04 Axioms for $\R$
We have looked at Dedekind cuts as an explicit construction of the set of real numbers. But how do we know that we have properly captured all of the properties that we would expect $\R$ to have? We need a list of properties that express exactly what it means to be "the real numbers." That list is given in the textbook as a set of fifteen axioms. The axioms are statements about a system consisting of a set together with operations of addition, multiplication, and comparison. It can be shown that any two systems that satisfy the fifteen axioms are essentially the same; that is, there is a one-to-one correspondence between the two sets that preserves all of the operations. Probably the most important thing to remember from Section 1.3 is simply what it means to have an axiomatic specification of the real numbers.
Just listing the axioms does not show that there is any system that actually satisfies them. But it can be shown that Dedekind cuts satisfy the fifteen axioms. So, we have come at the real numbers from two directions: We have a set of properties that completely characterize what we think the real numbers should be, and we have a particular mathematical object that has all of those properties. This allows us to say, in some sense, that we understand the real numbers (although we don't really understand them until we understand all the consequences of the axioms—a process that can never really be completed but that we will start in the next section).
The first eleven axioms say that the set of real numbers, together with the addition and multiplication operations, is a field. They define the algebra of the real number system. The requirement is that addition and multiplication are commutative and associative, that identities and inverses exist for both addition and multiplication, and multiplication distributes over addition. There are also closure axioms, which say that the sum and product of real numbers are also real numbers. Any system that satisfies the field axioms will automatically have many other algebraic properties. That is, it will have properties that can be proved from the axioms. For example, it can be proved that identities and inverses are unique.
The field axioms do not completely determine the real numbers. There are many other fields, including $\Q$, the set of rational numbers, and $\C$, the set of complex numbers. The set $\{a+b\sqrt{2}\,|\,a,b\in\R\}$ is a field. So is $\big\{\frac{p(\pi)}{q(\pi)}\,\big|\,p(x)$ and $q(x)$ are polynomials with rational coefficients and $q(x)$ is not the zero polynomial$\big\}.$ For all of the these sets, the addition and multiplication operations are the usual operations for numbers.
There are also finite fields. You are probably familiar with $\Z_n$, the set $\{0,1,\dots,n-1\}$ with the operations of addition and multiplication $mod\;\,n$. For a prime number $p$, $\Z_p$ is a field.
And there are algebraic systems that are not fields. The set $\Z$ of integers is not a field because not every integer has a multiplicative inverse in $\Z$. The set $\N$ of natural numbers is even further from being a field, because it is missing both additive and multiplicative inverses. The set $\Z_n$, where $n$ is not prime, is not a field. (Suppose $n$ factors as $n=ab$ where both $a$ and $b$ are strictly between 1 and $n.$ Then $b$ has no multiplicative inverse in $\Z_n$. For if $bx=1$ ($mod\;\,n$), then $0=0x=nx=abx=a1=a$ ($mod\;\,n$), but in fact $a\ne0$ ($mod\;\,n$) because $a$ is strictly between 1 and $n$.)
$\R$ is not just a field; it is an ordered field. There is an operator $<$ for comparing two real numbers. The properties of this operation are expressed by three axioms, which are actually properties of the set of positive numbers: There is a subset $P$ of $\R$ that is closed under addition and multiplication and $\R$ is a disjoint union of $P$, $\{0\}$, and $-P.$ Given the set $P$, the comparison operation is defined by saying $x<y$ if and only if $y-x\in P.$ The set $P$ is that set of positive elements of $\R$, and $-P=\{-a\,|\,a\in P\}$, the set of negative elements.
But there are other ordered fields besides $\R$, such as the rational numbers or indeed any field that is a proper subset of $\R$. The final piece to completely characterize the real numbers is completeness, which can be expressed by the least upper bound axiom. This axiom means, intuitively, that there are no "holes" in the real number line. As we have seen, it also implies the Archimedian property of the real numbers: There is no real number $x$ that is greater than every natural number. (Note that the field of rational numbers is Archimedian but is not complete, so the Archimedian property does not imply completeness.)
$\newcommand{\F}{{\mathbb F}}$
The Archimedian property might seem very natural to you, so it is interesting to look at an ordered field that is not Archimedian. For that, we can use rational functions $\frac{p(x)}{q(x)}$ where $p(x)$ and $q(x)$ are polynomials with real coefficients and $q(x)$ is not the zero polynomial. Note that we have to consider two rational functions $\frac{p_1(x)}{q_1(x)}$ and $\frac{p_2(x)}{q_2(x)}$ to be equivalent if they can be made equal by dividing out common factors from the numerator and denominator of each fraction. Every rational function is then equivalent to one of the form $\frac{p(x)}{q(x)}$ where $p(x)$ and $q(x)$ have no common factors of degree greater than zero and the leading coefficient of $q(x)$ is 1. [The leading coefficient of a polynomial is the constant factor in the term in which $x$ has the highest degree in that polynomial.] With that understanding, we let $\F$ be the set of rational functions with real coefficients, with the usual addition and multiplication of functions.
