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 make a small start in this 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.)
The axioms for the real numbers are a foundation that can be used to prove other properties of $\R.$ Section 1.3 includes some examples of using the field axioms to prove things that must be true in any field. But more important for us is using the order axioms to define and prove things about absolute value and distance. Application of the completeness axiom will begin in Section 1.4 with the Heine-Borel Theorem.
The field axioms can be used to prove things about arithmetic operations that you probably use automatically and regard as intuitive. The point here is that they are not things that have to be assumed separately; once we have the field axioms, these other "obvious" things follow logically.
There are several uniqueness results: 0 is the only additive identity, 1 is the only multiplicative identity, a real number $a$ has exactly one additive inverse, and a non-zero number $a$ has exactly one multiplicative inverse. This allows us to write $-a$ for the additive inverse of $a$ and to define subtraction by $b-a=b+(-a).$ Similarly, we can define $a^{-1}$ for a non-zero number $a$ and division by $\frac ba = b\cdot a^{-1}.$ The existence of inverses is what makes "cancellation" possible. For example,
Theorem: Let $a,$ $b,$ and $c$ be elements of some field. Suppose that $ab=ac$ and $a\ne 0.$ Then $b=c.$
Proof: Since $a$ is non-zero, it has a multiplicative inverse. So, we have $$\begin{array}{rll} ab\!\!\!\!&=ac &\mbox{(Assumption)}\\ a^{-1}(ab)\!\!\!\!&=a^{-1}(ac)&\mbox{($a$ has multiplicative inverse $a^{-1}$)} \\ (a^{-1}a)b\!\!\!\!&=(a^{-1}a)c &\mbox{(Associativity of multiplication)}\\ 1b\!\!\!\!&=1c &\mbox{(Multiplicative inverse: $a^{-1}a=1$)}\\ b\!\!\!\!&=c &\mbox{(1 is the multiplicative identity) }\qed\\ \end{array}$$
(The textbook justifies steps in some proofs by listing axiom numbers. Note that you do not have to memorize the axiom numbers, but you should remember the names of the properties.)
We can take most of the properties of addition, multiplication, and comparison as being well-known, but it would still be useful to look through their proofs in the textbook. We should spend a little more time on absolute value and distance.
The first order axiom is Trichotomy: For any real number $a,$ exactly one of the following is true: $a<0,$ $a=0,$ or $a>0.$ The other two order axioms state that for any two positive numbers $a$ and $b,$ $a+b$ and $ab$ are also positive. (Recall that we proved Trichotomy for $\R$ defined as the set of Dedekind cuts.) We can use the Trichotomy property to define absolute value, and then use absolute value to define distance.
Definition: For $a\in\R,$ we define the absolute value of $a,$ denoted $|a|,$ by $$|a|= \begin{cases} a&\mbox{if $a>0$}\\ 0&\mbox{if $a=0$}\\ -a&\mbox{if $a<0$} \end{cases}$$ For $a,b\in\R,$ we define the distance between $a$ and $b$ to be $|a-b|.$ (Note that $|a|$ is the distance from $a$ to 0.)
Of course, $|a|=a$ for all $a\ge0,$ but my definition makes it clear that it is the Trichotomy axiom that is being used. Some basic properties can be proved easily using the axioms.
Theorem: For $a,b\in\R,$ $|ab|=|a|\cdot|b|,$ $|-a|=|a|,$ $|a-b|=|b-a|,$ and $|a-b|=0$ if and only if $a=b.$
Proof: We first prove that $|ab|=|a|\cdot|b|.$ In the case $a=0$ or $b=0,$ both $|ab|$ and $|a|\cdot|b|$ are zero, so they are equal. In the case $a>0$ and $b>0,$ we have $ab>0,$ so $|ab|=ab=|a|\cdot|b|,$ In the case $a<0$ and $b<0,$ again $ab>0,$ so $|ab|=ab=(-a)(-b)=|a|\cdot|b|.$ In the case $a>0$ and $b<0,$ $ab<0,$ so $|ab|=-ab=a(-b)=|a|\cdot|b|.$ And the case $a<0$ and $b>0$ is similar.
