Real Number System

The real numbers carry three kinds of structure that play distinct roles in analysis: algebraic operations, an order compatible with those operations, and completeness. The first two are shared by every ordered field, whereas completeness distinguishes $\mathbb R$ from $$\mathbb Q$.$ We develop these three layers in turn. After deriving basic algebraic and order properties from the axioms, we use completeness to establish the Archimedean property of $$\mathbb R$,$ the density of $\mathbb Q$ in $$\mathbb R$,$ and the existence of positive $$n$-$ th roots.

Definition 1.1.1. Let $\mathcal L_{\mathrm{field}}$ be the first-order language $\mathcal L_{\mathrm{field}} = \lbrace 0, 1, +, -, \cdot\rbrace,$ where $$0$$ and $$1$$ are constant symbols, $$+$$ and $\cdot$ are binary function symbols, and $$-$$ is a unary function symbol.

Definition 1.1.2. Let $T_{\mathrm{field}}$ be the $$\mathcal L_{\mathrm{field}}$-$ theory which consists of the following axioms:

$\forall x,y,z \ ((x + y) + z = x + (y + z))$
$\forall x,y \ (x + y = y + x)$
$\forall x \ (x + 0 = x)$
$\forall x \ (x + (-x) = 0)$
$\forall x,y,z \ ((x \cdot y) \cdot z = x \cdot (y \cdot z))$
$\forall x,y \ (x \cdot y = y \cdot x)$
$\forall x \ (x \cdot 1 = x)$
$\forall x,y,z \ ((x \cdot y) + (x \cdot z) = x\cdot (y+z) )$
$0 \neq 1$
$\forall x \ (x \neq 0 \rightarrow \exists y \ (x \cdot y = 1))$

Definition 1.1.3. An $$\mathcal L_{\mathrm{field}}$-$ structure ${\mathcal F}=(F, 0, 1, +, -, \cdot)$ is called a field if and only if ${\mathcal F}\vDash T_{\mathrm{field}},$ and each $x\in F$ is called a scalar. $$0$$ and $$1$$ are called the additive identity and multiplicative identity, respectively. $$-x$$ is called the additive inverse of $$x.$$ If $x\ne 0$ and $$x\cdot y = 1$,$ then we write $$y=1/x$$ and call $$y$$ the reciprocal of $$x$.$

We shall often write

\[x-y,xy,\frac x y,2x,3x,x^2,x^3,\dots\]

in place of

\[x+(-y),x\cdot y,x\cdot \left(\frac 1 y \right),x+x,x+x+x,x\cdot x,x\cdot x\cdot x,\dots .\]

The field axioms encode only the fundamental laws of arithmetic. Familiar rules such as cancellation and the laws of signs are consequences of these axioms, as the following propositions show.

Proposition 1.1.4. For all elements $$x,y,z$$ of a field, the following hold:

if $$x+y=x+z$$ then $$y=z$;$
if $$x+y=x$$ then $$y=0$;$
if $$x+y=0$$ then $$y=-x$;$
$$-(-x)=x$.$

Proof. (1) Since $$x+y=x+z,$$ $$(x+y)-x=(x+z)-x.$$ Hence $$y=z.$$ (2) Since $$x+0=x,$$ we have $$x+y=x+0.$$ Hence $$y=0,$$ by (1). (3) Since $$x-x=0,$$ we have $$x+y=x-x.$$ Hence $$y=-x,$$ by (1). (4) Since $$(-x)+x=0,$$ $$-(-x)=x,$$ by (3). $\square$

Proposition 1.1.5. For all elements $$x,y,z$$ of a field, the following hold:

if $x\ne 0$ and $$xy=xz$,$ then $$y=z$;$
if $x\ne 0$ and $$xy=x$,$ then $$y=1$;$
$$xy=(-x)(-y)$;$
$$(-1)x=-x.$$

Proof. (1) Let $$xw=1.$$ Since $$(xy)w=(xz)w,$$ $$(xw)y=(xw)z.$$ Hence $$y=z.$$ (2) Since $x\cdot 1=x,$ $xy=x\cdot 1.$ Since $x\ne 0,$ by (1), $$y=1.$$ (3) Since $$xy+(-x)y=0,$$ we have $$xy=-((-x)y).$$ Similarly, $$(-x)y+(-x)(-y)=0$$ implies $$(-x)(-y)=-((-x)y).$$ Since the additive inverse of $$(-x)y$$ is unique, $$xy=(-x)(-y).$$ (4) Since $$x+(-1)x=(1-1)x=0,$$ by (3) in Proposition 1.1.4, $$(-1)x=-x.$$ $\square$

Algebraic structure alone does not provide a notion of positivity or comparison. To discuss inequalities, we equip a field with an order relation that is compatible with addition and multiplication.

