2 The Field and Order Axioms

Highlights

Two of the three axioms we will assume about the real number system: the field axiom which gives us the rules of arithmetic, and the order axiom which gives us the rules for inequalities.
We see examples of other number systems besides \(\R\) that also satisfy one or both of these axioms. In particular \(\Q\) also satisfies both.

2.1 Preparation

As discussed in the introduction, our approach to real analysis will be to start by laying out a set of axioms that we assume are true about the real number system \(\R\) and then prove everything else based upon this set of axioms.

In order to properly take this approach, we need a set of axioms that:

contains enough information to let us prove everything there is to know about \(\R\) (all of calculus, for example),
is specific enough so that it distinguishes \(\R\) from other number systems, and
is as simple as possible

The purpose of the first bullet point is clear enough: if I don’t assume anything, I won’t be able to prove anything.

The second bullet point about distinguishing \(\R\) from other number systems is related, but deserves some explanation. Consider the following proposition:

For every \(x>0\), there exists \(y\) such that \(y^2=x.\)

This is true if our universe is \(\R\) but false our universe is \(\Q.\) If our axioms for \(\R\) do not distinguish it from \(\Q\), we will have no hope of proving such a proposition. In short, I want a set of axioms that is true for \(\R\) but for no other number system.

Note

I am using the term number system a bit loosely. For our purposes, though, it’s fine to think of any set with binary operations corresponding to addition and multiplication—think \(\mathbb{N},\mathbb{Z},\mathbb{Q},\mathbb{R},\mathbb{C}.\)

Look ahead to the definition of a field (Definition 2.2) and definition of an ordered field (Definition 2.7). See if you can convince yourself the real number system does indeed satisfy both of these definitions. For example, every real number \(x\) does in fact have an additive inverse (a number \(-x\) that be added to \(x\) to obtain \(0\)); it is true that when you multiply two positive real numbers you get another positive real number.

Take another moment to scan some of the theorems that follow each of these axioms. None of these should be surprising—they are all well-known facts about how arithmetic and inequalities work in \(\R\)—but what may be surprising is that extremely simple facts like \(0x=0\) (Theorem 2.4) and \(1>0\) (Theorem 2.9) can be proven from even simpler assumptions.

For now, let’s focus on thinking about which other familiar number systems satisfy the field and order axioms.

Exercise 2.1 Think about some familiar number systems that aren’t the real numbers: \(\mathbb{N}\), \(\mathbb{Z}\), \(\mathbb{Q}\), and \(\mathbb{C}.\) You may rely on prior knowledge for this problem—no need to show all of your work.

Which of these are fields? For those that are not, which properties do they fail to have?
Which of these are ordered fields? Which of these satisfy the two conditions of Definition 2.7 despite not actually being a field?
(Exploratory) You should have found that one of the four number systems is a field and an ordered field. The Intermediate Value Theorem (for example) is true in \(\R\), but it is not true in any of these other four number systems. With that information, do you think one would be able to prove the Intermediate Value Theorem in \(\R\) if our only assumptions about \(\R\) are the field and order axioms?

2.2 Fields

For our purposes, a number system \(\mathbb{S}\) is a set with two binary operations \(+\) and \(\cdot.\) This is a nonstandard term, but we won’t use it much beyond this section.

Definition 2.2 A number system \(\mathbb{S}\) is said to be a field if it satisfies the following properties.

