Question

1√2 is a irrational number . Prove.

Answer 1

Answer:

Hey Mate!!

Let us consider, to the contrary, that 1√2 is rational.

••> 1√2 can be written in a form a/b where a and b are co-prime.

=> 1√2 = a/b

=> a = 1√2b

=> a^2 = (1√2b)^2

=> a^2 = 2b^2

=> a^2 ÷ 2 = b^2

As a^2 is divisible by 2, a is also divisible by 2.

Let a/2 = c

=> a = 2c

=> a^2 = (2c)^2

Substituting the value of a^2,

=> 2b^2 = 4c^2

=> b^2 = 2c^2

=> c^2 = b^2 ÷ 2

As b^2 is divisible by 2, b is also divisible by 2.

This means that a and b are divisible by 2.

However, this contradicts our assumption that a and b are co-prime.

This was due to our wrong assumption that 1√2 is rational.

Therefore, 1√2 is an irrational number.