1.4: Ordered Field Axioms
- Page ID
- 49095
In this book, we will start from an axiomatic presentation of the real numbers. That is, we will assume that there exists a set, denoted by \(\mathbb{R}\), satisfying the ordered field axioms, stated below, together with the completeness axiom, presented in the next section. In this way we identify the basic properties that characterize the real numbers. After listing the ordered field axioms we derive from them additional familiar properties of the real numbers. We conclude the section with the definition of absolute value of a real number and with several results about it that will be used often later in the text.
We assume the existence of a set \(\mathbb{R}\) (the set of real numbers) and two operations \(+\) and \(\cdot\) (addition and multiplication) assigning to each pair of real numbers \(x, y\), unique real numbers \(x+y\) and \(x \cdot y\) and satisfying the following properties:
(1a) \((x+y)+z=x+(y+z) \text { for all } x, y, z \in \mathbb{R}\).
(1b) \(x+y=y+x \text { for all } x, y \in \mathbb{R}\).
(1c) There exists a unique element \(0 \in \mathbb{R}\) such that \(x+0=x\) for all \(x \in \mathbb{R}\).
(1d) For each \(x \in \mathbb{R}\), there exists a unique element \(-x \in \mathbb{R}\) such that \(x+(-x)=0\).
(2a) \((x \cdot y) \cdot z=x \cdot(y \cdot z) \text { for all } x, y, z \in \mathbb{R}\).
(2b) \(x \cdot y=y \cdot x \text { for all } x, y \in \mathbb{R}\).
(2c) There exists a unique element \(1 \in \mathbb{R}\) such that \(1 \neq 0\) and \(x \cdot 1=x\) for all \(x \in \mathbb{R}\).
(2d) For each \(x \in \mathbb{R} \backslash\{0\}\), there exists a unique element \(x^{-1} \in \mathbb{R}\) such that \(x \cdot\left(x^{-1}\right)=1\). (We also write \(1 / x\) instead of \(x^{-1}\).)
(2e) \(x \cdot(y+z)=x \cdot y+x \cdot z \text { for all } x, y, z \in \mathbb{R}\).
We often write \(xy\) instead of \(x \cdot y\).
In addition to the algebraic axioms above, there is a relation \(<\) on \(\mathbb{R}\) that satisfies the order axioms below:
(3a) For all \(x, y \in \mathbb{R}\), exactly one of the three relations holds: \(x=y\), \(y<x\), or \(x<y\).
(3b) For all \(x, y, z \in \mathbb{R}\), if \(x<y\) and \(y<z\), then \(x<z\).
(3c) For all \(x, y, z \in \mathbb{R}\), if \(x<y\), then \(x+z<y+z\).
(3d) For all \(x, y, z \in \mathbb{R}\), if \(x<y\) and \(0<z\), then \(xz<yz\).
We will use the notation \(x \leq y\) to mean \(x<y\) or \(x=y\). We may also use the notation \(x>y\) to represent \(y<x\) and the notation \(x \geq y\) to represent \(y<x\) and the notation \(x \geq y\) to mean \(x>y\) or \(x=y\).
A set \(\mathbb{F}\) together with two operations \(+\) and \(\cdot\) and a relation \(<\) satisfying the 13 axioms above is called an ordered field. Thus the real numbers are an example of an ordered field. Another example of an ordered field is the set of rational numbers \(\mathbb{Q}\) with the familiar operations and order. The integers \(\mathbb{Z}\) do not form a field since for an integer \(m\) other than \(1\) or \(-1\), its reciprocal \(1 / m\) is not an integer and, thus, axiom 2(d) above does not hold. In particular, the set of positve integers \(\mathbb{N}\) does not form a field either. As mentioned above the real numbers \(\mathbb{R}\) will be defined as the ordered field which satisfies one additional property described in the next section: the completeness axiom.
From these axioms, many familiar properties of \(\mathbb{R}\) can be derived. Some examples are given in the next proposition. the proof illusrates how the given axioms are used at each step of the derivation.
