The Rational Numbers Fields - Department of Mathematics and Statistics

The Rational Numbers

Fields

The system TM of integers that we formally defined is an improvement algebraically on = (we can subtract in TM). But TM still has some serious deficiencies: for example, a simple equation like $B % oe # has no solution in TM. We want to build a larger number

system, the rational numbers, to improve the situation.

In Chapter 3, we introduced the idea of an algebraic structure called a field and we proved, for example, that TM: is a field iff : is a prime number.

The fields axioms, as we stated them in Chapter 3, are repeated here for convenience.

Definition Suppose J is a set with two operations (called addition and multiplication) defined inside J . J is called a field if the following "field axioms" are true.

1) There are elements ! - J and " - J (and ! ? "?

2? aB aC aD ?B C? D oe B ?C D? 2w? aB aC aD ?B C) D oe B ?C D? (addition and multiplication are associative)

3? aB aC B C oe C B

3w? aB aC B C oe C B

(addition and multiplication are commutative)

4? aB aC aD B ?C D? oe B C B D (the distributive law connects addition and multiplication)

5? aB B ! oe B

5w) aB ?B ? ! ? B " oe B?

(0 and 1 are "neutral" elements for addition and multiplication. ! is called

the additive identity element and " is called the multiplicative identity element in

J)

6? aB bC B C oe ! (C is called an additive inverse of B)

6w? aB ?B ? ! ? ?bC? B C oe "? ?for each B ? !, such a C is called a multiplicative inverse of B?

In any field J , multiplication "B C" is often written as just "BC".

The (still informal) systems , `, and , are other examples of fields. So is the collection ?+ ,?# ? +? , - ?, as you proved in a homework assignment.

Much of the material about fields in the next few pages is material you've seen before. It's just collected in a systematic way here, partly for review.

Any particular field such as TM:, and ` is called a model for the field axioms. There are significant differences between these models. For example,

? TM$ has only 3 elements, while TM& has 5 elements and and ` are infinite fields. ? In TM$, " " " oe !? in TM&? " " " " " oe !; in and `, no finite sum of "'s has ! for a sum. .

In other words, we cannot say that "all fields look alike." In that respect, the field axioms are quite different from the axioms P1-P5) for a Peano system: there, we were able to argue that "all Peano systems look alike" or, in more formal language, that any two Peano systems are isomorphic.

An axiom system for which "all models are isomorphic" is called a categorical axiom system. The field axioms are not categorical.

Some useful theorems can be proved just using the field axioms: these theorems therefore apply in all fields. Proving them in the abstract is efficient; it saves the effort of proving them over and over each time a new field comes up. For example, the statements in the following theorem are true in every field TM: ?: a prime)? ? `? ,? ??? .

Theorem 1 Suppose B? ?? @ are a members of a field J .

a) (Cancellation for addition) If B ? oe B @, then ? oe @. b) (Cancellation for multiplication) If B ? ! and B? oe B@, then ? oe @? c) The additive inverse of B is unique. d) If B ? !, then the multiplicative inverse of B is unique. e) The additive identity, !, is unique and the multiplicative identity, ", are

unique. f) aB - J ? B ! oe ! g) If ?@ oe !, then ? oe ! or @ oe !.

Proof a) Suppose B ? oe B @? By Axiom 5), B has an additive inverse C. Then

C ?B ?? oe C ?B @? ?C B? ? oe ?C B? @ ?B C? ? oe ?B C? @ !?oe!@ ?!oe@! ? oe @?

