What’s wrong with this proof that the set of natural numbers is uncountable? Suppose the set is countable, and can be arranged in a sequence. Then we construct a number whose nth digit is different than the nth digit of the nth number in the sequence, which means that it is not in the sequence.
 A: I guess your proof has been at least partially inspired by Cantor's diagonal argument, so I think I can see where your confusion stems from. In case you need a refresher, say we want to prove that the set of real numbers in $(0, 1)$ is uncountable. Suppose it isn't, then we can enumerate the set $\left\{s_0, s_1, s_2, ... \right\}$
Each $r \in (0, 1)$ can be represented in base 10 as $(0.a_0a_1a_2a_3...)_{10}$ where $a_i \in \left\{1,2,...,9\right\}$. We know that $r$ can be uniquely represented in base 2  as $(0.b_0b_1b_2b_3...)_2, \forall {b} \in \left\{0, 1\right\}$
From then on the proof is similar to what you proposed: We consider $s$ to be a number when the ith digit of it's binary representation is different to the ith digit of the binary representation of $s_i$. Since $s \in (0, 1)$ and $\nexists i, \ s_i = s \ $we can then conclude that $(0, 1)$ is uncountable.
Why doesn't the same argument work for $\mathbb N$? For one, we have no idea how we can even represent the members of $\mathbb N$. For example $(2)_{10} = (10)_2 = (.10000...)2^2$. But $(4)_{10} = (100)_2 = (0.100000...)2^3$, so you see that we can't uniquely define a number in the same way we did before.
I think at this point you can see why your argument fails: the nth digit isn't guaranteed to exist because you don't have a clear idea what the "nth digit" really is.
