Section 2.1 Ordered Fields
In the previous chapter we have discussed the rational numbers \({\mathbb Q}\) and said that they form an ordered field. We now take a different, and abstract, point of view. We extract some of the properties of the rational number, and list them as axioms. We will then say that any set that satisfies those properties is called an ordered field.
All the other properties of the ordered field must be consequences of these axioms. It is important to distinguish between axioms, and properties that derive from them (that we can call theorems). So for example \(x+0=x\) is one of the axioms, that cannot be proved. But instead \(x\cdot 0 =0\) is a theorem (of which we will soon see the proof).
Bear in mind that the elements of the set \(F\) we are about to introduce are not necessarily rational number. We think of them just as abstract elements of a set that satisfy a list of axioms. Later in the chapter we will add one more crucial axiom, and we will call the resulting set satisfying all those properties the Real Numbers.
Subsection 2.1.1 The field axioms
We start with a set \(F\text{.}\) Recall the definition of binary operation from the previous chapter: it is any function with domain \(F\times F\) and range \(F\) (recall that \(F\times F\) is the set of all pairs \((x,y)\) of elements of \(F\)). We now assume that our set \(F\) has two binary operations, called addition and multiplication. Given an input \((x,y)\text{,}\) the output of addition is denoted by \(x+y\) and the output of multiplication is denoted by \(xy\) (or sometimes \(x\cdot y\)). These operations satisfy the following axioms:
- Axiom A1 (Commutativity)
- \(x+y=y+x \hspace{5ex}\) for all \(x,y\in F\)
- Axiom A2 (Associativity)
- \(x+(y+z)=(x+y)+z \hspace{5ex}\) for all \(x,y,z \in F\)
- Axiom A3 (Additive identity)
- There is a unique element \(0\in F\text{,}\) called the additive identity, such that \(x+0=x
\hspace{1ex} \mbox{ for all } x\in F\)
- Axiom A4 (Additive inverse)
- For each \(x\in F\text{,}\) there is a unique element \(-x\in F\text{,}\) called the additive inverse, such that \(x+(-x)=0\text{.}\)
Any set \(F\) with a binary operation \(+\) that satisfies axioms A1-A4 is called an abelian group. So for example the integers \(\mathbb Z\) with the operation of addition are an abelian group. Of course a common name for the additive identity af axiom A3 is “zero”.
The multiplication operation satisfies the following axioms:
- Axiom M1 (Commutativity)
- \(xy=yx \hspace{1ex}\) for all \(x,y\in F\)
- Axiom M2 (Associativity)
- \(x(yz)=(xy)z \hspace{5ex}\) for all \(x,y,z \in F\)
- Axiom M3 (Multiplicative identity)
- There is a unique element \(1\in F\text{,}\) called the multiplicative identity, such that \(1\neq 0\) and \(1x=x
\hspace{1ex} \mbox{ for all } x\in F\)
- Axiom M4 (Multiplicative inverse)
- For each non-zero element \(x\in F\text{,}\) there is a unique element \(x^{-1}\in F\text{,}\) called the multiplicative inverse, such that \(xx^{-1}=1\text{.}\)
Note that the symmetry between the two operations is not perfect, because we require \(x\neq 0\) for the existence of the multiplicative inverse, but there is no restriction on the existence of the additive inverse. Another way to say this is that \(-1\) is defined, but \(0^{-1}\) is not. We now add one more axiom that relates the two operation, and further breaks the symmetry.
- Axiom D (Distributive law)
- \(x(y+z)=xy+xz \hspace{3ex} \) for all \(x,y,z\in F\)
Any set with two binary operations satisfying A1-A4, M1-M3, D is called a commutative ring. So the integers with the operations \(+\) and \(\cdot\) form a commutative ring. If axiom M4 is also satisfied, then \(F\) is called a field. So the rational numbers with the operations \(+\) and \(\cdot\) are an example of a field. But there are other examples. An important one is the set \({\mathbb Q}[i]=\{a+ib| a,b \in {\mathbb Q}, i^2=-1\}\text{,}\) discussed in the Exercises.
Subsection 2.1.2 Uniqueness questions
Note that in the definition of the additive identity \(0\text{,}\) the additive inverse \(-x\text{,}\) the multiplicative identity \(1\) and the multiplicative inverse \(x^{-1}\) we always included the word unique. This means that there is only one zero, only one additive inverse, and so on. While we have included unique in the axioms, in reality the fact that \(0\) is unique is a consequence of the other axioms.