For \(x, y, z \in \mathbb{R}\), the following hold:
- If \(x+y=x+z\), then \(y=z\);
- \(-(-x)=x\);
- If \(x \neq 0\) and \(xy=xz\), then \(y=z\);
- If \(x \neq 0\), then \(1 /(1 / x)=x\);
- \(0 x=0=x 0\);
- \(-x=(-1) x\);
- \(x(-z)=(-x) z=-(x z)\).
- If \(x>0\), then \(-x<0\); if \(x<0\), then \(-x>0\);
- If \(x<y\) and \(z<0\), then \(xz>yz\);
- \(0<1\).
- Proof
-
- Suppose \(x+y=x+z\). Adding \(-x\) (which exists by axiom (2d)) to both sides, we have
\[(-x)+(x+y)=(-x)+(x+z).\]
Then axiom (1a) gives
\[[(-x)+x]+y=[(-x)+x]+z.\]
Thus, again by axiom (2d), \(0+y=0+z\) and, by axiom (1c), \(y=z\).
- Since \((-x)+x=0\), we have (by uniqueness in axiom (2d)) \(-(-x)=x\).
The proofs of c. and d. are similar.
- Using axiom (2e) we have \(0 x=(0+0) x=0 x+0 x\). Adding \(-(0 x)\) to both sides (axiom (2d)) and using axioms (1a) and (1c), we get
\(0=-(0 x)+0 x=-(0 x)+(0 x+0 x)=(-(0 x)+0 x)+0 x=0+0 x=0 x\).
That \(0 x=x 0\) follows from axiom (2b).
- Using axioms (2c) and (2e) we get \(x+(-1) x=1 x+(-1) x=(1+(-1)) x\). From axiom (2d) we get \(1+(-1)=0\) and from part (e) we get \(x+(-1) x=0 x=0\). From the uniqueness in axiom (2d) we get \((-1) x=-x\) as desired.
- Using axioms (2e) and (1c) we have \(x z+x(-z)=x(z+(-z))=x 0=0\). Thus, using axiom (2d) we get that \(x(-z)=-(x z)\). The other equality follows similarly.
- From \(x>0\), using axioms (3c) and (1c) we have \(x+(-x)>0+(-x)=-x\). Thus, using axiom (2d), we get \(0>-x\). The other case follows in a similar way.
- Since \(z<0\), by part (h), \(-z>0\). Then by axiom (3d), \(x(-z)<y(-z)\). Combining this with part (g) we get \(-x z<-y z\). Adding \(xz+yz\) to both sides and using axioms (1a), (3c), (1b), and (1c) we get \(x y=(-x z+x z)+x y=-x z+(x z+x y)<-x y+(x z+x y)=-x y+(x y+x z)=(-x y+x y)+x z=x z\).
- Axiom (2c) gives that \(1 \neq 0\). Suppose, by way of contradiction, that \(1 < 0\). Then by part (i), \(1 \cdot 1>0 \cdot 1\). Since \(1 \cdot 1=1\), by axiom (2c) and \(0 \cdot 1=0\) by part (e), we get \(1>0\) which is a contradiction. It follows that \(1>0\). \(\square\)
Note that we assume that the set of all natural numbers is a subset of \(\mathbb{R}\) (and of any ordered field, in fact) by identifying the \(1\) in \(\mathbb{N}\) with the \(1\) in axiom (2c) above, the number \(2\) with \(1+1\), \(3\) with \(1+1+1\), etc. Futhermore, since \(0<1\) (from part (j) of the previous proposition), axiom (3c) gives, \(1<2<3\), etc (in particular all these numbers are distinct). In a similar way, can include \(\mathbb{Z}\) and \(\mathbb{Q}\) as subsets.
We say that a real number \(x\) is irrational if \(x \in \mathbb{R} \backslash \mathbb{Q}\), that is, if it is not rational.
Given \(x \in \mathbb{R}\), define the absolute value of \(x\) by
\[|x|=\left\{\begin{array}{ll}
x, & \text { if } x \geq 0 \text {;} \\
-x, & \text { if } x<0 \text {.}
\end{array}\right.\]
Figure \(1.1\): The absolute value function.
