Proof that rationals are uncountable Let's say I have a number $q \in \mathbb{Q}$, i.e.
 $$\forall a \in \mathbb{Z}. \forall b \in \mathbb{N}^+. q = \frac{a}{b}$$
Now I look at $a$ as an endless sequence $(a_0, a_1, a_2, \ldots)$ where $a_i$ can be determined by the expression $a = \sum_{i=0}  a_i \cdot 10^{i} $
and I look at $b$ as an endless sequence $(b_0, b_1, b_2, \ldots)$, where $b_i$ can be determined by the expression $b = \sum_{i=0} b_i \cdot 10^{i}$
Now I can combine these two sequences into one sequence, that looks like $(a_0, b_0, a_1, b_1, \ldots)$. Note that this mapping of a rational number to an endless sequence is a bijection. For every rational number, there is an endless sequence.
Now let's say I take all existing rational numbers and created endless sequences like this, and I would write them like this:
$$ 
\begin{align} 
a_{0_{q0}} b_{0_{q0}} a_{1_{q0}} b_{1_{q0}} \ldots \\
a_{0_{q1}} b_{0_{q1}} a_{1_{q1}} b_{1_{q1}} \ldots \\
a_{0_{q2}} b_{0_{q2}} a_{1_{q2}} b_{1_{q2}} \ldots \\
\vdots
\end{align}$$
Could I use this structure to apply Cantor's diagonal theorem and prove rationals to be uncountable? If not, where is the contradiction?
 A: The error is here:

Now I look at $a$ as an endless sequence $(a_0,a_1,a_2, \dots)$ where $a_i$ can be
  determined by the expression $a=\sum_{i=0}a_i\cdot 10^i$ 
  and I look at $b$ as an endless sequence $(b_0,b_1,b_2,\cdots)$, where $b_i$ can be
  determined by the expression $b=\sum_{i=0}b_i\cdot 10^i$.

While these "endless sequences" are, I suppose, technically "endless", after finitely many terms the sequences will become all zeros.  This is because $a$ and $b$ are both integers, and therefore have only finitely-many digits in their decimal expansion.
Likewise, when you interlace the sequences of $a$s and $b$s, you get a sequence of $c$s that eventually becomes all zeros.
When you apply the Cantor diagonalization trick, you end up with a new sequence that, by construction, ends with infinitely-many nonzero entries.  While it's true that that sequence of digits does not already exist in the list, it does not correspond to a rational number, because the way you have set up your correspondence, a rational number corresponds to a sequence that eventually becomes all zeros.
