Constructing Number Sets 2
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This is continuation of the last post, where we worked on building some important number sets and endowing them with properties.
Rational Numbers
Integers introduced us to subtraction, which enabled us to create an infinite ring structure. The rationals will expand the integers to allow for division, and create a field structure. There are a lot of similarities between the constructions of the rationals and the integers.
Consider the following equivalence relation on \(\bbZ\times\bbZ\setminus\{0\}\): \((a,b)\sim(c,d)\) iff \(ad=bc\). This is an equivalence relation because
- Reflexivity: \(ab=ba\) implies \((a,b)\sim(a,b)\)
- Symmetry: \(ad=bc\) implies \(cb=da\)
- Transitivity: \(ad=bc\) and \(cf=de\) are given. So, we have \(fad=fbc\), and thus \(fad=bde\). Since \(d\neq0\), by multiplication being almost injective, we have \(af=be\).
There's a key difference between how we define integers and rationals. For the integers, we defined the equivalence relation on \(\bbN\times\bbN\) since addition is injective. For the rationals, we defined the relation on \(\bbZ\times\bbZ\setminus\{0\}\) since multiplication is injective on non-zero values.
The intuition on how to interpret \([(a,b)]\in\bbQ\) is very similar to that of the integers. In the integers, it was interpreted as \(a-b\) (although at that point, we hadn't defined subtraction yet). Here, we'll interpret it as \(a/b\) (again, this is not defined right now, but this is solely for your intuition about the coming definitions).
For any \(a,b,c\in\bbZ\), if \(c\neq0\) and \([(a,c)]=[(b,c)]\), then \(a=b\).
This in particular implies that the map \(\bbZ\to\bbQ\) defined by \(z\mapsto[(z,1)]\) is injective. We call this the canonical embedding of the integers into the rationals. Thus, we can call \([(z,1)]\) just \(z\), and with this interpretation, \(\bbZ\subset\bbQ\). As a lot of these proofs are very similar to those of the integers, I won't be providing too much commentary on them.
Let \(+\) be a binary operation on \(\bbQ\) defined by \([(a,b)]+[(c,d)]=[(ad+bc,bd)]\). This operation is well-defined on \(\bbQ\)
\([(a,b)],[(c,d)],[(e,f)]\in\bbQ\),$$[(a,b)]+([(c,d)]+[(e,f)])=([(a,b)]+[(c,d)])+[(e,f)]$$
For all \([(a,b)]\), \([(a,b)]=0\) iff \(a=0\)
For \([(a,b)]\in\bbQ\), we define \(-[(a,b)]\) to be \([(-a,b)]\). In particular, \([(a,b)]+(-[(a,b)])=0\). We define subtraction as adding the negation, i.e., \([(a,b)]-[(c,d)]=[(a,b)]+(-[(c,d)])\)
\(0\) is the only identity element
We call a rational number \([(a,b)]\) positive iff \(a\cdot b>0\). Likewise, we call a rational negative if \(a\cdot b\lt0\). It should be clear from the previous results that a rational number is exactly one of positive, negative, or zero. We say \([(a,b)]\geq[(c,d)]\) iff \([(a,b)]-[(c,d)]\) is not negative. This is a valid order definition.
- Well-defined: Given \([(a,b)],[(c,d)],[(e,f)],[(g,h)]\in\bbQ\) such that \([(a,b)]=[(c,d)]\) and \([(e,f)]=[(g,h)]\), if \([(a,b)]\geq[(e,f)]\), then $$(af-be)bf\geq0$$Since \(ad=bc\) and \(eh=gf\), and as in the transitivity proof, we can assume \(b,d,f,h\in\bbN\setminus\{0\}\), we have $$afdh-bedh\geq0$$$$(bcfh-bdgf)dh\geq0$$$(ch-dg)dh$$$
- Reflexivity: Since \([(a,b)]-[(a,b)]=0\) and \(0\) is not negative, \([(a,b)]\leq[(a,b)]\)
- Anti-Symmetry: For \([(a,b)],[(c,d)]\in\bbQ\), if \((ad-bc)bd>0\), then \([(c,d)]-[(a,b)]=[(-(ad-bc),bd)]\), and thus \(-(ad+bc)bd\lt0\). So, \([(c,d)]<[(a,b)]\). If \((ad+bc)bd\lt0\), then \([(a,b)]<[(c,d)]\) by the same logic. So if neither are true, then \((ad-bc)bd=0\), and since \(bd\neq0\), \(ad-bc=0\), so \([(a,b)]=[(c,d)]\)
- Transitivity: If \([(a,b)],[(c,d)],[(e,f)]\in\bbQ\), we can assume without loss of generality that \(b,d,f\in\bbN\setminus\{0\}\) since for all \([(x,y)]\in\bbQ\), \([(-x,-y)]=[(x,y)]\). Thus, if \((ad-bc)bd\geq0\) and \((cf-de)df\geq0\), we have \(ad-bc\geq0\) and \(cf-de\geq0\). So, \(fad-fbc+bcf-bde\geq0\), and thus \(fad-bde\geq0\). Since \(d>0\), \(af-be\geq0\).
