Real and complex number systems: axioms and proof consequences

Learning goals

Identify what is assumed about the real numbers, prove consequences of those axioms, and explain why complex numbers extend arithmetic but not the real order.

Earlier constructions embed natural numbers into integers and integer fractions into rationals. Here we take an ordered complete field $\mathbb R$ as the real-number system, and then build complex numbers from real pairs. A construction, an axiom and a theorem derived from an axiom are three different kinds of statement.

1. What an ordered complete field assumes

A field has associative and commutative addition and multiplication, additive identity zero, multiplicative identity one with $0\ne1$, an additive inverse for every element and a multiplicative inverse for every nonzero element, and distributivity of multiplication over addition. These rules imply cancellation where the relevant inverse exists; cancellation by zero is not a field law.

An ordered field also has a total order compatible with arithmetic: adding the same number preserves strict inequalities, and multiplying two positive numbers gives a positive number. Its order has $0<1$. Completeness is the additional least-upper-bound axiom: every nonempty real subset bounded above has a real supremum.

Rationals satisfy the ordered-field algebra but not completeness. Thus completeness cannot be inferred merely from the ability to add, multiply or compare numbers. The empty set and an unbounded-above set are excluded from the real supremum axiom; we do not assign them finite real suprema by convention.

Proposition: squares and inverse signs

Every real square is nonnegative, and is positive for a nonzero input. A positive real has positive reciprocal, and if $0<a<b$ then $0<1/b<1/a$.

Proof. If $x>0$, the positive-product axiom gives $x^2>0$. If $x<0$, then $-x>0$ and distributivity gives $(-x)^2=x^2>0$. Zero has square zero. For $a>0$, its reciprocal is nonzero; if negative, $-a^{-1}>0$ and $a(-a^{-1})=-1$ would be positive, contradicting $-1<0$. Thus $a^{-1}>0$. Finally $1/a-1/b=(b-a)/(ab)>0$, since numerator and denominator are positive. The sign hypotheses justify every comparison.

These algebraic consequences also prove that the equation $x^2=-1$ has no real solution. This is a limitation of the ordered real system, not an error in the field operations.

2. Completeness supplies a square root

For a real $A>0$, let $S=\{x\ge0:x^2<A\}$. It is nonempty and bounded above by $A+1$. It also contains a positive number: take $\min(A/2,1/2)$. Completeness gives $r=\sup S>0$.

If $r^2<A$, choose $0<\delta<\min(1,(A-r^2)/(2r+1))$. Then $(r+\delta)^2-r^2=2r\delta+\delta^2\le(2r+1)\delta<A-r^2$, so $r+\delta\in S$, contradicting that $r$ is an upper bound.

If $r^2>A$, choose $0<\delta<\min(r/2,(r^2-A)/(2r))$. Then $r-\delta>0$ and $(r-\delta)^2\ge r^2-2r\delta>A$. Every $x\in S$ must be smaller than $r-\delta$, since larger nonnegative numbers have larger squares. Thus $r-\delta$ is a smaller upper bound, contradicting leastness. Hence $r^2=A$.

Uniqueness among nonnegative roots is separate: if $r,s\ge0$ and $r^2=s^2=A>0$, then $(r-s)(r+s)=0$. The positive factor $r+s$ is nonzero, so $r=s$. This proves existence and uniqueness using completeness and algebra, not an unproved radical symbol.

For $A=2$, the resulting root is irrational: if $r=a/b$ in lowest integer terms with $b>0$, then $a^2=2b^2$ makes $a$ even, and substitution makes $b$ even too, contradicting coprimality.

The standard counterexample is

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

This set is not empty, because $1\in S$. It is also bounded above in $Q$; for instance $2$ is an upper bound.

What would happen if $Q$ were complete? Then $S$ would have a rational supremum. The whole argument below shows that no rational number can play that role.

Theorem

Q is not complete

The ordered set $(Q,\le)$ is not complete.

