How do we know that the diagonal number produced in Cantor's argument for the uncountability of real numbers isn't periodic? As for example stated in the answer to this question: Why does Cantor's diagonal argument not work for rational numbers?
 A: There's a common misconception that Cantor's diagonal argument produces a "special real." That's really not what's going on. Let's look at exactly what Cantor's argument shows.
The argument shows that, if $L$ is any list of real numbers (formally: $L$ is a function from natural numbers to reals - think of the $i$th entry in the list as being $L(i)$), then there is some real number $r$ not in $L$ (formally: not in the range of $L$). Call the real built in this way "$Diag(L)$."
So "the real we get from Cantor's argument" isn't unique: it depends on what list we feed it. There are some lists $L$ such that $Diag(L)$ is rational (a good exercise), but these are lists that don't contain every rational number. If $L$ is a list that does contain every rational number, then since $Diag(L)\not\in L$, we know $Diag(L)$ is not rational.

Note that Cantor's argument is usually presented as a proof by contradiction. I think this obscures what's actually going on: the diagonal method gives a perfectly constructive procedure for, given a list of reals $L$, outputting a real not in $L$. This is really the "meat" of the argument. EDIT: As Mitchell points out below, there is no constructive method to go from a countable set of reals (that is, unlisted) to a real not in that set. (Precisely: ZF ( = set theory without choice) doesn't even prove that there is any function $d$ from $\{$countable sets of reals$\}$ to $\mathbb{R}$ with $d(X)\not\in X$.) The ordering is a crucial ingredient here. 
One way to phrase the conclusion which I've found helpful is 

"Any countable set of real numbers, isn't all of $\mathbb{R}$."

Of course, this just means "the reals are uncountable," but I think it makes things less mysterious.
