Number-system depth: constructing integers without hidden subtraction

Assumptions and goals

Writing an integer pair as “a minus b” before negative numbers have been constructed would assume the object we intend to build. We instead start from pairs of natural numbers and use only natural addition, multiplication, order and cancellation in the construction proofs. The goal is not a construction of the natural numbers themselves.

Assume $\mathbb N=\{0,1,2,\ldots\}$ with its usual commutative associative addition and multiplication, distributivity, identities zero and one, and additive cancellation. Assume its total order is compatible with addition: if $a\ge b$, there is a unique $n\in\mathbb N$ with $a=b+n$. Also assume sums or products of positive naturals are positive. These are stated prior facts, not consequences of already-existing negative integers.

1. The equivalence relation records a formal difference

In the following shared proof, $N$ denotes $\mathbb N$ and $\sim_Z$ denotes $\sim$.

Definition

The equivalence relation for the integers

Work in the set $N^2$ of ordered pairs of natural numbers.

Define a relation $\sim_Z$ by

$$(a,b) \sim_Z (c,d) \quad \Longleftrightarrow \quad a+d=b+c.$$

The integers are defined to be the equivalence classes of this relation.

Theorem

The relation really is an equivalence relation

The relation $\sim_Z$ on $N^2$ is reflexive, symmetric, and transitive, so it is an equivalence relation.

Reflexivity, symmetry, and transitivity ensure that these classes form a quotient set; each step uses only natural-number arithmetic.

Proof: $\sim_Z$ is an equivalence relation

For reflexivity, every pair satisfies

$$a+b=b+a,$$

so $(a,b)\sim_Z(a,b)$.

For symmetry, if $(a,b)\sim_Z(c,d)$, then $a+d=b+c$. The same equality can be read backwards as $c+b=d+a$, so $(c,d)\sim_Z(a,b)$.

For transitivity, suppose

$$(a,b)\sim_Z(c,d) \quad \text{and} \quad (c,d)\sim_Z(e,f).$$

Then

$$a+d=b+c, \qquad c+f=d+e.$$

Adding these equalities and cancelling $c+d$ on both sides gives

$$a+f=b+e,$$

which means $(a,b)\sim_Z(e,f)$.

Let $Z$ be the quotient set and write its classes $[a,b]$. The notation suggests the formal difference of $a$ and $b$, but the class, not an assumed integer difference, is the definition. Every diagonal shift preserves a class: $(a+k,b+k)\sim(a,b)$ by commutativity.

2. Addition and negation are well-defined

Define

$$[a,b]+[c,d]=[a+c,b+d],\qquad -[a,b]=[b,a].$$

Proof: integer addition is well-defined

If $(a,b)\sim_Z(a',b')$ and $(c,d)\sim_Z(c',d')$, then

$$a+b'=b+a', \qquad c+d'=d+c'.$$

Adding these equalities and rearranging in $N$ gives

$$(a+c)+(b'+d')=(b+d)+(a'+c').$$

This is exactly $(a+c,b+d)\sim_Z(a'+c',b'+d')$, so the addition formula descends from pairs to integer classes.

For negation the required equality $b+a'=a+b'$ is the original equality reversed, so it also respects classes.

The element $0_Z=[0,0]$ is an additive identity. The sum $[a,b]+[b,a]=[a+b,a+b]$ equals $[0,0]$, proving that the proposed negative really is an additive inverse. Addition is associative and commutative because its coordinates are. Therefore the construction has an abelian additive group. Once inverses are established, subtraction inside $Z$ can be defined as addition of the negative; it was not assumed in proving the equivalence relation.

3. Multiplication needs a less obvious representative check

Define

$$[a,b]\,[c,d]=[ac+bd,ad+bc].$$

Theorem

The multiplication formula must be well-defined

If $(a,b)\sim_Z(a',b')$ and $(c,d)\sim_Z(c',d')$, then

$$(ac+bd,ad+bc)\sim_Z(a'c'+b'd',a'd'+b'c').$$

Therefore the product class does not depend on the representatives chosen.

