It is well known that there are more irrational numbers than there are rational ones, and a typical prove for that is the Dedekind cut. Thinking about alternative proves I came up with the following idea:
- Interpret any $n\in\mathbb N$ as a hexadecimal number and convert it to a string (using e.g. UTF-8 encoding).
- Of all the strings obtained this way, only consider those that provide plain English instructions via which one or more real numbers can be obtained.
- All numbers obtained this way are called "describable", the set of them shall be denoted $\mathbb D$.
Now my problem: Since $\mathbb R$ is uncountable, this means there must be an uncountable amount of "indescribable" real numbers $\mathbb I := \mathbb R\backslash \mathbb D$. But using the "describability" defined above, $\mathbb N$ can provide infinitely many descriptions of arbitrary lengths (thus $\mathbb D$ is countable as well), and I fail to see how the countability of $\mathbb N$ can imply the uncountability of $\mathbb I$ (other than already asserting the uncountability of $\mathbb R$). So can the uncountability of $\mathbb I$ be proven?
Since no requirement is made about finite evaluation of the description of any $d\in\mathbb D$, I think $\mathbb D$ is a superset of the computable numbers, though I'm not sure.