened due to our wrong assumption.

Hence, √2 is irrational.