Why no rational number can be sup(S)

Any upper bound $s$ must satisfy $s\ge1$, since $1\in S$. In particular, $s\le0$ cannot be an upper bound. We may therefore restrict the candidate supremum to positive rational numbers before comparing squares.

The two rational perturbations

Suppose $s\in Q$ and $s\gt 0$. If $s^2\lt 2$, let $M=2-s^2$ and choose

$$h=\frac12\min\left\{1,\frac{M}{2s+1}\right\}\gt 0.$$

This is an explicit rational choice. Since $h\lt 1$ and $h\lt M/(2s+1)$,

$$(s+h)^2=s^2+h(2s+h)\lt s^2+M=2.$$

Thus $s+h\in S$ and $s+h\gt s$, so $s$ is not an upper bound.

If $s^2\gt 2$, let $M=s^2-2$ and choose

$$h=\frac12\min\left\{s,\frac{M}{2s}\right\}\gt 0.$$

Now $r=s-h\gt 0$, and

$$r^2=s^2-2sh+h^2\gt s^2-2sh\gt 2.$$

Every $x\in S$ satisfies $x\lt r$: otherwise $x\ge r\gt 0$ would imply $x^2\ge r^2\gt 2$. Consequently $r$ is an upper bound smaller than $s$, contradicting leastness. Positivity of $r$ is essential; a negative rational whose square exceeds $2$ is not an upper bound of $S$.

Finally, $s^2=2$ is impossible for rational $s$ by the earlier parity proof. These cases exhaust the positive candidates, so $S$ has no supremum in $Q$. All quantities used here are rational; no real square root or completeness principle was assumed.

The same two perturbations prove that $S_+=\{q\in\mathbb Q:q\ge0,\ q^2<2\}$ has no rational supremum: $1\in S_+$, and both perturbed bounds and the points used in the argument are positive.

3. Complex numbers are constructed real pairs

Define $\mathbb C=\mathbb R^2$ with

$$(a,b)+(c,d)=(a+c,b+d),\qquad (a,b)(c,d)=(ac-bd,ad+bc).$$

The embedded real $a$ is $(a,0)$, and $i=(0,1)$ satisfies $i^2=(-1,0)$. Thus $(a,b)$ is written $a+bi$. The symbols express the defined pair operations, not an assumed real solution of $x^2=-1$.

The pair formulas inherit additive group laws from real coordinates. Multiplication is commutative. Associativity is checked by expansion: the real coordinate of either triple product is $ace-bde-adf-bcf$, and the imaginary coordinate is $acf-bdf+ade+bce$. Distributivity follows by coordinate expansion as well. Zero is $(0,0)$ and one is $(1,0)$.

For nonzero $z=a+bi$, define $\overline z=a-bi$ and $|z|=\sqrt{a^2+b^2}$. Since $a,b$ are not both zero, $a^2+b^2>0$. Direct multiplication gives $z\overline z=a^2+b^2$, so

$$z^{-1}=\frac{a-bi}{a^2+b^2}.$$

This proves every nonzero complex number has a multiplicative inverse and completes the field verification. In particular every quotient should first exclude a zero denominator; the conjugate trick does not manufacture an inverse for zero.

Conjugation preserves sums and products by expansion. Hence

$$|zw|^2=(zw)\overline{(zw)}=(z\overline z)(w\overline w)=|z|^2|w|^2.$$

Both sides' square roots are nonnegative, giving $|zw|=|z||w|$. The modulus is a nonnegative real measure of size, not an order relation between complex numbers.

4. Why the complex field cannot inherit an ordered-field structure

Suppose a total order on $\mathbb C$ satisfied the ordered-field rules. The square of every nonzero element would be positive by the same sign proof used over the reals. In particular $i^2=-1$ would be positive. But $1>0$ makes $-1<0$, a contradiction. Therefore there is no ordered-field order on $\mathbb C$.

One may sort complex pairs lexicographically for bookkeeping, or compare their moduli, but neither gives the real-compatible ordered-field structure needed by the supremum axiom. Equal moduli do not imply equal numbers: $1,-1,i,-i$ all have modulus one. The real completeness axiom is an ordered-field assertion and cannot simply be copied word for word to all complex numbers.

