Completeness has been assumed; now it starts paying. The first thing it buys is a fact that looks too obvious to need a proof: no real number is larger than every natural number. It needs one because nothing said so far forbids a number sitting above the whole of the naturals — an ordered field can have such elements — and completeness is exactly what rules them out.

A staircase with steps of equal height reaches any ceiling, however high, provided the steps keep coming. The claim is not that some particular step is high enough; it is that no ceiling can outrun the whole staircase. Completeness is what stops a ceiling hanging just above every step at once.

However far out $x$ lies, some natural number $n$ is past it. The proof does not find that $n$ by counting; it shows that a real number above every natural would force the naturals to have a least upper bound, and then walks back one step from that bound to reach a contradiction.

The Archimedean property states that for every real $x$ there is a natural $n$ with $n>x$. Two rewritings are used constantly: for every $\varepsilon>0$ there is a natural $n$ with , and for every positive and there is a natural with . Each follows from the first by dividing, and each is the form in which the property appears in a later proof.

• $\forall\varepsilon>0\ \exists n\in\mathbb{N}\ \big(\tfrac1n<\varepsilon\big)$: apply the property to . This is the form that makes in chapter four.

Suppose the claim fails. Then some real number bounds the whole of the naturals above, so by completeness the naturals have a least upper bound. Step back from that bound by one: the result is smaller, so it is not an upper bound, so some natural number beats it — and adding one to that natural number produces a natural number above the least upper bound itself. That is impossible, so the supposition fails.

Assume the denial: some real number bounds the naturals above.

The naturals are non-empty and bounded above, so completeness supplies a least upper bound.

Anything below the least upper bound fails to be an upper bound.

Add one to both sides; the result is still a natural number.

A natural number above the supremum of the naturals is impossible, so the supposition was false.

The property is a statement about the naturals as a whole, not about any particular one.

The one move a least upper bound allows: anything below it is not an upper bound, so something in the set beats it.

• No real number is at or above every natural number.
• Equivalent forms: some $\tfrac1n$ is below any given tolerance, and steps of any fixed size pass any target.
• The proof spends completeness, and the property fails in ordered fields that are not complete.
• It is the tool that picks the N in every epsilon–N proof.

Great work! You now know how to:

• State the Archimedean property and its two working forms
• Produce an n making a reciprocal smaller than a given tolerance
• Say which step of the proof spends completeness
• Keep the order of the quantifiers straight