The chain of number sets \(\mathbb{N}\subseteq\mathbb{N}_0\subseteq\mathbb{Z}\subseteq\mathbb{Q}\subseteq \mathbb{R}\subseteq\mathbb{C}\) starts with counting. This discussion builds the natural numbers from the Peano axioms, derives addition, multiplication, and order from them, and then constructs the integers as the smallest extension of the natural numbers in which subtraction always works, closing with why even the integers are not yet enough, which is exactly where the rational numbers pick up.
The Natural Numbers
The natural numbers \(\mathbb{N}=\{1,2,3,\dots\}\) can be pictured as evenly spaced points on a ray, starting at \(1\) and extending indefinitely. Giuseppe Peano gave this picture a small, precise list of rules, the Peano axioms:
\[ \begin{array}{rl} \text{P1.} & 1\in\mathbb{N}.\\ \text{P2.} & \text{Every } n\in\mathbb{N} \text{ has a successor } S(n)\in\mathbb{N}.\\ \text{P3.} & 1 \text{ is not the successor of any natural number.}\\ \text{P4.} & S(m)=S(n) \implies m=n.\\ \text{P5.} & \text{If } M\subseteq\mathbb{N} \text{ contains } 1 \text{ and contains } S(n) \text{ whenever it}\\ & \text{contains } n, \text{ then } M=\mathbb{N}. \end{array} \]
Axiom P5 is the induction axiom, and it is what makes \(\mathbb{N}\) exactly the numbers reachable from \(1\) by repeatedly taking successors, with nothing extra mixed in. Restated as a proof technique, this is the principle of mathematical induction: to prove a statement \(A(n)\) for every \(n\in\mathbb{N}\), it suffices to prove the base case \(A(1)\) and the inductive step \(A(n)\implies A(n+1)\); taking \(M=\{n\in\mathbb{N}:A(n)\text{ holds}\}\) in P5 then gives \(M=\mathbb{N}\). This is the same technique used, for instance, to prove Bernoulli's inequality.
Addition and Multiplication
Writing \(n+1\) for \(S(n)\), addition is defined by induction on the second argument:
\[ a+1 := S(a), \qquad a+S(n) := S(a+n). \]
Multiplication is then defined by induction in the same way, using addition:
\[ a\cdot 1 := a, \qquad a\cdot S(n) := a\cdot n + a. \]
Both operations can be shown, by induction on these two definitions, to be commutative and associative, and to satisfy the distributive law \(a(b+c)=ab+ac\); the arguments are routine but involve checking several cases, so they are omitted here.
Order and the Well-Ordering Principle
Order on \(\mathbb{N}\) is defined through addition:
\[ m<n :\iff \exists k\in\mathbb{N}: m+k=n. \]
This relation totally orders \(\mathbb{N}\): for any \(m,n\in\mathbb{N}\), exactly one of \(m<n\), \(m=n\), \(n<m\) holds. A stronger fact, equivalent to induction, is the well-ordering principle: every nonempty subset of \(\mathbb{N}\) has a least element.
To see this from induction, suppose \(S\subseteq\mathbb{N}\) is nonempty but has no least element, and let \(M\) be the set of \(n\in\mathbb{N}\) such that none of \(1,\dots,n\) lie in \(S\). Then \(1\in M\), since \(1\in S\) would trivially make \(1\) the least element of \(S\). If \(n\in M\), then \(n+1\notin S\) either: otherwise, since \(1,\dots,n\notin S\), the element \(n+1\) would itself be the least element of \(S\), contradicting the assumption. So \(n+1\in M\), and induction gives \(M=\mathbb{N}\), meaning no natural number lies in \(S\) at all — contradicting that \(S\) is nonempty. Hence \(S\) must have a least element.
From Natural Numbers to Integers
Closure Under Subtraction
The natural numbers are closed under addition and multiplication, but not under subtraction: there is no \(n\in\mathbb{N}\) with \(3+n=2\), so "\(2-3\)" has no natural-number answer. Extending \(\mathbb{N}\) to a number system where subtraction always works requires adding \(0\) and the negative numbers.
A Formal Construction
The same equivalence-class idea used to construct the rational numbers as fractions applies one level down, to construct the integers as differences. On pairs \((a,b)\in\mathbb{N}_0\times\mathbb{N}_0\), each pair standing for the not-yet-defined value \(a-b\), define
\[ (a,b)\sim(c,d) :\iff a+d=b+c, \]
using only addition on \(\mathbb{N}_0\), which is already available. This relation is reflexive, symmetric, and transitive, so it partitions \(\mathbb{N}_0\times\mathbb{N}_0\) into equivalence classes, and
\[ \mathbb{Z} := (\mathbb{N}_0\times\mathbb{N}_0)/{\sim}. \]
Every class contains exactly one pair with \(a=0\) or \(b=0\): reducing \((a,b)\) by cancelling the common part of \(a\) and \(b\) leaves either \((n,0)\) or \((0,n)\) for some \(n\in\mathbb{N}_0\), which are identified with \(n\) and \(-n\) respectively, recovering the familiar picture \(\mathbb{Z}=\{\dots,-2,-1,0,1,2,\dots\}\).