Proof: integer multiplication is well-defined

Assume the hypotheses

$$a+b'=b+a', \qquad c+d'=d+c'.$$

To prove that the two proposed product representatives are equivalent, the definition of $\sim_Z$ asks us to show

$$(ac+bd)+(a'd'+b'c')=(ad+bc)+(a'c'+b'd').$$

First change only the representative of the first factor. Multiplying $a+b'=b+a'$ by $c$ and by $d$ gives

$$ac+b'c=bc+a'c, \qquad ad+b'd=bd+a'd.$$

Therefore, using only associativity and commutativity of addition in $N$,

$$\begin{aligned} (ac+bd)+(a'd+b'c) &=(ac+b'c)+(bd+a'd)\\ &=(bc+a'c)+(ad+b'd)\\ &=(ad+bc)+(a'c+b'd). \end{aligned}$$

This proves equivalence with the product formed from $(a',b')$ and $(c,d)$. Now keep $(a',b')$ fixed and apply $c+d'=d+c'$, multiplying by $a'$ and $b'$; the same natural-number rearrangement proves equivalence with the product formed from $(c',d')$. Transitivity gives the claimed output equivalence. No subtraction or integer multiplication is used in this proof.

The product is commutative by natural commutativity. Its identity is $1_Z=[1,0]$. For associativity, with a third pair $(e,f)$, the first coordinate on either association expands to $ace+bde+adf+bcf$, and the second to $acf+ade+bce+bdf$, up to reordering. Thus the representative pairs agree literally. Distributivity likewise follows by expanding each coordinate: for example the first coordinate of $[a,b]([c,d]+[e,f])$ is $ac+ae+bd+bf$, matching the sum of the two first product coordinates. These checks prove the integer ring laws rather than merely assuming formal negative-number algebra to obtain them.

4. Natural numbers embed and every class has a normal form

Define $j:\mathbb N\to Z$ by $j(n)=[n,0]$. If $j(n)=j(m)$, the equivalence relation gives $n=m$, so it is injective. The coordinate formulas show $j(n+m)=j(n)+j(m)$ and $j(nm)=j(n)j(m)$; zero and one are preserved.

For a representative $(a,b)$, total natural order gives either $a\ge b$ or $b\ge a$. In the first case, write $a=b+n$ to get $[a,b]=[n,0]$. In the second write $b=a+n$ to get $[a,b]=[0,n]=-j(n)$. These are the nonnegative and nonpositive normal forms.

Each form is unique: $[n,0]=[m,0]$ forces $n=m$, and similarly for the negative forms. If $[n,0]=[0,m]$, the defining relation gives $n+m=0$, forcing $n=m=0$. Thus a class is either zero, uniquely positive, or uniquely negative. This justifies the ordinary signed-integer interpretation after constructing it, without using that interpretation to establish the operations.

Worked checks. The pair $(7,4)$ belongs to $[3,0]$ because $7+0=4+3$. The product of $[0,2]$ and $[0,3]$ is $[6,0]$, so the product of the constructed negatives of two and three is the embedded positive six. This familiar sign rule is now a consequence of the class formula.

5. Order can also be defined without representatives

Define $x\le_Z y$ when $y=x+j(k)$ for some $k\in\mathbb N$. This refers to already well-defined elements and operations in $Z$. It is reflexive using $k=0$ and transitive by adding the nonnegative increments. For antisymmetry, if $y=x+j(k)$ and $x=y+j(\ell)$, group cancellation gives $j(k+\ell)=0_Z$. Injectivity of $j$ gives $k+\ell=0$, so both increments are zero and $x=y$.

To prove totality, the already constructed difference $y-x$ has one of the two normal forms $j(n)$ or $-j(n)$. In the first case $x\le_Z y$, and in the second $y\le_Z x$. Adding the same element preserves order by the defining equation. Products of nonnegative embedded elements remain nonnegative by preservation of natural multiplication. The order therefore recovers the expected signed arithmetic consistently with the construction.

6. Why this construction is a template

There are four logically separate stages: define representatives; prove an equivalence relation; check that proposed operations respect it and produce valid representatives; prove the structural laws and the normal form. Only then interpret the classes as familiar objects. Rational construction repeats the pattern with integer numerator/nonzero-denominator pairs, but its equivalence is cross multiplication rather than the additive relation used here. Copying one relation or operation formula into the other construction without a fresh proof is not justified.

The construction also explains uniqueness of additive inverses without coordinate calculations. In any associative system with commutative addition and identity, if $x+y=0=x+z$, then $y=y+0=y+(x+z)=(y+x)+z=0+z=z$. The inverse is a uniquely determined object even though its representing pair is not unique.