But how do we prove that something is unique? The general strategy to prove that some mathematical object is unique is to assume that there are two such objects, and then prove that they have to be the same. So suppose we want to prove that \(0\) is unique. Then we assume that there is a second zero, that we denote by \(0'\text{.}\) So this means that \(x+0'=x\) must be true for all \(x \in F\text{.}\) Using this with \(x=0\text{,}\) we then find \(0+0'=0\text{.}\) But we can also use \(x=0'\) in Axiom A3, to find \(0'+0=0'\text{.}\) So we conclude that \(0'=0'+0=0\text{.}\) In a similar way we can prove that \(1, -x,x^{-1}\) are unique.
Subsection 2.1.3 Axiomatic proofs
We stress once more that a field is defined by the nine axioms A1-A4, M1-M4, D. This means that we must be able to prove any statement we make about a field using only these nine axioms. This may be hard to do at first, because we routinely use rules from algebra and arithmetic, as they are taught in school, that are not found on the list of the nine axioms. In fact, any such rules can be broken down into steps that use only the nine axioms. An example will illustrate this.
Example 2.1.1.
Suppose that \(a, b\in F\text{,}\) and \(a\neq 0\text{.}\) Solve the equation for \(x\text{:}\)
\begin{equation*}
ax+b=0.
\end{equation*}
A typical solution, showing all steps, will be as follows. Subtract \(b\) from both sides, and get
\begin{equation*}
ax+b - b = 0-b.
\end{equation*}
Now simplify, to get
\begin{equation*}
ax=-b.
\end{equation*}
Divide by \(a\text{:}\)
\begin{equation*}
\frac{ax}{a}=\frac{-b}{a}
\end{equation*}
and simplify, to find the answer:
\begin{equation*}
x=-\frac{b}{a}
\end{equation*}
But none of these steps is found in any of the nine axioms! In fact, we never even mentioned subtraction and division in the axioms. At this point, we don’t know what those operations are. In spite of this, the steps we took really follow directly from the axioms, and our solution is, of course, correct. We will now solve the equation again using only the field axioms, but following the steps of our “normal” solution as a guide. We will refer to such a proof as an axiomatic proof. We start with the given equation:
\begin{equation*}
ax+b=0
\end{equation*}
The first step was “subtract \(b\)”. We don’t know what that means, but we have Axiom A4 that says that \(b\) has an additive inverse, denoted by \(-b\text{.}\) We can then add it to both sides of the equation, because we do have the addition operation. But we need to be careful. The left side of the equation is \(ax+b\text{,}\) so we need to add the \(-b\) to this, not just to \(b\text{.}\) This means we need to use parentheses:
\begin{equation*}
(ax+b)+(-b) = 0+(-b).
\end{equation*}
The next step should be clear: we want the \(-b\) to be added to the \(b\text{,}\) so we use Axiom A2 to re-arrange the parentheses, and on the right side we use the property of the additive identity (aka 0), Axiom A3:
\begin{equation*}
ax+(b+(-b))=-b.
\end{equation*}
Now we use Axiom A4 (the property of the additive inverse):
\begin{equation*}
ax+0=-b
\end{equation*}
and then Axiom A3 again:
\begin{equation*}
ax=-b.
\end{equation*}
The next step should be “divide by \(a\)”, but again, we don’t know what that means. But we do have Axiom M4, that says that \(a\text{,}\) since it is not zero, has a multiplicative inverse, that is denoted by \(a^{-1}\text{.}\) So we multiply both sides of the equation by \(a^{-1}\text{,}\) again carefully using parentheses:
\begin{equation*}
a^{-1}(ax) =a^{-1}(-b).
\end{equation*}
Now we can use Axiom M2:
\begin{equation*}
(a^{-1}a)x = a^{-1}(-b)
\end{equation*}
then the definition of multiplicative inverse, Axiom M4:
\begin{equation*}
1x=a^{-1}(-b)
\end{equation*}
and finally the definition of multiplicative identity, Axiom M3:
\begin{equation*}
x=a^{-1}(-b).
\end{equation*}
This is the solution of our equation. We almost found the same solution as before. As we will soon see, we can now consider \(\frac{-b}{a}\) just as another way to write \(a^{-1}(-b)\text{.}\) But the solution we found the “normal” way was \(-\frac{b}{a}\text{.}\) Is this the same as \(\frac{-b}{a}\text{?}\) Of course it is, but this fact is not found in the axioms. So it must be proved from the axioms, and we will soon do it.
The previous example shows that if we insist on using only the axioms to prove our statements, then even solving an equation as simple as \(ax+b=0\) becomes a fairly complex task. What about all the algebra skills and rules learned in many years of school? Have they now become useless? To the contrary, we will need to often use them. When solving exercises, unless the directions explicitly say “use only the field axioms”, you should continue to use all the algebra rules you have used so far.
