Until now I have studied and understood Cantor's proof. My problem comes when looking at binary representation:
integer binary representation encoding for diag proof
1 1 10000...
2 10 01000...
3 11 11000...
4 100 00100...
5 101 10100...
So far so good. We have now represented all integers uniquely by their binary representation.
Now let's apply the diagonalisation and generate the number K:
K = 0011111111111111....
I know that after 1 and 2, the representation will be 1 because we move right faster than the 1's do.
So according to Cantor, this binary number will not be present in our enumeration, so what is this number?
K = 0*1 + 0*2 + 1*4 + 1*8 + ...
K = 4 * (1 + 2 + 4 + ... )
K = 4 * (3 + K)
K = -4
Clearly something funny is happening here, I have two questions:
- Has this demonstrated a binary number not present in our enumeration? If not, why not?
- How have we enumerated a negative binary number?
$K$ is not a real number, since $\sum_{n=2}^\infty 2^n$ does not converge. So no, you did not find an integer which is missing from your enumeration. Your proof for $K=-4$ is similar to the "proof" that $\sum_{n=1}^\infty n=-\frac{1}{12}$ in that both apply facts about convergent series to a non-convergent series.
Usually if you write down this kind of diagonal encoding you're trying to prove that the reals are uncountable. So you should encode the reals (or at least a nicely encodable subset like $[0,1)$), not the integers. If you do that, the string $a_1a_2a_3\dots$ encodes the series $\sum_{n=0}^\infty a_n2^{-n}$, which does converge, since its dominated by the convergent geometric series. So there you will actually find a not yet encoded real number, since the series then actually corresponds to a number.