Construction of the Real Numbers via Dedekind Cuts

Reflexivity: for any $(a,b)$ we have $a+b=b+a$, so $(a,b)\sim (a,b)$.

Symmetry: suppose $(a,b)\sim (c,d)$, i.e. $a+d=b+c$. Then $c+b=d+a$, which is the defining equation for $(c,d)\sim (a,b)$.

Suppose $(a,b)\sim (c,d)$ and $(c,d)\sim (e,f)$, i.e. $a+d=b+c$ and $c+f=d+e$.

Adding the two equations: $(a+d)+(c+f)=(b+c)+(d+e)$.

Regrouping both sides: $(a+f)+(c+d)=(b+e)+(c+d)$.

Cancelling $(c+d)$: $a+f=b+e$, so $(a,b)\sim (e,f)$; the relation is transitive.

The relation is reflexive, symmetric, and transitive, hence an equivalence relation.

By hypothesis $a+b^{\mathrm{\prime }}=b+a^{\mathrm{\prime }}$.

By hypothesis $c+d^{\mathrm{\prime }}=d+c^{\mathrm{\prime }}$.

Adding the two equations: $(a+b^{\mathrm{\prime }})+(c+d^{\mathrm{\prime }})=(b+a^{\mathrm{\prime }})+(d+c^{\mathrm{\prime }})$.

Commuting the equation of step 1.1: $b+a^{\mathrm{\prime }}=a+b^{\mathrm{\prime }}$, which is the defining equation for $(b,a)\sim (b^{\mathrm{\prime }},a^{\mathrm{\prime }})$; negation is well defined.

Regrouping both sides: $(a+c)+(b^{\mathrm{\prime }}+d^{\mathrm{\prime }})=(b+d)+(a^{\mathrm{\prime }}+c^{\mathrm{\prime }})$, which is the defining equation for $(a+c,b+d)\sim (a^{\mathrm{\prime }}+c^{\mathrm{\prime }},b^{\mathrm{\prime }}+d^{\mathrm{\prime }})$; addition is well defined.

Both operations are independent of representatives.

By hypothesis $a+b^{\mathrm{\prime }}=b+a^{\mathrm{\prime }}$; write $s$ for this common value.

Regrouping and factoring: $(ac+bd)+(a^{\mathrm{\prime }}d+b^{\mathrm{\prime }}c)=c(a+b^{\mathrm{\prime }})+d(b+a^{\mathrm{\prime }})$.

Regrouping and factoring: $(ad+bc)+(a^{\mathrm{\prime }}c+b^{\mathrm{\prime }}d)=d(a+b^{\mathrm{\prime }})+c(b+a^{\mathrm{\prime }})$.

By step 1.1 both right-hand sides equal $cs+ds$, so $(ac+bd)+(a^{\mathrm{\prime }}d+b^{\mathrm{\prime }}c)=(ad+bc)+(a^{\mathrm{\prime }}c+b^{\mathrm{\prime }}d)$.

That equation is precisely the defining relation $(ac+bd,ad+bc)\sim (a^{\mathrm{\prime }}c+b^{\mathrm{\prime }}d,a^{\mathrm{\prime }}d+b^{\mathrm{\prime }}c)$: the product class is unchanged when the first factor's representative changes.

The product formula is symmetric in its two arguments: swapping $(a,b)\leftrightarrow (c,d)$ sends $(ac+bd,ad+bc)$ to $(ca+db,cb+da)$, the same pair. Hence, by the argument of steps 1.1–3.1 applied to the second factor, the product class is also unchanged when $(c,d)$ is replaced by $(c^{\mathrm{\prime }},d^{\mathrm{\prime }})$.

Replacing first $(a,b)$ by $(a^{\mathrm{\prime }},b^{\mathrm{\prime }})$ and then $(c,d)$ by $(c^{\mathrm{\prime }},d^{\mathrm{\prime }})$: $[(a,b)]\cdot [(c,d)]=[(a^{\mathrm{\prime }},b^{\mathrm{\prime }})]\cdot [(c,d)]=[(a^{\mathrm{\prime }},b^{\mathrm{\prime }})]\cdot [(c^{\mathrm{\prime }},d^{\mathrm{\prime }})]$; multiplication is well defined.

