4.4 Axioms for the reals and first approximations

What must a larger ordered number system provide?

The rational numbers support addition, subtraction, multiplication and division by nonzero elements, with an order compatible with those operations. Yet the rational set $S=\{q\in Q:q^2\lt2\}$ is nonempty and bounded above without having a least upper bound in $Q$. Preserving rational arithmetic is therefore only part of the problem: a larger system must also supply the missing bounds.

A complete ordered field states both requirements. Defining this target does not prove that a model exists. In the arguments below, $R$ denotes an assumed model of these axioms; the Dedekind construction will establish existence using only rational data.

The formal target

Definition 4.4.1

(Model of the real numbers)

A set $R$ is a model of the real numbers if it is equipped with:

• elements $0,1\in R$,
• binary operations $+$ and $\cdot$,
• a total order $\le$,

such that:

• $(R,0,1,+,\cdot,\le)$ is an ordered field;
• $(R,\le)$ is complete.

Completeness requires a least upper bound for every nonempty subset bounded above, and equivalently a greatest lower bound for every nonempty subset bounded below. These bounds belong to the ambient field, not necessarily to the subset.

Theorem 4.4.2

(Uniqueness of the real numbers)

Any two complete ordered fields are isomorphic as ordered fields. Thus different valid constructions can be treated as models of the same $R$.

The uniqueness theorem is stated here without proof. It explains why we can move between different constructions of $R$. Dedekind cuts and Cauchy sequences look different internally, but once both constructions satisfy the complete ordered field axioms, they represent the same real number system up to the unique structure that matters here: addition, multiplication, and order.

Worked example 4.4.3

(Why Q is not yet a model of the reals)

The rational numbers already satisfy the ordered-field part. What fails is completeness.

The set

$$S=\{x\in Q\mid x^2\lt 2\}$$

is nonempty and bounded above in $Q$, but it has no rational supremum. So $Q$ is close to the target, yet still not the real-number system.

Why decimal expansions are not the full story

An infinite decimal supplies rational truncations, so it is a plausible way to record a number. It does not yet define equality or arithmetic on these records. For example, the familiar identity $0.9999\ldots=1$ suggests that distinct digit strings may need to represent one object; its formal justification comes after the construction. Moreover, choosing base ten is a choice of notation, not a property that the ordered field should depend on.

The approximation idea

If a model $R$ and a decimal representation have already been supplied, we may write, for illustration,

$$r=10.234890234809234\ldots$$

Then we can approximate $r$ from below and above by rational numbers:

$$10\lt r\lt 11,$$ $$10.2\lt r\lt 10.3,$$ $$10.23\lt r\lt 10.24,$$

and so on.

Each additional digit narrows a rational interval. The eventual construction must explain why all these finite comparisons determine one boundary.

Worked example 4.4.4

(Lower and upper rational fences)

For a decimal expansion such as $r=3.1415\ldots$, the first few rational fences look like

$$3\lt r\lt 4,\qquad 3.1\lt r\lt 3.2,\qquad 3.14\lt r\lt 3.15.$$

Each extra digit gives a narrower rational interval containing $r$.

Dividing the rationals into two camps

In such a model, a number $r$ determines two sets of rationals:

$$A=\{q\in Q\mid q\lt r\}, \qquad B=\{q\in Q\mid q\ge r\}.$$

These sets retain every rational comparison with $r$, not merely a finite collection of approximations. But their definition still uses $r$, so they cannot by themselves construct it.

The direction can be reversed when the comparison rule uses rational arithmetic alone. For the gap above, the set

$$C=\{q\in Q:q\le0\text{ or }q^2\lt2\}$$

records the entire rational lower side without referring to an existing real root. The nonpositive rationals must be included: $S$ itself is not downward closed. The Dedekind construction will make such lower sets into numbers and define their arithmetic. The following consequences describe what any complete ordered-field model must satisfy; they are not premises for the existence of that model.

