6.4 Constructing number systems

In this section we shall see how it is possible to begin with $\mathbb{N}$, the simplest number system we know, and from it build larger and larger ones, at each stage in an attempt to overcome a particular shortcoming of the system so far created. We shall present three such constructions; equivalence relations will be very useful in the first two, but first we need a new concept.

We are now ready to begin our constructions. For the first one we start with $\mathbb{N}$; here we have addition and multiplication, but we cannot solve the equation $x+5=3$, for example. We define a relation $\sim$ on $\mathbb{N}^{2}$ by

$$(a,b)\sim(c,d)\quad\hbox{if}\quad a+d=b+c.$$

Thus for example $(3,1)\sim(4,2)$, as $3+2=1+4$. We show that this is an equivalence relation.

1. (i)

   For all $(a,b)\in\mathbb{N}^{2}$ we have $a+b=b+a$, i.e., $(a,b)\sim(a,b)$; so $\sim$ is reflexive.

2. (ii)

   If $(a,b)\sim(c,d)$ then $a+d=b+c$, so $c+b=d+a$, i.e., $(c,d)\sim(a,b);$ so $\sim$ is symmetric.

3. (iii)

   If $(a,b)\sim(c,d)$ and $(c,d)\sim(e,f)$ then $a+d=b+c$ and $c+f=d+e,$ so $a+d+f=b+c+f=b+d+e$; cancelling $d$ from both sides, we obtain $a+f=b+e$, i.e., $(a,b)\sim(e,f)$; so $\sim$ is transitive.

Thus we have an equivalence relation, and so equivalence classes; we may identify the class $\widehat{(a,b)}$ with the integer $a-b$ (which may be positive, negative or zero), because if $\widehat{(a,b)}=\widehat{(c,d)}$ then $a+d=b+c$, so $a-b=c-d$. There is a copy of $\mathbb{N}$ in the new system ($n$ corresponds to $\widehat{(n+1,1)}$), and we can define addition and multiplication in it by

$$\widehat{(a,b)}+\widehat{(c,d)}=\widehat{(a+c,b+d)}\qquad\hbox{and}\qquad\widehat{(a,b)}\cdot\widehat{(c,d)}=\widehat{(ac+bd,ad+bc)}$$

(the second of these arises from considering what $(a-b)(c-d)$ should be). Moreover, setting $-\widehat{(a,b)}=\widehat{(b,a)}$, we have negation and thus subtraction; so we can solve all equations $x+b=a$ because we can always form $x=a-b$. After checking that everything works, we have thus built the number system $\mathbb{Z}$ from $\mathbb{N}$.

The construction above no doubt seems very artificial; certainly we would never write integers as equivalence classes of ordered pairs. The point is that the construction can be done, using only what is available in $\mathbb{N}$.

Now let us turn to our second construction, which proceeds in a rather similar fashion to the first, but should end up seeming less unnatural than it. This time we start with the newly-created $\mathbb{Z}$ (but written by now in the usual manner); here we have addition, subtraction and multiplication, but we cannot solve an equation like $5x=3$. Define $\mathbb{Z}^{*}=\mathbb{Z}\setminus\{0\}$, and consider the direct product $\mathbb{Z}\times\mathbb{Z}^{*}$; its elements are pairs of integers $(a,b)$ with $b\neq 0$. We define a relation $\sim$ on $\mathbb{Z}\times\mathbb{Z}^{*}$ by

$$(a,b)\sim(c,d)\quad\hbox{if}\quad ad=bc.$$

Thus for example $(2,3)\sim(6,9)$ as $2\cdot 9=3\cdot 6$. We show this is an equivalence relation.

1. (i)

   For all $(a,b)\in\mathbb{Z}\times\mathbb{Z}^{*}$ we have $ab=ba$, i.e., $(a,b)\sim(a,b)$; so $\sim$ is reflexive.