By [L3], each axiom may be verified on arbitrary fixed representatives $(a,b),(c,d),(e,f)$.

Associativity of $+$: $((a,b)+(c,d))+(e,f)$ and $(a,b)+((c,d)+(e,f))$ both equal $(a+c+e,b+d+f)$.

Commutativity of $+$: $(a+c,b+d)=(c+a,d+b)$.

Additive identity: $(a,b)+(0,0)=(a+0,b+0)=(a,b)$.

Additive inverses: $(a,b)+(b,a)=(a+b,b+a)\sim (0,0)$, since $(a+b)+0=(b+a)+0$.

Commutativity of $\cdot$: swapping $(a,b)\leftrightarrow (c,d)$ sends $(ac+bd,ad+bc)$ to $(ca+db,cb+da)$, the same pair.

Multiplicative identity: $(a,b)\cdot (1,0)=(a\cdot 1+b\cdot 0,a\cdot 0+b\cdot 1)=(a,b)$.

Associativity of $\cdot$: expanding, both $((a,b)(c,d))(e,f)$ and $(a,b)((c,d)(e,f))$ equal $(ace+adf+bcf+bde,acf+ade+bce+bdf)$.

Distributivity: $(a,b)\cdot ((c,d)+(e,f))=(a(c+e)+b(d+f),a(d+f)+b(c+e))=(ac+bd,ad+bc)+(ae+bf,af+be)=(a,b)(c,d)+(a,b)(e,f)$.

Steps 1.2–1.9 verify all axioms: $(\mathbb{Z},+,\cdot ,0,1)$ is a commutative ring with identity and additive inverses.

Suppose $(a,b)\sim (a^{\mathrm{\prime }},b^{\mathrm{\prime }})$ and $(c,d)\sim (c^{\mathrm{\prime }},d^{\mathrm{\prime }})$, i.e. $a+b^{\mathrm{\prime }}=b+a^{\mathrm{\prime }}$ and $c+d^{\mathrm{\prime }}=d+c^{\mathrm{\prime }}$.

Suppose $a+d\le b+c$.

Totality: in $\mathbb{N}$ either $a+d\le b+c$ or $b+c\le a+d$, so $x\le y$ or $y\le x$.

Antisymmetry: $x\le y$ and $y\le x$ give $a+d\le b+c$ and $b+c\le a+d$, hence $a+d=b+c$, which is $x=y$ as classes.

Transitivity: from $a+d\le b+c$ and $c+f\le d+e$, adding gives $(a+d)+(c+f)\le (b+c)+(d+e)$; regrouping, $(a+f)+(c+d)\le (b+e)+(c+d)$; cancelling, $a+f\le b+e$, i.e. $x\le z$.

Compatibility with addition: $x+z\le y+z$ reads $(a+e)+(d+f)\le (b+f)+(c+e)$, which regroups to $(a+d)+(e+f)\le (b+c)+(e+f)$, equivalent to $a+d\le b+c$, i.e. $x\le y$.

Suppose $0<x$ and $0<y$, i.e. $b<a$ and $d<c$.

Adding $(a^{\mathrm{\prime }}+d^{\mathrm{\prime }})$: $(a+d)+(a^{\mathrm{\prime }}+d^{\mathrm{\prime }})\le (b+c)+(a^{\mathrm{\prime }}+d^{\mathrm{\prime }})$.

Regrouping with the equations of step 1.1: $(b+c)+(a^{\mathrm{\prime }}+d^{\mathrm{\prime }})=(b+a^{\mathrm{\prime }})+(c+d^{\mathrm{\prime }})=(a+b^{\mathrm{\prime }})+(d+c^{\mathrm{\prime }})=(a+d)+(b^{\mathrm{\prime }}+c^{\mathrm{\prime }})$.

Write $a=b+k$ and $c=d+m$ with nonzero naturals $k,m$.

So $(a+d)+(a^{\mathrm{\prime }}+d^{\mathrm{\prime }})\le (a+d)+(b^{\mathrm{\prime }}+c^{\mathrm{\prime }})$; cancelling $(a+d)$ gives $a^{\mathrm{\prime }}+d^{\mathrm{\prime }}\le b^{\mathrm{\prime }}+c^{\mathrm{\prime }}$. Applying the same argument with primed and unprimed representatives exchanged proves the converse implication, so the relation is independent of representatives.