But it is important to bear in mind that any such rule is really a theorem that can be proved from the field axioms. In fact, our first theorem will be the all-important property of \(0\) that we already mentioned, and that is not an axiom, but that you surely would use when solving a problem without stopping to wonder why it is true.
Theorem 2.1.2.
In any field \(F\text{,}\) \(x\cdot 0 = 0\) for all \(x\in F\text{.}\)
Hint.
Use the fact that \(0+0=0\) and the distributive law.
Axiomatic proof Using A3 with \(x=0\text{,}\) we find \(0=0+0\text{.}\) Then using the distributive law, we have
\begin{equation*}
x\cdot 0=x\cdot (0+0)= x\cdot 0+x\cdot 0.
\end{equation*}
Now add \(-(x\cdot 0)\) to both sides to get
\begin{equation*}
x\cdot 0 +-(x\cdot 0) = (x\cdot 0+x\cdot 0)+-(x\cdot 0).
\end{equation*}
Use the definition of additive inverse on the left and the associative law A2 on the right:
\begin{equation*}
0=x\cdot 0+(x\cdot 0+-(x\cdot 0))
\end{equation*}
Now use the definition of additive inverse again:
\begin{equation*}
0=x\cdot 0+0
\end{equation*}
and finally use A3 to find
\begin{equation*}
0=x\cdot 0.
\end{equation*}
The next theorem can be used to see that the two solutions
\(\frac{-b}{a}\) and
\(-\frac{b}{a}\) of
Example 2.1.1 are the same (see
Exercise 2.3.2 1.).
Theorem 2.1.3.
For each \(x\in F\text{,}\) \((-1)x=-x.\)
Hint.
Re-write \(x\) as \(1x\) and add it to \((-1)x\text{.}\)
Proof (Axiomatic proof). We need to show that
\((-1)x\) is the additive inverse of
\(x\text{.}\) This means that if we add it to
\(x\text{,}\) we should get 0. We find:
| \(x+(-1)x\) |
\(=1x+(-1)x\) |
By Axiom M3 |
|
\(=(1+(-1))x\) |
By Axiom D |
|
\(=0x\) |
By Axiom A4 |
|
\(=0\) |
By Theorem 2.1.2
|
Since
\((-1)x\) added to
\(x\) is 0, it must be
\(-x\text{.}\) Note that here we are using the fact that
\(-x\) is
unique.
Subsection 2.1.4 Subtraction and division
The operation of subtraction is a binary operation defined by:
\begin{equation*}
(x,y) \mapsto x+(-y).
\end{equation*}
Unfortunately, the symbol for this operation (the minus sign, \(-\)) is the same as the symbol used for the additive inverse. So \(x-y\) means \(x+(-y)\text{.}\) But the two concepts (additive inverse, and subtraction) are quite different. In fact, calculators have two different keys for the two meanings of the symbol \(-\text{.}\)
The subtraction operation is neither commutative, nor associative, as can be easily checked.
If \(y\neq 0\text{,}\) we can define the operation of division by
\begin{equation*}
(x,y)\rightarrow xy^{-1}.
\end{equation*}
This time, fortunately we have not just one, but two different and commonly used symbols for this operation: either \(x\div y\) or \(\frac{x}{y}\text{.}\) They both mean the same thing: \(xy^{-1}\text{.}\) In particular, since \(1\) is the multiplicative identity,
\begin{equation*}
\frac{1}{x}=1\cdot x^{-1}=x^{-1}\text{.}
\end{equation*}
Like subtraction, division is neither commutative nor associative.
Subsection 2.1.5 The order axioms
We now introduce four order axioms. Remember from the previous chapter that a binary relation on a set \(F\) is just any subset \(R\) of \(F\times F\text{.}\) We assume that there is a binary relation \(R\) on \(F\text{,}\) and we will write \(x\lt y\) just as a convenient way to say that \((x,y)\in R\text{.}\) So we will have no further use of the symbol \(R\text{,}\) and we will refer to the binary relation by the symbol \(\lt\text{.}\) We assume that the binary relation \(\lt\) satisfies the axioms:
- Axiom O1 (Total order)
- For all \(x,y\in F\text{,}\) either \(x\lt y\text{,}\) or \(y\lt x\text{,}\) or \(x=y\text{.}\)
- Axiom O2 (Transitivity)
- If \(x\lt y\) and \(y\lt z\text{,}\) then \(x\lt z\text{.}\)
- Axiom O3 (Preserves addition)
- If \(x\lt y\text{,}\) then \(x+z\lt y+z\) for all \(z\in F\text{.}\)
- Axiom O4 (Preserves multiplication by positive elements)
- If \(x\lt y\) and \(0\lt z\text{,}\) then \(xz\lt yz\text{.}\)
So the rational numbers together with the usual ordering form an ordered field. But there is no way to define an order satisfying the four order axioms on the field \({\cal Q}[i]\text{.}\) Axiom O1 is often referred to by saying that the relation \(\lt \) is a total order, and Axiom O2 says that \(\lt \) is transitive.
