Question

Prove that √2 is a irrational number

Answer 1

let √2 = a / b wher a,b are integers b ≠ 0
we also let that a / b is written in the simplest form
⇒ √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 notion that a/b is written in the simplest form
as our notion is wrong
∴ √2 is irrational number. 

Answer 2

Let √2 is rational and represented in the simplest form of a/b. (HCF=1)

√2 =a/b
√2b =a
by squ. both the sides
2b² = a² .....(i)
2 is the factor of a², in this way we can also say that 2 is factor of a.

Now, put a=2c
2b²=4c²
b²=2c² .....(ii)

2 is tge factor of b².
Also, 2 is the factor of b.
From (i) & (ii) , 2 is also the factor of a and b which contradicts the fact .

Hence,
√2 is irrational.