(using Axiom 2) (using Axiom 3) (since C is an additive inverse for B) (using Axiom 3) (using Axiom 5)

b) If B ? !? then B has a multiplicative inverse C (Axiom 6w). So if B? oe B@, then

C?B?? oe C?B@? ?CB?? oe ?CB)@ ?BC?? oe ?BC?@ "?oe"@

?"oe@" ?oe@

(using Axiom 2w) (using Axiom 3w)

(since C is a multiplicative inverse

for B) (using Axiom 3w) (using Axiom 5w)

c) Suppose C? Cw - J . If B C oe ! and B Cw oe ! both are true, then B C oe B Cw? Adding C to both sides, we get

C ?B C? oe C ?B Cw? ?C B? C oe ?C B? Cw ?B C? C oe ?B C? Cw ! C oe ! Cw

C ! oe Cw ! C oe Cw

(using Axiom 2) (using Axiom 3) (because C is an additive inverse for B) (using Axiom 3) (using Axiom 5).

Since B has a unique additive inverse, we can talk about the additive inverse of B and give it a name: ? B?? Then B ? B? oe ! and (using Axiom 3) ? B? B oe !?

The last equation says that B is the additive inverse for B ? that is ? B? oe B? This is not some profound fact; it's just a consequence of the notation we chose for the additive inverse.

d) Suppose B ? !, that C? Cw - J . If BC oe " and BCw oe " both are true, then BC oe BCw? Multiplying both sides by C, we get

C?BC? oe C?BCw? ?CB?C oe ?CB?Cw ?BC?C oe ?BC?Cw " C oe " Cw

C " oe Cw " C oe Cw

(using Axiom 2w)

(using Axiom 23)

(since C is a multiplicative inverse for B) (using Axiom 3w) (using Axiom 5w)

If B ? !, then B has a unique multiplicative inverse, so we can talk about the multiplicative inverse of B and give it a name: B"? Then B B" oe " and (using Axiom 3w) B" B oe "?

The latter equation says that B is the multiplicative inverse for B" ? that is ?B"?" oe B?

e) Suppose D - J and that D is an additive identity, that is, aB B D oe B oe B !? Using part a) to cancel the B's, we get D oe !? .

Suppose A - J and that A is a multiplicative identity, that is, aB ? !? BA oe B oe B "? Using part b) to cancel the B's, we get A oe "?

f) Suppose B - J ? 0 is the additive identity in J , so ! ! oe !?

Therefore

B ?! !? oe B !

B!B!oeB!

(using Axiom 4)

Let C be the additive inverse of B !, that is C oe ?! B?

?! B? ?B ! B !? oe ?! B? B ! ? ?! B? B 0) B ! oe ?! B? B !

!B!oe ! B!oe!

Part g) is left as an exercise. ?

In a field, we can also define subtraction and division.

Definition Suppose B and C are members of a field J .

a) We define the difference B C oe B ? C?? So subtraction is defined in terms of addition (adding the additive inverse).

b) If B ? !, we define the quotient C B oe CB"? So division is defined in terms of multiplication (multiplying by the multiplicative inverse).

Example Consider the field TM( oe ?!? "? #? $? %? &? '? ( where "? #? ???? ' are abbreviations for the equivalence classes ?"?? ?#?? ???? ?'?)

a) The additive inverse of % is 3 because % $ oe ! (and only $ has this property). Therefore we write % oe $.

b) Subtraction: # % oe # ? %? oe # $ oe &?

c) The multiplicative inverse of $ is & because $ & oe " (and only & has this property). Therefore we write $" oe & (and &" oe $??

d) Division: # $ oe # $" oe # & oe $? " $ oe " $" oe " & oe &?

Example If +? ,? - are members of any field J and + ? !, then the equation +B , oe has a unique solution in J . We can simply use the "algebra of fields" contained in the axioms and theorems to solve the equation (some detailed justifications like references to the repeated use of associativity and commutativity in TM are omitted).

+B , oe +B , ? ,? oe - ? ,?

+B ! oe - ? ,? +B oe - , +"?+B? oe +"?- ,?

" B oe +"?- ,? B oe +"?- ,?

Using the additive inverse of ,

+ ? ! so + has a multiplicative inverse

For a more specific example, we solve the equation $B % oe 2 in TM(. (See the preceding example.)

$B % oe # $B % % oe # % $B % $ oe # $ $B oe & & $B oe & & "Boe% B oe %?

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