We have already seen the axioms for Archimedean ordered fields and how such fields contain a copy of the rational numbers, $\mathbb{Q}$. This web page discusses some beautiful and very powerful consequences of these facts all describing how the rationals are embedded in an Archimedean ordered field such as the reals.

All the results here concern an Archimedean ordered field and the copy $\mathbb{Q}$ of the set of rationals embedded in it. To help your intuition (and to spell out the most important case) we shall call our field $\mathbb{R}$ but the results here apply to a general Archimedean ordered field $F$ too.

We start with a useful equivalent formulation of the Archimedean propery.

Proposition.

Let $\epsilon >0$ be an element of $\mathbb{R}$. Then there is $N\in \mathbb{N}$ such that $0<\frac{1}{N}<\epsilon $.

**Proof.**

Since $\epsilon >0$, $\frac{1}{\epsilon}$ is defined and positive. Let $N\in \mathbb{N}$ be such that $\frac{1}{\epsilon}<N$. (This $N$ exists by the Archimedean Property.) Then $\epsilon >\frac{1}{N}$ by the proposition on reciprocals in ordered fields.

The next result is rather beautiful and gets used a lot in more advanced work.

Density of the rationals in the reals.

Let $x,y\in \mathbb{R}$ with $x<y$. Then there is some $q\in \mathbb{Q}$ with $x<q<y$.

**Proof.**

Using the previous proposition, let $N\in \mathbb{N}$ satisfy $\frac{1}{N}<y-x$. Let $k\in \mathbb{N}$ be the least natural number such that $\frac{k}{N}>x$. (Note that $k$ with $k>Nx$ exists by the Archimedean Property and we may take the least such by the least number principle.) Then $x<\frac{k}{N}<y$ for if $\frac{k}{N}\u2a7ey$ we have

By a similar sort of argument one can prove much more than this:
that every real number is the limit of a sequence of rational
numbers. This too is a powerful result that is discussed in other
topics and courses in analysis. It also shows something important
about the rational numbers, that although $\mathbb{Q}$ is
closed under addition, multiplication, etc., it is not closed
under taking limits of rational sequences. (We will say that
the field $\mathbb{Q}$ is not complete

.)

Definition.

A sequence $\left({a}_{n}\right)$ is monotonic nondecreasing if ${a}_{n+1}\u2a7e{a}_{n}$ for all $n$. It is monotonic nonincreasing if ${a}_{n+1}\u2a7d{a}_{n}$ for all $n$. It is monotonic if it is either monotonic nondecreasing or monotonic nonincreasing.

Theorem.

For any $x\in \mathbb{R}$ there is a monotonic nondecreasing sequence of rationals converging to $x$.

**Proof.**

We start by defining a sequence $\left({a}_{n}\right)$ using induction on $n$. Start with any ${a}_{0}<x$ in $\mathbb{Q}$. (Such ${a}_{0}$ can actually be found in $\mathbb{Z}$ by the Archimedean Property.)

Inductively assume that ${a}_{n}$ is defined and ${a}_{n}<x$. If ${a}_{n}+\frac{1}{n}>x$ let ${a}_{n+1}={a}_{n}$. Otherwise, let ${a}_{n+1}={a}_{n}+\frac{k}{n}$ where $k\in \mathbb{N}$ is largest such that ${a}_{n}+\frac{k}{n}<x$. Such $k$ exists by a combination of the Archimedean Property and the least number principle, as in the previous proof.

By construction, $\left({a}_{n}\right)$ is a nondecreasing sequence of rationals and bounded above by $x$. We show that it converges to $x$.

Let $\epsilon >0$ be arbitrary. It will suffice to show that
there is $N\in \mathbb{N}$ such that ${a}_{N}>x-\epsilon $,
for this means $x-\epsilon <{a}_{N}\u2a7d{a}_{n}\u2a7dx$ so
$\left|{a}_{n}-x\right|<\epsilon $ for all $n\u2a7eN$ by the fact
that the sequence is nondecreasing. If not, we
have that *all*
${a}_{n}\u2a7dx-\epsilon $. We show that this
leads to a contradiction.

So, given that all ${a}_{n}\u2a7dx-\epsilon $, let $n\in \mathbb{N}$
be such that $\frac{1}{n}<\epsilon $, by the Archimedean Property. Since
${a}_{n}\u2a7dx-\epsilon $ we have ${a}_{n}+\frac{1}{n}<x$.
Thus in the construction, ${a}_{n+1}={a}_{n}+\frac{k}{n}$
where $k$ is largest so that ${a}_{n+1}<x$.
But, by the assumption we are making,
we also have ${a}_{n+1}\u2a7dx-\epsilon $ so
${a}_{n+1}+\frac{1}{n}<x$. This means
${a}_{n}+\frac{k}{n}+\frac{1}{n}={a}_{n}+\frac{k+1}{n}<x$
contradicting the definition of $k$, since it seems that $k+1$
is a larger natural number that would have worked. This is our contradiction
and we conclude that *some*
${a}_{n}>x-\epsilon $, as required.

Exercise.

Either by modifying the above proof, or by using the result it shows, show that every real number is the limit of a nonincreasing sequence of rationals.

The rationals form a dense

subset of $\mathbb{R}$,
or indeed of any Archimidean ordered field. We've seen this
illustrated in three different ways: firstly near 0

,
then globally

, using the $<$ relation to
explain what this density

means, and finally by showing that
every real number is the limit of a nondecreasing
sequence of rationals.