The Archimedean property of the real numbers can be stated as follows: for any real number x, there exists an integer greater than x. (In other words, the set of integers is not bounded above.) A corollary of this fact, which is itself sometimes called the "Archimedean property", is that for any positive reals x and y, there exists an integer n such that n x is greater than y.

This property of the reals can be used to prove that there is a rational number between any two reals.

Archimedes' property is a consequence of the completeness axiom (or least upper bound property) for the real numbers, which states that any non-empty subset of the reals which is bounded above has a supremum. A proof of the property runs as follows: assume that there is some real number x such that no integer is greater than x. Then x is an upper bound for the integers, so by the completeness axiom the set of integers has a supremum--call it s. Because s is a least upper bound, s-1 cannot be an upper bound for the integers; hence there is an integer n which is greater than s-1. But then n+1 is an integer greater than s, a contradiction.

"No matter how long a stick you have, you can always make it longer."

This is one of the more obvious properties of the real numbers, attributed to Archimedes. The natural numbers are spread about the reals like police on socialist street-corners; it just makes sense that no matter how far down the road you walk, you'll run into one on the next corner in a bit.

But this is mathematics, and mathematics has to be proven.

The Toolbox

If you believe the real numbers exist, there are some axioms you have to accept. Many of these are derivable from some model of the real numbers (like Dedekind cuts), but that is intractable for me. So I put what little faith I have in the axioms of the reals. Among these axioms is the Completeness Axiom (which see) that guarantees that every subset of the reals with an upper bound has a supremum (which see, a.k.a. least upper bound). The CA may seem a bit odd, but take a look at this picture:

-(--]---(---------)----*---
            M          b

Here we have a subset of the reals, M, and an upper bound, b. Common sense says that there has to be a number that is just beyond the reach of M, but not so great that there could be something between it and M. Like the creepy stalker guy that lets nothing get between him and his girl, the least upper bound of M sticks to it like glue.

The Method

In technical terms, the Archimedes' Property says that you can always find a natural number greater than any given real number. In more technical terms, (Ax in R)(En in N)(x<n).

∀ and ∃ are ugly, so I'm borrowing an alternative notation. (Ax in R) means specifically "for all real numbers x", and (En in N) means "there exists a natural number n"

The classic proof in analysis is a proof by contradiction, which happens to work in this case. The best way to answer to a problem is to know the answer. So, we begin thus:

BEGIN PROOF
Assume that you can't always find     ¬AP : ¬(Ax in R)(En in N)(x < n)
a natural number greater than a
given real number.

That means there's a real number      (Ex in R)(An in N)(x ≥ n)
out there that's greater than any
natural number.

That means (contrary to everything    x is an upper bound for N.
we already believe) that the natural
numbers are bounded!

Hmmm.  N is bounded, and we know      1 in N
it's not empty (1 is a natural        By CA, (Ey in R)(y = sup N).
number, after all).  So we throw
the Completeness Axiom at it.

The important thing is that the       y-1 < y
supremum of N is the least upper      So y-1 is not an upper bound for N.
bound.  Subtract one from it, and
it won't be an upper bound.

If it isn't an upper bound, there     (Em in N)(y-1 < m)
must be a natural number greater than
it.

So we add one to it.                   y < m+1

But you make the natural numbers       m+1 in N.
by adding one to the previous natural
number.  So we've made a natural number that's  greater than the 
'supremum' guaranteed by the Completeness Axiom.  That can't be right.

Our assumption must be wrong.          ¬AP is false.

So its opposite must be right!         AP is true!
END OF PROOF

If you've read this far, you may be asking yourself how such an ordinary looking property could be of any use. The Archimedes' property is good for proving limits of sequences (in that delta epsilion proof manner of speaking) and appears so many times in a discussion about analysis that, if it weren't possible to prove, it would probably be taken axiomatically.

But we can prove it. Obviously. So we don't have to take it as an axiom. The fewer axioms, the better!

My head hurts.

R --\    /-- Q is dense, R\Q is dense
     \  /
CA -- AP --- lim (sequences)
     /  \
MI -/    \-- sup, inf (proofs)