Expanding both components of $xy=[(ac+bd,ad+bc)]$: $ac+bd=(b+k)(d+m)+bd=(bd+bm+kd+bd)+km$ and $ad+bc=(b+k)d+b(d+m)=bd+kd+bd+bm$, so $ac+bd=(ad+bc)+km$.

Since $k,m$ are nonzero, $0<k$ and $0<m$ by [L4]. Hence [L5] gives $0=0\cdot m<km$, and step 3.2 yields $ad+bc<ac+bd$, which is $0<xy$.

The order is well defined, total, compatible with addition, and positives are closed under multiplication; together with [L6], this makes $\mathbb{Z}$ a totally ordered commutative ring.

By trichotomy, each of $x$ and $y$ is positive or negative; split on the sign pattern.

Case both positive: $0<xy$ by closure of positives, so $xy\mathrm{\ne }0$.

Case exactly one negative, say $y<0$ (the other case is identical by commutativity): $0<-y$, so $0<x(-y)=-(xy)$, hence $xy<0$ and $xy\mathrm{\ne }0$.

Case both negative: $0<-x$ and $0<-y$, so $0<(-x)(-y)=xy$, and $xy\mathrm{\ne }0$.

In every case $xy\mathrm{\ne }0$: the product of nonzero integers is nonzero.

Cancellation: if $xz=yz$ with $z\mathrm{\ne }0$, then $(x-y)z=xz-yz=0$; were $x-y\mathrm{\ne }0$, the product would be nonzero, so $x-y=0$, i.e. $x=y$.

Injectivity: $[(m,0)]=[(n,0)]$ means $m+0=0+n$, i.e. $m=n$.

Addition: $\iota (m)+\iota (n)=[(m+n,0+0)]=\iota (m+n)$.

Multiplication: $\iota (m)\iota (n)=[(mn+0\cdot 0,m\cdot 0+0\cdot n)]=\iota (mn)$.

Order: $\iota (m)\le \iota (n)$ reads $m+0\le 0+n$, i.e. $m\le n$.

Image: if $0\le [(a,b)]$ then $b\le a$, so $a=b+k$ and $(a,b)\sim (k,0)$, giving $[(a,b)]=\iota (k)$; conversely $0\le \iota (k)$ for every $k$.

$\iota$ embeds $\mathbb{N}$ into $\mathbb{Z}$, preserving arithmetic and order, with image the nonnegative integers.

Reflexivity: $ab=ba$, so $(a,b)\sim (a,b)$.

Symmetry: if $ad=cb$ then $cb=ad$, which is the defining equation for $(c,d)\sim (a,b)$.

Suppose $(a,b)\sim (c,d)$ and $(c,d)\sim (e,f)$, i.e. $ad=cb$ and $cf=ed$.

Multiplying the first equation by $f$ and the second by $b$: $adf=cbf$ and $cfb=edb$.

Chaining: $(af)d=adf=cbf=cfb=edb=(eb)d$.

Cancelling the nonzero $d$: $af=eb$, so $(a,b)\sim (e,f)$; the relation is transitive.

The relation is reflexive, symmetric, and transitive, hence an equivalence relation.

By hypothesis $a{b}^{\mathrm{\prime }}=a^{\mathrm{\prime }}b$ and $c{d}^{\mathrm{\prime }}=c^{\mathrm{\prime }}d$.

Product: $(ac)(b^{\mathrm{\prime }}{d}^{\mathrm{\prime }})=(a{b}^{\mathrm{\prime }})(c{d}^{\mathrm{\prime }})=(a^{\mathrm{\prime }}b)(c^{\mathrm{\prime }}d)=(a^{\mathrm{\prime }}{c}^{\mathrm{\prime }})(bd)$, so $(ac,bd)\sim (a^{\mathrm{\prime }}{c}^{\mathrm{\prime }},b^{\mathrm{\prime }}{d}^{\mathrm{\prime }})$.