- Trichotomy: Given by the fact that all rationals are positive, negative, or zero, and the negation of a negative number is positive.
For integers \(a,b\in\bbZ\), \([(a,1)]\leq[(b,1)]\) iff \(a\leq b\). In particular, the naturals are the non-negative integers
If \([(a,b)]\leq[(c,d)]\) and \([(e,f)]\leq[(g,h)]\), then $$[(a,b)]+[(e,f)]\leq[(c,d)]+[(g,h)]$$
$$[(c,d)]+[(g,h)]-([(a,b)]+[(e,f)])=([(c,d)]-[(a,b)])+([(g,h)]-[(e,f)])$$The right side is a sum of two non-negative rationals, so all we need to show is that for two non-negative rationals \([(v,w)],[(x,y)]\), \([(vy+xw,wy)]\) is non-negative. We can assume without loss of generality that \(w,y\geq0\), which implies \(v,x\geq0\), so the claim is true.
Let \(\cdot\) be the binary operation on \(\bbQ\) which is defined by \([(a,b)]\cdot[(c,d)]=[(ac,bd)]\). It is well-defined on \(\bbQ\).
For all \([(a,b)],[(c,d)]\), \([(a,b)]\cdot[(c,d)]=[(c,d)]\cdot[(a,b)]\)
For all \([(a,b)],[(c,d)],[(e,f)]\),$$[(a,b)]\cdot([(c,d)]\cdot[(e,f)])=([(a,b)]\cdot[(c,d)])\cdot[(e,f)]$$
For all \([(a,b)],[(c,d)],[(e,f)]\),$$[(a,b)]\cdot([(c,d)]+[(e,f)])=[(a,b)]\cdot[(c,d)]+[(a,b)]\cdot[(e,f)]$$
All non-zero rationals have multiplicative inverses. We denote the inverse of \([(a,b)]\) by \([(a,b)]^{-1}\)
Given rational \([(a,b)]\) and non-zero rational \([(c,d)]\), we define division as follows: $$\frac{[(a,b)]}{[(c,d)]}=[(a,b)]\cdot[(c,d)]^{-1}$$
If \([(a,b)]\leq[(c,d)]\) and \([(e,f)]\geq0\), $$[(a,b)]\cdot[(e,f)]\leq[(c,d)]\cdot[(e,f)]$$And, if \([(e,f)]\leq0\), $$[(a,b)]\cdot[(e,f)]\geq[(c,d)]\cdot[(e,f)]$$
For all \([(a,b)],[(c,d)]\) such that \([(a,b)]<[(c,d)]\), there exists \([(e,f)]\) such that $$[(a,b)]<[(e,f)]<[(c,d)]$$
In summary, the rationals are an ordered field containing the integers and respecting their operations.
Real Numbers
Finally, we arrive at the real numbers. The jump from the rationals to the reals is more complicated than any of the previous constructions (barring perhaps the natural numbers), since as we will see in future posts, the real numbers are a larger space than any of the previous sets.
We'll define the reals numbers axiomatically for a few reasons. 1. It'll motivate some of the logical discussions we will have later in the course. 2. It is far more convenient to work with than a constructive approach. 3. It shows the naturality of the definition of the reals.
In any ordered set \(P\) with \(S\subset P\), \(x\) is an upper bound of \(S\) iff for all \(y\in S\), \(y\leq x\). \(z\) is a least upper bound of \(S\) iff for all upper bounds \(x\) of \(S\), \(z\leq x\).
The Real Numbers \(\bbR\) are a set that satisfy the following properties.
- \(\bbQ\subset\bbR\)
- \(\bbR\) is an ordered field, with its order, addition, and multiplication structures the same as the rational ones on the rationals.
- For all \(x,y,z\in\bbR\),
- \(x\geq y\) implies \(x+z\geq y+z\)
- \(x,y\geq0\) implies \(xy\geq0\)
- For all \(S\subset\bbR\), if there exists an upper bound \(X\), there exists a least upper bound \(x\).
This definition, if you haven't seen it before, should confuse you. First, how do we know such a set exists? What if there are multiple sets that satisfy the conditions? Are they both the real numbers?
This is where the notion of an isomorphism becomes useful. Loosely speaking, an isomorphism is a bijection between sets that respects their structure. In this case, if spaces \((X,\lt_X,+_X,\cdot_X),\;(Y,\lt_Y,+_Y,\cdot_Y)\) both satisfy the conditions, an isomorphism would be a bijection \(f:X\to Y\) such that for all \(x,y\in X\),
- \(\;f(x)+_Yf(y)=f(x+_Xy)\)
- \(\;f(x)\cdot_Yf(y)=f(x\cdot_Xy)\)
- \(\;x\lt_Xy\) iff \(f(x)\lt_Yf(y)\)
The beauty of isomorphisms is that spaces with isomorphisms between them, which we call isomorphic spaces, can be considered to be the same set, as the structures behave identically, with the only difference between the spaces being the names assigned to each element. Think of this the same way as we identified \(n\) with \([(n,0)]\) in the integers, and \(z\) with \([(z,1)]\) in the rationals. It is an easy exercise to show that composing isomorphisms gives us an isomorphism.
