代数章节翻译

2Integers and rational numbers2.1 The integersThe integers are familiar to us from elementary arithmetic, but here we shall want to express that familiarity in precise terms. We do this by writing down a list of properties which the integers posses and on which we shall base all our deductions. Later, in Vol.2, we shall see that all the properties listed here can actually be deduced from quite a brief list of axioms, but this is immaterial at present. We denote the set of positive integers (also called natural numbers) 1,2,3.by N and write Z for the set of all integers, positive, negative, and zero. Here N stands for number and Z for Zahl, the German for number; both abbreviations are generally used in mathematics (see Appendix 2 ).The set z admits three operations: addition, x+y, subtraction, x-y,and multiplication, x.y, or xy.Often it is convenient to express subtraction by adding the negative:x-y=x+(-y). These operations are connected by the following laws.Z.1 Associative law: (x+y)+z=x+(y+z), (xy)z=x(yz).Z.2 Commutative law: x+y=y+x, xy=yx.Z.3 Existence of neutral element:x+0=x, x1=x.Z.4 Existence of (additive) inverse: x+(-x)=0. The number 0 is said to be neutral for addition because adding it to any number x leaves x unchanged; likewise 1 is neutral for multiplication. Every number x has the additive inverse -x (which undoes the effort of adding x), but apart from 1 and -1, no integer has a multiplicative inverse. However we shall find such inverse once we come to consider rational numbers in 2.4.In addition to the above laws, there is a further law, relating addition and multiplication:Z.5 Distributive law: x(y+z)=xy+xz.A set R with two operations x+y,xy and a negative -x, satisfying Z.1-5 is called a ring;more precisely it is commutative ring (because the multiplication is commutative,cf.Ex.(1) ,6.1). Thus the set Z of all integers is a commutative ring.However, these laws are not yet sufficient to determine Z; in Ch.6 we shall give a general definition of a ring and we shall find that there are many different types.We now look at some consequence of the above laws. It follows from the distributive law that x0=0 for all x. By the associative law, the sum of any number of terms is independent of the way in which brackets are placed, and by the commutative law the order of the terms is immaterial. A similar remark applies to multiplication; for the present we shall accept this without proof and return to this point in Ch.3 to give a general proof.Thus the sum of numbers a1,.,an may be written a1+.+an. Often one abbreviates this expression by writing down the general term an with a capital sigma, S,to show that the sum is to be taken, with some indiction of the range over which the terms are to be summed (unless this is clear from the context). So instead of a1+.+an we may write or or or simply , where in each case v is a dummy variable(cf. 1.1). When n=0, the sum written here is empty and ,by convention, this is taken to be 0. This notation is not only briefer; it can also help to make our formula more perspicuous as well as more accurate. For instance,in the expression1+2+.+n,the reader is expected to guess that he is dealing with an arithmetic progression; the expressionremoves all doubt. Thus the formula for the sum of the first n natural numbers may be written =n(n+1). We observe that for n=0 the right-hand side reduces to 0, so with our convention about empty sums,this formula still holds for n=0.For another example consider the distributive law. This has a generalized version which reads (cf. Ex.(2)(a1+.+ar)(b1+.+bs)=a1b1+a1b2+.+arbs,or in abbreviated from Sau .Sbv=Su,vaubv. Here we have not indicated the precise range of summation u,v.A similar abbreviation exists for repeated products, using capital pi,P, in place of S.Thus instead of a1a2.an we write or or or simply .Integers and rational numbersFor example, the factorial function may be defined as n!=. An empty product is taken to be 1; thus empty sums and products are neutral for addition and multiplication respectively.It is an important property of the integers that the product of two non-zero integers is never zero:Z.6 For any integers a, b, if a0 and b0,then a*b0; morever 10. This has the following useful consequence:Cancellation law For a, b, cZ, if ca=cb and c0,then a=b.This asserts that multiplication by a non-zero integer is an injective mapping of Z into itself. To prove it, suppose that ab, then a-b0 and hence(by Z.6) c(a-b)0,therefore ca-cb=c(a-b)0.Besides the operations on Z We have an order relation, i.e. an ordering on Z: xy or yx. If xy but xy, We write xy or also yx. This relation satisfies the requirements for a total ordering (see 1.5) and is related to the operations of Z by the following rules:Z.7 If x1x2 and y1y2, then x1+y1x2+y2.Z.8 If xy and z0, then zxzy.The presence of these rules means that Z is a totally ordered ring. Using the ordering we can describe the set N of positive integers asN=xZx0. Later we shall see how to reconstruct Z from N; for the moment we note that, for every xZ, either x=0 or xN or xN and that these three possibilities are mutually exclusive. In fact, this is true in any totally ordered ring, taking N to be defined by (1). For we know that just one of the following holds (because we have a total order): x=0 or x0 or x0. Now x+ (-x) =0, hence, if x0, then 00 or x0, as asserted.In order to fix Z completely, we use the following condition on the set N of positive integers:I (principle of induction): Let S be a subset of N such that 1S and n+1S whenever nS. Then S=N.This principle forms the basis of the familiar method of proof by induction. Let P(n) be an assertion about a positive integer n, e.g. ,P(n) might bethe sum of the first n positive integers is n(n+1)/2 .Suppose we wish to prove P(n) for all n, i.e. (n)P(n). Then by I it will be enough to prove () P(1) and ()(n)(P(n)=P(n+1). For this means that the set S of all n for which P(n) holds contains 1 and contains n+1 whenever it contains n.2.1 The integersHence by I,S=N, i.e. P(n) holds for all nN.There are two alternative forms of I that are often useful.I Let S be a subset of N such that 1S and nS whenever mS for all mI=I=I,I=I. Let S be such that 1S and nS whenever mS for all mI. Let S be a set of positive integers without a least element;we shall show that S is empty. Denoting the complement of S by S, we must show that S=N.Now, since S has no element,1S,so 1S, moreover,if mS for all mI. 1 is the least element of N(for if the least element r satisfied r1,then r2P(n+1)。这意味着对所有n构成的集合S,当P(n)包含1和n+1成立时，P(n)包含n同样成立。 2.1 整数因为通过I,S=N，举例来说，对于所有nN，P（n)成立。I有两个可供选择的形式通常很有用。I令S是N的一个子集，这样1S且nS，当mI=I=I。I=I。令S满足1S且nS，则当mI。令S为一个无最小元素的正整数集合，可看出S为空。用S表示S的补集，我们必须证明S=N。已知S没有最小元素，1S，此外，当mI。1是N中的最小元素，（因为如果最小元素满足r1，那么r2r，矛盾）。令S是N的一个子集，这样有1S且n+1S，当nS则N中集合S的补集S没有最小元素。对于1S且如果nS，则n-1S，因此，通过I，S=，S=N得证。用关于通过推导证明的实际性的评论来结束本节内容。通常定理比较容易证明，假设越强结论越弱。但是在一项推导证明中，如果结论被强化，第n步的结论变成等n+1步的假设，定理将更容易得证。以4.3中引理4为例。我们习以为常从0（取代1）开始推导，显然这并不影响其正确性。