I'm wondering if there's a way to symbolically (or is there a more lose constraint?) represent ANY irrational number in a systematic way.
You can represent any rational number as two integers and I know you can represent some irrational numbers as sums of some infinite series. Is there any representation that can surely be found for any irrational number?
Now, I'm pretty sure there's an interesting observation that since the set of all possible English sentences is countable, there has to be a great set of numbers that can't be expressed as an English sentence. What I'm looking for is perhaps a way to loosen that rule to allow an uncountable infinity of numbers to be expressed.
My previous research into similar topics has usually led to people saying a question like this is ambiguous because a "representation" is not a very well defined idea. So take your best guess at what the definition might be so that this question makes sense (if there are any).
Best Answer
Every irrational number $x$ has a single unique decimal expansion, in a straightforward way. There is a non-negative integer $n$ and a sequence $d_n, d_{n-1}, d_{n-2}, \ldots, d_0, d_{-1}, d_{-2}\ldots$ of decimal digits for which $$x = \sum_{i\le n} d_i 10^i.$$
Every irrational number has a single unique continued fraction expansion: there is a sequence of positive integers $c_0, c_1, c_2,\ldots$ for which $$x = c_0 +\cfrac1{c_1 + \cfrac1{c_2+\cdots}}$$
Both of these are useful. "Unique" here means that not only do different irrationals have different representations, but also that each irrational has only one representation in each system. (This is not true for rational numbers.)
But each of these representations gives every irrational number an infinite representation, in terms of the infinite sequence of $d_i$ in the first case, or the $c_i$ in the second case. There is no method for giving every irrational number a finite representation, because it can be shown that for any such system, the family of such representations is countably infinite, and so there must be uncountably many irrationals with no finite representation. (This has been discussed many times on this site; try a search for "finite strings countable".)