[Entangled]

> The cardinality of the rationals is the same as the integers.

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
    :
    N. ARNR ?

[sykhi]

It might be worth your time to consider the possibility that it is likely that the whole of the mathematics profession is not wrong in this matter. Are there any professional mathematicians that agree with you? If pretty much the whole profession thinks you are wrong about something then it’s quite likely you are indeed wrong.

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.

[dooglius]

It has been formally verified: http://us.metamath.org/mpegif/canth2.html

So, if you don't accept the proof, you have to reject some axiom used. Which one do you have a problem with?

[mcphage]

> If we use fruits as an example, taking a diagonal from their letters won't form a fruit either.

It will, however, form a sequence of characters. The diagonalization argument requires all possible sequences to be valid, which isn't true for fruits.