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Construction of Numbers
We have used the notion of natural numbers and real numbers informally in previous chapters. In this chapter, we shall construct the number system using the axioms of set theory. The 10th axiom, which is the infinity axiom, will be formally introduced, completeing our axiom list.
It is also possible to first construct the real number system by a set of axioms for the real numbers and work from them. Then the set of natural numbers is a subset of the set of real numbers. We will also use this axiomatic approach as well as the constructive approach.
Contents
Contents
1. * Natual Numbers
1.1. Definition
1.2. Induction and Recursion
1.2.1. Peano's Postulates
1.2.2. Cardinal number of a finite set
1.3. Addition and Multiplication
2. Integers and Rational Numbers
2.1. Integers
2.2. Rational Numbers
3. Continued on Page Two
4. Real Numbers
5. Complex Numbers
1. * Natual Numbers
Informally, the set \mathbb N of natural numbers is the set
\[\{0,1,2,\dots\}.\]
In set theory, the notation \omega is used more frequently. We will see why soon enough.
In the context of modern mathematics, it seems more common to include 0 in the set \mathbb N, while in elementary mathematics it is more likely to be excluded. Therefore we go with the following option: use \mathbb N to include zero and \mathbb N^\ast to exclude it.
There are a variey of ways to construct specific sets that serve perfectly well as numbers, such as Zermelo's construction: \varnothing, \{\varnothing\}, \{\{\varnothing\}\}\dots A more standard one is due to von Neumann, which is to be introduced in the following.
If we want to define a natural number as a specific set, then the ordinal number of a finite set is a good candidate. (It is suggested to review Chapter 4 here.) If A is a finite set and isomorphic to an ordinal number: A\cong \alpha, then we expect \alpha to be a natural number.
For instance, suppose A=\{a,b,c\} and a<b<c. Then A is a well-ordered set. Let F:A\to\alpha be the isomorphism from A onto the ordinal number \alpha. We can calculate what \alpha=\{F(a),F(b),F(c)\} is like.
\begin{align*}
F(a)&=\{F(x)\mid x<a\}=\varnothing,\\
F(b)&=\{F(x)\mid x<b\}=\{\varnothing\},\\
F(c)&=\{F(x)\mid x<c\}=\{\varnothing,\{\varnothing\}\}.
\end{align*}
This suggests that
\[3:=\{\varnothing,\{\varnothing\},\{\{\varnothing\}\}\},\]
if we want 3 to be the set \alpha.
1.1. Definition
We define 0 to be the least ordinal number, \varnothing:
\[0:=\varnothing.\]
For a number n, we define its successor n^+ by
\[ n^+=n\cup\{n\}.\]
By Theorem thm443, n^+ is an ordinal number provided n is an ordinal number. Actually this is the least ordinal number larger than n. Clearly n\in n^+, so n^+ is larger than n. Now suppose n\in\alpha for an ordinal number \alpha. By transitivity of \alpha, n\subseteq\alpha, indicating n^+\subseteq\alpha. Since \alpha\notin\alpha, we have \alpha\notin n^+, so n^+=\alpha or n^+\in\alpha. This shows that n^+ is the least ordinal number larger than n.
We define 1,2,\dots to be the successors of 0,1,\dots:
\[1:=0^+=\varnothing^+,\quad 2:=0^{++},\quad 3:=0^{+++},\dots\]
Specifically,
\begin{align*}
0&=\varnothing,\\
1&=\{0\}=\{\varnothing\},\\
2&=\{0,1\}=\{\varnothing,\{\varnothing\}\}\\
3&=\{0,1,2\}=\{\varnothing,\{\varnothing\},\{\{\varnothing\}\}\},\\
&\dots
\end{align*}
They are the least ordinal numbers and
\[0\in 1\in2\in\dots,\quad 0\subseteq1\subseteq2\subseteq\dots\]
And then the set \mathbb N should be the set of them all.
