Question

prove underroot 2 in contradiction method

Answer 1

Sol:
Given √2 is irrational number.
Let √2 = a / b wher 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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