The Archimedean consequence without circularity

The approximation picture must not quietly assume the Archimedean property as an unexplained fact about $R$. In the complete ordered-field target, it is derived from completeness. Suppose, for contradiction, that the natural numbers were bounded above in $R$. Completeness would give

$$s=\sup\{0,1,2,\ldots\}.$$

Because $s-1$ is smaller than $s$, it cannot be an upper bound. Hence some natural number $n$ satisfies $s-1\lt n$, and then $n+1\gt s$, contradicting that $s$ was an upper bound. Therefore the natural numbers are unbounded above. For every real $x$ there is a natural $n$ with $n\gt x$; applying this to $1/\varepsilon$ gives the usual small-mesh estimates.

Theorem 4.4.5

(Archimedean approximation principle)

For every $x\in R$ there is $n\in N$ with $n\gt x$. Equivalently, for every $\varepsilon\gt 0$ there is $n\in N$, $n\ge1$, such that $1/n\lt \varepsilon$. The proof uses completeness and does not assume that a decimal expansion has already constructed the real line.

Worked example 4.4.6

(Choosing a rational fence width)

Given $\varepsilon\gt 0$, choose $n\gt 1/\varepsilon$. Then $1/n\lt \varepsilon$. If an approximation interval has width at most $1/n$, its width is therefore less than $\varepsilon$. This is the quantitative reason that successive decimal or rational fences can be made arbitrarily narrow.

Common mistake

(Do not use the conclusion as the construction)

Saying “choose a decimal expansion of the real number” before proving that the fences converge assumes the very boundary that the construction is meant to produce. First establish the order and completeness principles; then use the Archimedean consequence to control the approximation width.

Density and rational fences

The Archimedean property does more than assert abstract unboundedness: it lets us place rational numbers around any real number with a prescribed accuracy. The complete mesh argument is written explicitly in the next section. It uses a least-integer choice and then divides by a positive integer; the resulting rational fence works in every base and does not depend on decimal notation.

Theorem 4.4.7

(Rational density with a requested error)

For every $x\in R$ and every $\varepsilon\gt 0$, there is $q\in Q$ such that $|x-q|\lt \varepsilon$. Apply the interval argument below to $x-\varepsilon$ and $x+\varepsilon$.

Worked example 4.4.8

(A rational inside a specified interval)

Choose $n=1000$, so $1/n=0.001\lt 1.42-1.41$. The least integer strictly above $1000(1.41)=1410$ is $m=1411$. Hence $m/n=1.411$ lies strictly inside $(1.41,1.42)$. Choosing $n=100$ would give the right endpoint $1.42$, so the strict mesh inequality is essential.

Worked example 4.4.9

(Why completeness is still needed)

Between any two distinct rationals there is another rational, their midpoint. Nevertheless, $S=\{q\in Q:q^2\lt2\}$ has no rational supremum. Thus the existence of points between given endpoints does not supply a least upper bound for every nonempty set bounded above.

Trichotomy, absolute value, and error control

The total-order axiom gives a three-way decision for every pair of real numbers. Exactly one of $x\lt y$, $x=y$, or $y\lt x$ holds. In particular, every real number is positive, zero, or negative, and the sign of a difference determines which side of an interval it occupies. The three alternatives are exhaustive and mutually exclusive, so later arguments may split into cases without leaving an unclassified possibility.

Theorem 4.4.10

(Triangle inequality for rational error terms)

For $u,v\in Q$ (and, once the ordered-field axioms are established, for $u,v\in R$),

$$|u+v|\le |u|+|v|.$$

Consequently, if $|x-q|\lt\varepsilon$ and $|q-r|\lt\delta$, then $|x-r|\lt\varepsilon+\delta$.

To prove the first statement, if $u+v\ge0$, then $|u+v|=u+v\le|u|+|v|$ because $u\le|u|$ and $v\le|v|$. If $u+v\lt0$, then $|u+v|=-(u+v)=(-u)+(-v)\le|u|+|v|$, since $-u\le|u|$ and $-v\le|v|$. The two sign cases cover every possibility. For the consequence, write $x-r=(x-q)+(q-r)$ and apply the inequality:

$$|x-r|\le|x-q|+|q-r|\lt\varepsilon+\delta.$$

This is why a mesh estimate is useful: errors add in a controlled way rather than merely becoming “visually small.”