So here's the claim: there exists a set that satisfies the definition of the real numbers, and any two sets that satisfy the conditions are isomorphic. We'll show this by constructing one example of a set that satisfies these conditions and then constructing an isomorphism to any other set satisfying the conditions. With both of these results, we can, without any reservation, refer to the real numbers as a space.
First, let's construct an example of a set that satisfies the conditions.
A Dedekind Cut is a set \(S\subsetneq\bbQ\) that satisfies the following conditions.
- \(S\neq\varnothing\)
- For all \(x\in S\) and \(y\in\bbQ\), if \(y\leq x\), \(y\in S\).
- For all \(x\in S\), there exists \(y\in S\) such that \(x\lt y\).
A cut should be thought of as the set of all rationals less than the number the cut is representing.
For all \(q\in\bbQ\), the set \(q^*=\{r\in\bbQ\mid r\lt q\}\) is a dedekind cut. And, if \(q\neq r\), \(q^*\neq r^*\).
- \(q\notin q^*\), so \(q^*\neq\bbQ\)
- For all \(q\in\bbQ\), \(q-1\lt q\), so \(q-1\in q^*\)
- For \(x,y,q\in\bbQ\), if \(y\in q^*\), \(y\lt q\), and if \(x\lt y\), \(x\lt q^*\) by transitivity, so \(x\in q^*\).
- For \(x,q\in\bbQ\), if \(x\in q^*\), \(x\lt q\), so by density of rationals, there exists \(y\in\bbQ\) such that \(x\lt y\lt q\), so \(y\in q^*\).
This is useful since we can define an injection \(f:\bbQ\to D\) that sends \(q\mapsto q^*\). As we've done a few times now, we can now identify \(q\) with \(q^*\), so that \(\bbQ\subset D\).
Consider the binary operation \(+\) on \(D\) that is defined by \(A+B=\{a+b\mid a\in A,\;b\in B\}\). This operation is well-defined and is the same as rational addition on the rationals.
To show that it is well defined, we need to show that for all \(A,B\in D\), \(A+B\in D\).
- There exists \(x\in\bbQ\setminus A\), \(y\in \bbQ\setminus B\). Thus, \(x\) is an upper bound for \(A\) (otherwise, \(x\in A\)), and likewise, \(y\) is an upper bound for \(B\). Thus, for any \(c\in A\) and \(d\in B\), \(c\lt x\) and \(d\lt y\), so \(c+d\lt x+y\). Thus, \(x+y\notin A+B\).
- There exists some \(x\in A\) and \(y\in B\), so \(x+y\in A+B\)
- If \(a\in A\), \(b\in B\), and \(y\lt a+b\), then \(y-a\lt b\) and thus \(y-a\in B\). So, \(a+(y-a)\in A+B\), implying \(y\in A+B\).
- If \(a\in A\), \(b\in B\), there exists \(c\in A\), \(d\in B\) such that \(a\lt c\) and \(b\lt d\), and thus \(a+b\lt c+d\), \(c+d\in A+B\)
Commutativity and associativity of this definition are given just by the commutativity and associativity of addition in the rationals. This also implies that no non-zero element is an identity element.
For all \(A\in D\), \(0+A=A\). Moreover, no other element is an identity element.
If \(a\in A\), there exists \(b\in A\) such that \(b>a\). Thus, \(a-b\lt0\), and thus, \(a-b\in 0\). So, \((a-b)+b\in 0+A\). If \(x+y\in 0+A\), then since \(x\lt 0\), \(x+y\lt y\), so \(x+y\in A\). To see that no other element can be an identity, note that for any two identity elements \(e_1,e_2\), \(e_1+e_2=e_1=e_2\).
For any \(A\in D\), let \(-A\) be defined as follows: $$-A=\begin{cases}(-q)^*&\exists q\in\bbQ,\;A=q^*\\\bbQ\setminus\{-a\mid a\in A\}&\text{otherwise}\end{cases}$$For all \(A\), \(-A\in D\), and \(A+(-A)=0\).
If \(A=q^*\) for some \(q\in\bbQ\), this is obvious. So assume not. First, let's show that \(-A\) is a cut.
- There exists \(a\in A\), so \(-a\notin -A\), so \(-A\neq\bbQ\)
- There exists \(b\notin A\), so \(-b\in -A\).
