(diagonal argument)
Suppose there is a bijection between N and R.
i.e.
1 <=> R1
2 <=> R2
3 <=> R3 .... etc.

Then there must exist a real number different to all Rn. This is evident because we can construct a number, say Rprime that has a different nth digit to every Rnth number (for any n, Rprime is different to all Rp: p<=n).

Therefore there is no bijection.

(my dumbshit argument)
Suppose there is a bijection between N and N.
i.e.
1 <=> N1
2 <=> N2
3 <=> N3 .... etc.

Then there must exist a natural number different to all Nn. This is evident because we can construct a number, say Nprime that has a different quantity to every Nnth number (for any n, Nprime is different to all Np: p<=n). This number is constructed by ordering the set of all Nn, starting at N1, choose a natural number different to N1 (i.e. N1 + 1) then check N2 (add 1 until Nprimeconstruction =/= N1 or N2) then check N3 etc.

Therefore there is no bijection.

What is wrong with my argument?