As further examples of applications of the axioms, we deduce some corollaries to Axioms VII-IX. They apply to any ordered field.
Corollary 7. If \(x\) is a positive element of an ordered field \(F\), then \(-x\) is negative; and if \(x\) is negative, \(-x\) is positive.
Proof. Given \(x>0\), we may add \((-x)\) to both sides, by Axiom IX, \begin {align*} x + (-x) >0 +(-x).\;\; \text {i.e.,}\;\; 0>-x, \end {align*}
as required. Similarly it can be shown that \(x<0\) implies \(-x>0\). □
Corollary 8 (Addition and Multiplication of inequalities). If \(a,b,x,y\) are elements of an ordered field \(F\), such that \(a<b\) and \(x<y\), then \begin {align*} a+x<b+y \end {align*}
(i.e., we may always add two inequalities).
If, further, \(a,b,x,y\) are positive , then \(a<b\) and \(x<y\) implies \(ax<by\).
(i.e., the inequalities may be multiplied)
Proof. Suppose \(a<b\) and \(x<y\), with \(a,b,x,y\) positive. Then, multiplying the first inequality by \(x\), and the second by \(b\) (Axiom IX), we have, \begin {align*} ax<bx\;\;\text {and}\;\; bx<by. \end {align*}
Hence, by transitivity, \(ax<bx<by\), i.e., \(ax<by\), as required. □
Corollary 9. All nonzero elements of an ordered field have positive squares. That is, if \(a\neq 0\), then \(a^2=a\cdot a >0\). (Hence \(1= 1^2>0\).)
Proof. As \(a\neq 0\), we have, by trichotomy (Axiom VII), either \(a>0\) or \(a<0\).
If \(a>0\), then we may multiply by \(a\), obtaining, \(a\cdot a > a\cdot 0\), by Axiom IX, i.e., \(a^2>0\).
If \(a<0\), then by corollary 7, \(-a>0\); so we can multiply the inequality \(a<0\) by \((-a)\), using again
Axiom IX, we then obtain \begin {align*} a(-a)<0\cdot (-a) =0 \end {align*}
i.e., \(-a^2<0\), whence \(a^2>0\), as required. □
Definition 1.3.1 (Absolute value).
Given an element \(x\) of an ordered field \(F\), we define its absolute value, denoted as \(|x|\), as follows:
If \(x\geq 0\), then \(|x|=x\); if, however, \(x<0\), then \(|x|=-x\).
In particular, \(|0|=0\). It follows that \(|x|\) is always nonnegative. In fact, if \(x\geq 0\), then \(|x|=x\); and if \(x<0\), then by corollary 7, \(-x>0\); and here \(-x = |x| >0\). Moreover, we always have \begin {align} -|x| \leq x \leq |x|. \end {align}
For, if \(x \geq 0\), then, \(|x|=x\) by definition, and \(-|x| =-x \leq 0 \leq x\). If, however, \(x<0\), then \(|x|>x\), since \(|x|\) is positive, while \(x\) is negative and \(x=-|x|\). Thus, this holds for both the cases.
Proof. Suppose first that \(|x|<y\). Then by formula (1.1), we have \(x \leq |x| <y\), whence \(x<y\). It remains to prove that \(-y<x\). This is certainly true if \(x\) is non negative (for \(-y\) is negative here). If, however, \(x\) is negative, then by definition, \(-x =|x|\), whence \(-x<y\); that is \(-y<x\). Thus, in all cases, \(|x|<y\) implies \(-y<x<y\).
The converse is proven in a similar way, by distinguishing two cases: \(x\geq 0\) and \(x<0\). Suppose that \(-y < x <y\). For the case where \(x \geq 0\), by definition, \(|x| =x <y\), but when \(x<0\), by definition, we have \(|x|=-x\), because \(-y<x\), \(-x<y\), which means that \(|x| =-x <y\). Thus, in all cases, \(-y<x<y\) implies \(|x|<y\). Proving the equivalence. □
Corollary 11. For any elements \(a\) and \(b\) of an ordered field \(F\), we have \begin {align*} |ab| = |a|\cdot |b|. \end {align*}
If, further \(b\neq 0\), then \begin {align*} \frac {|a|}{|b|} = \left | \frac {a}{b} \right |. \end {align*}
Proof. For the proof, consider the four possible cases:
(1) \(a\geq 0,\: b \geq 0\); (2) \(a\geq 0,\: b<0\); (3) \(a<0,\: b\geq 0\); (4) \(a<0,\: b<0\)
(1)
Because, \(a\geq 0,\: b \geq 0\), \(|a| =a\), and \(|b| =b\), by the definition of absolute value. Additionally, by Axiom IX,
we have \(ab\geq 0\), Hence, \(|ab|=ab = |a|\cdot |b|\), from corollary 2. If \(b\neq 0\), the same reasoning applies but with
replacing \(b\) with \(b^{-1}\).
