Is there an uncomputable number between any two real numbers?

Question

I know close to nil about uncomputable numbers, so perhaps it doesn't even make sense to ask this question. All the information I can find about them is unaccessible with my level of education, but I have heard that "most numbers are uncomputable."

Answer 1

Well since computable numbers are numbers that can be calculated by a finite computer program (in an infinite amount of time), now a finite computer program can be expressed as a finite string of charachters such as a,b,c...1,2....,+,-,*... etc, the space is also a character, so "hello, how are you?" would be a string of 19 characters, let's say there are $n$ characters in total, so the amount of finite strings of $m$ characters would be $n^m$, a finite number lets define the set of strings of $m$ characters as $Str(m)$, it is clear that any finite computer program would belong to $\bigcup_{m\geq 1}Str(m)$, but this is the countable union of finite sets, so the set of computable numbers is countable, since any interval $(a,b)$ is uncountable there would exist an uncomputable number between $a$ and $b$. In fact since the Lebesgue measure of a countable set is $0$ you can say that almost every number is uncomputable.

Answer 2

Assume $\ a<b\ $ are real numbers and define $\ d:=b-a\ $ .

Case $1$ : $\ a\ $ is uncomputable

Choose positive integer $\ n\ $ so large that $\ \frac{1}{n}<d\ $ holds, then $\ a+\frac{1}{n}\ $ is an uncomputable number in the interval $\ (a,b)\ $.

Case $2$ : $\ a\ $ is computable

Let $\ r\ $ be an arbitary positive non-computable number and choose $\ n\ $ so large that $\ \frac{r}{n}<d\ $ holds. Then, $\ a+\frac{r}{n}\ $ is an uncomputable number in the interval $\ (a,b)\ $.