Question

Prove that √2 is an irrational number​

Answer 1

let √2 is 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 (squaring both the sides)

⇒ 2 = a square/b square

⇒ 2b square = a square

⇒ a square is divisible by 2

⇒ a is divisible by 2

∴ let a = 2c

a square= 4c square

⇒ 2b square= 4c square

⇒ b square = 2c square

∴ 2c sqaure is divisible by 2

∴ b square 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.

Answer 2

Let us assume that, √2 is a rational number

√2 = $\dfrac{a}{b}$

Here, a and b are co-prime numbers.

• Squaring on both sides, we get

=> (√2)² = $( { \dfrac{a}{b}) }^{2}$

=> 2 = $\dfrac{ {a}^{2} }{ {b}^{2} }$

=> 2b² = a² ________( eq 1)

Clearly;

a² is divisible by 2.

So, a is also divisible by 2.

Now, let some integer be c.

=> a = 2c

=> a² = 4c²

=> 2b² = 4c² [From (eq 1)]

=> b² = 2c² _______(eq 2)

This means that, 2 divides b², and so 2 divides b also.

2 divides b² and 2 divide a² also.

So, our assumption is wrong.

√2 is irrational number.

________________________________