- If \(x\in -A\), if \(y\lt x\), then \(-y\gt-x\). If \(-y\in A\), then \(-x\in A\), a contradiction. Thus, \(-y\notin A\), and \(-y\in -A\)
- If \(x\in -A\) is such that there is no \(y\in -A\) with \(y\gt x\), then \(-x\notin A\) but for all \(y\lt -x\), \(y\in A\). I claim that \(A=(-x)^*\). We already showed that \((-x)^*\subset A\). And, if there exists \(z\in A\) such that \(z\geq -x\), then \(-x\in A\), a contradiction. So, \(A=(-x)^*\). However, we assumed that this wasn't the case, so this is a contradiction.
- For any \(x\in0\), \(-x\gt0\). I claim that there exists some \(a\in A\) such that \(a-x\notin A\). Suppose not. Since \(A\) is non-empty, there exists \(b\in A\). Let \(k\in\bbN\) be the smallest natural for which \(b+k\cdot(-x)\notin A\). \(k\neq0\) by the given, so \(k-1\) is natural, and thus \(b+(k-1)\cdot(-x)\in A\). But, by our assumption, this implies \(b+k\cdot(-x)\in A\), a contradiction. Thus, \(b+k\cdot(-x)\in A\) for all \(k\in\bbN\). Next, for any \(q\in\bbQ\), either \(q\leq b\), in which case \(q\in A\), or \(z=\frac{q-b}{-x}\) is positive since the numerator and denominator are. Thus, \(z=[(v,w)]\), where \(v,w\) are positive naturals. Thus, \((v+1)\gt z\). So, we have $$b+(v+1)\cdot(-x)\gt b+z\cdot(-x)=q$$Since \(b+(v+1)\cdot(-x)\in A\), \(q\in A\). This implies \(A=\bbQ\), a contradiction. Thus, there exists some \(a\in A\) such that \(a+(-x)\notin A\). So, \(x-a\in -A\), so \((x-a)+a=x\in (-A)+A\).
For the other side, note that if \(a\in A\), \(b\in -A\) such that \(a+b=0\), then \(b=-a\), and thus \(b\notin -A\), a contradiction. If \(a+b\gt0\), then \(b\gt-a\), so \(-a\in -A\), a contradiction. This completes the proof.
We define \(A-B\) as \(A+(-B)\).
Now we introduce order, which is surprisingly easy to define.
We define the order relation on the reals as follows: \(A\leq B\) iff \(A\subset B\). This is an order, and it is identical to the rational order on the rationals.
- Reflexivity: Obvious
- Anti-Symmetry: Obvious
- Transitivity: Obvious
- Trichotomy: If there exists \(x\in A\) such that \(x\notin B\), then \(x\) is an upper bound for \(B\) (if not, \(x\in B\)). This implies that for any \(y\in B\), \(y\lt x\), so \(y\in A\). Thus \(B\subset A\).
We say a number is positive if it is greater than zero, and negative if it is less than zero.
If \(x\lt 0\), then \(-x\gt0\), and vice versa.
If \(x\geq y\), then for any \(z\), \(x+z\geq y+z\)
Now, we prove the completeness of the reals, via the least upper bound property.
Given a non-empty set \(S\subset D\), \(\bigcup S=\bbQ\) or \(\bigcup S\in D\).
- If \(\bigcup S=\bbQ\), we are done.
- Since there exists \(X\in S\subset D\), and \(X\neq\varnothing\), \(X\subset\bigcup S\neq\varnothing\)
- If \(x\in\bigcup S\), there exists \(X\in S\) such that \(x\in X\). Thus, for any \(y\lt x\), \(y\in X\), so \(y\in\bigcup S\).
- If \(x\in\bigcup S\), \(x\in X\) as in the previous bullet, so there exists \(y\in X\) such that \(y\gt x\), and thus \(y\in\bigcup S\).
Given a non-empty set \(S\subset D\), if there exists an upper bound \(Y\in D\) for \(S\), there exists a least upper bound \(X\in D\) for \(S\).
I claim that \(X=\bigcup S\). Note that \(Y\) is not in any of the elements of \(S\), so \(Y\notin X\), and thus, \(X\neq\bbQ\). By the previous theorem, \(X\) is a cut. It is an upper bound for \(S\) since if \(X\in Z\) for any \(Z\in S\), then \(X\in\bigcup S\), and thus \(X\in X\), which is false by the axiom of regularity. And, for any \(Z\in D\), if \(Z\lt X\), there exists \(a\in X\) such that \(a\notin Z\). Thus, there exists \(A\in S\) such that \(a\in A\), which implies that \(Z\lt A\), meaning that \(Z\) is not an upper bound of \(S\). Thus, \(X\) is the least upper bound of \(S\).
Given a non-empty set \(S\subset D\), if there exists a lower bound \(Y\in D\) for \(S\), there exists a greatest lower bound \(X\in D\) for \(S\).
Now we define multiplication.
