近世代数答案(一、二三章).doc

Chapter 1 1、proof Let A,B,C be sets .Suppose that x∈B,we get x∈A∩B or ,and or since and .so x∈C and .Similarly ,we have and so B=C . 2、proof ① First,consider .Then or ,but .This implies if x is not an element of A ,then .Hence and . Conversely, if ,then by definition , or . This generates two cases: (a1) If ,clearly ; (b2) If ,then either or not . i.e.,either and or but , in either case, we have . Hence . Therefore =. ② Suppose that .Then but and . So -B and and by definition . Hence . Converssely, Assume that ,then -B and ,and we have but and .Hence , , i.e., . Thereforeand,so= 3.(a) surjective (b) bijective (c) bijective 4.proof if f: XY and g: YZ are functions,then their composite denoted by gf, is the function XZ given by gf: Xg(f(x)) suppose that (gf)(a)= (gf)(b), where a,b X. we have g(f(a))=g(f(b)) by definition, and f(a)=f(b) since g is injective, similarly, a=b for f is injective. Therefore, gf is injective. For each ZZ, there is y Y with g(y)=z since g is surjective, and for each y Y, there exists a x with f(a)=y since f is surjective. So for z Z, there is a x with (gf)(a)=g(f(a))=g(y)=z. which implies gf is surgective. 5. proof clearly, :RR is a function. Suppose that (a)= (b) where a, b R are distinct. Then , cross multiplying yields , which simplifies to and hence ,so is injective. for given y R, from,we get equation , which can be solved for x, i.e .for each y R,there is at least x x such that .whic implies is surjective. Therefore is bijective. 6、(a) R is reflexive, symmetric, transitive. (b) R is reflexive, not symmetric, transitive. (c ) R is reflexive, symmetric, transitive. (d) R is reflexive, symmetric, transitive. 7、proof (1) For every ∈R-{0}，we have , and so (2) If , where , i.e., then , i.e., , (3) If , where , i.e. , then .i.e.,.Therefore, the relation ～ is an equivalence relation . 8、 There are 1,3,5,15 equivalence relations on a set S with 1,2,