A proposition...

For all x in R, there exists a sequence a_n in Q such that the limit of a_n as n->+inf = x

This seems trivially false to me just from cardinality arguments (e.g. if you assume it's true then the cardinality of Q being less than the cardinality of R immediately implies that there are two different x in R that must have the same limit in Q), but I'll be damned if I can prove it rigorously.  Is there an easier way?

>>23
There are no "right ones"; any two elements will do. I glossed over this in >>16 since I figured anyone who had graduated from junior high would be able to fill int he details. Since you apparently are still struggling through 7th grade, I clarified it with >>22. With regards to whether or not it is possible to choose two elements, it is painfully trivial with ZFC. Without axiom of choice, it is significantly harder, but still entirely possible. Essentially it is possible to prove that for any non-empty set S, there exists a function f from {S} to S, and f(S) is some element of S. Having chosen one element of S, we remove it from S and repeat the process to get a second element.