(2)
From earlier, \(|a|=a\), but \(|b|=-b\), by the definition of absolute value. From Corollary 6, we
have \(a(-b) = -(a\cdot b)\), hence we have, \(|a|\cdot |b|= a(-b)=-(a\cdot b)\). For \(|a\cdot b|\), because \(a\geq 0\) and \(b<0\), \(a\cdot b \leq 0\), if \(a\neq 0\), then \(a\cdot b<0\), by the monotonicity of
inequality (Axiom IX). Hence, \(|a\cdot b| = -(a\cdot b).\) Thus, (2) is also proven.
(3)
Trivial, see (2).
(4)
Because \(a<0\) and \(b<0\), \(|a| =-a\) and \(|b|=-b\), by the definition of absolute value. From Corollary 6, we
again have \(|a|\cdot |b| = (-a)\cdot (-b) =ab = a \cdot b\). Because \(-a>0\) and \(-b>0\), using Corollary 8, we get: \((-a)\cdot (-b) > 0 \cdot 0\). \(0\cdot 0= 0\) from Corollary 5,
Thus, \((-a)\cdot (-b)=ab >0\) and \(|ab|=ab\), giving the desired result. For division, again, replace, \(b\) with
\(b^{-1}\) □
Corollary 12 (Triangular Inequalities). For any elements \(a\) and \(b\) of an ordered field \(F\), we have:
Proof. Inequality (i) can be proved simply by using the definition of absolute value and Corollary 10. We have: \begin {align*} -|a| \leq a\leq |a| \;\;\text {and}\;\; -|b|\leq b \leq |b| \end {align*}
Adding the inequalities, we have: \begin {align*} -(|a|+|b|) \leq a+b \leq |a|+|b| \end {align*}
By Corollary 10, \(|a+b|\leq |a|+|b|\), as required.
To prove (ii), let \(x =a-b\). By (i), \(|x+b| \leq |x|+|b|\), i.e., \begin {align*} |(a-b) +b| \leq |a-b| +|b| \end {align*}
whence, \(|a|\leq |a-b|+|b|\), or \(|a|-|b| \leq |a-b|\). Similarly, interchanging \(a\) and \(b\): \begin {align*} |x+a|\leq |x|+|a|\\ \implies |(b-a)+a| \leq |b-a|+|a|\\ \implies -|a-b| \leq |a| -|b| \end {align*}
Hence, by Corollary 10, \(||a|-|b||\leq |a-b|.\) □
Corollary 13 (Density of an Ordered Field). Given any two elements \(a\) and \(b\) of an ordered field \(F\), there always is an element \(x \in F\) such that \(a<x<b\). (This element is said to lie between \(a\) and \(b\))
This is an important proposition. This is often expressed by saying that every ordered field is densely ordered.
Proof. Because we have \(a<b\), we have \(0<b-a\). Assume we have some element \(0<c<1\), where \(c \in F\) , by the monotonicity of multiplication (Axiom IX), \(0<c\cdot (b-a)\), additionally, because \(c<1\), again by the monotonicity of multiplication (Axiom IX), \(c\cdot (b-a)< 1\cdot (b-a)=b-a\), by property of multiplicative identity (Axiom IV). Thus we have this key inequality: \begin {align*} 0 <c\cdot (b-a) < b-a \end {align*}
Now, we add \(a\) to inequality, giving: \begin {align*} a+0< a+(c\cdot (b-a)) < a+(b-a) \end {align*}
By applying associativity, commutativity and additive zero properties, we have the desired result: \begin {align*} a<a+(c\cdot (b-a)) < b. \end {align*}
Hence, \(x= a+(c\cdot (b-a))\) is between \(a\) and \(b\), where \(0<c<1\). Indeed, this process can be repeated an infinite amount of times, hence this proposition can be strengthened to say that there are an infinite number of elements between \(a\) and \(b\). □