5. Embeddings and genuinely new elements

The standard chain $\mathbb N\subset\mathbb Z\subset\mathbb Q\subset\mathbb R\subset\mathbb C$ is shorthand for specified injective structure-preserving maps, not a claim that every construction uses literally identical underlying sets. Each inclusion adds new objects: negative one, a half, the positive square root of two, and $i$ distinguish successive stages. The inherited arithmetic agrees along the embeddings because those maps preserve the operations.

A property may be lost or gained at a stage: rationals and reals are ordered fields, but only reals are complete in the stated order sense; complexes remain a field but cannot be an ordered field. Larger does not mean retaining every earlier property in its original form.

Graded proof and diagnosis tasks

1. Order proof. Prove that $a^2=b^2$ over $\mathbb R$ implies $a=b$ or $a=-b$. State where the field zero-product property enters.
2. Sign audit. Explain why $a<b$ alone does not determine whether $1/b<1/a$ without excluding zero and checking the signs. Give all same-sign and opposite-sign cases.
3. Completeness. In the square-root argument, why do nonemptiness, boundedness, positivity of the supremum and both strict-square cases each need justification?
4. Complex inverse. Derive the inverse of $a+bi$ from its pair product, and identify precisely when the formula is undefined.
5. Order impossibility. Prove the impossibility of an ordered-field structure on $\mathbb C$ without merely saying complex numbers are two-dimensional.
6. Construction audit. A student says “rationals have the same algebra as reals, so every bounded rational set has a rational supremum.” Diagnose the missing axiom and give a concrete bounded nonempty set witnessing failure.

Full solutions

1. Factor $a^2-b^2=(a-b)(a+b)=0$. In a field, a product of two elements is zero only if at least one is zero: if one factor is nonzero, multiply by its inverse. Thus $a-b=0$ or $a+b=0$, giving the two possibilities. No order assumption is needed for this factorization conclusion.
2. First require $a,b\ne0$. The difference $1/a-1/b=(b-a)/(ab)$ has positive numerator. If $a,b$ have the same sign, $ab>0$ and $1/b<1/a$. If $a<0<b$, then $ab<0$ and instead $1/a<1/b$, consistent with one reciprocal negative and the other positive. A case with $a<b$ and $a>0>b$ is impossible. If either is zero, one reciprocal is undefined.
3. The supremum axiom applies only to a nonempty bounded-above set. Positivity of $r$ permits the downward perturbation and a positive denominator $2r$. Excluding $r^2<A$ proves it is not too small; excluding $r^2>A$ proves it is not too large. Upper boundedness alone or only one exclusion would not establish $r^2=A$. Uniqueness still needs the later separate factorization argument.
4. Multiplication by the conjugate gives $(a+bi)(a-bi)=a^2+b^2$. For a nonzero pair this is positive, so division by that real number yields $(a-bi)/(a^2+b^2)$ and product one. The denominator vanishes exactly when both coordinates are zero, so the formula excludes precisely the complex zero.
5. In any ordered field every nonzero square is positive, by considering the sign of its base. Thus $i^2$ would be positive, but $i^2=-1$ and $1>0$ force it negative. This contradiction concerns order compatibility with multiplication; dimensionality by itself is not a proof.
6. Ordered-field algebra does not include the least-upper-bound axiom. The set $\{q\in\mathbb Q:q\ge0,q^2<2\}$ is nonempty and bounded in $\mathbb Q$ but has no rational supremum: any proposed rational supremum with square below or above two admits the corresponding rational perturbation contradiction, while equality would make the square root of two rational. It is real completeness, not field algebra, that supplies its real supremum.

Summary and terms

An ordered field combines field arithmetic with compatible total order. Completeness supplies suprema of nonempty bounded sets and is an extra axiom. An embedding preserves structure injectively between number systems. A complex conjugate reflects the second coordinate and supplies the inverse formula, while a modulus measures size without imposing an ordered-field order. State the logical status of each fact before using it in a construction or proof.