We will often write \(x\gt y\) to mean \(y\lt x\text{,}\) and we call the elements \(x\) of \(F\) such that \(x\gt 0\) positive. We also write \(x\leq y\) to mean that either \(x\lt y\) is true, or \(x=y\) is true, and similarly for \(x\geq y\text{.}\)
Axiom O3 is described by saying that \(\lt \) preserves addition, and Axiom O4 by saying that it preserves multiplication by positive elements. The next two theorems derive basic properties of the ordering relation that we routinely use, but that, once again, must be shown to derive directly only from the axioms. More properties are derived in the exercises.
Theorem 2.1.4.
Let \(F\) be an ordered field, and suppose \(x,y\in F\text{.}\) Then \(x\lt y\) is equivalent to \(-y\lt -x\text{.}\)
Hint.
To prove that \(x\lt y \Rightarrow -y\lt -x\text{,}\) use Axiom O3 to add \((-x)+(-y)\) to both sides of the equation, then use the axioms to simplify.
Proof (Axiomatic proof). Suppose that
\(x\lt y\text{.}\)
| \(x+((-x) +(-y)) \lt y+((-x) +(-y))\) |
By Axiom O3 |
| \((x+(-x)) +(-y) \lt y+((-y) +(-x))\) |
By Axioms A1 and A2 |
| \(0 +(-y) \lt (y+(-y)) +(-x)\) |
By Axioms A4 and A2 |
| \(-y \lt 0 +(-x)\) |
By Axioms A3 and A4 |
| \(-y \lt -x\) |
By Axiom A3 |
The implication
\(-y\lt -x\Rightarrow x\lt y\) is similar.
Theorem 2.1.5.
Suppose \(x,y,z \in F\text{,}\) \(x\lt y\text{,}\) and \(z\lt 0\text{.}\) Then \(zx \gt zy\)
Hint.
Use the previous theorems to prove that \(-z\gt 0\text{.}\) Then use Axiom O4.
We conclude this section with an all-important theorem. It is one of the simplest examples of what Analysis results are. We could say that while in Algebra we typically try to show that two quantities are equal, in Analysis most of the time we try to prove that one is smaller than the other. Instead of working in an abstract ordered field \(F\text{,}\) this time we state our theorem for the rational numbers \({\mathbb Q}\text{.}\) Then we can use all the properties and rules we already know for rational numbers. In this theorem, we also make our first encounter with the Greek letter \(\epsilon\) (pronounced epsilon), that is commonly used in Analysis to denote a generic positive number (and often thought of as very small).
Theorem 2.1.6.
Suppose that \(x,y \in {\mathbb Q}\text{,}\) and suppose that the following is true:
\begin{equation*}
\left(\forall \epsilon \gt 0\right) \left(x\leq y+\epsilon\right).
\end{equation*}
Then \(x\leq y\text{.}\)
Hint.
Prove the contrapositive.
Proof If
\({\cal P}: \left(\forall \epsilon \gt 0\right) \left(x\leq y+\epsilon\right)\) and
\({\cal Q}: x\leq y\text{,}\) we want to prove
\({\cal P} \Rightarrow {\cal Q}\text{.}\) The contrapositive is
\(\sim {\cal Q}\Rightarrow \sim {\cal P}\text{.}\)
| The negation of \({\cal Q}\) is |
\(\sim {\cal Q}: x\gt y\) |
| The negation of \(\cal P\) is |
\(\sim{\cal P}: \left(\exists \epsilon \gt 0\right) \left(x\gt y+\epsilon\right) \) |
So assume that
\(x\gt y\text{.}\) This means that
\(x-y\gt 0\text{.}\) Let
\(\epsilon=\frac{x-y}{2}\text{.}\) Then
\(\epsilon \gt 0\text{,}\) and
\begin{align*}
y+\epsilon\amp =y+\frac{x-y}{2}\\
\amp = y+\frac{x}{2}-\frac{y}{2}\\
\amp =\frac{y}{2}+\frac{x}{2}\\
\amp \lt \frac{x}{2}+\frac{x}{2}=x.