The problem is, however, we cannot prove the existence of this set with the axioms so far. (In Chapter 3 we simply assumed it exists.) Currently we can only form finite sets such as using pairing axiom. To justify the existence of \mathbb N, we adopt the infinity axiom.
We call a set A to be inductive if \varnothing\in A and for any a\in A we have a^+\in A.
- Infinity Axiom: There exists an inductive set, i.e.,
\[(\exists A)\, (\varnothing\in A\wedge(\forall a\in A)\, a^+\in A).\]
By infinity axiom, there exists an inductive set S. By subset axiom, there is a set \mathbb N=\{x\in S\mid P(x)\}, where P(x) means x is a natural number. Then n\in \mathbb N if and only if n\in S and n belongs to every inductive set. Since S itself is an inductive set, n\in \mathbb N if and only if n belongs to every inductive set, so \mathbb N is a set consisting of all natural numbers.
It is easy to check \mathbb N is inductive and a subset of any inductive set. It is reasonable to say \mathbb N is the smallest inductive set.
Hence, by definition, 0\in\mathbb N. It follows that 1=0^+\in\mathbb N, and so forth. Then we can form a subset \{0,1,2,\dots\}\subseteq \mathbb N consisting of all elements finitely reachable from 0. This is a set containing zero and the successors of each element, so it is inductive and \mathbb N\subseteq\{0,1,2,\dots\}. We have
\[\mathbb N=\{0,1,2,\dots\}.\]
1.2. Induction and Recursion This method is the induction principle for \mathbb N. We state it formally:
Induction principle for \mathbb N: Any inductive subset of \mathbb N coincides with \mathbb N.
It is not diffifult to see that the transfinite induction is a gerneralization of this induction principle (and the strong induction principle to be stated soon). An easy application: We can use it to prove every natural number except 0 is the successor of another natural number.
The set \mathbb N is a set of ordinal numbers. Suppose \alpha\in\beta\in\mathbb N. Then \beta is a natural number. It is easy to show by induction that a natural number is a set all of whose are natural numbers. We then have \alpha\in\mathbb N, and therefore \mathbb N is a transitive set.
By Theorem thm443, the set \mathbb N itself is an ordinal number, commonly denoted by a lowercase Greek \omega. (We might use both notations interchangeably.) We have the following theorem and recursion theorem on \omega, which is a corollary of transfinite recusrion.
Strong Induction Principle for \omega: Let A\subseteq \omega, and assume that for any n\in\omega, "every number less than n is in A" implies n\in A. Then A=\omega.
Recursion Theorem on \omega: Let g:A\to A be a mapping. There exists a unique mapping F:\omega\to A such that F(0)=a for a given a\in A and F(n^+)=g(F(n)) for all n\in\omega.
Remark: The recursion theorem is not so obvious actually. There is a fallacious proof of it: First F(0)=a, so F is defined on 0. And whenever F is defined on n, then F(n^+)=g(F(n)) is defined. Therefore on n^+, F is defined as well. By induction F is defined on all n\in\omega.
There are serveral problems with this short "proof". First, it uses induction to prove the domain of F is just \omega, but we cannot refer to the domain of an undefined mapping. Second, it fails to utilize the structure of \omega. For example, if 3^+=0 or 3^+=1, then such a mapping F does not necessarily exist. Third, if in the view of mapping extension, we cannot define a mapping by an infinite sequence of extensions.
1.2.1. Peano's Postulates In late 19\textsuperscript{th} century, Peano proposed an axiomatic approach to the natural numbers. The axioms are called Peano's postulates, although he attributed the formulation to Dedekind. We will show Peano's postulates become theorems under the von Neumann construction, and also show that the number system satisfying Peano's postulates is unique.
We define a Peano system as a set N with a function S such that:
- There is a "zero" element e\in N;
- For any n\in N, S(n)\in N;
- For any n\in N, S(n)\neq e;
- The function S is injective.
