Solutions To Mathematics Textbooks/Proofs and Fundamentals/Chapter 3

Exercise 3.2.1 3, namely and Exercise 3.2.2 3∈(3,5]. — False. (3,5] is open at 3, so 3 is not a member. Counterexample: 3∈/(3,5]. 10∈(−∞,2​]. — False. 2​≈1.414, so the interval (−∞,2​] contains only numbers ≤1.414; 10 is much larger. 7∈{2,3,4,…,11}. — True. The set {2,3,4,…,11} lists the integers from 2 through 11; 7 is one of them. π∈(2,∞). — True. π≈3.14159>2, so π lies in (2,∞). −1.3∈{…,−3,−2,−1}. — False. The set {…,−3,−2,−1} contains only integers; −1.3 is not an integer (not equal to −1), so it is not a member. Counterexample: −1.3∈/{…,−3,−2,−1}. [1,2]⊆{0,1,2,3}. — False. [1,2] contains non-integer numbers like 1.5, but {0,1,2,3} does not. Counterexample: 1.5∈[1,2] but 1.5∈/{0,1,2,3}. {−1,0,1}⊆[−1,1). — False. [−1,1) contains −1 and 0 but not 1. Counterexample: 1∈{−1,0,1} but 1∈/[−1,1). [5,7]⊆(4,∞). — True. Every x∈[5,7] satisfies x≥5>4, so all are in (4,∞). {2,4,8,16,…}⊆[2,∞). — True. The set is (powers of 2 starting at 2) and every element is ≥2, hence lies in [2,∞). Exercise 3.2.3 1 The set of even integers 2 The set of composite numbers 3 The set of all rational numbers. Exercise 3.2.4 (1) {x∈P∣x∈M and x has a child} → All men who have at least one child. (2) {x∈P∣∃y,z∈P such that y is a child of x and z is a child of y} → All people who have at least one grandchild (i.e. they have a child who in turn has a child). (3) {x∈P∣∃m∈F such that x is married to m} → All people who are married to a woman. (4) {x∈P∣∃q∈P such that x and q have the same mother} → All people who have at least one other person (a sibling) with the same mother. (5) {x∈P∣∃h∈P such that h is older than x} → All people for whom there exists someone older — i.e. everyone except the very oldest person(s). (6) {x∈P∣∃n∈M such that x is the child of n and x is older than n} → All people who are the child of some man and who are older than that man — i.e. people older than (one of) their fathers. Exercise 3.2.5 1 2 there exist such that 3 there exist such that 4 {n^3|n is an integer and -5<n<5} 5 there exist such that Exercise 3.2.6 Exercise 3.2.7 Exercise 3.2.8 Exercise 3.2.9 A = {1,2}, B = {1,2,{1,2}} Exercise 3.2.10 Using the definition of a subset: For any x ∈ A, then x ∈ B, and because x ∈ B, x ∈ C. The same goes for any y ∈ B or any z ∈ C. Given A⊆B B⊆C C⊆A We need to prove: A=B=C. Step 1: From the assumptions get all inclusions. We have directly: A⊆B. B⊆C. C⊆A. From A⊆B and B⊆C, by transitivity of “⊆”: A⊆C. Combine with C⊆A: A⊆C and C⊆A ⇒ A=C. So we already know A=C. Step 2: Show B=A as well. We already have: A⊆B. B⊆C and C=A ⇒ B⊆A. So A⊆B and B⊆A ⇒ A=B. Step 3: Conclude equality of all three sets. We have A=B and A=C. Therefore A=B=C. Boxed result A=B=C Exercise 3.2.11 Exercise 3.2.12 False. Counterexample. Let A be a set of even integers and B a set of odd integers.Then A and B are not equal, and A is not a subset of B, and B is not a subset of A. A and B are disjoint. Exercise 3.2.13 Exercise 3.2.14 Exercise 3.2.15 1 2 Exercise 3.2.16 (1) false (2) true (3) true (4) true (5) false (6) true (7) false (8) false (9) true