There is an arithmetic of cardinal numbers and it is well understood if you accept the axiom of choice. For instance, using your notation, N*N = N and N^2 = N. You can read more here
If you combine N as all possible numerators with N as denominators you get that cardinality of Q = N * N
Also, I don't accept the diagonal argument as proof. Given all possible combinations of numbers, any given number will occur in that set no matter what. If you add special rules of course it falls apart and Cantor's argument is just a special rule.
If we use fruits as an example, taking a diagonal from their letters won't form a fruit either.
1. [A]PPLE
2. O[R]ANGE
3. MA[N]GO
4. CHE[R]RY
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N. ARNR ?It’s worth pointing out that your logic on Q = N times N is a bit faulty too. Since you are counting things like 4/4 as different than 1/1. Even so you are correct that the cardinality of Q is N times N. This is because N times N = N.
So, if you don't accept the proof, you have to reject some axiom used. Which one do you have a problem with?
It will, however, form a sequence of characters. The diagonalization argument requires all possible sequences to be valid, which isn't true for fruits.