- Inductive property holds: For a formula \varphi, \varphi(e)\wedge[(\forall n\in N)\, \varphi(n)\Rightarrow \varphi(S(n))] implies (\forall n\in N)\, \varphi(n).
The last of the conditions is the Peano induction postulate. Alternatively, any subset A of N that contains e and is closed under S equals N itself.
This is easy to prove. Only (4) needs verification: We shall show \alpha=\beta if \alpha^+=\beta^+ for \alpha,\beta\in\omega. If \alpha\neq\beta, then \alpha\in\beta or \beta\in\alpha. By symmetry assume \beta\in\alpha. Then \beta^+=\alpha\in\alpha^+ or \beta^+\in\alpha\in\alpha^+, indicating \beta^+\in\alpha^+. Hence \beta\notin\alpha. We conclude \alpha=\beta.□
This theorem shows \mathbb N is the unique system satisfying Peano's postulates, up to isomorphism.
The recursion theorem provides us with a mapping F:\mathbb N\to N such that F(n^+)=S(F(n)) and F(0)=e. This F is clearly surjective: e is in the range of F and for any F(n) in the range, S(F(n)) is also in the range, so by the inductive property N is all in the range of F.
We show F is injective by induction. Suppose F(m)=F(0)=e for some m\in\mathbb N. If m\neq0, then since we have known a nonzero natural number is the successor of another natural number k, e=F(m)=F(k^+)=S(F(k)), contradicting the postulate that e is not in the range of S. Therefore m=0, so F(m)=F(0) implies m=0.
We now prove F(m)=F(n^+) implies m=n^+, with the assumption F(m)=F(n)\Rightarrow m=n for any m\in\mathbb N. Then F will be injective by induction. Suppose F(m)=F(n^+)=S(F(n)). Then m\neq0, otherwise e=S(F(n)), contradicting the postulates. Therefore there is a ntural number k\in\mathbb N such that m=k^+, so S(F(n))=F(m)=F(k^+)=S(F(k)). By injectivity of S, F(n)=F(k). By assumption, n=k, so m=k^+=n^+. The injectivity of F follows.□
1.2.2. Cardinal number of a finite set We have defined a natural number as an ordinal number. In Chapter 4, we also defined the cardinal numer of a general set as an ordinal number, while the cardinal number of a finite set is a natural number n if the set is equinumerous to \mathbb N _ {<n}=\{0,1,\dots,n-1\}. Now it can be seen that for a finite set the definitions agree with each other.
Let A be a finite set equinumerous to \mathbb N _ {<n}. Under the von Neumann construction, \mathbb N _ {<n} is simply the set n, so A\sim n and A is not equinumerous to any smaller ordinal, since the smaller
ordinals are just the natural numbers in n. Hence |A|=n.
1.3. Addition and Multiplication We should define some basic operations such as addition +:\mathbb N\times \mathbb N\to\mathbb N, multiplication \times:\mathbb N\times \mathbb N\to\mathbb N and exponentiation on \mathbb N. There are two definitions that are equivalent. The first is natural and uses the recursion theorem, and the second is general and uses the cardinal numbers to define.
Suppose m\in\mathbb N. We want to define a function such that the image of n is the number m+n. By recursion theorem there exists a unique function A _ m:\mathbb N\to\mathbb N such that A _ m(0)=m and A _ m(n^+)=A _ m(n^+) for n\in\mathbb N. Then we define addition by m+n=A _ m(n). With this definition, we have
\begin{align*}
m+0&=m,\\
m+n^+&=(m+n)^+.
\end{align*}
Similarly we define multiplication and exponentiation such that
\[
\begin{aligned}
m\cdot0&=0,\\
m\cdot n^+&=m\cdot n+m,
\end{aligned}\qquad
\begin{aligned}
m^0&=1,\\
m^{n^+}&= m^{n}\cdot m.