Addition, Negation, and Multiplication
Addition and negation are defined directly through representatives:
\[ [(a,b)]+[(c,d)] := [(a+c,\,b+d)], \qquad -[(a,b)] := [(b,a)]. \]
Well-definedness of addition follows immediately from the definition of \(\sim\): if \((a,b)\sim(a',b')\) and \((c,d)\sim(c',d')\), adding \(a+b'=b+a'\) and \(c+d'=d+c'\) gives \((a+c)+(b'+d')=(b+d)+(a'+c')\), which is exactly \((a+c,b+d)\sim(a'+c',b'+d')\). Every element has an additive inverse, since
\[ [(a,b)]+(-[(a,b)]) = [(a+b,\,b+a)] = [(0,0)] = 0, \]
using \(a+b=b+a\) and \((0,0)\sim(a+b,\,a+b)\). Multiplication mirrors the formal expansion \((a-b)(c-d)=(ac+bd)-(ad+bc)\):
\[ [(a,b)]\cdot[(c,d)] := [(ac+bd,\,ad+bc)]. \]
Subtraction is then defined as addition of the negation, \(a-b := a+(-b)\), exactly the operation \(\mathbb{Z}\) was built to provide. Addition and multiplication on \(\mathbb{Z}\) are commutative, associative, and distributive, and restrict to the original operations on \(\mathbb{N}\) under the embedding \(n\mapsto[(n,0)]\).
Order, and the Failure of Well-Ordering
Order extends from \(\mathbb{N}_0\) to \(\mathbb{Z}\) by
\[ [(a,b)] < [(c,d)] :\iff a+d < b+c, \]
again totally ordering \(\mathbb{Z}\) and agreeing with the order on \(\mathbb{N}\) under the embedding above. Unlike \(\mathbb{N}\), however, \(\mathbb{Z}\) has no well-ordering principle: the well-ordering proof for \(\mathbb{N}\) relied on there being a smallest natural number to start the induction from, but \(\mathbb{Z}\) itself, or any set of negative integers, has no least element at all — for any candidate \(k\), the integer \(k-1\) is smaller and still in the set.
Absolute Value and Distance
One more notion built directly on the order above is worth making precise before collecting \(\mathbb{Z}\) into a ring, since it is used implicitly whenever a proof needs to compare how far two integers are from each other or from \(0\). For \(a\in\mathbb{Z}\), the absolute value is
\[ |a| := \begin{cases} a, & a\geq 0,\\ -a, & a<0. \end{cases} \]
Because the order on \(\mathbb{Z}\) is total, exactly one branch applies to any given \(a\), so \(|a|\) is well-defined, and three facts follow immediately: \(|a|\geq0\) always, with equality exactly when \(a=0\); \(|-a|=|a|\), since negating \(a\) swaps which branch is active without changing the resulting value; and \(a\leq|a|\) as well as \(-a\leq|a|\), since whichever branch equals \(|a|\), the other expression is no larger than it.
Multiplicativity, \(|ab|=|a||b|\), reduces to a case split on signs: if \(a\) and \(b\) have the same sign then \(ab\geq0\) and \(|ab|=ab=|a||b|\); if their signs differ, say \(a<0\leq b\), then \(ab=-((-a)b)\leq0\), so \(|ab|=-(ab)=(-a)b=|a||b|\). The same case split gives the triangle inequality
\[ |a+b|\leq|a|+|b|, \]
most directly by adding \(a\leq|a|\) and \(b\leq|b|\) to get \(a+b\leq|a|+|b|\), and separately adding \(-a\leq|a|\) and \(-b\leq|b|\) to get \(-(a+b)\leq|a|+|b|\); since \(|a+b|\) is whichever of \(a+b\) or \(-(a+b)\) is nonnegative, both bounds together give \(|a+b|\leq|a|+|b|\).
Absolute value turns subtraction into a distance on \(\mathbb{Z}\), \(d(a,b):=|a-b|\), which is symmetric since \(a-b=-(b-a)\) gives \(|a-b|=|b-a|\), and satisfies \(d(a,c)\leq d(a,b)+d(b,c)\) by applying the triangle inequality to \(a-c=(a-b)+(b-c)\). This is exactly the notion of distance that carries over unchanged once \(\mathbb{Z}\) is embedded into \(\mathbb{Q}\) and \(\mathbb{R}\).
The Ring of Integers
Collecting the structure above, \((\mathbb{Z},+,\cdot)\) has an additive identity \(0\), an additive inverse \(-a\) for every \(a\), a multiplicative identity \(1=[(1,0)]\), and commutative, associative, distributive addition and multiplication: this combination is called a commutative ring. Unlike a field, however, multiplicative inverses are rare in \(\mathbb{Z}\): if \(ab=1\) for integers \(a,b\), then \(|a||b|=1\), which forces \(|a|=|b|=1\), so only \(1\) and \(-1\) have a multiplicative inverse in \(\mathbb{Z}\). This is precisely the closure gap that motivates constructing the rational numbers next, using the very same equivalence-class technique demonstrated above, now applied to pairs of integers instead of pairs of natural numbers.