Foundations of the Real Numbers for Analysis

A field is a set $F$ equipped with two binary operations $+$ (addition) and $\cdot$ (multiplication) and two distinguished elements $0\mathrm{\ne }1$ of $F$, satisfying:

• (A) $(F,+)$ is an abelian group with identity $0$: addition is associative and commutative, $x+0=x$ for all $x$, and every $x$ has an additive inverse $-x$ with $x+(-x)=0$.
• (M) Multiplication is associative and commutative on all of $F$, and $x\cdot 1=x$ for every $x\in F$; moreover $(F\setminus \{0\},\cdot )$ is an abelian group with identity $1$, so every $x\mathrm{\ne }0$ has a multiplicative inverse $x^{-1}$ with $x\cdot x^{-1}=1$.
• (D) Multiplication distributes over addition: $x\cdot (y+z)=x\cdot y+x\cdot z$.

We write $x-y:=x+(-y)$ and $x\mathrm{/}y:=x\cdot y^{-1}$ (for $y\mathrm{\ne }0$), and abbreviate $x\cdot y$ as $xy$.

• The rationals $\mathbb{Q}$ (The rationals form a field) and both constructions of the reals (The reals form a field, The Dedekind reals form a field) are fields. This definition states the field axioms once, abstractly, so that every field fact proved from them applies to all of these at once.
• The requirement $0\mathrm{\ne }1$ excludes the trivial one-element structure.
• Why (M) quantifies over all of $F$ explicitly (amended 2026-07-28, owner approved). An earlier wording headed (M) with the group $(F\setminus \{0\},\cdot )$ alone, and under that strict reading the axioms do not pin down the products $0\cdot y$: take ${\mathbb{F}}_2$ and redefine $0\cdot y:=y$, which leaves (A) untouched, leaves $(F\setminus \{0\},\cdot )$ the trivial group, and satisfies (D) in all eight instances, yet has $0\cdot 1=1$ — making Multiplication by zero: $0\cdot a=0$ false in a purported model of the axioms. Associativity, commutativity and $x\cdot 1=x$ are therefore asserted on all of $F$, the element $0$ included. Commutativity on all of $F$ is also what turns (D) into its right-hand form $(y+z)\cdot x=y\cdot x+z\cdot x$, which Multiplication by zero: $0\cdot a=0$ uses.
• The identities $0,1$ and the inverses $-x,x^{-1}$ are unique (Identities and inverses in a field are unique ↗), so writing them as single-valued notation, and the abbreviations $x-y:=x+(-y)$ and $x\mathrm{/}y:=x{y}^{-1}$, are legitimate.

An ordered field is a field $F$ (Field) together with a subset $P\subseteq F$, the set of positive elements (the positive cone), satisfying:

• (O1) Trichotomy. For each $x\in F$, exactly one of the following holds: $x\in P$, or $x=0$, or $-x\in P$.
• (O2) Closure. If $x,y\in P$ then $x+y\in P$ and $x\cdot y\in P$.

We write $x>0$ for $x\in P$ (read "$x$ is positive"), and define the order by

$a<b:⟺b-a\in P,a\le b:⟺a<b\text{ or }a=b\mathrm{.}$

Thus $a>b$ means $b<a$, and $a\ge b$ means $b\le a$. An element $x$ with $-x\in P$ (equivalently $x<0$) is called negative.

• By trichotomy applied to $b-a$, for any $a,b$ exactly one of $a<b$, $a=b$, $a>b$ holds; this makes $\le$ a total order.
• (O2) says the positives are closed under addition and multiplication: sums and products of positives are positive.
• The rationals (The rationals form a totally ordered field) and both constructions of the reals (The reals form a totally ordered field, The Dedekind reals form a totally ordered field) are ordered fields, so every fact proved here from (O1)-(O2) holds in each of them.