Definition 1.1.6. Let $\mathcal{L}_{\mathrm{of}} = \mathcal{L}_{\mathrm{field}} \cup \{<\},$ where $$<$$ is a binary relation symbol, and let $T_{\mathrm{of}}$ be the $$\mathcal{L}_{\mathrm{of}}$-$ theory that consists of the field axioms together with the following five axioms:

$\forall x \ \neg(x < x)$
$\forall x,y,z\ ((x < y \land y < z) \rightarrow x < z)$
$\forall x,y\ (x\ne y \rightarrow (x < y \lor y < x))$
$\forall x,y,z\ (x < y \rightarrow x + z < y + z)$
$\forall x,y,z\ ((0 < x \land 0 < y) \rightarrow 0 < x \cdot y)$

An $$\mathcal{L}_{\mathrm{of}}$-$ structure $\mathcal{F} = (F, 0, 1, +, -, \cdot, <)$ is called an ordered field if and only if $\mathcal{F} \vDash T_{\mathrm{of}}.$

The ordered-field axioms yield the familiar rules for signs and inequalities.

Proposition 1.1.7. For all elements $$w,x,y,z$$ of an ordered field, the following hold:

if $$0<x$,$ then $$-x<0$;$
if $$x<0$$ and $$y<0$,$ then $$0<xy$;$
if $$x<y$$ and $$z<w$,$ then $$x+z<y+w$;$
$0\le x^2.$

Proof. (1) From (OF14) in Definition 1.1.6, for all $$x,y,u,v,$$ if $$x<y$$ then $$u<v$$ where $$v-u = y-x.$$ Thus $$0<x$$ implies $$-x<0.$$ (2) Since $$0<-x$$ and $$0<-y,$$ $$0<(-x)(-y).$$ Thus, by (3) of Proposition 1.1.5, $$0<xy.$$ (3) Since $$x<y,$$ it follows that $$x+z<y+z.$$ Similarly, since $$z<w,$$ we have $$z+y<w+y.$$ Since the order relation is transitive, $$x+z<w+y.$$ (4) Since the order is comparable if $0\ne x$ then $$0<x$$ or $$x<0.$$ If $$0<x$$ then $$0x<x^2,$$ so that $$0<x^2.$$ If $$x<0$$ then $$0<x^2.$$ The only case left is that $$x=0.$$ If $$x=0$$ then $$x^2=0x=0.$$ Therefore $0\le x^2.$ $\square$

The ordered-field axioms are satisfied by both $\mathbb Q$ and $$\mathbb R$,$ so they do not capture what is distinctive about the real numbers. The additional ingredient is completeness. For $$\mathbb R$,$ completeness is expressed by the least-upper-bound property: every nonempty subset of $\mathbb R$ that is bounded above has a least upper bound.

Theorem 1.1.8. The real number system $\mathcal R=(\mathbb R,0,1,+,-,\cdot,<)$ is an ordered field and has the least-upper-bound property.

We henceforth identify $$\mathbb N$,$ $$\mathbb Z$,$ and $\mathbb Q$ with their respective images $$\mathbb N^{\ast}$,$ $$\mathbb Z^{\ast}$,$ and $\mathbb Q^{\ast}$ in $\mathbb R.$ Under this identification, we have

\[\mathbb N \subseteq \mathbb Z \subseteq \mathbb Q \subseteq \mathbb R.\]

In particular, the rational field $\mathcal Q=(\mathbb Q,0,1,+,-,\cdot,<)$ is a subfield of $\mathcal R.$

We now derive several fundamental consequences of completeness. The first is the Archimedean property: integer multiples of any positive real number eventually exceed any prescribed real number.

Theorem 1.1.9. For all $x\in \mathbb R$ and all $y\in \mathbb R^{+},$ there exists $n\in \mathbb N^{+}$ such that

\[x<ny.\]

Proof. Let $x\in \mathbb R$ and $y\in\mathbb R^{+}$ be given, and let $$S$$ be the set of all $$ny,$$ where $$n$$ runs through the positive integers. That is,

\[S = \lbrace ny : n\in \mathbb N^{+}\rbrace.\]

We suppose, for contradiction, that $$S$$ is bounded above. Then $$S$$ has the supremum in $\mathbb R,$ by Theorem 1.1.8. Put $\alpha = \sup S.$

Lemma 1.1.10. Let $A\subseteq \mathbb R$ be bounded above, and let $$\alpha=\sup A$.$ Then, for any $$\varepsilon>0$,$ there exists $x\in A$ such that

\[\alpha-\varepsilon<x.\]

Subproof. Since $\alpha$ is least upper bound for $$A$,$ $\alpha-\varepsilon$ is not an upper bound for $$A$.$ Hence there exists $x\in A$ such that $\alpha-\varepsilon<x.$ $\blacksquare$

Lemma 1.1.10 guarantees the existence of an element $$my$$ such that $\alpha - y < my.$ But then we have $\alpha < (m+1)y\in S,$ which contradicts that $\alpha$ is an upper bound for $$S.$$ Therefore it is concluded that the Theorem 1.1.9 is in fact true. $\square$

The Archimedean property of $\mathbb R$ leads to the density of $\mathbb Q$ in $$\mathbb R$:$ between any two distinct real numbers there is a rational number.