For \(x,y\in D\), $$x\cdot y=\begin{cases}0\cup\{p\cdot q\mid p\in x,\;q\in y,\;p,q\geq0\}&x,y\geq0\\-((-x)\cdot y)&x\lt0\leq y\\-(x\cdot(-y))&y\lt0\leq x\\(-x)\cdot(-y)&\text{otherwise}\end{cases}$$This definition produces cuts and behaves identically to rational multiplication on the rationals.
First, we show that this operation always produces cuts: let \(D\ni x,y\geq0\). If \(x=0\) or \(y=0\), then the product is just \(0\) by definition, which is a cut. Assume \(x,y\gt0\)
- If \(0\lt r\lt p\) and \(0\lt s\lt q\), \(r\cdot s\lt p\cdot q\), so for some \(p\notin x\), \(q\notin y\), \(p\cdot q\notin x\cdot y\).
- \(-1\in x\cdot y\)
- If \(z\in x\cdot y\) and if \(v\lt z\), if \(v\lt 0\), then \(v\in x\cdot y\) trivially. Otherwise, there exists \(a\in x\), \(b\in y\) such that \(a\cdot b=z\) with \(a,b\gt0\). Note that \(\frac vz\lt1\), so \(\frac{av}z\lt a\), meaning that \(\frac{av}z\in x\). Thus, we have \(v=\frac{av}z\cdot b\in x\cdot y\)
- If \(z\in x\cdot y\), since \(x,y\gt 0\), there exists \(r\in x\), \(s\in y\) such that \(r,s\gt0\), so \(r\cdot s\gt0\). Thus, if \(z\leq 0\), \(z\lt r\cdot s\). If \(z\gt 0\), there exists \(a\in x\), \(b\in y\) such that \(a\cdot b=z\). Since \(x,y\) are cuts, there exists \(\overline{a}\in x\), \(\overline{b}\in y\) such that \(\overline{a}\gt a\), \(\overline{b}\gt b\). Thus, \(x\cdot y\ni \overline{a}\cdot\overline{b}\gt a\cdot b=z\).
By definition, we have \(x,y\geq0\) implies \(xy\geq0\) since \(0\subset xy\). Some important results for multiplication.
Given \(x,y\in D\), \(x\cdot y=y\cdot x\)
First assume \(x,y\geq0\). If \(r\in x\cdot y\), then either \(r\lt 0\) or there exists \(a\in x\), \(b\in y\) such that \(r=a\cdot b\). In the first case, \(r\in y\cdot x\) since \(0\subset y\cdot x\). In the second case, \(r=a\cdot b=b\cdot a\), implying \(r\in y\cdot x\). Since \(x,y\) arbitrary, we are done with this case.
If \(x,y\lt0\), $$x\cdot y=(-x)\cdot(-y)=(-y)\cdot(-x)=y\cdot x$$If \(x\lt 0\leq y\), $$x\cdot y=-((-x)\cdot y)=-(y\cdot(-x))=y\cdot x$$The final case is parallel.
If \(x,y\lt0\), $$x\cdot y=(-x)\cdot(-y)=(-y)\cdot(-x)=y\cdot x$$If \(x\lt 0\leq y\), $$x\cdot y=-((-x)\cdot y)=-(y\cdot(-x))=y\cdot x$$The final case is parallel.
Given \(x,y,z\in D\), \(x\cdot (y\cdot z)=(x\cdot y)\cdot z\).
Assume first that \(x,y,z\geq0\). If \(r\in x\cdot(y\cdot z)\), either \(r\lt0\), in which case \(r\in(x\cdot y)\cdot z\), or there exists \(a\in x,y\in b,z\in c\) such that $$r=a\cdot(b\cdot c)=(a\cdot b)\cdot c\in(x\cdot y)\cdot z$$To handle the general case, we need casework:
- \(x\lt0\), \(y,z\geq0\): $$x\cdot(y\cdot z)=-((-x)\cdot(y\cdot z))=-(((-x)\cdot y)\cdot z)=(-((-x)\cdot y)\cdot z)=(x\cdot y)\cdot z$$
- \(x,y\lt0\leq z\): $$x\cdot(y\cdot z)=-((-x)\cdot-((-y)\cdot z))=(-(x\cdot(-y))\cdot z)=(x\cdot y)\cdot z$$
- \(x,y,z\lt0\):$$x\cdot(y\cdot z)=-((-x)\cdot((-y)\cdot(-z)))=-((x\cdot y)\cdot(-z))=(x\cdot y)\cdot z$$
\(1\) is the sole multiplicative identity of \(D\)
For \(x\in D\), let \(y\in x\). There exists \(z\neq0\in x\) such that \(y\lt z\lt x\). Thus, \(\frac yz\lt 1\). Thus, \(y=z\cdot\frac yz\in x\cdot 1\). And, if \(a\cdot b\in x\cdot 1\), with \(a\lt x\), \(b\lt 1\), then \(a\cdot b\lt a\), so \(a\cdot b\in x\). This proves that \(x\cdot 1=x\).
To show that this is a unique identity, note that for any two identities \(e_1,e_2\), \(e_1\cdot e_2=e_1=e_2\).