\end{align*}
This last line says that \(\sim{\cal P}\) is true. So we are done.
Subsection 2.1.6 Absolute value and inequalities
You have surely already used the absolute value function, and you know that it’s the function that “turns negative numbers into positive numbers”. In this section, we will give a precise definition of \(|x|\) that uses only the ordered field axioms. Then we discuss its meaning from a geometric point of view, and show how to re-write inequalities involving the absolute value.
Definition 2.1.7.
Let \(F\) be an ordered field, and \(x\in F\text{.}\) We define:
\begin{equation*}
|x|=\begin{cases} x \amp \mbox{ if } x\geq 0\\
-x \amp \mbox{ if } x\lt 0
\end{cases}
\end{equation*}
So we can think of \(|x|\) as the “size” of \(x\text{,}\) disregarding the sign. Another way to think of it is that \(|x|\) is the distance between \(x\) and \(0\text{.}\) See the picture, illustrating that \(3=|3|=|-3|\) is the distance from \(0\) of both \(3\) and \(-3\)
The next theorem summarizes the important properties of the absolute value function, that will often be used throughout this course.
Theorem 2.1.8.
\(|x|\geq 0\) for all \(x\in F\) .
Let
\(a\in F\text{,}\) \(a\geq 0\text{.}\) Then
\begin{equation*}
|x|\leq a \Longleftrightarrow -a\leq x \leq a.
\end{equation*}
Let
\(a,x_0\in F\text{,}\) \(a\geq 0\text{.}\) Then
\begin{equation*}
|x-x_0|\leq a \Longleftrightarrow x_0-a\leq x \leq x_0+ a.
\end{equation*}
For all \(x,y\in F\text{,}\) \(|xy|=|x||y|\text{.}\)
(Triangle inequality) For all \(x,y\in F\) \(|x+y|\leq |x|+|y|\text{.}\)
(Reverse triangle inequality) For all \(x,y\in F\text{,}\) \(||x|-|y||\leq |x-y|\text{.}\)
Note that combining e. and f.in
Theorem 2.1.8, and using
\(|-y|=|y|\text{,}\) we find that for all
\(x,y\in F\text{,}\)
\begin{equation*}
||x|-|y||\leq |x\pm y|\leq |x|+|y|.
\end{equation*}
Hint.
For part b., consider the two cases \(x\geq 0\) and \(x\lt 0\text{.}\) For part c., use part b. For part d., consider the cases \(xy\geq 0\) and \(xy\lt 0\text{.}\) For part e., note that \(-|x|\leq x \leq |x|\) and \(-|y|\leq y \leq |y|\) and add the two inequalities. For part f., note that \(|x|=|x-y+y|\leq |x-y|+|y|\text{,}\) and so \(|x|-|y|\leq |x-y|\text{,}\) then do the same switching \(x\) and \(y\text{.}\)
Subsection 2.1.7 Intervals
Definition 2.1.9.
An interval in an ordered field \(F\) is a subset \(I\) of \(F\) with the property that if \(x,y \in I\) with \(x\lt y\) and \(x\lt z \lt y\text{,}\) then \(z\in I\text{.}\)
Intuitively, an interval
\(I\) is just one connected piece, without gaps or holes. This is in accordance with the familiar notion of intervals as “stretches” of the real line. We will use the usual notation and write
\([a,b]\) to for the “stretch” that starts at
\(a\) and ends at
\(b\text{,}\) and replace the brackets
\([\text{ } \text{ } ]\) with the parentheses
\((\text{ } )\) if the endpoints are excluded. The next definition makes a list of all the subsets of that form, and in the next section we will see that in fact the list consists of exactly all the possible types of intervals. But
Definition 2.1.9 should be remembered as the one property that is common to all of them, and defines what an interval is.
Definition 2.1.10.
If \(a\in F\text{,}\) we define:
\begin{align*}
(-\infty,a)\amp=\{x\in F: x\lt a\}\\
(-\infty,a]\amp=\{x\in F: x\leq a\}\\
(a,\infty)\amp=\{x\in F: x\gt a\}\\
[a,\infty)\amp=\{x\in F: x\geq a\}.
\end{align*}
If \(a,b\in F\) and \(a\lt b\text{,}\) we define
\begin{align*}
(a,b)\amp=\{x\in F: a\lt x \lt b\}\\
(a,b]\amp=\{x\in F: a\lt x \leq b\}\\
[a,b)\amp=\{x\in F: a\leq x \lt b\}\\
[a,b]\amp=\{x\in F: a\leq x \leq b\}.
\end{align*}
We also write \((-\infty,\infty)=F.\)
