Google is being a little useless, but I have it on decent authority that one can prove, by contradiction, that there exists a well ordered relation for ℜ. In other words; that there is a minimal real number. That about that for a second.
(context switch) You can look at something like metamath and find a axiom system, on top of which a sizeable amount of maths is built. Generally, all maths should be done this way but it's too tedious, so generally people don't bother being that precise in proofs, but the general idea is that they could, if they wished.
Now you can think of the axiom system as a tree with n roots (one for each axiom). You can move from any point in a number of different ways by using a rule of inference the make a new theorem. (Ian and Will can stop shouting GEB at this point). This axiom tree branches into the space of all possible theorems.
Now, we all hope like hell that our axiom tree only ever hits true statements. But we know that it doesn't hit all true statements (by Gödel). But when people prove by contradiction they start from a theorum and reach ⊥ (false). Since we assume that, starting from a point on the tree, we cannot reach a false statement like ⊥, they then assume that the original statement is false. Which is rubbish. What they have showed is that it isn't on the tree, and thus that it's not-provable (using that system).