# Density results for the rationals

## 1. Introduction

We have already seen the axioms for Archimedean ordered fields and how such fields contain a copy of the rational numbers, $ℚ$. 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.

## 2. Density results

All the results here concern an Archimedean ordered field and the copy $ℚ$ 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 $ℝ$ 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 $ε > 0$ be an element of $ℝ$. Then there is $N ∈ ℕ$ such that $0 < 1 N < ε$.

Proof.

Since $ε > 0$, $1 ε$ is defined and positive. Let $N ∈ ℕ$ be such that $1 ε < N$. (This $N$ exists by the Archimedean Property.) Then $ε > 1 N$ 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 $x , y ∈ ℝ$ with $x < y$. Then there is some $q ∈ ℚ$ with $x < q < y$.

Proof.

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

$k - 1 N = k N - 1 N ⩾ y - 1 N > y - ( y - x ) = x ⁢$

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 $ℚ$ is closed under addition, multiplication, etc., it is not closed under taking limits of rational sequences. (We will say that the field $ℚ$ is not complete.)

Definition.

A sequence $( a n )$ is monotonic nondecreasing if $a n + 1 ⩾ a n$ for all $n$. It is monotonic nonincreasing if $a n + 1 ⩽ a n$ for all $n$. It is monotonic if it is either monotonic nondecreasing or monotonic nonincreasing.

Theorem.

For any $x ∈ ℝ$ there is a monotonic nondecreasing sequence of rationals converging to $x$.

Proof.

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

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

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

Let $ε > 0$ be arbitrary. It will suffice to show that there is $N ∈ ℕ$ such that $a N > x - ε$, for this means $x - ε < a N ⩽ a n ⩽ x$ so $| a n - x | < ε$ for all $n ⩾ N$ by the fact that the sequence is nondecreasing. If not, we have that all $a n ⩽ x - ε$. We show that this leads to a contradiction.

So, given that all $a n ⩽ x - ε$, let $n ∈ ℕ$ be such that $1 n < ε$, by the Archimedean Property. Since $a n ⩽ x - ε$ we have $a n + 1 n < x$. Thus in the construction, $a n + 1 = a n + k n$ where $k$ is largest so that $a n + 1 < x$. But, by the assumption we are making, we also have $a n + 1 ⩽ x - ε$ so $a n + 1 + 1 n < x$. This means $a n + k n + 1 n = a n + k + 1 n < x$ 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 $a n > x - ε$, 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.

## 3. Summary

The rationals form a dense subset of $ℝ$, 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.