Determining the equality of two real numbers Can the equality of two real numbers always be determined? 
Let us say that we have derived an expression for a real number X.
We also have obtained an (entirely different) expression for a real number Y.
Now one mathematician claims that X and Y must necessarily be exactly equal.
Another mathematician claims that X and Y can not be equal, however their difference is extremely small; say 0.5^N with N some unknown but very large number.
To settle the dispute the mathematicians decide to use a Turing machine.
As input the machine takes the binary expansions x(n) for X [n=1,2,3,4...] and y(n) for Y.
If the machine finds that for some n the pair of digits x(n) and y(n) are unequal, then it concludes that X and Y are unequal and the program halts. However if X and Y are equal, the machine will continue to check digits "ad infinitum" without ever reaching a conclusion. So in the case of equality (X = Y) there appears to be no halting criterium.
In the example that I described, the Turing machine is of no great help. The machine just continues to run "forever". Thus it remains unresolved whether this is because X and Y are exactly equal, or whether they are unequal but the first difference in their digits occurs at some unattainable high index N (required CPU time > age of the universe).
 A: The way you stated the problem (with real numbers given as infinite sequences of binary digits) there is no way to determine if two numbers are equal, the argument for which is simple (and similar to the one for halting problem):
For simplicity, assume that we're considering the question for numbers in $[0,1]$.
Suppose, by contradiction, that we have an algorithm which could solve the question of any two numbers. Such an algorithm would necessarily finish after reading only finitely many digits (we can assume without loss of generality that it reads them in sequence, and reads the same number of digits from each).
Let $n$ be the number of digits the algorithm reads before reaching the conclusion that, in fact, $0=0$. Then, when supplied with $0$ and $2^{-n-1}$, it would read the exact same input as in the $0=0$ case, so it would also say that they're equal, which is a contradiction.
As shown by Robert Israel, the question heavily depends on the permitted presentations of the numbers. Note that while there are only countably many numbers that can be expressed in the language of real closed fields (which somewhat limits the scope of Robert's answer), any real number (even undefinable) has a binary expansion, and no two numbers share one (even though it is not, in general, computable, even for definable numbers).
A: It depends on what kind of "expressions" are allowed. The theory of real closed fields is decidable, so for expressions using only +,-,*,/, RootOf and explicit integers the equivalence can be decided algorithmically.  If other operations are allowed, it may be
undecidable.  See e.g. http://en.wikipedia.org/wiki/Richardson%27s_theorem
and  http://dl.acm.org/citation.cfm?id=190429