Note that a real number $a$ can be considered to be the rational function $\frac{p(x)}{q(x)}$ where $p(x)=a$ and $q(x)=1$. So we can consider $\R$ to be a subset of $\F$. Since $\frac a1\cdot \frac b1 = \frac{ab}1$ and $\frac a1 + \frac b1 = \frac{a+b}1$, addition and multiplication of real numbers are the same in $\F$ as in $\R.$ That is, $\R$ is actually a subfield of $\F$.
Theorem:The field of rational functions, $\F$, is a non-Archimedian ordered field.
Proof: We give a brief outline of the proof. It is not hard to show that $\F$ satisfies the eleven field axioms. To make it into an ordered field, we need to define the set of positive elements in $\F$. We define $P=\big\{\frac{p(x)}{q(x)}\,\big|\,$the leading coefficient of $q(x)$ is 1 and $p(x)$ is non-zero and has a positive leading coefficient$\big\}$. We have to check that $P$ satisfies the three ordering axioms. Recalling that any rational function can be written as a fraction in which the denominator has leading coefficient 1, it is clear that any rational function $r(x)$ satisfies exactly one of $r(x)\in P$, $r(x)=0$, or $-r(x)\in P$. To see that $P$ is closed under addition and multiplication, suppose $r(x)=\frac{ax^n+\cdots}{x^m+\cdots}$ and $s(x)=\frac{bx^k+\cdots}{x^\ell+\cdots}$ where $a$ and $b$ are positive and in each case "$\dots$" represents terms of lower degree. Then $$r(x) \cdot s(x) = \frac{ax^n+\cdots}{x^m+\cdots} \cdot \frac{bx^k+\cdots}{x^\ell+\cdots} = \frac{abx^{n+k}+\cdots}{x^{m+\ell}+\cdots}$$ and so $r(x)\cdot s(x)\in P$ since $ab>0$. For addition, $$\begin{align*} r(x) + s(x) &= \frac{ax^n+\cdots}{x^m+\cdots} + \frac{bx^k+\cdots}{x^\ell+\cdots} \\[8pt] &= \frac{(x^\ell+\cdots)(ax^n+\cdots)+(x^m+\cdots)(bx^k+\cdots)}{(x^m+\cdots)(x^\ell+\cdots)}\\[8pt] &= \frac{ax^{\ell+n}+b^{k+m}+\cdots}{x^{m+\ell}+\cdots} \end{align*}$$ and so $r(x)+s(x)\in P$ because the leading coefficient in the numerator is either $a$ or $b$ or $a+b$ (depending on whether $\ell+n$ is greater than or less than or equal to $k+m$), and all of these are positive.
To see that $\F$ is not Archimedian, consider the rational function $\frac x1$ (where the numerator, $x$, is the polynomial $p(x)=x$). We claim that for every natural number $n$, $\frac n1<\frac x1$. (In $\F$, the natural number $n$ becomes the rational fraction $\frac n1$.) By definition, $\frac n1 <\frac x1$ means that $\frac x1 - \frac n1 \in P$. But $\frac x1 - \frac n1 = \frac{x-n}{1}$, and the fact that the leading coefficient of the numerator is positive means that $\frac{x-n}{1}\in P$. We have shown that $\frac x1$ is an element of $\F$ that is greater than every natural number, which means $\F$ is non-Archimedian. $\qed$
It is also true in $\F$ that $\frac 1x < \frac 1n$ for any integer $n$, so there are elements of $\F$ that are greater than zero but less than $\frac 1n$ for every integer $n$.
You can try to imagine all the elements of $\F$ laid out in a linear order, with $\R$ as a small bit in the middle. Any rational function in which the degree of the numerator is greater than the degree of the denominator lies to the left or to the right of $\R$. Any rational function in which the degree of the numerator is smaller than the degree of the denominator lies within $\R$, infinitely close but not equal to zero. I will leave you to find the rational functions in which the degrees of the numerator and denominator are equal. We started our investigation of the real numbers by identifying $\R$ as "the real line." But we had only some intuitions about what "the real line" should mean. Is it possible that $\F$ would be a better model for the real numbers, instead of the set of Dedekind cuts? Maybe so, but it would mean giving up the least upper bound axiom, which is the basis of most of the things that we will prove about real numbers in this course.