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2.101
Example.
The rational numbers
![$(\mbox{{\bf Q}},\mbox{${\mbox{{\bf Q}}}^{+}$})$](img573.gif)
form an ordered field, where
![$\mbox{${\mbox{{\bf Q}}}^{+}$}$](img574.gif)
denotes
the familiar set of positive rationals.
2.102
Notation (
.)
Let
![$(F,F^+)$](img576.gif)
be an ordered field. We let
We call
![$F^-$](img575.gif)
the set of
negative elements in
![$F$](img406.gif)
. Thus
and
We can restate the Trichotomy axiom as: For all
![$x\in F$](img489.gif)
, exactly one of the
statements
is true.
2.103
Theorem.
Let
be an ordered field. Then for all
,
.
Proof: Since
, we know
or
. Now
and
2.104
Corollary.
In any ordered field,
.
2.106
Remark.
The method used in the second proof above shows that none of the fields
![$\mbox{{\bf Z}}_n$](img363.gif)
are orderable.
2.107
Definition (
)
Let
![$(F,F^+)$](img576.gif)
be an ordered field, and let
![$a,b\in F$](img476.gif)
. We define
2.108
Remark.
In any ordered field
![$(F,F^+)$](img576.gif)
:
2.109
Exercise.
Let
![$(F,F^+)$](img576.gif)
be an ordered field, and let
![$a,b\in F$](img476.gif)
. Show that exactly one of
the statements
is true.
2.110
Theorem (Transitivity of
.)
Let
be an ordered field. Then for all
,
Proof: For all
we have
2.111
Exercise (Addition of inequalities.)
Let
![$(F,F^+)$](img576.gif)
be an ordered field, and let
![$a,b,c,d\in F$](img605.gif)
. Show that
and
2.112
Exercise.
Let
![$(F,F^+)$](img576.gif)
be an ordered field, and let
![$a,b\in F$](img476.gif)
. Show that
2.113
Notation.
Let
![$(F,F^+)$](img576.gif)
be an ordered field, and let
![$a,b,c,d\in F$](img605.gif)
. We use notation like
![\begin{displaymath}
a\leq b<c=d
\end{displaymath}](img609.gif) |
(2.114) |
to mean
![$(a\leq b)$](img610.gif)
and
![$(b<c)$](img611.gif)
and
![$(c=d)$](img612.gif)
, and similarly we write
![\begin{displaymath}
a>b=c\geq d
\end{displaymath}](img613.gif) |
(2.115) |
to mean
![$a>b$](img614.gif)
and
![$b=c$](img615.gif)
and
![$c\geq d$](img616.gif)
. By transitivity of
![$=$](img143.gif)
and of
![$<$](img602.gif)
, you can
conclude
![$a<c$](img617.gif)
from (
2.114), and you can conclude
![$b\geq d$](img618.gif)
and
![$a>c$](img619.gif)
from
(
2.115). A chain of inequalities involving both
![$<$](img602.gif)
and
![$>$](img620.gif)
shows bad
style, so you should not write
2.116
Exercise (Laws of signs.)
Let
![$(F,F^+)$](img576.gif)
be an ordered field, and let
![$a,b\in F$](img476.gif)
. Show that
and
and
and
These laws together with the axiom
are called the
laws of signs.
2.117
Notation.
Let
![$F$](img406.gif)
be an ordered field, and let
![$a,b$](img511.gif)
be non-zero elements of
![$F$](img406.gif)
. We say
![$a$](img24.gif)
and
have the same sign if either (
![$a,b$](img511.gif)
are both in
![$F^+$](img568.gif)
) or (
![$a,b$](img511.gif)
are both in
![$F^-$](img575.gif)
). Otherwise we say
and
have opposite signs.
2.118
Corollary (of the law of signs.)Let
be an ordered field and let
. Then
2.119
Notation.
I will now start to use the convention that `` let
![$F$](img406.gif)
be an ordered field" means ``let
![$\left(F, F^+\right)$](img630.gif)
be an ordered field''; i.e.,
the set of positive elements of
![$F$](img406.gif)
is assumed to be called
![$F^+$](img568.gif)
.
2.120
Exercise.
Let
![$F$](img406.gif)
be an ordered field and let
![$a,b,c\in F$](img504.gif)
. Prove that
2.121
Theorem (Multiplication of inequalities.)Let
be an ordered field and let
be elements of
. Then
Proof: By the previous exercise we have
Hence, by transitivity of
,
2.122
Exercise.
Let
![$F$](img406.gif)
be an ordered field, and let
![$a\in F\backslash\{0\}$](img635.gif)
. Show that
![$a$](img24.gif)
and
![$a^{-1}$](img636.gif)
have the same sign.
2.123
Exercise.
A
Let
![$F$](img406.gif)
be an ordered field, and let
![$a,b\in F\backslash\{0\}$](img628.gif)
.
Under what conditions (if any) can you say that
Under what conditions (if any) can you say that
2.124
Definition (Square root.)
Let
![$F$](img406.gif)
be a field, and let
![$x\in F$](img489.gif)
. A square root for
![$x$](img85.gif)
is any element
![$y$](img86.gif)
of
![$F$](img406.gif)
such that
![$y^2=x$](img639.gif)
.
2.125
Examples.
In
![$\mbox{{\bf Z}}_5$](img392.gif)
, the square roots of
![$-1$](img33.gif)
are
![$2$](img63.gif)
and
![$3$](img640.gif)
.
In an ordered field
, no element in
has a square root.
In
, there is no square root of
. (See theorem 3.45
for a proof.)
2.126
Theorem.
Let
be an ordered field and let
. If
has a square root, then it
has exactly two square roots, one in
and one in
, so if
has a square
root, it has a unique positive square root.
Proof: Suppose
has a square root
. Then
, since
. If
is any square root of
, then
, so, as we saw in theorem 2.95,
or
. By trichotomy, one of
is in
, and the other is in
.
Proof: Let
be elements of
. Then
, unless
,
so
. Hence
2.129
Remark.
The implication (
2.128) is also true when
![$<$](img602.gif)
is replaced by
![$\leq$](img654.gif)
in both
positions. I'll leave this to you to check.
Next: 2.7 Absolute Value
Up: 2. Fields
Previous: 2.5 Subtraction and Division
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