It then follows that $|-a|=|-1|\cdot|a|=1\cdot|a|=|a|.$ From that, it follow that $|a-b|=|-(b-a)|=|b-a|.$ Finally, by definition of absolute value, $|a-b|=0$ if and only if $a-b=0,$ which is true if and only if $a=b.$
The last part of this theorem is the important fact that the distance between $a$ and $b$ is zero if and only if $a$ is equal to $b.$ And the next-to-last part says that distance is symmetric: the distance from $a$ to $b$ is the same as the distance from $b$ to $a.$ One of the most important properties of distance is the triangle inequality, which says that for any $a,b,c\in\R,$ $$|a-c|\le|a-b|+|b-c|$$ That is, going from $a$ to $c$ directly cannot be longer than going from $a$ to $b$ and then from $b$ to $c.$ The triangle inequality can also be expressed as a fact about absolute value, which is the version given in the book: For any $a,b\in\R,$ $|a|+|b|\ge|a+b|.$ This is intuitively clear since when $a$ and $b$ are both positive or both negative, $|a|+|b|=|a+b|,$ but when one is positive and one is negative, you get some cancellation on the right side of the inequality but not on the left (for example, $|3+(-5)| = |-2| = 2,$ but $|3|+|-5|=3+5=8$). The triangle inequality for distance follows from the triangle inequality for absolute value: Suppose $a,b,c\in\R.$ Let $x=a-b$ and $y=b-c.$ Then $$\begin{align*} |a-b|+|b-c| &= |x|+|y|\\ &\ge|x+y|\\ &=|(a-b)+(b-c)|\\ &=|a + (-b+b) - c|\\ &=|a + 0 -c|\\ &=|a-c| \end{align*}$$
The triangle inequality will be one of the most useful tools for proofs in the rest of this course. Some related facts about absolute value and distance will also be useful, including Theorem 1.3.12 from the textbook:
Theorem: For any $a,b\in \R$ with $b\ge0,$ $|a|\le b$ if and only if $-b\le a\le b.$
This is used in the book's proof of the triangle inequality. I want to use it to prove two additional facts that will be useful in future proofs. The first fact is useful when we want to show that some quantity, $a,$ is not too close to zero. If we can show that $a$ is close to some non-zero number, $m,$ then $a$ can't be too close to zero. In particular, if $a$ is within $\frac{|m|}{2}$ of $m,$ then $a$ cannot be closer to zero than $\frac{|m|}{2}.$
Theorem: Suppose $a,m\in \R$ and $|a-m|\le\frac{|m|}{2}.$ Then $|a|\ge\frac{|m|}{2}.$
Proof: Suppose that $|a-m|\le\frac{|m|}{2}.$ By the previous theorem, $-\frac{|m|}{2}\le a-m \le \frac{|m|}{2}.$
Consider the case $m\ge0,$ so that $|m|=m.$ Then the first inequality, $-\frac{|m|}{2}\le a-m,$ becomes $-\frac{m}{2}\le a-m$ and adding $m$ to both sides gives $\frac{m}{2}\le a.$ That is, $a\ge\frac{m}{2}=\frac{|m|}{2}.$ And because $\frac{m}{2}\ge0,$ we also have $a\ge0,$ so $a=|a|.$ So we get $|a|\ge\frac{|m|}{2}.$
Turing to the case $|m|<0,$ we have $|m|=-m$ and the inequality $a-m \le \frac{|m|}{2}$ becomes $a-m\le -\frac{m}{2}.$ Adding $m$ to both sides gives $a\le\frac{m}{2}.$ Multiplying by $-1$ then gives $-a\ge=-\frac{m}{2}.$ Since $\frac{m}{2}<0,$ we see that $a$ is also negative, so the inequality $-a\ge=-\frac{m}{2}$ can be written $|a|\ge\frac{|m|}{2}.$ $\qed$
If the proof is confusing, this diagram shows that the geometry is clear:
Sometimes, we need to bound $a.$ That is, we want to find some limit on how big $|a|$ can be. If we can show that $a$ is close to some number $m,$ then we can show that $|a|$ cannot be too much bigger than $|m|.$
Theorem: Suppose $a,m\in \R$ and $\eps>0.$ If $|a-m|\le\eps,$ then $|a|\le|m|+\eps.$
Proof: Suppose that $|a-m|\le\eps.$ Then $-\eps\le a-m\le \eps.$ Adding $m$ gives $m-\eps\le a\le m+\eps.$
Consider the case where $m\ge0.$ Then the inequality $a\le m+\eps$ becomes $a\le |m|+\eps.$ [If $\eps$ is small compared to $m,$ then $a$ will have the same sign as $m,$ so in this case, $a$ would be positive, and $a=|a|.$ This means we already have $|a|\le |m|+\eps;$ however, to cover all possibilities, we need to consider the general case where $a$ can be negative.] Now, for any $m,$ $-|m|\le m,$ and so from the inequality $m-\eps\le a,$ we get $-(|m|+\eps)=-|m|-\eps\le m-\eps\le a.$ Combining that with the previous inequality, we have $-(|m|+\eps)\le a\le (|m|+\eps),$ which is equivalent to $|a|\le|m|+\eps.$
Turning to the case $m<0,$ from $|m-a|<\eps,$ we get $|(-m)-(-a)|=|-(m-a)|=|m-a|<\eps.$ Since $-m>0,$ we can apply the previous case to $|(-m)-(-a)|<\eps,$ showing that $|-a|≤|-m|+\eps.$ But $|-a|=|a|$ and $|-m|=|m|,$ so we have in fact $|a|<|m|+\eps.$ $\qed$
And here is a diagram showing the geometry, in the case where $\eps$ is small compared to $|m|:$