Sum: $(ad+cb)(b^{\mathrm{\prime }}{d}^{\mathrm{\prime }})=(a{b}^{\mathrm{\prime }})(d{d}^{\mathrm{\prime }})+(c{d}^{\mathrm{\prime }})(b{b}^{\mathrm{\prime }})=(a^{\mathrm{\prime }}b)(d{d}^{\mathrm{\prime }})+(c^{\mathrm{\prime }}d)(b{b}^{\mathrm{\prime }})=(a^{\mathrm{\prime }}{d}^{\mathrm{\prime }}+c^{\mathrm{\prime }}{b}^{\mathrm{\prime }})(bd)$, so $(ad+cb,bd)\sim (a^{\mathrm{\prime }}{d}^{\mathrm{\prime }}+c^{\mathrm{\prime }}{b}^{\mathrm{\prime }},b^{\mathrm{\prime }}{d}^{\mathrm{\prime }})$.

Negation: from $a{b}^{\mathrm{\prime }}=a^{\mathrm{\prime }}b$, $(-a)b^{\mathrm{\prime }}=(-a^{\mathrm{\prime }})b$, so $(-a,b)\sim (-a^{\mathrm{\prime }},b^{\mathrm{\prime }})$.

All three operations are independent of representatives.

By hypothesis $(a,b)\sim (a^{\mathrm{\prime }},b^{\mathrm{\prime }})$, that is $a{b}^{\mathrm{\prime }}=a^{\mathrm{\prime }}b$ in $\mathbb{Z}$.

Commuting each product by [L1], $b^{\mathrm{\prime }}a=b{a}^{\mathrm{\prime }}$, hence $b{a}^{\mathrm{\prime }}=b^{\mathrm{\prime }}a$.

The equation $b{a}^{\mathrm{\prime }}=b^{\mathrm{\prime }}a$ is exactly the defining relation $(b,a)\sim (b^{\mathrm{\prime }},a^{\mathrm{\prime }})$, and since $a,a^{\mathrm{\prime }}\mathrm{\ne }0$ the pairs $(b,a),(b^{\mathrm{\prime }},a^{\mathrm{\prime }})$ are legal rational representatives; therefore $[(a,b){]}^{-1}=[(b,a)]=[(b^{\mathrm{\prime }},a^{\mathrm{\prime }})]=[(a^{\mathrm{\prime }},b^{\mathrm{\prime }}){]}^{-1}$, so the reciprocal is well-defined on $\mathbb{Q}\setminus \{0\}$.

By [L1], each axiom may be verified on fixed representatives.

Associativity of $+$: both $((a,b)+(c,d))+(e,f)$ and $(a,b)+((c,d)+(e,f))$ have representative $(adf+cbf+ebd,bdf)$.

Commutativity of $+$: $(ad+cb,bd)$ and $(cb+ad,db)$ are equal pairs.

Additive identity: $(a,b)+(0,1)=(a\cdot 1+0\cdot b,b\cdot 1)=(a,b)$.

Additive inverses: $(a,b)+(-a,b)=(ab-ab,b^2)=(0,b^2)\sim (0,1)$.

Associativity and commutativity of $\cdot$: $((ac)e,(bd)f)=(a(ce),b(df))$ and $(ac,bd)=(ca,db)$.

Multiplicative identity: $(a,b)(1,1)=(a,b)$; and $1\mathrm{\ne }0$ since $1\cdot 1\mathrm{\ne }0\cdot 1$.

Distributivity: $(a,b)((c,d)+(e,f))$ has representative $(a(cf+ed),b(df))$, while $(a,b)(c,d)+(a,b)(e,f)$ has representative $(acbf+aebd,b^{2}df)=(b(acf+aed),b(bdf))$; cancelling the common nonzero factor $b$ these are equivalent.

Inverses: $[(a,b)]=0$ iff $a\cdot 1=0\cdot b$ iff $a=0$; for $a\mathrm{\ne }0$, $(a,b)(b,a)=(ab,ba)\sim (1,1)$.

All field axioms hold: $\mathbb{Q}$ is a field.

Since $b\mathrm{\ne }0$, by trichotomy [L1] either $b>0$ or $b<0$.

If $b>0$, the representative $(a,b)$ already has positive denominator.

If $b<0$, then $-b>0$ by [L1], and $a(-b)=(-a)b$ by [L2] is exactly the defining relation $(a,b)\sim (-a,-b)$; so $(-a,-b)$ represents the same class and has positive denominator $-b$.

In either case the class has a representative with positive denominator; hence the order Order on the rationals, stated on such representatives, is defined for every rational.