Let $F$ be an ordered field (Ordered field). The absolute value of $x\in F$ is

$\mathrm{\mid }x\mathrm{\mid }:=\{\begin{array}{ll}x& \text{if }x\ge 0,\\ -x& \text{if }x<0.\end{array}$

• The distance between $x$ and $y$ is $\mathrm{\mid }x-y\mathrm{\mid }$.
• The basic algebraic properties ($\mathrm{\mid }x\mathrm{\mid }\ge 0$, $\mathrm{\mid }x\mathrm{\mid }=0⟺x=0$, $\mathrm{\mid }xy\mathrm{\mid }=\mathrm{\mid }x\mathrm{\mid }\mathrm{\mid }y\mathrm{\mid }$, and the characterisation $\mathrm{\mid }x\mathrm{\mid }<c⟺-c<x<c$) are established in Basic properties of the absolute value; the triangle inequality is The triangle inequality.

Let $F$ be an ordered field (Ordered field). For a natural number $n\ge 1$, write $n\cdot 1_F:=\underset{n}{\underbrace{1_F+\cdots +1_F}}$ for the $n$-fold sum of the multiplicative identity, and $0\cdot 1_F:=0$. These are the canonical natural numbers of $F$.

$F$ is Archimedean if for every $x\in F$ there is a natural number $n$ with

$x<n\cdot 1_F\mathrm{.}$

Equivalently, the canonical naturals $(n\cdot 1_F{)}_{n\ge 1}$ are cofinal: no single element of $F$ is an upper bound for all of them.

• Equivalently (applying the definition to $1\mathrm{/}\epsilon$): for every $\epsilon >0$ in $F$ there is $n$ with $1\mathrm{/}(n\cdot 1_F)<\epsilon$, so the canonical fractions $1\mathrm{/}n$ are arbitrarily small.
• That the canonical naturals are well-defined, positive, and strictly increasing is Canonical naturals are positive and strictly increasing. Every complete ordered field is Archimedean (Every complete ordered field is Archimedean); an ordered field need not be (Not every ordered field is Archimedean).

Let $F$ be an ordered field (Ordered field) and $S\subseteq F$.

• $u\in F$ is an upper bound of $S$ if $s\le u$ for all $s\in S$; $S$ is bounded above if it has an upper bound.
• $u\in F$ is a least upper bound (or supremum, $\sup S$) of $S$ if $u$ is an upper bound of $S$ and $u\le u^{\mathrm{\prime }}$ for every upper bound $u^{\mathrm{\prime }}$ of $S$.

$F$ is a complete ordered field (equivalently, $F$ has the least-upper-bound property, or is Dedekind complete) if every nonempty $S\subseteq F$ that is bounded above has a least upper bound in $F$.

• A least upper bound, if it exists, is unique (two least upper bounds are each $\le$ the other, so equal by antisymmetry of the order).
• Applying the property to $\{-s:s\in S\}$ yields the dual greatest lower bound (infimum) property, so the two are equivalent.
• The Dedekind-cut reals have this property by construction (Dedekind completeness: the least-upper-bound property); the Cauchy-sequence reals acquire it via The Cauchy-sequence reals have the least-upper-bound property. This definition is the target for the uniqueness theorem Uniqueness of the complete ordered field: $\mathbb{R}$ up to a unique isomorphism.

Let $F$ and $G$ be fields (Field). A field homomorphism $\phi :F\to G$ is a map satisfying, for all $x,y\in F$,

$\phi (x+y)=\phi (x)+\phi (y),\phi (xy)=\phi (x)\phi (y),\phi (1_F)=1_G\mathrm{.}$

An embedding (or monomorphism) is an injective field homomorphism. An isomorphism is a bijective field homomorphism.

