Proof that rationals are uncountable

Question

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?

Answer 1

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.