The following properties of absolute value follow directly from the definition.
Let \(x, y, M \in \mathbb{R}\) and suppose \(M>0\). The following properties hold:
- \(|x| \geq 0\);
- \(|-x|=|x|\);
- \(|x y|=|x||y|\);
- \(|x|<M\) if and only if \(-M<x<M\). (The same holds if \(<\) is replaced with \(\leq\).)
- Proof
-
We prove (d) and leave the other parts as an exercise.
(d) Suppose \(|x|<M\). In particular, this implies \(M>0\). We consider the two cases separately: \(x \geq 0\) and \(x<0\). Suppose first \(x \geq 0\). Then \(|x|=x\) and, hence, \(-M<0 \leq x=|x|<M\). Now suppose \(x<0\). Then \(|x|=-x\). Therefore, \(-x<M\) and, so \(x>-M\). It follows that \(-M<x<0<M\).
For the converse, suppose \(-M<x<M\). Again, we consider different cases. If \(x \geq 0\), then \(|x|=x<M\) as desired. Next suppose \(x<0\). Now \(-M<x\) implies \(M>-x\). Then \(|x|=-x<M\). \(\square\)
Given \(x, y \in \mathbb{R}\),
\[|x+y| \leq|x|+|y|.\]
- Proof
-
From the observation above, we have
\[-|x| \leq x \leq|x|\]
\[-|y| \leq y \leq|y|.\]
Adding up the inequalities gives
\[-|x|-|y| \leq x+y \leq|x|+|y|.\]
Since \(-|x|-|y|=-(|x|+|y|)\), the conclusion follows form Proposition 1.4.2 (d). \(\square\)
For any \(x, y \in \mathbb{R}\),
\[|| x|-| y|| \leq|x-y|.\]
The absolute value has a geometric interpretation when considering the numbers in an ordered field as points on a line. the number \(|a|\) denotes the distance from the number \(a\) to \(0\). More generally, the number \(d(a, b)=\mid a-b\) is the distance between the points \(a\) and \(b\). It follows easily from Proposition 1.4.2 that \(d(x, y) \geq 0\), and \(d(x, y)=0\) if and only if \(x=y\). Moreover the triangle inequality implies that
\[d(x, y) \leq d(x, z)+d(z, y),\]
for all numbers \(x,y,z\).
Exercise \(\PageIndex{1}\)
Prove that \(n\) is an even integer if and only if \(n^{2}\) is an even integer. (Hint: prove the "if" part by contraposition, that is, prove that if \(n\) is odd, then \(n^{2}\) is odd.)
- Answer
-
Add texts here. Do not delete this text first.
Exercise \(\PageIndex{2}\)
Prove parts c. and d. of Proposition 1.4.1
- Answer
-
Add texts here. Do not delete this text first.
Exercise \(\PageIndex{3}\)
Let \(a, b, c, d \in \mathbb{R}\). Suppose \(0<a<b\) and \(0<c<d\). Prove that \(a c<b d\).
- Answer
-
Add texts here. Do not delete this text first.
Exercise \(\PageIndex{4}\)
Prove parts a., b., and c. of Proposition 1.4.2.
- Answer
-
Add texts here. Do not delete this text first.
Exercise \(\PageIndex{5}\)
Prove Corollary 1.4.4.
- Answer
-
Add texts here. Do not delete this text first.
Exercise \(\PageIndex{6}\)
Given two real numbers \(x\) and \(y\), prove that
\[\max \{x, y\}=\frac{x+y+|x-y|}{2} \text { and } \min \{x, y\}=\frac{x+y-|x-y|}{2}.\]
- Answer
-
Add texts here. Do not delete this text first.
Exercise \(\PageIndex{7}\)
Let \(x, y, M \in \mathbb{R}\). Prove the following
- \(|x|^{2}=x^{2}\).
- \(|x|<M\) if and only if \(x<M\) and \(-x<M\).
- \(|x+y|=|x|+|y|\) if and only if \(x y \geq 0\).
- Answer
-
Add texts here. Do not delete this text first.