Are there real numbers that cannot be uniquely expressed with a finite number of symbols? Are there real numbers that cannot be uniquely expressed with a finite number of symbols? Is this the same thing as an uncomputable number? 
I can show that if there is one such number then there are infinitely many, and the set of all these numbers is not compact if it is nonempty, but I don't know if any such numbers exist. 
Suggestions for tags are welcome. 
Edit: I want to add that by "symbols" I mean numbers, mathematical symbols, or English language descriptions (or any language really). So for π, you could write it as "the ratio of a circle's circumference to its diameter" and that would count. 
 A: Internally, most of them cannot be so expressed, assuming you have a finite alphabet. The set of all finite strings from a finite alphabet is countable, but the reals are uncountable. Consequently, every function from the set of strings to the set reals is missing uncountably many reals.
Externally, it is possible (but not necessary) that every real number can be uniquely expressed in the same metalanguage you use to describe the set theory you're using. Here is a related question asking whether, not just each individual real number, but every set can be defined.
A: Yes.
Just note that there are uncountably many real numbers, while there are only countably many finite strings of symbols (unless you are willing to use uncountably many symbols, but then I guess the question is moot, as you might as well have one symbol for each real), and most of them don't even describe a real number.
The numbers you are asking about are called definable numbers. They do include computable numbers, but there are definable numbers which are not computable.
A: I'd appreciate it if a set-theorist chimes in to shred this answer but here is a half-serious attempt:
Apply a well-ordering to the real numbers. 
Define $$E = \{x \in \mathbb R : \text{$x$ cannot be uniquely expressed with a finite number of symbols}\}.$$
Assume $E \not= \emptyset$. Then $E$ has a least element $e$. Thus $e$ cannot be uniquely expressed with a finite number of symbols, yet we can unambiguously state 

$e$ is the least element of $\mathbb R$ (unique, under the given well-ordering) that cannot be uniquely expressed with a finite number of symbols

which uniquely expresses $e$ with a finite number of symbols. This is a contradiction which leads to $E = \emptyset$.
p.s. My own guess is that the argument fails because the well-ordering itself is not expressible with finitely many symbols.
