# 8.4: The Real Numbers

• Bob Dumas and John E. McCarthy
• University of Washington and Washington University in St. Louis

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We complete our construction of the real numbers (we have the irrational numbers remaining) with the objective of proving the ordercompleteness of the real numbers, and deriving some important consequences of completeness. Many of the most powerful and interesting results of calculus depend on this property of the real numbers. If you have been asked to accept some of these theorems on faith, now it is time to reward your trust.

There are a couple of different ways to construct the real numbers from the rational numbers. One approach is to define real numbers as convergent sequences of rational numbers. The other common approach is to characterize real numbers as subsets of rational numbers that satisfy certain conditions.

DEFINITION. Dedekind cut A Dedekind cut $$L$$ is a nonempty proper subset of $$\mathbb{Q}$$ that has no maximal element and satisfies $(\forall a, b \in \mathbb{Q})[a \in L \wedge b<a] \Rightarrow[b \in L] .$ Let $$L$$ be a Dedekind cut. Then there is some rational number $$a \in L$$, and therefore all rational numbers less then $$a$$ are in $$L$$. Let $$R=\mathbb{Q} \backslash L$$. Since $$L \neq \mathbb{Q}$$, there is $$c \in R$$ and every rational number greater than $$c$$ is in $$R$$. It is clear that $$\{L, R\}$$ is a partition of $$\mathbb{Q}$$ and that every element of $$L$$ is less than every element of $$R$$. So Dedekind cuts "split" the rational numbers. We shall associate each Dedekind cut with a real number located at the split on the real number line.

REMARK. To help our mental picture of what is going on, we think of $$L$$ as all rational numbers to the left of some fixed real number $$\alpha$$, i.e. as $$(-\infty, \alpha) \cap \mathbb{Q}$$, and $$R$$ as the rational numbers to the right, $$[\alpha, \infty) \cap \mathbb{Q}$$. Of course we don’t yet know what exactly we mean by "the real number $$\alpha$$ ", but this is the idea to keep in mind. Note that $$R$$ will have a least element iff $$\alpha$$ is rational.

To understand how Dedekind cuts relate to numbers we construct an injection from the rational numbers to the Dedekind cuts. Let $$\mathcal{D}$$ be the set of Dedekind cuts. We define an injection $$i: \mathbb{Q} \rightarrow \mathcal{D}$$ by $i(a)=\{b \in \mathbb{Q} \mid b<a\} .$ The function $$i$$ is a well-defined injection that informs us of how $$\mathbb{Q}$$ fits into $$\mathcal{D}$$.

We shall define order and operations on $$\mathcal{D}$$ so that they agree with the usual linear ordering and operations on $$\mathbb{Q}$$ that are inherited in $$i[\mathbb{Q}]$$. That is, we shall define the linear order, addition and multiplication on $$\mathcal{D}$$ so that for $$a, b \in \mathbb{Q}$$, \begin{aligned} {[a \leq b] } & \Longleftrightarrow[i(a) \leq i(b)] \\ i(a+b) &=i(a)+i(b) \\ i(a \cdot b) &=i(a) \cdot i(b) \end{aligned} If we can do this, we can think of $$\mathcal{D}$$ as an extension of $$\mathbb{Q}$$. How do we do it?

For $$L, K \in \mathcal{D}$$, we define the relation $$\leq$$ in $$\mathcal{D}$$ by $[L \leq K] \Longleftrightarrow[L \subseteq K] .$ You should confirm that $$\leq$$ is a linear ordering of $$\mathcal{D}$$ and that the relation $$\leq$$ on $$i[\mathbb{Q}]$$ satisfies (1). If $$L \in \mathcal{D}$$ and $$L<i(0)$$ we say that $$L$$ is negative. If $$L>i(0)$$, we say that $$L$$ is positive. With a similar objective in mind we define addition and multiplication on $$\mathcal{D}$$. That is, we want the operations to satisfy certain properties of addition and multiplication and we want the operations defined on $$i[\mathbb{Q}]$$ to agree with the operations on $$\mathbb{Q}$$.

If $$L, K \in \mathcal{D}$$, then $L+K:=\{a+b \mid a \in L \text { and } b \in K\} .$ Verify that $$L+K$$ is a Dedekind cut, and that (2) holds.

Multiplication takes a bit more effort to define. (Why can’t we let $$L \cdot K=\{a b \mid a \in L, b \in K\}$$ ?) If $$L$$ or $$K$$ is $$i(0)$$, then $L \cdot K:=i(0) .$ If $$L, K \in \mathcal{D}$$ are both positive, then $L \cdot K=\{a \cdot b \mid a \in L, b \in K, a>0 \text { and } b>0\} \cup\{c \in \mathbb{Q} \mid c \leq 0\} .$ Verify that $$L \cdot K$$ is a Dedekind cut, and that (3) holds for $$a, b>0$$.

How do we define multiplication by "negative" Dedekind cuts? Let’s start with defining multiplication by $$-1$$. Let $$L \in \mathcal{D}$$ and $$R=$$ $$\mathbb{Q} \backslash L$$. We define $$-L$$ by $-L:=\{c \in \mathbb{Q} \mid(\exists r \in R)-c>r\} .$ Now we can define multiplication on arbitrary elements of $$\mathcal{D}$$ to satisfy the properties we desire. If $$L, K \in \mathcal{D}$$ and both are negative, then $L \cdot K:=(-L \cdot-K) .$ If exactly one of $$L$$ and $$K$$ is negative, then $L \cdot K:=-(-L \cdot K) .$ DEFINITION. Real numbers, $$\mathbb{R}$$ The real numbers are the Dedekind cuts, with addition, multiplication and $$\leq$$ defined as above. We denote the real numbers by $$\mathbb{R}$$ when we do not need to think of them explicitly as Dedekind cuts.

We have defined the real numbers as sets of rational numbers. Since the rational numbers were defined using basic ideas about sets, functions and relations, so are the real numbers. The properties of the real numbers that we discussed at the beginning of this section are satisfied by the Dedekind cuts. For every rational number $$a$$, we identify $$a$$ with the Dedekind cut $$i(a)$$.

THEOREM 8.2. The real numbers as defined above satisfy:

(i) Addition and multiplication are both commutative and associative.

(ii) $$(\forall L \in \mathcal{D}) L+0=L, L \cdot 1=L$$.

(iii) $$(\forall L \in \mathcal{D}) L+(-L)=0$$.

(iv) $$(\forall L \in \mathcal{D} \backslash\{0\})(\exists K \in \mathcal{D}) L \cdot K=1$$.

(v) $$(\forall L, K, J \in \mathcal{D}) L \cdot(K+J)=L \cdot K+L \cdot J$$.

Proof. Exercise.

This page titled 8.4: The Real Numbers is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by Bob Dumas and John E. McCarthy via source content that was edited to the style and standards of the LibreTexts platform.