2. (ii)

   If $(a,b)\sim(c,d)$ then $ad=bc$, so $cb=da$, i.e., $(c,d)\sim(a,b)$; so $\sim$ is symmetric.

3. (iii)

   If $(a,b)\sim(c,d)$ and $(c,d)\sim(e,f)$ then we have $ad=bc$ and $cf=de$, so $adf=bcf=bde$; cancelling $d$ from both sides (which we may because $d\neq 0$), we obtain $af=be$, i.e., $(a,b)\sim(e,f)$; so $\sim$ is transitive.

Thus we have an equivalence relation, and so equivalence classes; we may identify the class $\widehat{(a,b)}$ with the rational number $a/b$, because if $\widehat{(a,b)}=\widehat{(c,d)}$ then $ad=bc$ and $b,d\neq 0$, so $a/b=c/d$.

There is a copy of $\mathbb{Z}$ in the new system ($n$ corresponds to $\widehat{(n,1)}$), and we can define addition, multiplication and negation (and therefore subtraction) in it by

$$\widehat{(a,b)}+\widehat{(c,d)}=\widehat{(ad+bc,bd)},\qquad\widehat{(a,b)}\cdot\widehat{(c,d)}=\widehat{(ac,bd)}\qquad\hbox{and}\qquad-\widehat{(a,b)}=\widehat{(-a,b)}.$$

(The first of these arises from considering what $(a/b)+(c/d)$ should be.) Moreover, setting

$${\widehat{(a,b)}\,}^{-1}=\widehat{(b,a)}$$

if $a\neq 0$, we have inversion of non-zero elements and thus division (since $x/y=xy^{-1}$); so we can solve all equations $bx=a$ (as long as $b\neq 0$) because we can always form $x=a/b$. After checking that everything works, we have thus built the number system $\mathbb{Q}$ from $\mathbb{Z}$.

This time, although the notation is a little strange, things look reasonably familiar; after all, we do denote rational numbers by pairs of integers, the second of which is non-zero, except that instead of writing them horizontally, separated by a comma, we write them vertically, separated by a line. Moreover, we are quite used to the idea that a given rational number may be expressed in different ways, e.g.,

$$\frac{2}{3}=\frac{4}{6}=\frac{6}{9}=\dots,$$

corresponding to different choices of representative pair within the equivalence class $\widehat{(2,3)}$. Again, we observe that the construction has only used what is available in $\mathbb{Z}$.

Note: The remainder of this chapter is not examinable.

The next step would be to begin with $\mathbb{Q}$ (where we have addition, subtraction, multiplication and division, but cannot solve the equation $x^{2}=2$, for example) and build $\mathbb{R}$ from it. This is, however, significantly harder than the previous two constructions, reflecting the fact that the system $\mathbb{R}$ is considerably more complicated than its predecessors; it requires material which you will only meet later this year, namely convergence of infinite sequences. It does use an equivalence relation on an appropriate set, though.

Instead, our final construction will start from $\mathbb{R}$; here we have addition, subtraction, multiplication and division, and can solve many equations, but we cannot solve $x^{2}=-1$ since the square of any real number is non-negative.

We form the set $\mathbb{R}^{2}$; this time, for a change, we shall not require an equivalence relation. Nevertheless, we proceed much as before. There is a copy of $\mathbb{R}$ in the new system ($x$ corresponds to $(x,0)$), and we can define addition, negation (and therefore subtraction), multiplication and inversion of non-zero elements (and therefore division) in it by

$$\begin{array}[]{rlrl}(a,b)+(c,d)&\!\!\!=(a+c,b+d),&-(a,b)&\!\!\!=(-a,-b),\\ (a,b)\cdot(c,d)&\!\!\!=(ac-bd,ad+bc),&(a,b)^{-1}&\!\!\!=\biggl(\displaystyle{\frac{a}{a^{2}+b^{2}},-\frac{b}{a^{2}+b^{2}}}\biggr).\end{array}$$

(The third and fourth of these are not at all obvious, but note that $a^{2}+b^{2}>0$ for $(a,b)\neq(0,0)$, so the formula for inversion does at least make sense.) The formula for multiplication shows that

$$(0,1)^{2}=(0,1)\cdot(0,1)=(-1,0).$$

In other words, $x=(0,1)$ solves the equation $x^{2}=-1$. After checking that everything works, we have thus built a new number system from $\mathbb{R}$; we call it $\mathbb{C}$, the complex numbers.

