I'm currently reading Roger Penrose's book Shadows of the Mind, in which (at pp.72-77) he gives a simple, somewhat preliminary I guess, proof for Gödel's incompleteness theorem by using turing machines. However, I'm having trouble understanding the argument.
The argument
Suppose that we list all the possible computations that can be carried out on a certain input number, $n$
\begin{equation} \label{firstList} C_0(n), C_1(n), C_2(n), C_3(n), C_4(n), C_5(n), \ldots, \end{equation}
In other words; we're listing all the turing machines that can operate on a certain input number, $n$.
Now, further assume that we have some computational procedure, $A$, which tries to predict whether a certain turing machine will stop when working with a certain input number. This procedure won't always be able to discern if a certain turing machine stops when it's fed with a certain input number, but it is sound, meaning that it never will say that it stops if it doesn't.
So, formally, $A$ takes two inputs: $q$ and $n$. $q$ represents a certain turing machine and $n$ represents the input to this turing machine. $A$ signals that a certain turing machine will not stop by stopping itself. This then has the consequence that
\begin{equation} \text{If } A(q, n) \text{ stops, then } C_q(n) \text{ does not stop.} \end{equation}
Now, Cantor's diagonal slash is introduced, in which one always choses a $q$ and $n$ that are the same. This means that we can write
\begin{equation} \text{If } A(n, n) \text{ stops, then } C_n(n) \text{ does not stop.} \end{equation}
Penrose here states (I'm citing him because I'm not entirely sure what the essential components of these steps are)
We now notice that $A(n, n)$ depends upon just one number $n$, not two, so it must be one of the computations $C_0, C_1, C_2, C_3, \ldots$ (as applied to $n$), since this was supposed to be a listing of all the computations that can be performed on a single natural number $n$. Let us suppose that it is in fact $C_k$, so we have:
\begin{equation} A(n, n) = C_k(n) \end{equation}
Now examine the particular value n=k. [...] We have [from the equation above]:
\begin{equation} A(k, k) = C_k(k) \end{equation}
Now, substituting $n$ with $k$ in the statement before the quotes we get
\begin{equation} \text{If } A(k, k) \text{ stops, then } C_k(k) \text{ does not stop.} \end{equation}
By then taking advantage of $A(k, k) = C_k(k)$ above, we get
\begin{equation} \text{If } C_k(k) \text{ stops, then } C_k(k) \text{ does not stop.} \end{equation}
Penrose then states that (not important for my question but just providing some context for readers not familiar with what this proof leeds to)
From this, we must deduce that the computation $C_k(k)$ does not in fact stop. (For if it did then it does not, according to [the equation above]. But $A(k, k)$ cannot stop either, since by [one of the equations above], it is the same as C_k(k). Thus, our procedure $A$ is incapable of ascertaining that this particular computation $C_k(k))$ does not stop even though it does not.
Moreover, if we know that $A$ is sound, then we know that $C_k(k)$ does not stop. Thus, we know something that $A$ is unable to ascertain. It follows that $A$ cannot encapsulate our understanding.
What I don't get
Now, what I don't get here is why the diagonal slash is necessary for the argument. Why can't I just choose two numbers; one for the turing machine, $x$; and one for the input number, $y$; and then proclaim that $A(x, y)$ has to correspond to a certain computation, $C_x(y)$? From that point on I should be able to say that if $C_x(y)$ stops, then $C_x(y)$ does not stop. An important part of the argument seems to be that "$A(n, n)$ depends upon just one number $n$, not two", but I don't understand why.