11.5 Extending \(\mathbb N\) to \(\mathbb Z\)

Now we have a basic set of axioms and a model for \(\mathbb N\). Can we do something similar for \(\mathbb Z\), the set of all integers? Rather than start again but use \(\mathbb Z\) rather than \(\mathbb N\), we instead ‘extend’ the set of natural numbers \(\mathbb N\) to form the set of integers \(\mathbb Z\) and then ‘extend’ the basic properties, such as commutativity, associativity etc, to the integers. We do this by using equivalence relations. Let \(\mathbb N\) be the set of natural numbers (as outlined above) and consider the relation \(\sim\) defined on the set \(\mathbb N\times\mathbb N\), of ordered pairs of natural numbers, as follows \[ (a,b)\sim(c,d)\text{ if and only if }a+d=b+c. \] First we prove
Theorem 11.14

The relation \(\sim\) is an equivalence relation on \(\mathbb N\times\mathbb N\).

Proof. We prove that \(\sim\) is reflexive, symmetric and transitive.

  1. Let \((a,b)\in\mathbb N\times\mathbb N\). Then since \(a+b=b+a\) it follows that \((a,b)\sim(a,b)\) and so \(\sim\) is reflexive.
  2. Let \((a,b)\sim(c,d)\). Then \(a+d=b+c\) and so \(c+b=d+a\) and hence \((c,d)\sim(a,b)\) and \(\sim\) is symmetric.
  3. Let \((a,b)\sim(c,d)\) and \((c,d)\sim(e,f)\). Then \(a+d=b+c\) and \(c+f=d+e\). Consequently, \((a+f)+(c+d) = (a+d)+(c+f) = (b+c) + (d+e) = (b+e) + (c+d)\) and so \(a+f = b+e\) by Theorem 11.5 and so \((a,b)\sim(e,f)\) and \(\sim\) is transitive.

Now let \(a,b\in\mathbb N\) and consider the equivalence class \[ [a,b] = \{(c,d)\in\mathbb N\times\mathbb N|(c,d)\sim(a,b)\}, \] and let \[ Z=\{[a,b]|a,b\in\mathbb N\}. \] Define an operation of addition on \(Z\) by \[ [a,b]+[c,d] = [a+c,b+d]. \] Then this operation is well-defined since if \([a,b] = [a',b']\) and \([c,d]=[c',d']\) then \((a,b)\sim(a',b')\) and \((c,d)\sim(c',d')\). Hence, \(a+b'=b+a'\) and \(c+d'=d+c'\) and so \((a+c)+(b'+d') = (a+b')+(c+d') = (b+a')+(d+c') = (a'+c') + (b+d)\) or in other words \((a+c,b+d)\sim(a'+c',b'+d')\). Hence \([a,b]+[c,d] = [a',b']+[c',d']\).

In a similar way we can define and operation of subtraction on \(Z\) by \[ [a,b]-[c,d] = [a+d,b+c] \] and again we can justify that this is well-defined. Finally we can define a well-defined operation of multiplication on \(Z\) by \[ [a,b]\ast[c,d] = [a\ast c+b\ast d,a\ast d+b\ast c]. \] Notice that all these operations are defined only in terms of \(+\) and \(\ast\) on \(\mathbb N\).

Theorem 11.15

Let \(Z\) be the set of all equivalence classes \(Z=\{[a,b] |a,b\in\mathbb N\}\) and define a function \(f:Z\to\mathbb Z\) by \(f([a,b]) = a-b\). Then \(f\) is a bijection which preserves the operations of addition, subtraction and multiplication.

Proof. First we need to show that \(f\) is well-defined. By this we mean that if \([a,b] = [a',b']\) then we should be able to deduce that \(f([a,b]) = f([a',b'])\). This is easy, as \([a,b]=[a',b']\) is equivalent to \(a+b'=a'+b\). But then \(a-b=a'-b'\) as required.

Suppose that \(f([a,b]) = f([c,d])\). Then \(a-b=c-d\) and so \(a+d=b+c\) and so \((a,b)\sim(c,d)\). Hence \([a,b]=[c,d]\) and \(f\) is an injection. Let \(z\in\mathbb Z\). If \(z\ge0\) then \(f([z,0]) = z-0=z\) while if \(z<0\) then \(f([0,-z]) = 0-(-z) = z\) and so \(f\) is a surjection and hence a bijection.

You should check that the operations of addition, subtraction and multiplication are preserved (actually they were defined in such a way as to make sure they were preserved under this bijection).

So the set \(Z\) given above, acts as a model for \(\mathbb Z\) in a similar way to von Neumann’s model for \(\mathbb N\).

Notice that in this model, the subset of the natural numbers are represented by the classes \([n,0]\) while the negative integers, \(-n,n\in\mathbb N\), by the classes \([0,n]\). Zero is represented by \([0,0] = [1,1] = [2,2] = \ldots\).

As an exercise you may wish to prove that some of the previous theorems concerning \(\mathbb N\), extend in an appropriate way to this model for \(\mathbb Z\), e.g. \([a,b]+[c,d]=[c,d]+[a,b]\) for all \([a,b],[c,d]\in Z\).