The rational mesh written without a hidden choice

The density argument can be made completely explicit. Given $x\lt y$ in $R$, first choose $n\in N$, $n\ge1$, with $1/n\lt y-x$. By the Archimedean property, the set of integers strictly greater than $nx$ is nonempty. It has a least element $m$ after translating by a sufficiently large integer and applying the well-ordering of $N$, an induction consequence whose proof appears in the later order discussion. Minimality gives

$$m-1\le nx\lt m.$$

Since $n>0$, division preserves the inequalities. Thus

$$x\lt\frac mn\le x+\frac1n\lt y.$$

The number $m/n$ is rational and lies strictly in the requested interval. The strict inequality $1/n\lt y-x$ is essential: if one used merely $1/n\le y-x$, the constructed fraction could land exactly at $y$.

Worked example 4.4.11

(A three-part approximation question)

Suppose $x\in R$ and $\varepsilon>0$. Choose $n>2/\varepsilon$ and apply the mesh construction to the interval $(x-\varepsilon/2,x+\varepsilon/2)$. The resulting rational $q$ satisfies $|x-q|\lt\varepsilon/2\lt\varepsilon$. If a second rational $r$ satisfies $|q-r|\lt\varepsilon/2$, the triangle inequality gives $|x-r|\lt\varepsilon$. Thus the same proof simultaneously produces a rational approximation and a rigorous tolerance for replacing one approximation by another.

Worked example 4.4.12

(A trichotomy check around a rational fence)

Suppose a real $x$ satisfies $1.41\lt x\lt1.42$. Trichotomy says that exactly one of $x\lt1.411$, $x=1.411$, or $1.411\lt x$ holds. If the third case holds, then $q=1.411=1411/1000$ is a rational strictly below $x$ and still inside the original interval. If the first case holds, use $1.41$ as the lower fence and repeat with a finer mesh. The equality case is not an error: it means the rational fence has already reached $x$. Keeping all three cases prevents a strict inequality from being silently substituted for a non-strict one.

The dependence is completeness $\Rightarrow$ the Archimedean property $\Rightarrow$ rational density. The last step also uses least-integer choice and division by a positive integer. These results control approximations inside an assumed model. Constructing that model still requires objects with defined equality, arithmetic and a complete order; a decimal notation alone does not establish those properties.

Quick checks

Quick check

(Which property does Q fail, so that it cannot be a model of the real numbers?)

Compare Q with the formal definition above.

Solution · Answer

$Q$ fails completeness. It is an ordered field, but a nonempty subset of $Q$ bounded above need not have a supremum in $Q$; dually, a nonempty subset bounded below need not have an infimum in $Q$.

Quick check

(What information do the inequalities $10.23<r<10.24$ give you about r?)

State it in terms of rational approximation.

Solution · Answer

They place $r$ inside a narrow rational interval. In other words, they give a lower rational approximation and an upper rational approximation that trap $r$.

Exercises

Quick check

(Let $x\in R$ and $\varepsilon>0$. Find rational approximations $q,r$ so that $|x-q|<\varepsilon/3$ and $|q-r|<2\varepsilon/3$, then prove $|x-r|<\varepsilon$.)

Apply rational density twice and keep track of the two error budgets.

Solution · Model solution

Rational density gives $q\in Q$ with $|x-q|<\varepsilon/3$. Apply it again to $q$ with tolerance $2\varepsilon/3$ to obtain $r\in Q$. The triangle inequality then gives

$$|x-r|\le|x-q|+|q-r|<\frac{\varepsilon}{3}+\frac{2\varepsilon}{3}=\varepsilon.$$

The tolerances can be unequal; what matters is that their sum stays within the requested error budget. No particular decimal expansion is needed.

Quick check

(Suppose we try to construct real numbers from infinite digit strings. Besides assigning a boundary to each string, what equality and arithmetic checks must the construction establish?)

Use $1.25000\ldots=1.24999\ldots$ to test the meaning of equality.

Solution · Model solution

The construction must identify different strings that describe the same boundary. Addition, multiplication, and order must be independent of which representative string is chosen. Finally it must verify the ordered-field laws and completeness. An approximation notation is useful, but does not establish these structural properties on its own.

Related notes

Read this after 4.3 Completeness and gaps in Q. Then continue with 4.5 Dedekind cuts and the embedding of Q, where the informal left/right split of $Q$ becomes the first fully rigorous construction of the real numbers.