To show that this is a unique identity, note that for any two identities \(e_1,e_2\), \(e_1\cdot e_2=e_1=e_2\).
Given \(a,b,c\in D\), \(a\cdot(b+c)=a\cdot b+a\cdot c\).
Assume first that \(a\geq0\), \(b+c\geq0\). Let \(x\in a\cdot(b+c)\). There are two cases
- \(x\lt0\): Let \(y\in0\) such that \(y\notin x\). If \(y^*=x\), then replace \(y\) with \(\frac y2\), as \(y\lt\frac y2\lt0\). Thus, \(x\lt y^*\lt0\). Thus, \(y^*\in a\cdot b\) and \(x-y^*\in a\cdot c\), so \(x=y^*+(x-y^*)\in a\cdot b+a\cdot c\)
- There exists \(0\leq i\in a\), \(j\in b\), \(k\in c\) such that \(x=i\cdot(j+k)=i\cdot j+i\cdot k\). If \(j\leq0\), then \(i\cdot j\leq0\), so \(i\cdot j\in a\cdot b\). If \(i=0\), then the relation is obvious. And, if \(0\lt i\in a\), \(0\lt j\in b\), then \(i\cdot j\in a\cdot b\). Parallel logic shows \(i\cdot k\in a\cdot c\), proving the desired relation.
- \(a\lt0\), \(b+c\geq0\): Thus, $$a\cdot(b+c)=-((-a)\cdot(b+c))=-(-(a\cdot b)-(a\cdot c))=a\cdot b+a\cdot c$$ \(a\gt0\), \(b+c\lt0\)$$a\cdot(b+c)=-(a\cdot(-b-c))=-(-a\cdot b-a\cdot c)=a\cdot b+a\cdot c$$
- \(a\lt0\), \(b+c\lt0\):$$a\cdot(b+c)=(-a)\cdot(-b-c)=a\cdot b+a\cdot c$$
For any \(x,y\in D\) such that \(x\lt y\), there exists \(q\in\bbQ\) such that \(x\lt q^*\lt y\).
For all \(x,y\in D\), if \(x\cdot y=0\), \(x=0\) or \(y=0\).
Suppose there exists \(x,y\) such that the theorem is false. Then, \(x\cdot(-y)=-(x\cdot y)=0\) and \((-x)\cdot y=-(x\cdot y)\), so we can assume \(x,y\gt0\) without loss of generality. Thus, there exists \(p\in x\) and \(q\in y\) such that \(p,q\gt0\). But, \(p\cdot q\gt0\) since the rationals are an integral domain, and \(p\cdot q\in x\cdot y\). Thus, \(x\cdot y\neq0\).
For any \(x\neq0\), there exists \(y\) such that \(x\cdot y=1\). We call \(y\) \(x^{-1}\), or the reciprocal of \(x\).
Let \(x\gt0\). Let \(S\) be the set of cuts \(y\) such that \(x\cdot y\leq1\). \(0\in S\), so \(S\) is non-empty. And, since \(x\gt0\), there exists \(q\in x\) such that \(q\gt0\). Note that for any \(A\in D\) with \(A\geq(q^{-1})^*\), \(x\cdot A\geq x\cdot(q^{-1})^*\ni 1\), so \(A\notin S\). Thus, \((q^{-1})^*\) is an upper bound for \(S\). By the least upper bound property, there exists a cut \(Y\) that is the least upper bound of \(S\). I claim that \(x\cdot Y=1\).
- If \(x\cdot Y\lt0\), then \(Y\lt0\), and thus \(Y\) is not an upper bound since \(0\in S\)
- If \(0\leq x\cdot Y\lt 1\), by density of the rationals, there exists \(r\in\bbQ\) such that \(x\cdot Y\lt r^*\lt 1\). By density again, there exists a rational \(s\) such that \(r\lt s\lt1\).
Thus, \(\frac sr-1\gt0\) and \(\left(\frac sr-1\right)Y\gt0\) since \(D\) is an integral domain and \(x,y\geq0\) implies \(x\cdot y\geq0\). So, \(\frac srY\gt Y\), and \(\frac sr(r-x\cdot Y)\gt0\)$$x\cdot \frac srY\lt x\cdot \frac srY+\frac sr(r-x\cdot Y)=\frac sr(x\cdot Y+r-x\cdot Y)=s<1$$This implies that \(\frac srY\in S\), but \(\frac srY\gt Y\) as we showed, contradicting the fact that \(Y\) is an upper bound. - If \(x\cdot Y\gt1\), by density of the rationals, there exists \(r,s\in\bbQ\) such that \(1\lt s^*\lt r^*\lt x\cdot Y\). Thus, \(\frac sr-1\lt0\), and thus \(\left(\frac sr-1\right)Y\lt0\). This implies \(\frac srY\lt Y\). Moreover, $$x\cdot \frac srY=\frac sr(x\cdot Y)\gt s\gt1$$This implies that \(\frac srY\) is an upper bound for \(S\), contradicting the least upper bound nature of \(Y\).