• From additivity, $\phi (0_F)=0_G$ and $\phi (-x)=-\phi (x)$; from multiplicativity, $\phi (x^{-1})=\phi (x{)}^{-1}$ for $x\mathrm{\ne }0$. These read off because $\phi (-x)$ is an additive inverse of $\phi (x)$ and $\phi (x^{-1})$ a multiplicative inverse of $\phi (x)$, and inverses are unique (Identities and inverses in a field are unique).
• Every field homomorphism is automatically injective: its kernel $\{x:\phi (x)=0\}$ is an ideal of $F$, and a field has only the ideals $\{0\}$ and $F$; since $\phi (1_F)=1_G\mathrm{\ne }0$, the kernel is $\{0\}$. This injectivity is used explicitly in Homomorphisms out of a complete ordered field are order-preserving.
• Order-compatible homomorphisms are the subject of Ordered-field isomorphism.

Let $F$ and $G$ be ordered fields (Ordered field). An ordered-field isomorphism $\phi :F\to G$ is a bijective field homomorphism (Field homomorphism and embedding) that is order-preserving in both directions:

$a<b⟺\phi (a)<\phi (b)\text{for all }a,b\in F\mathrm{.}$

Two ordered fields are isomorphic if there exists an ordered-field isomorphism between them; we write $F\cong G$.

• Equivalently, $\phi$ is a field isomorphism carrying the positive cone of $F$ onto that of $G$ ($\phi (P_F)=P_G$); the inverse ${\phi }^{-1}$ is then also an ordered-field isomorphism.
• Because a field homomorphism preserves all of $+,-,\cdot ,{}^{-1},0,1$, an ordered-field isomorphism identifies $F$ and $G$ as ordered fields completely: every field-theoretic and order-theoretic statement transfers across it.
• For homomorphisms out of a complete ordered field, order-preservation is automatic (Homomorphisms out of a complete ordered field are order-preserving); this is what makes the isomorphism in Uniqueness of the complete ordered field: $\mathbb{R}$ up to a unique isomorphism unique.

In a field $F$ (Field) the additive identity, the multiplicative identity, each additive inverse, and each multiplicative inverse are unique. Hence the notations $0$, $1$, $-x$, and $x^{-1}$ denote well-defined elements, as the field definition and its consequences assume.

In any field $F$ (Field), for every $a\in F$ we have $0\cdot a=0$.

In any field $F$ (Field), for all $a,b\in F$ we have $(-a)b=-(ab)$ and $(-a)(-b)=ab$.

In any field $F$, the identity $(-1)(-1)=1$ holds.

In any field $F$ (Field), if $a,b\in F$ satisfy $ab=0$, then $a=0$ or $b=0$.

In any field $F$ (Field), if $a,b,c\in F$ with $c\mathrm{\ne }0$ and $ac=bc$, then $a=b$.

Let $F$ be an ordered field (Ordered field) with positive cone $P$, and let $a,b,c,d\in F$.

1. Translation invariance. If $a<b$ then $a+c<b+c$.
2. Adding inequalities. If $a<b$ and $c<d$ then $a+c<b+d$.

Let $F$ be an ordered field (Ordered field) with positive cone $P$, and let $a,b,c\in F$.

1. If $a>0$ and $b>0$ then $ab>0$.
2. If $a>0$ and $b<0$ then $ab<0$.
3. If $a<0$ and $b<0$ then $ab>0$.
4. If $c>0$ then $a<b⟺ac<bc$.
5. If $c<0$ then $a<b⟺bc<ac$ (multiplication by a negative reverses order).

Let $F$ be an ordered field (Ordered field) with positive cone $P$. For every $a\in F$ with $a\mathrm{\ne }0$ we have $a^2>0$, where $a^2=a\cdot a$. In particular, every nonzero square is positive.

In any ordered field $F$ with positive cone $P$, the multiplicative identity satisfies $0<1$; that is, $1\in P$.

Let $F$ be an ordered field (Ordered field) with positive cone $P$, and let $a,b\in F$.

1. If $a>0$ then $a^{-1}>0$.
2. If $0<a<b$ then $0<b^{-1}<a^{-1}$.