I was reading another question on mse about cantors proof and I'm curious about a number that could be defined from it. Well there could be a whole heap of them, but one for now.
Define $A=\Bbb{Q} \cap (0,1)$. $A $ is countable so it can be enumerated. Do so as follows $A=\{\frac 12, \frac 13, \frac 23, \frac 14 , \frac 34, \ldots\}=\{a_1, a_2,\ldots \}$. I don't know a slick formula for the terms in this sequence. Now suppose we write the sequence as a list in decimal form with the convention of using recurring zeros rather than recurring nines for fractions which are multiples of $2$ or $5$:
$\frac 12 =0.50000000\ldots$
$\frac 13 =0.33333333\ldots$
$\frac 23=0.66666666\ldots$
$\vdots$
Now let $a_{m, n}$ be the $n$'th digit of the decimal expansion (not including the whole number part) of $a_m$. Define $b_i=a_{i, i}$ and $b= \sum\limits_{i=1}^\infty \frac{b_i}{10^i}$.
I would assume based on nothing concrete at all, that $b$ is irrational but I have no idea on how to go about that.
I think this is a neat question.
The key seems to me to be that you will have recurring decimals e.g. $\frac 19, \frac 1{90}, \frac 1{900}\dots$ infinitely often (note these have non-zero digits in the recurring part, and give non-zero values for the relevant places beyond $9, 90, 900 \dots$.
You will also have $\frac 12, \frac 14, \frac 34, \frac 18, \frac 38, \frac 58, \frac 78 \dots$ which will give arbitrarily long sequences of zeros.
Those two facts mean that you can't have a recurring decimal, or a terminating decimal, so the number you have defined will not be rational (rationals terminate or recur).