So, \(x\cdot Y=1\). If \(x\lt0\), let \(Y\) be such that \((-x)\cdot Y=1\). Thus, $$x\cdot (-Y)=-(x\cdot Y)=(-x)\cdot Y=1$$This completes the proof.
We define division as follows: For \(x,y\in D\), \(\frac xy\) = \(x\cdot y^{-1}\)
These rather tedious proofs show that \(D\) satisfies all of the conditions of the real numbers. Now, we show that any other set that does so is isomorphic to \(D\).
For any space \(X,\lt_X,+_X,\cdot_X\) satisfying the real conditions, for all \(x\in X\), there exists some \(n\in\bbN\) such that \(x\lt_Xn\).
In any space \(X,\lt_X,+_X,\cdot_X\) with the real conditions, for any \(x\lt_Xy\), there exists \(q\in\bbQ\) such that \(x\lt_Xq\lt_Xy\).
If \(x\lt0\lt y\), the claim is obvious (\(0\in\bbQ\)). If \(x\lt y\leq0\), apply this theorem on \(0\leq (-y)\lt (-x)\) to get \(0\leq(-y)\lt q\lt(-x)\). Thus, \(x\lt-q\lt y\leq0\). So, if we show this theorem for \(0\leq x\lt y\), we have shown the result for all \(x,y\).
By the archimedean property, let \(n\in\bbN\) be such that \(n\gt_X2(y-x)^{-1}\gt0\). Thus, \(ny-nx\gt(y-x)\cdot2(y-x)^{-1}=2\). By the well-ordering principle, let \(m\in\bbN\) be the least natural greater than \(ny-1\) (which exists thanks to the archimedean principle). Clearly, \(m\lt ny\), since if not, \(m-1\gt ny-1\), contradicting the minimality of \(m\). And, $$m\gt ny-1\gt ny-(ny-nx)=nx$$So, \(nx\lt m\lt ny\). Since \(n\gt0\), we have \(x\lt \frac mn\lt y\) as desired.
By the archimedean property, let \(n\in\bbN\) be such that \(n\gt_X2(y-x)^{-1}\gt0\). Thus, \(ny-nx\gt(y-x)\cdot2(y-x)^{-1}=2\). By the well-ordering principle, let \(m\in\bbN\) be the least natural greater than \(ny-1\) (which exists thanks to the archimedean principle). Clearly, \(m\lt ny\), since if not, \(m-1\gt ny-1\), contradicting the minimality of \(m\). And, $$m\gt ny-1\gt ny-(ny-nx)=nx$$So, \(nx\lt m\lt ny\). Since \(n\gt0\), we have \(x\lt \frac mn\lt y\) as desired.
Any space \((X,\lt_X,+_X,\cdot_X)\) that satisfy the real conditions is isomorphic to \((D,\lt,+,\cdot)\).
I claim that the map \(f:X\to D\) defined by \(A\mapsto\{q\in\bbQ\mid q\lt_X A\}\) is an isomorphism.
- Well-defined: We need to show that the output of this function is always a cut. By the archimedean property, let \(n_A\in\bbN\) be such that \(n_A\gt_X A\) for all \(A\in X\)
- The archimedean property implies that \(n_A\in\bbQ\setminus f(A)\).
- \(n_{-A}\gt_X-A\), so \(-n_{-A}\lt_XA\), so \(-n_{-A}\in f(A)\).
- If \(q\in f(A)\) and \(r\lt_X q\), then \(r\lt_X q\lt_X A\), so \(r\in f(A)\)
- If \(q\in f(A)\), then \(q\lt_X f(A)\), so by the density of the rationals, there exists \(r\in\bbQ\) with \(q\lt_X r\lt_X A\), so \(r\in f(A)\)
- Injective: For any \(A,B\in X\) with \(A\lt_XB\), by the density of the rationals, there exists \(q\in\bbQ\) with \(A\lt_X q\lt_XB\), so \(q\in f(B)\), \(q\notin f(A)\)
- Surjective: For any cut \(S\in D\), observe that \(S\subset X\) has an upper bound in \(\bbQ\) (if not, all elements of \(\bbQ\) would be in \(S\), a contradiction). \(S\) is also non-empty, so \(S\) has a least upper bound \(Y\) in \(X\). I claim that \(f(Y)=S\).
