Next, we discuss the set QQ of rational numbers, which we ordinarily think of as quotients k/nk/n of integers. Of course, we do not allow the “second” element nn of the quotient k/nk/n to be 0. Also, we must remember that there isn't a 1-1 correspondence between the set QQ of all rational numbers and the set of all such quotients k/n.k/n. Indeed, the two distinct quotients 2/32/3 and 6/96/9 represent the same rational number. To be precise, the set QQ is a collection of equivalence classes of ordered pairs (k,n)(k,n) of integers, for which the second component of the pair is not 0. The equivalence relation among these ordered pairs is this:
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(1)We will not dwell on this possibly subtle definition, but will rather accept the usual understanding of the rational numbers and their arithmetic properties. In particular, we will represent them as quotients rather than as ordered pairs, and, if rr is a rational number, we will write r=k/n,r=k/n, instead of writing rr as the equivalence class containing the ordered pair (k,n).(k,n). As usual, we refer to the first integer in the quotient k/nk/n as the numerator and the second (nonzero) integer in the quotient k/nk/n as the denominator of the quotient. The familiar definitions of sum and product for rational numbers are these:
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(3)Addition and multiplication of rational numbers satisfy the three basic algebraic relations of commutativity, associativity and distributivity stated earlier.
We note that the integers ZZ can be identified in an obvious way as a subset of the rational numbers Q.Q. Indeed, we identify the
integer kk with the quotient k/1.k/1. In this way, we note that QQ contains the two numbers 0≡0/10≡0/1 and 1≡1/1.1≡1/1. Notice that any other quotient k/nk/n that is equivalent to 0/10/1 must satisfy k=0,k=0, and any other quotient k/nk/n that is equivalent to 1/11/1 must satisfy k=n.k=n.
Remember, k/n≡k'/n'k/n≡k'/n' if and only if kn'=k'n.kn'=k'n.
The set QQ has an additional property not shared by the set of integers Z.Z. It is this: For each nonzero element r∈Q,r∈Q, there exists an element r'∈Qr'∈Q for which r×r'=1.r×r'=1. Indeed, if r=k/n≠0,r=k/n≠0, then k≠0,k≠0, and we may define r'=n/k.r'=n/k. Consequently, the set QQ of all rational numbers is what is known in mathematics as a field.
- Definition 1:
A field is a nonempty set FF on which there are defined two binary operations, addition (++) and multiplication (××), such that the following six axioms hold:
- Both addition and multiplication are commutative and associative.
- Multiplication is distributive over addition; i.e.,
x×(y+z)=x×y+x×zx×(y+z)=x×y+x×z
(4) - There exists an element in F,F, which we will denote by 0,0,
that is an identity for addition;
i.e.,
x+0=xx+0=x for all x∈F.x∈F.
- There exists a nonzero element in F,F, which we will denote by 1,1,
that is an identity for multiplication; i.e.,
x×1=xx×1=x for all x∈F.x∈F.
- If x∈F,x∈F, then there exists a unique element
y∈Fy∈F such that x+y=0.x+y=0.
This element yy is called the additive inverse of xx
and is denoted by -x.-x.
- If x∈Fx∈F and x≠0,x≠0, then there
exists a unique element y∈Fy∈F such that
x×y=1.x×y=1.
This element yy is called the multiplicative inverse of xx
and is denoted by x-1.x-1.
REMARK. There are many examples of fields.
(See Exercise 1.) They all share certain arithmetic properties, which can be derived from the axioms above. If xx is an element of a field F,F, then according to one of the axioms above,
we have that 1×x=x.1×x=x. (Note that this “1” is the multiplicative identity of the field FF and not the natural number 1.)
However, it is tempting to write x+x=2×xx+x=2×x in the field F.F.
The “2” here is not à priori an element of F,F,
so that the equation x+x=2×xx+x=2×x is not really justified.
This is an example of a situation where a careful recursive definition can be useful.