Addition is commutative (for all \(a,b\in\mathbb{S}\), \(a+b=b+a\)).
Addition is associative (for all \(a,b,c\in\mathbb{S}\), \((a+b)+c=a+(b+c)\)).
There is an element \(0\in\mathbb{S}\), called the additive identity such that for all \(a\in\mathbb{S}\), \(a+0=a.\)
Every \(a\in\mathbb{S}\) has an additive inverse \(-a\) with the property that \(a+(-a)=0.\)
Multiplication is commutative (for all \(a,b\in\mathbb{S}\), \(a\cdot b=b\cdot a\)).
Multiplication is associative (for all \(a,b,c\in\mathbb{S}\), \((a\cdot b)\cdot c=a\cdot(b\cdot c)\)).
There is an element \(1\in\mathbb{S}\) different from 0, called the multiplicative identity such that for all \(a\in\mathbb{S}\), \(a\cdot 1=a.\)
Every \(a\in\mathbb{S}\setminus\{0\}\) has an multiplicative inverse \(a^{-1}\) with the property that \(a\cdot(a^{-1})=1.\)
The distributive rule holds: for all \(a,b,c\in\mathbb{S}\), \(a(b+c)=a\cdot b+a\cdot c\).

Collectively, these properties tell us how arithmetic works for fields. And the first axiom we make about the real number system is that it is indeed a field.

Axiom 2.3 \(\R\) is a field.

Note that there is no reference to the operations of subtraction and division in the definition of a field, but these can be defined in terms of the established properties: \(a-b\) is defined to be \(a+(-b)\) (after all, \(-b\) is defined above), and if \(b\neq0\), we define \(\frac{a}{b}\) to be \(a\cdot b^{-1}\) (after all, \(b^{-1}\) is defined above).

What is somewhat remarkable is that all of the other rules of arithmetic for the real numbers can be derived from the field axiom. For example, you may have supposed that the following theorem would be an assumption, but one can prove it from the even more basic assumptions in the definition of a field.

Theorem 2.4 For any \(x\in\R\), \(0x=0.\)

Exercise 2.5 Prove Theorem 2.4.

Hint

Note that Theorem 2.4 involves both multiplication and, implicitly, addition (since it involves the additive identity). Thus its proof must involve the one part of the field axiom that combines addition and multiplication: the distributive property.

Any facts that we can prove about \(\R\) that use only the assumption that \(\R\) is a field, will in fact be true for any field. And there are some weird fields.

Exercise 2.6 Consider the number system \(\mathbb{F}_2=\{\alpha,\beta\}\) consisting of just two elements, where the addition and multiplication operations are defined by \[\alpha+\alpha=\beta \qquad \alpha+\beta=\alpha \qquad \beta+\alpha=\alpha \qquad \beta+\beta=\beta\] and \[\alpha\cdot\alpha=\alpha \qquad \alpha\cdot\beta=\beta \qquad \beta\cdot\alpha=\beta \qquad \beta\cdot\beta=\beta.\]

It turns out that \(\mathbb{F}_2\) is a field. Which element of \(\mathbb{F}_2\) is \(0?\) Which is \(1?\) Which is \(-1?\)

Hint

Asking which element is \(0\) means asking which element is the additive identity—the thing that can be added to any element and get that same element back. Similarly for the multiplicative identity \(1.\) And asking which element is \(-1\) means asking which element is the additive inverse of \(1\) (the thing you add to \(1\) and get back \(0\)).

So the field axiom alone fails miserably at the goal of distinguishing \(\R\) from other number systems—it cannot even distinguish \(\R\) from a number system with just two elements! So we need more axioms capturing more of the structure of \(\R.\)

2.3 Order and Inequalities

Another important feature of the real number system is the behavior of inequalities and how they interact with arithmetic. This is captured by the next axiom.

Definition 2.7 A field \(\mathbb{F}\) is called an ordered field if it has a relation \(>\) with the following properties.

For all \(a\in\mathbb{F}\), exactly one of the following is true:
- \(a>0\),
- \(a=0\), or
- \(-a>0.\)
If \(a>0\) and \(b>0\), then \(a+b>0\) and \(a\cdot b>0.\)
\(b>a\) if and only if \(b-a>0.\)

We can then define relations \(<,\leq,\geq\) in the usual way based on the relation \(>\)¹.

Axiom 2.8 \(\R\) is an ordered field.