In practice, we do not use exactly this notation for the elements of $\mathbb{C}$. Instead we write $\mathrm{i}$ for $(0,1)$, and recalling that we have identified the real numbers $a$ and $b$ with $(a,0)$ and $(b,0)$, respectively, we find

$\displaystyle a+b\,\mathrm{i}=(a,0)+(b,0)\cdot(0,1)=(a,0)+(b\cdot 0-0\cdot 1,b\cdot 1+0\cdot 0)=(a,0)+(0,b)=(a,b).$

The formulæ above then become

	$\displaystyle(a+b\,\mathrm{i})+(c+d\,\mathrm{i})$	$\displaystyle=(a+c)+(b+d)\,\mathrm{i},$	$\displaystyle-(a+b\,\mathrm{i})$	$\displaystyle=(-a)+(-b)\,\mathrm{i},$
	$\displaystyle(a+b\,\mathrm{i})\cdot(c+d\,\mathrm{i})$	$\displaystyle=(ac-bd)+(ad+bc)\,\mathrm{i},$	$\displaystyle(a+b\,\mathrm{i})^{-1}$	$\displaystyle=\frac{a}{a^{2}+b^{2}}-\frac{b}{a^{2}+b^{2}}\,\mathrm{i}.$

Note that

$$(a+b\,\mathrm{i})\biggl(\frac{a}{a^{2}+b^{2}}-\frac{b}{a^{2}+b^{2}}\,\mathrm{i}\biggr)=\frac{aa-b(-b)}{a^{2}+b^{2}}+\frac{a(-b)+ba}{a^{2}+b^{2}}\,\mathrm{i}=1,$$

so the formula for $(a+b\,\mathrm{i})^{-1}$ is correct.

The formula for the product of two complex numbers looks complicated. Fortunately, there is no need to memorize it, as there is an easier way to multiply complex numbers: first expand the brackets, then use the fact that $\mathrm{i}^{2}=-1$, and finally gather together like terms. For instance, we have

	$\displaystyle(2-3\mathrm{i})(-1+2\mathrm{i})$	$\displaystyle=2(-1)+2\cdot 2\mathrm{i}+(-3)(-1)\mathrm{i}+(-3)2\mathrm{i}^{2}$
		$\displaystyle=-2+4\mathrm{i}+3\mathrm{i}+(-6)(-1)=4+7\mathrm{i}.$

Likewise, there is a trick to avoid memorizing the formula for inversion: multiply the fraction $1/(a+b\,\mathrm{i})$ by $a-b\,\mathrm{i}$ in both numerator and denominator, and expand the brackets; the fact that

$$(a+b\,\mathrm{i})(a-b\,\mathrm{i})=a^{2}-(b\,\mathrm{i})^{2}=a^{2}+b^{2}$$

ensures that the denominator is real and non-zero (because $a$ and $b$ are not both zero). The following example illustrates this procedure:

$$\frac{1}{2-3\mathrm{i}}=\frac{2+3\mathrm{i}}{(2-3\mathrm{i})(2+3\mathrm{i})}={\frac{2+3\mathrm{i}}{2^{2}-(3\mathrm{i})^{2}}}={\frac{2+3\mathrm{i}}{4+9}}={\frac{2}{13}+\frac{3}{13}\,\mathrm{i}.}$$

There is of course much more to be said about $\mathbb{C}$, but that will be covered elsewhere. We conclude this section by mentioning that, perhaps surprisingly, we find there is in fact no need to build any larger system to enable us to solve equations; the process stops here.