\end{aligned}
\]
For example, we can compute 2+1=2+0^+=(2+0)^+=2^+=3.
The usual properties are satisfied, which can be proved by induction.
- Associative law: (m+n)+k=m+(n+k) and (m\cdot n)\cdot k=m\cdot (n\cdot k);
- Commutative law: m+n=n+m and m\cdot n=n\cdot m;
- Distributive law: m\cdot (n+k)=m\cdot n+m\cdot k;
- For exponentiation:
\[m^{n+k}=m^n\cdot m^k,\quad (m\cdot n)^k=m^k\cdot n^k,\quad (m^n)^k=m^{n\cdot k}.\]
Only (m+n)+k=m+(n+k) is proved for illustration and the others are similar. We shall show the set A=\{k\in\mathbb N\mid (m+n)+k=m+(n+k)\} is inductive for given numbers m,n. First 0\in A. Suppose k\in A. Then
\begin{align*}
(m+n)+k^+&=((m+n)+k)^+=(m+(n+k))^+\\
&=m+(n+k)^+=m+(n+k^+).
\end{align*}
Hence k^+\in A.
We now turn to the another definition of these operations. The idea is actually simple.
- If A\cap B=\varnothing, then \alpha+\beta:=|A\cup B|.
- \alpha\cdot\beta:=|A\times B|.
- \beta^\alpha=|B^A|.
We have actually defined the exponentiation in Definition def37 and here we simply repeat it again. The addition and multiplication are also well-defined (but we will not verify it here). For example, if A\cap B\neq\varnothing, we can replace A,B with A\times\{0\} and B\times\{1\} respectively. The cardinal numbers are unchanged but the new sets are disjoint. If |A _ 1|=|A _ 2|=\alpha and |B _ 1|=|B _ 2|=\beta, then A _ 1\cup B _ 1\sim A _ 2\cup B _ 2 so they have the same cardinal number.
The properties of arithmetic are derived from those of set operations, such as A\cup B=B\cup A. The properties of exponentiation of sets are listed in the following theorem.
- A^{B\cup C}\sim A^B\times A^C;
- (A\times B)^C\sim A^C\times B^C;
- (A^B)^C\sim A^{B\times C}.
The proof of this theorem is omitted, as well as the proof of the following theorem, (which can be proved by induction).
The second definition is provided because it still works for general cardinal numbers, not just for cardinal numbers in \omega. For infinite cardinal numbers, some new unusual properties appear. There is a useful absorption law in cardinal arithmetic, (which will not be proved here): Let \alpha,\beta be two cardinal numbers, the larger of which is infinite and the smaller of which is nonzero. Then
\[
\alpha+\beta=\alpha\cdot\beta=\max(\alpha,\beta).
\]
2. Integers and Rational Numbers Our next step is naturally to define the set \mathbb Z of integers, and the set \mathbb Q of rational numbers.
2.1. Integers A negative number can be obtained by using two natural numbers and a "substraction" operation. We know, say, 0-1=1-2, but (0,1)\neq(1,2). A simple definition using only an ordered pair does not work, but we can define an equivalence relation such that (0,1)\sim(1,2), and then -1 is the equivalence class of it. Since we have no substraction yet, the equivalence relation should be expressed in terms of addition.
Then the set \mathbb Z of integers is the quotient set (\mathbb N\times \mathbb N)\mathbin/\sim.
The relation \sim can be easily checked to be an equivalence relation. The definition is well-defined. A quick example, the integer -2 is just the equivalence class [(0,2)]=\{(0,2),(1,3),\dots\}.
We then can define the addition operation on \mathbb Z: [(m,n)]+[(m',n')]=[(m+m',n+n')]. Of course, we should check that the result is independent of the choice of the representatives (m,n) and (m',n'). Suppose (m,n)\sim (m',n') and (p,q)\sim(p',q'). Then it can be checked that (m+p,n+q)\sim(m'+p',n'+q'). (Recall Theorem thm224.)
