One of the very first things we learn at higher math education is the different kinds of number sets, e.g. the naturals, rationals and reals. These are however just taken for granted, and we merely assume that these actually exist and are well-defined. I’ll spend the next few blog posts to characterize the rationals and the reals from facts known from the natural numbers alone. This post will be dedicated to proving the existence of the rationals, i.e. the theorem:

Theorem.There exist a countable dense linear ordering without endpoints.

*Proof.* Define and the relation given by .

* Claim 1. * is an equivalence relation.

*Proof of claim.* Reflexivity and symmetry follows directly from commutativity of in . Transitivity: Assuming and , meaning and by definition. Now if then by necessity as well and we’re done. Assume thus , which then implies , where the assumption was used at .

Now define for the equivalence class of and define . These will resemble our intuition of fractions, such that for instance 1/2=2/4, which is stated as . We now want to define order on our rationals:

**Claim**** 2.** defines a relation on .

*Proof of claim.* We need to show that the relation is independent of the choice of representatives. Let thus be such that and . We have to show that . Since then and likewise . Then . Now it is seen that the bi-implication holds, since e.g. assuming then by necessity since otherwise .

Having defined the order, we show that it fits the DLO requirements and *almost* satisfying having no endpoints:

**Claim 3.**** ** is a countable dense linear ordering with no right endpoint and as left endpoint.

*Proof of claim. *Since , is countable. We show that is a DLO. Reflexivity and symmetry follows directly from commutativity of in .

Transitive: Assume and , meaning and . Then .

Linearity: Assume . It has to be shown that . By definition, and . Thus by totality of <, .

Density: Assume . Since both and are even numbers, there exist an odd number satisfying . Then as well, implying and thus and likewise to show that .

Lastly we have to prove the facts about the endpoints. It has no right endpoint since given any then since . Assume now that is *not* the left endpoint. Then there is satisfying , meaning , but , contradiction.

Now define and let . Following intuition, this should fix our problem with the left endpoint – and indeed, our intuition is correct, as the following claim shows:

**Claim 4.**** ** is a countable dense linear ordering with no endpoints.

*Proof of claim. *Clearly it is a DLO with no right endpoint as removing a single element doesn’t change these properties. Now given we have since , due to .

Now technically we’ve proven our stated theorem, but we’d like to show our usual idea of the rationals also exists, which includes “the negative fractions”. This requires a reversed order, which motivates the following claim:

**Claim 5.**** ** is a countable dense linear ordering with no endpoints.

*Proof of claim. *Only the order is reversed, so countability and DLO properties still hold. Only change is that is now the right endpoint instead of left, but an argument analogous to that in claim 4 shows that it has no endpoints as well.

Now finally define the well-known rationals , which is the two orderings concatenated after each other (juxtaposition). We show the final claim that this in fact fits the requirement:

* Claim 6. * is a countable dense linear ordering with no endpoints.

*Proof of claim. *Since , is countable due to . Since is a DLO without endpoints by claim 5 and is a DLO without a right endpoint and as left endpoint by claim 3, then has no endpoints and is a DLO as well.

And there we have it! The rationals as we know them along with their well-known properties actually exist. But the question is now if there are different “kinds” of rationals, or are they in fact unique? We already saw another DLO without endpoints in claim 4 of the proof, but is this version equivalent to our rationals? This will be the focus of the next post!

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