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# 2.7 Absolute Value

2.130   Definition (Absolute value.) Let be an ordered field, and let . Then we define

2.131   Remark. It follows immediately from the definition that
1. for all
2. for all .
3. .

2.132   Theorem. Let be an ordered field. Then for all ,
 (2.133)

Proof: If then (2.133) becomes , which is true. If then . If then . Hence (2.133) holds in all cases.

2.134   Exercise. Let be an ordered field. Prove that for all and for all .

2.135   Exercise (Product formula for absolute value.) A Prove that for all ,

2.136   Theorem. Let be an ordered field, let , and let with . Then
 (2.137)

and
 (2.138)

Proof: We first show that

 (2.139)

Case 1.
If , then

Case 2.
If , then

Case 3.
If , then is true, so (2.139) is true. Hence (2.139) is valid in all cases.

Next we show that

 (2.140)

Case 1.
If , then

Case 2.
If , then

Case 3.
If then is true, so (2.140) is true. Hence (2.140) is true in all cases.
We have proved (2.137).

Since is true if and only if is true,

i.e.,

This is 2.138.

2.141   Remark. I leave it to you to check that (2.137) holds when is replaced by , and (2.138) holds when and are replaced by and , respectively.

2.142   Theorem (Triangle inequality.)Let be an ordered field. Then
 (2.143)

Proof: The obvious way to prove this is by cases. But there are many cases to consider, e.g. ( and and ). I will use an ingenious trick to avoid the cases. For all , we have

and

By adding the inequalities, we get

By theorem 2.136 it follows that .

2.144   Exercise. A Let be an ordered field. For each statement below, either prove the statement, or explain why it is not true.
a)
for all , .
b)
for all , .

2.145   Exercise (Quotient formula for absolute value.) A Let be an ordered field. Let with . Show that
a)
,
b)
.

2.146   Definition (Distance.) Let be an ordered field, and let . We define the distance from to to be .

2.147   Remark. If is the ordered field of real or rational numbers, represents the familiar notion of distance between the points on the real line (or the rational line).

2.148   Exercise. Let be an ordered field. Let with . Show that
 (2.149)

HINT: Use theorem 2.136. Do not reprove theorem 2.136.

2.150   Remark. We can state the result of exercise 2.148 as follows. Let , and let . Then the set of points whose distance from is smaller than , is the set of points between and .

2.151   Definition (Intervals and endpoints.) Let be an ordered field. Let with . Then we define

A set that is equal to a set of any of these nine types is called an interval. Note that and , so the empty set is an interval and so is a set containing just one point. Sets of the first four types have endpoints and , except that has no endpoints. Sets of the second four types have just one endpoint, namely . The interval has no endpoints.

2.152   Examples. Let be an ordered field. By exercise 2.148 the set of solutions to is

I can read this result from the figure by counting 4 units to the left and right of 3. This method is just a way of remembering the result of theorem 2.148.

Now suppose I want to find the solutions in to

 (2.153)

Here, thinking of as the distance from to , I want to find all elements that are nearer to 2 than to 5. From the picture I expect the answer to be . Although this picture method is totally unjustified by anything I've done, it is the method I would use to solve the inequality in practice. If I had to use results we've proved to solve (2.153), I'd say (since )

which agrees with my answer by picture.

2.154   Exercise. A Let be an ordered field, let with . Show that

Interpret the result geometrically on a number line.

2.155   Exercise. Let be an ordered field. Express each of the following subsets of as an interval, or a union of intervals. Sketch the sets on a number line.
a)
b)
c)
d)

2.156   Note. Girolamo Cardano (1501-1576), in an attempt to make sense of the square root of a negative number, proposed an alternate law of signs in which the product of two numbers is negative if at least one factor is negative. He concluded that plus divided by plus gives plus'', and minus divided by plus gives minus'', but plus divided by minus gives nothing'' (i.e. zero), since both of the assertions plus divided by minus gives plus'' and plus divided by minus gives minus'' are contradictory.[40, p 25]

I believe that our axioms for an ordered field are due to Artin and Schreier in 1926 [6, page 259].

Systems satisfying various combinations of algebraic and order axioms were considered by Huntington [28] in 1903.

The notation for absolute value was introduced by Weierstrass in 1841[15, vol.2, page 123]. It was first introduced for complex numbers rather than real numbers.

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