Square root of two, an irrational number: how to prove it?

√2=1.41421356237309504880168872420969807856967187537694807317667973799...

And it continues!!
What is a rational number?
It is a number which can be expressed in the form a/b, a fraction in its lowest terms, where a and b are two integers (whole numbers), and b different from 0 and 1.
And expressing √2 in this form is not obvious!
So now we're going to prove that it's irrational, by contradiction, that means:
If it was a rational number, then

√2=a/b

2=a²/b²
2b²=a²
Now we know that a² is even, because it is a multiple of 2, and we can deduce that a is also even! Why?
If a is even, then it equals to a multiple of two: a=2 x k
a²=a x a

substituting a by 2k
a²=2k x 2k
which gives us

a²=4k²
a² is also even!

If the product of 2 whole numbers has to be even, then at least one of those numbers has to be even.So if the square of a number (which is a multiplication of the number by itself) is even, that number is also even.We deduce that if a² is even than a is even.

When it comes to an odd number, the product of 2 numbers which are not multiples of 2, the product can't be even too.

So now that we know that a is even if √2 is rational, it equals to a/b and we know that a equals to 2k, so we can say that:

√2 = 2k/b
then

2 = (2k)²/b²

2b² = 4k²

b² = 2k²

Being a multiple of 2, b² is even, in the same way, b is also even.
But we have started the process, saying that a/b is in its lowest terms but they are both even, which means that they can be simplified by two.

So we can deduce that √2 is an irrational number.
By ADY Canimozhi and SIVAPRAGASSAM Aswin