A more complete proof that the square root of 2 is irrational.

I don't know about you but I never quite liked the usual proof by contradiction that the square root of 2 is irrational. It seems incomplete in some way. I never felt convinced by it. Here's what I feel is the missing piece of the puzzle.

Assume that the square root of 2 is rational. So,
√2 = a/b
=>
2 = (a/b)^2
=>
2 = a^2 / b^2
=>
2 b^2 = a^2

So far so good. The usual proof continues with the following statement:

Since a^2 is equal to a natural number multiplied by 2, a^2 is an even number. But for a^2 to be even, a must be even too (see proof in the appendix at the end). So that means that there is a natural number k where a = 2k.

Since a = 2k and 2 b^2 = a^2,
2 b^2 = a^2
=>
2 b^2 = (2k)^2
=>
2 b^2 = 4 k^2
=>
b^2 = 2 k^2

Just like for a, b must also be an even number.

The proof usually ends right there, claiming that since a and b are both even numbers, then the fraction a/b is not simplified and irreducible, contradicting that a/b exists. But let's see where the proof takes us if we just keep on going.

If both a and b are even, then the fraction a/b can be simplified by dividing both a and b by 2, that is, if a = 2k and b = 2l, then we can say that √2 = k/l. But after doing this we can reapply the same reasoning on k and l and we'll discover that k and l are also both even numbers, and we can do it again and again ad infinitum.

So, which natural numbers can be divided by 2 infinitely? Only 1 number can do that, zero. But replacing zero for both a and b will not make their quotient a real number, or if you want to define 0/0, it will not result in a number whose square equals 2. So there is no fraction a/b which gives √2.

So there you have it, a proof that goes on till the end.

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APPENDIX
Now on to the proof that an even square can only come from an even number squared:

Let a^2 be an even number.

a can either be even or odd, that is there must exist an n where
a = 2n or a = 2n + 1
If a = 2n, a^2 = (2n)^2 = 4 n^2 = 2(2 n^2), which is an even number
If a = 2n + 1, a^2 = (2n + 1)^2 = 4 n^2 + 4n + 1 = 2(2 n^2 + 2n) + 1, which is an odd number

So an even number squared will give an even number and an odd number squares will give an odd number. Hence, a square even number can only come from an even number squared.