Question

prove that √2 is irrational by method of contradiction

Answer 1

We prove this by contradiction. Assume it to be a rational number, so it can be expressed in the form of a/b where a & b are co-prime integers(Having 1 as only common factor). 
root 2 = a/b 
squaring both sides 
2 = square of a/ square of b. 
2 is an integer, whereas square of a/ square of b is a fraction, as a & b are co-prime integers so is square of a & square of b. 
Thus we arrive at the wrong conclusion. 
Hence our assumption is wrong. 
Thus square root of 2 is an irrational number.

Answer 2

Let's suppose √2 is a rational number. Then we can write it √2  = a/b where a, b are whole numbers, b not zero.

We additionally assume that this a/b is simplified to lowest terms, since that can obviously be done with any fraction. Notice that in order for a/b to be in simplest terms, both of a and b cannot be even. One or both must be odd. Otherwise, we could simplify a/b further.

From the equality √2  = a/b it follows that 2 = a2/b2,  or  a2 = 2 · b2.  So the square of a is an even number since it is two times something.

From this we know that a itself is also an even number. Why? Because it can't be odd; if a itself was odd, then a · a would be odd too. Odd number times odd number is always odd. Check it if you don't believe me!

Okay, if a itself is an even number, then a is 2 times some other whole number. In symbols, a = 2k where k is this other number. We don't need to know what k is; it won't matter. Soon comes the contradiction.

If we substitute a = 2k into the original equation 2 = a2/b2, this is what we get:

2=(2k)2/b22=4k2/b22*b2=4k2b2=2k2

This means that b² is even, from which follows again that b itself is even. And that is a contradiction!!!

 Therefore √2 cannot be rational.