Question

proof that square of root 2 is irrational.

Answer 1

The famous contradiction for root 2 being rational is this.......

Assume that root 2 is rational, thus it can be written as a numerator over a denominator. We factor out any common factors that the numerator and denominator have, to write is as a fraction a/b where a and b share no common factors. Then:

root 2 = a/b

2 = (a^2) / (b^2)

2 * b^2 = a^2

Therefore a^2 is an even number, thus a is an even number. Define c as the integer which is half of a.

2 * b^2 = (2c)^2 = 4 * c^2

b^2 = 2 * c^2

Therefore b^2 is an an even number, thus b is an even number. With a and b both being even numbers, this contradicts our original assertion that they have no common factors, thus our initial assumption must be incorrect.

Now, substituting root 4 instead of root 2...

root 4 = a/b where a and b have no common factors.

4 = (a^2) / (b^2)

4 * b^2 = a^2

Therefore a^2 is an even number, thus a is an even number. I'm guessing this is where you are going wrong: you might also be thinking that because a^2 is a multiple of four, then a is a multiple of four... but that's not true. 36 is a multiple of four, yet 6 is not.
Hope it helps

Answer 2

Hey friend

Here is your answer

We know that irrational no. Are the decimals which are neither terminating nor recurring.

√2= 0.4142......

So, root 2 is neither terminating nor recurring.

So, root 2 is an irrational no.

Hope it helps you