The familiar properties of addition follow easily from the corresponding properties of addition of natural numbers.
Inverses of the group provide us with a subtraction operation: b-a=b+(-a).
We can also define the multiplication operation on \mathbb Z by
\[[(m,n)]\cdot[(m',n')]=[(mp+nq,mq+np)].\]
It can be checked that the result is independent of the choice of the representatives.
The order on \mathbb Z can be defined by: [(m,n)]<[(m',n')] if and only if m+q<p+n. It can be checked this is well-defined.
An integer greater than [(0,0)] is called a positive integer. If it is less than [(0,0)], it is called a negative integer.
The following theorem shows that the inclusion mapping is an isomorphism from \mathbb N to a subset of \mathbb Z, which preserves addition, multiplication, and order. Its proof is straightforward and omitted.
Thus, it is reasonable to denote the integers [(n,0)] simply by n.
2.2. Rational Numbers For natural numbers, there is not necessarily a solution for n+x=0, i.e., n may not have an additive inverse. But after we embed \mathbb N into \mathbb Z, we obtain an additive group and can talk about substraction. Now suppose n\in\mathbb Z, it may not have a multiplicative inverse: n\cdot x=1. With the same idea, we extend the integers to the rationals. We are, for the most part, retracing our previous steps.
Since a/b=c/d if and only if ad=bc, we choose to define \sim as follows. Let \mathbb Z^\ast=\mathbb Z\setminus\{0\} be the set of nonzero integers.
Then the set \mathbb Q of rational numbers is the quotient set (\mathbb Z\times\mathbb Z^\ast)\mathbin/\sim.
Here only the transitivity is verified. Suppose [(a,b)]\sim[(c,d)] and [(c,d)]\sim[(e,f)]. Then ad=bc and cf=de, so adf=bcf=bde. By assumption d is nonzero, so we can apply cancellation law to get af=be, implying [(a,b)]\sim[(e,f)].
Next we define the operations on \mathbb Q. We hope the sum of a/b and c/d is (ad+bc)/(bd). Therefore we define [(a,b)]+[(c,d)]=[(ad+bc,bd)]. Since b\neq0 and d\neq0, bd\neq0, so [(ad+bc,bd)]\in\mathbb Q. Moreover, if (a,b)\sim(a',b') and (c,d)\sim(c',d'), then (ad+bc,bd)\sim(a'd'+b'c',b'd'). (The detail of the verification is omitted.) The definition of addition is well-defined.
It can be proved that (\mathbb Q,+) is also an Abelian group. For a rational r, the unique inverse of addition is denoted by -r. It can be proved the inverse of [(a,b)] is [(-a,b)].
For multiplication, we define [(a,b)]\cdot[(c,d)]=[(ac,bd)]. It can be verified that if (a,b)\sim(a',b') and (c,d)\sim(c',d'), then (ac,bd)\sim(a'c',b'd'). (The detail of the verification is omitted.) The definition of addition is well-defined.
The nonzero rationals with multiplication form an Abelian group. For a rational number r, the unique inverse of multiplication is denoted by r^{-1}. It can be proved the inverse of [(a,b)] is [(b,a)]. Inverses of the multiplicative group provide us with a division operation: b/a=b\cdot(a^{-1}).
The order on \mathbb Q can be defined by [(a,b)]<[(c,d)] (b,d>0) if and only if ad<cb. Since [(a,b)]=[(-a,-b)], every rational number can always be represented by a fraction with a positive denominator. It can be proved this order relation is well-defined.
Thus \mathbb Q is an ordered field.
There is also an isomorphism from \mathbb Z to a subset of \mathbb Q, which preserves addition, multiplication and order. Its proof is straightforward and omitted.
Thus, it is reasonable to denote the rationals [(n,1)] simply by n.
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