Theorem 1.1.11. For all $x,y\in \mathbb R,$ if $$x<y$$ then there exists $p\in \mathbb Q$ such that

\[x<p<y.\]

Proof. Let $$x$$ and $$y$$ be reals such that $$x<y.$$ There exists $n\in \mathbb N^{+}$ such that $$1<n(y-x),$$ by Theorem 1.1.9. Hence we can choose such a number and denote it by $$n.$$ Then we consider the set

\[S = \lbrace k \in \mathbb Z : nx < k\rbrace.\]

Theorem 1.1.9 implies that $$S$$ is nonempty. It is also bounded below, so $$S$$ has the least element; denote it by $$m.$$ Then we have

\[nx < m \le nx + 1<ny.\]

Since $$n>0$$ it follows that

\[x < \frac m n < y.\]

This proves Theorem 1.1.11. $\square$

Completeness also guarantees the existence of positive $$n$-$ th roots.

Theorem 1.1.12. For all $y\in \mathbb R^{+}$ and all $n\in \mathbb N^{+},$ there exists the unique $x\in \mathbb R^{+}$ such that

\[x^n = y.\]

Proof. It is clear that there is at most one such $$x,$$ since $$0<x_1<x_2$$ implies $$x_1^n<x_2^n.$$ We consider the set

\[S = \lbrace x\in \mathbb R^{+}: x^n<y\rbrace.\]

If $$x=y/(1+y)$$ then $$0<x<1.$$ Since $$x^n<x<y,$$ $x\in S.$ Thus $$S$$ is nonempty. If $$x>1+y$$ then $$y<x<x^n,$$ so that $x\notin S.$ Thus $$1+y$$ is an upper bound for $$S.$$ Therefore there exists $\alpha$ such that $\alpha = \sup S.$ We suppose, for contradiction, that $\alpha^n<y$ or $y<\alpha^n.$ We first assume $\alpha^n<y.$ Choose $$h$$ so that

\[0< h< \min\left\lbrace 1,\frac{y-\alpha^n}{n(\alpha+1)^{n-1}}\right\rbrace.\]

Then

\[(\alpha + h)^n - {\alpha}^n < hn(\alpha + h)^{n-1}<hn(\alpha + 1)^{n-1} < y-{\alpha}^n.\]

Thus $(\alpha + h)^n < y,$ so that $\alpha + h \in S.$ Since $\alpha<\alpha+h,$ this result contradicts the fact that $\alpha$ is an upper bound for $$S.$$ We now assume $y<\alpha^n.$ Choose

\[h = \frac{\alpha^n-y}{n\alpha^{n-1}}.\]

Then $0<h<\alpha.$ If $\alpha - h \le x,$ we conclude that

\[\alpha^n - x^n \le \alpha^n - (\alpha - h)^n<hn\alpha^{n-1} = \alpha^n - y.\]

Thus $$y<x^n$$ so that $x\notin S.$ Since $\alpha -h<\alpha,$ this results contradicts the fact that $\alpha$ is the supremum for $$S.$$ Therefore $\alpha^n=y,$ and this prove Theorem 1.1.12. $\square$

The existence and uniqueness established in Theorem 1.1.12 make the following notation well-defined.

Definition 1.1.13. Let $x\in \mathbb R$ and $n\in \mathbb N^{+}.$ We define $$n$-$ th root of $$x$$ as follows:

\[x^{1/n} = \begin{cases} \text{the unique }y>0\text{ such that }y^n=x, & \text{if } x>0, \\ 0, & \text{if } x=0.\\ \end{cases}\]

The uniqueness of positive $$n$-$ th roots gives the expected product rule.

Proposition 1.1.14. For all $x,y\in \mathbb R^{+}$ and $n\in \mathbb N^{+},$

\[(xy)^{1/n}=x^{1/n}y^{1/n}.\]

Proof. Let $x,y\in \mathbb R^{+}$ be given. We let $a=(xy)^{1/n}, b=x^{1/n},$ and $c=y^{1/n}.$ Then we have $$a^n=xy, b^n=x,$$ and $$c^n=y.$$ Hence $$a^n=b^nc^n=(bc)^n,$$ so that

\[\left(a^n\right)^{1/n}=\left((bc)^n\right)^{1/n}\]

by the uniqueness assertion of Theorem 1.1.12. Since

\[\left(a^n\right)^{1/n}=a\quad \text{and} \quad \left((bc)^n\right)^{1/n}=bc,\]

it follows that $$a=bc.$$ By definition of $$a,b,$$ and $$c,$$ therefore,

\[(xy)^{1/n}=x^{1/n}y^{1/n}.\tag*{$\square$}\]

References

Rudin, W. (1976). Principles of mathematical analysis (3rd ed.). McGraw-Hill Education.
Munkres, J. R. (2000). Topology (2nd ed.). Pearson.

External

1.1 Real Number System

References