- If \(q\in S\), then by cut properties, there exists \(r\in S\) such that \(q\lt r\). Since \(Y\) is an upper bound of \(S\), \(q\lt_Xr\leq_XY\), so \(q\in f(Y)\)
- If \(q\in f(Y)\), then \(q\lt_X Y\). If there is no rational \(r\in S\) such that \(q\lt_X r\), then \(q\) is an upper bound of \(S\), contradicting the minimality of \(Y\). Thus, there exists such a \(r\in S\), and since \(S\) is a cut, \(q\in S\)
- \(\;x\leq_Xy\) implies \(f(x)\leq f(y)\): If \(q\in f(y)\), then \(q\lt_Xx\), so \(q\lt_Xy\), and thus \(q\in f(y)\)
- \(\;f(x+_Xy)=f(x)+f(y)\): If \(q\in f(x+_Xy)\), then \(q\lt_Xx+_Xy\). By density of the rationals, there exists \(r\in\bbQ\) such that \(q-y\lt_Xr\lt_Xx\) since \(x-(q-y)=x+y-q\gt_X0\). Thus, \(q-r\lt_X q-(q-y)=y\), so \(q=r+(q-r)\in f(x)+f(y)\). If \(q\in f(x)+f(y)\), there exists \(\bbQ\ni a\lt_Xx\), \(\bbQ\ni b\lt_Xy\) such that \(q=a+b\). But, \(q=a+b\lt_Xx+y\) by order properties.
- \(\;f(x\cdot_Xy)=f(x)f(y)\): Assume first that \(x,y\geq0\). If \(x=y=0\), then the claim is obvious. If \(q\in f(x\cdot_Xy)\), then \(q\lt x\cdot_Xy\). If \(q\leq_X0\), then \(q\in f(x)f(y)\) by definition. If \(q\gt0\), then \(x,y\gt0\).
In this case, \(x\cdot_Xy\gt_Xq\) implies \(x\gt_X\frac qy\). Thus, by density of rationals, let \(r\in\bbQ\) such that \(\frac qy\lt_Xr\lt_Xx\). Thus, \(\frac qr\lt_Xq\cdot\left(\frac qy\right)^{-1}=y\). So, \(q=r\cdot_X\frac qr\in f(x)f(y)\). If \(q\in f(x)f(y)\), there exists \(0\lt_Xr\lt_Xx\) and \(0\lt_Xs\lt_Xy\) such that \(q=rs\). By order properties, \(rs\lt_Xx\cdot_Xy\), implying that \(q\in f(x\cdot_Xy)\).
If \(x\lt0\leq y\), then $$f(x\cdot_Xy)=-f(-x\cdot_Xy)=-f(-x)f(y)=f(x)f(y)$$If \(y\lt0\leq x\), then$$f(x\cdot_Xy)=-f(x\cdot_X(-y))=-f(x)f(-y)=f(x)f(y)$$If \(x,y\lt0\), $$f(x\cdot_Xy)=f(-x\cdot_X-y)=f(-x)f(-y)=f(x)f(y)$$
These proofs show that our characterization of the real numbers via axioms is well-defined, since there is a space that satisfies the axioms, and any two spaces that do so are equivalent in a very strong sense. So why bother with other constructions of the real numbers? The answer is simple: perspective. Alternate constructions of the real numbers will emphasize different properties of them, and make proving certain theorems easier. For example, the following construction of the reals allows for a far easier definition of order, addition, and multiplication. These exercises are not too difficult (much less difficult than the proofs above) and will give you some intuition on how the reals work.
Given two rationals \(a,b\), the distance \(d(a,b)\) between the two rationals is defined as the larger of \(a-b\) and \(b-a\).
Given an infinite sequence of rationals \(q_0,q_1,q_2,\ldots\), we call the sequence Cauchy if for all rationals \(r\gt0\), there exists a natural \(n\in\bbN\) such that for all \(\bbN\ni i,j\geq n\), \(d(q_i,q_j)\lt r\). In other words, a Cauchy sequence is one in which all of the terms progressively get closer together.
Let \(X\) be the space of all Cauchy sequences. Let sequences \(q_0,q_1\ldots\) and \(r_0,r_1,\ldots\) be equivalent iff \(q_0,r_0,q_1,r_1,\ldots\) is Cauchy. Prove that this is an equivalence relation on \(X\). Define \(\bbR\) as the partition associated with this equivalence relation.
Let the sequence \([q,q,q,\ldots]\) be identified with \(q\) so that \(\bbQ\subset\bbR\). Show that no two such sequences are in the same equivalence class.
Given \([q_0,\ldots],[r_0,\ldots]\in\bbR\), let \([r_0,\ldots]\gt[q_0,\ldots]\) if \(r_i-q_i\gt0\) for all but finitely many \(i\). Show that this is an order relation on \(\bbR\), and that it is identical to the rational order on the rationals.
Given \([q_0,\ldots],[r_0,\ldots]\in\bbR\), let $$[q_0,\ldots]+[r_0,\ldots]=[q_0+r_0,q_1+r_1,\ldots]$$Show that this definition has all of the properties of addition in the real conditions.
Given \([q_0,\ldots],[r_0,\ldots]\in\bbR\), let $$[q_0,\ldots]\cdot[r_0,\ldots]=[q_0r_0,q_1r_1,\ldots]$$Show that this definition has all of the properties of multiplication in the real conditions
Show that \(\bbR\) satisfies all of the remaining real conditions