Again, the properties listed in Definition 2.7 are not surprising; we already knew they were true for \(\R.\) But more surprising is the fact the everything else we know about how inequalities behave and interact with arithmetic can be prove from Axiom 2.3 and Axiom 2.8.

Theorem 2.9 \(1>0.\)

Exercise 2.10 Prove Theorem 2.9 using only the Field and Order Axioms.

You are probably familiar with the basics of how inequalities and arithmetic interact, and henceforth, we can use them at will without citation. In this class, being fluent with the rules for inequalities will be essential, though, so we collect a list of “inequality rules” (all of which could in principle be proven from Axiom 2.3 and Axiom 2.8) in Section 2.4.

Worth reflecting on now is where we stand on our stated goals in creating the axioms. The Axiom 2.8 was successful in distinguishing \(\R\) from \(\mathbb{F}_2\) because there is no ordering on \(\mathbb{F}_2\) satisfying Definition 2.7 (you will prove this in Exercise 2.11 below).

However, we can note that \(\mathbb{Q}\) is also an ordered field since it satisfies Axiom 2.8. This means that anything we can prove directly from Axiom 2.3 and Axiom 2.8 will apply just as well if our universe were \(\mathbb{Q}\) as it would if our universe were \(\mathbb{R}.\) In particular, we would have no hope of proving, say, the Intermediate Value Theorem, which is true in \(\mathbb{R}\) but not in \(\mathbb{Q}.\)

This all means that we need another assumption—an axiom that will distinguish the real number system from the rationals in a way that will allow us to prove everything else we’ll need in order to do calculus.

Exercise 2.11 Prove that the following fields are not ordered.

\(\mathbb{F}_2\) (defined in Exercise 2.6)
The complex numbers \(\mathbb{C}\)

Hint

This is a good opportunity to use a proof by contradiction. What contradiction can you create by assuming there is an ordering on \(\mathbb{F}_2\) satisfying the conditions of Definition 2.7? What about in \(\mathbb{C}\)?

The following exercise says something very simple, but ends up being very useful in practice.

Exercise 2.12 Suppose \(r\geq0\) and that for all \(\epsilon>0\), \(r\leq \epsilon.\) Prove that \(r=0.\)

Note

Now that you’ve finished this section, you can use whatever facts you want about arithmetic and order in \(\R\) without citing theorems or axioms.

2.3.1 Review Questions

Give some examples of number systems that are fields. Give some examples of number systems that are not fields.
Give some examples of fields that are ordered fields. Give some examples of fields that are not ordered fields.
Summarize for yourself how we can be certain that it is impossible to prove a statement like “every positive number in \(\R\) has a square root” building off of only the field and order axioms.

2.4 Inequality Rules

\(a\) is positive \(\Leftrightarrow\) \(-a\) is negative
\(1>0\)
\(x\leq y\) and \(y\leq z\) \(\Rightarrow\) \(x\leq z\)
\(a>b\) and \(c\geq d\) \(\Rightarrow\) \(a+c>b+d\)
\(a>b>0\) and \(c\geq d>0\) \(\Rightarrow\) \(ac>bd\)
\(a>b\) and \(c<0\) \(\Rightarrow\) \(ac<bc\)
\(a<0\) and \(b<0\) \(\Rightarrow\) \(a+b<0\)
\(a<0\) and \(b<0\) \(\Rightarrow\) \(ab>0\)
\(a>0\) and \(b<0\) \(\Rightarrow\) \(ab<0\)
\(a>0\) \(\Leftrightarrow\) \(\frac{1}{a}>0\)
\(a>1\) \(\Rightarrow\) \(a^2>a\)
\(0<a<1\) \(\Rightarrow\) \(a^2<a\)
\(a>b>0\) \(\Rightarrow\) \(a^2>b^2\)

That is, we define \(a<b\) to mean \(b>a\), \(a\leq b\) to mean \(a<b\) or \(a=b\), and \(a\geq b\) to mean \(a>b\) or \(a=b.\)↩︎