- Definition 2:
If xx is an element of a field F,F,
define inductively elements n·x≡nxn·x≡nx of FF by
1·x=x,1·x=x, and, if k·xk·x is defined,
set (k+1)·x=x+k·x.(k+1)·x=x+k·x.
The set SS of all natural numbers nn for which
n·xn·x is defined is therefore, by the axiom of mathematical induction, all of N.N.
Usually we will write nxnx instead of n·x.n·x.
Of course, nxnx is just the element of FF obtained by adding xx to itself nn times:
nx=x+x+x+...+x.nx=x+x+x+...+x.
- Justify for yourself that the set QQ of all
rational numbers is a field.
That is, carefully verify that all six of the axioms hold.
- Let F7F7 denote the seven elements
{0,1,2,3,4,5,6}.{0,1,2,3,4,5,6}. Define addition and multiplication on F7F7
as ordinary addition and multiplication mod 7.
Prove that F7F7 is a field.
(You may assume that axioms (1) and (2) hold.
Check only conditions (3)–(6).)
Show in addition that 7x=07x=0 for every x∈F7.x∈F7.
- Let F9F9 denote the set consisting of the nine
elements {0,1,2,3,4,5,6,7,8}.{0,1,2,3,4,5,6,7,8}.
Define addition and multiplication on F9F9 to be ordinary addition
and multiplication mod 9. Show that F9F9 is not a field.
Which of the axioms fail to hold?
- Show that the set NN of natural numbers is not a field.
Which of the field axioms fail to hold?
Show that the set ZZ of all integers is not a field.
Which of the field axioms fail to hold?
Let FF be any field. Verify that the
following arithmetic properties hold in F.F.
- 0×x=00×x=0 for all x∈F.x∈F.
HINT: Use the distributive law and the fact that 0=0+0.0=0+0.
- If xx and yy are nonzero elements of F,F, then
x×yx×y is nonzero.
And, the multiplicative inverse of x×yx×y satisfies (x×y)-1=x-1×y-1.(x×y)-1=x-1×y-1.
- (-1)×x=(-x)(-1)×x=(-x) for all x∈F.x∈F.
- (-x)×(-y)=x×y(-x)×(-y)=x×y for all x,y∈F.x,y∈F.
- x×x-y×y=(x-y)×(x+y).x×x-y×y=(x-y)×(x+y).
- (x+y)×(x+y)=x×x+2·x×y+y×y.(x+y)×(x+y)=x×x+2·x×y+y×y.
- Definition 3:
Let FF be a field, and let
xx be a nonzero element of F.F.
For each natural number n,n, we define inductively an
element xnxn in FF as follows:
x1=x,x1=x, and, if xkxk is defined, set xk+1=x×xk.xk+1=x×xk.
Of course, xnxn is just the product of nnxx's.
Define x0x0 to be 1.1.
For each natural number n,n, define x-nx-n to be the multiplicative inverse (xn)-1(xn)-1 of the element xn.xn.
Finally, we define 0m0m to be 0 for every positive integer m,m,
and we leave 0-n0-n and 0000 undefined.
We have therefore defined xmxm for every nonzero xx and every integer m∈Z.m∈Z.
Let FF be a field.
Derive the following laws of exponents:
- xn+m=xn×xmxn+m=xn×xm for all
nonzero elements x∈Fx∈F and all integers nn and m.m.
HINT: Fix x∈Fx∈F and m∈Zm∈Z and use induction
to derive this law for all natural numbers n.n.
Then use the fact that in any field (x×y)-1=x-1×y-1.(x×y)-1=x-1×y-1.
- xn×m=(xm)nxn×m=(xm)n for all nonzero x∈Fx∈F
and all n,m∈Z.n,m∈Z.
- (x×y)n=xn×yn(x×y)n=xn×yn for all nonzero x,y∈Fx,y∈F
and all n∈Z.n∈Z.
From now on, we will indicate multiplication in a field
by juxtaposition; i.e., x×yx×y will be denoted simply as xy.xy.
Also, we will use the standard fractional notation to indicate
multiplicative inverses. For instance,
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(5)
