昆士兰大学MATH1061 Assignment 3课业解析题意：完成5到数学题解析：第一题： Prove that for all sets A, B, and C
A × (B ∩ C) = (A × B) ∩ (A × C)
证明：（1）
任取(x,y)∈A×(B∩C) 则x∈A,y∈B∩C 由y∈B∩C得y∈B,且y∈C 由x∈A,y∈B得(x,y)∈(A×B) 由x∈A,y∈C得(x,y)∈(A×C) 所以(x,y)∈(A×B)∩(A×C) 所以A×(B∩C) 包含于 (A×B)∩(A×C) （2）
任取(x,y)∈(A×B)∩(A×C)
则(x,y)∈(A×B) ,且(x,y)∈(A×C) 由(x,y)∈(A×B) 得x∈A,y∈B 由(x,y)∈(A×C) 得x∈A,y∈C 由y∈B及y∈C得y∈(B∩C) 又因为x∈A
所以(x,y)∈A×(B∩C) 由（1）（2）得
A×(B∩C) = (A×B)∩(A×C)涉及知识点：更多可加微信讨论微信号：IT_51zuoyejunpdf
MATH1061 Assignment 3 Due 10am Monday 23 September 2019This Assignment is compulsory, and contributes 5% towards your final grade. It must be submittedby 10am on Monday 23 September, 2019. In the absence of a medical certificate or other valid documented excuse, assignments submitted after the due date will not be marked.Submission You will receive a coversheet for this assignment by email. Print that coversheet, stapleit to the front of your assignment (which may be handwritten) and submit your assignment using theassignment submission system which is located in the corridor between buildings 69 and 62.1. (4 marks) Prove that for all sets A, B, and C,A × (B ∩C) = (A × B) ∩ (A × C):2. (5 marks) Consider the function f : Z× Z ! Z where f((x; y)) = 3x + 5y for all (x; y) 2 Z× Z.(a) Is the function f one-to-one? Prove your answer.(b) Is the function f onto? Prove your answer.3. (8 marks) Let S = f(a1; a2; : : : ; an) j n ≥ 1; ai 2 Z≥0 for i = 1; 2; : : : ; n; an 6= 0g. So S is the setof all finite ordered n-tuples of nonnegative integers where the last coordinate is not 0. Find abijection from S to Z+.4. (6 marks) Define a relation r on R as follows. For all a; b 2 R, a r b if and only if ab- a- b < 0.(a) Is r reflexive? Explain your answer.(b) Is r symmetric? Explain your answer.(c) Is r transitive? Explain your answer.5. (9 marks) Define the relation τ on Z by a τ b if and only if there exists x 2 f1; 4; 16g such thatax ≡ b (mod 63).(a) Prove that τ is an equivalence relation.(b) Prove that there exists an integer n with 1 ≤ n ≤ 62 such that the equivalence class of n isfm 2 Z j m ≡ n (mod 63)g.