Construction and verification exercises

1. Noncircular proof. Prove transitivity of $\sim$ using only natural-number equalities and cancellation. Identify the line in a proposed proof “a-b=c-d=e-f” that would be circular at this construction stage.
2. Generating shifts. If $(a,b)\sim(c,d)$, prove they admit a common diagonal extension, obtained by adding the same natural number to both coordinates of each pair. Deduce that a function on pairs invariant under every diagonal shift is constant on equivalence classes.
3. Target matters. Is $[a,b]\mapsto a+b$ a natural-valued function? Is $[a,b]\mapsto[a+b]_2$ a function into classes modulo two? Prove both answers.
4. Operation check. Compute $[2,5][4,1]$ directly from the representative formula, reduce it to a signed normal form, and verify the result using a different equivalent representative for the first factor.
5. Uniqueness. Prove additive inverse uniqueness abstractly from the group laws established here. Explain why different representative pairs of that inverse do not violate uniqueness.
6. Challenge: extension of arithmetic. Let $R$ be a commutative ring with identity, and let $h:\mathbb N\to R$ preserve addition, multiplication, zero and one. Prove that $H([a,b])=h(a)-h(b)$ is the unique addition/multiplication/identity-preserving map $Z\to R$ extending $h$. Subtraction is allowed in $R$ because its additive group is part of the declared hypothesis.

Full solutions

1. From $a+d=b+c$ and $c+f=d+e$, adding gives $a+d+c+f=b+c+d+e$. Reorder and cancel the common natural summand $c+d$ to obtain $a+f=b+e$. Writing natural pairs as already-existing integer differences before constructing $Z$ assumes both subtraction and its equality laws. It may be a later interpretation, but cannot serve as the foundational proof here.
2. Shift $(a,b)$ by $d$, getting $(a+d,b+d)$, and shift $(c,d)$ by $b$, getting $(c+b,d+b)$. Their first coordinates agree by equivalence and their second by commutativity, so the pairs coincide. If a pair function is invariant under every such shift, its value at each original pair equals its value at the shared extension, proving class constancy. This is a genuine proof that these shifts generate the relation, not a general permission to test only special representative changes in arbitrary quotients.
3. The natural-valued proposal fails because $[0,0]=[1,1]$ but their coordinate sums are zero and two. For the modulo-two target, $a+d=b+c$ implies $a+b\equiv c+d\pmod2$: move terms in the already established congruence arithmetic, where a negative equals its positive modulo two. Equivalently add $b+d$ to both sides to get $a+b+2d=c+d+2b$. Thus the output classes agree, so the second proposal is well-defined.
4. The product pair is $[2\cdot4+5\cdot1,2\cdot1+5\cdot4]=[13,22]=[0,9]$. Replace the first factor by $[3,6]$, which is equivalent by a diagonal shift. The new product is $[18,27]=[0,9]$. Both match the signed interpretation of negative three times positive three, obtained only after normal forms have been justified.
5. If $x+y=0=x+z$, associativity and commutativity give $y=y+0=y+(x+z)=(y+x)+z=z$. Hence the inverse element is unique in $Z$. Pairs such as $(b,a)$ and $(b+k,a+k)$ name that same element; uniqueness of an element does not require uniqueness of its written representative.
6. If $a+d=b+c$, apply $h$ to get $h(a)+h(d)=h(b)+h(c)$; additive rearrangement in $R$ gives $h(a)-h(b)=h(c)-h(d)$. Thus $H$ is well-defined. Its sum formula follows by additivity of $h$, and multiplying differences in $R$ matches $h(ac+bd)-h(ad+bc)$ by distributivity and multiplicativity, so it preserves products. It takes $[1,0]$ to one and extends $h$ on $[n,0]$. Every class is $j(a)-j(b)$, so any additive map extending $h$ must take it to $h(a)-h(b)$. This proves uniqueness of the extension as well as existence.

Summary and terms

A representative construction defines new objects as classes of existing data. A normal form is a chosen canonical description proved to exist and be unique, unlike an arbitrary representative. An embedding is an injective structure-preserving map. An extension map preserves an earlier arithmetic interpretation on the embedded subset. Keep construction, representative checks, structural laws and interpretation in that logical order to avoid circular use of the number system being built.