Solutions to Math 109 Homework: Equivalence Relations and Functions, Assignments of Introduction to Sociology

Download Solutions to Math 109 Homework: Equivalence Relations and Functions and more Assignments Introduction to Sociology in PDF only on Docsity! Solutions of Homework 8, Math 109 Warning: Below, you will find only one possible proof among a plethora of possibilities (The word plethora might be a little strong, but take it as a metaphor!). So, if your solution is not exactly as the one presented here, it does not mean that it is wrong. Another remark is that these solutions might not be the optimal ones, and most of the time they will not be. And I cannot guarantee that these solutions are free of mistakes or typos. Moreover, these are formal proofs, so you do not find the work done previously on a scratch paper. Nevertheless, we emphasize the fact that this work is crucial for writing a proof. It is really hard to write a formal proof right from the start without previously throwing ideas on a piece of scratch paper. Finally, pages refer to the book by Eccles. Solutions Problem 17, page 273 Part(i) It is not reflexive since 0  0. It is symmetric. It is also transitive. Part(ii) It is reflexive, because a2 ≥ 0 no matter what a is. It is also clearly symmetric, but it is not transitive. In- deed −2 ∼ 0 and 0 ∼ 2, but −2  2. Therefore, it is not an equivalence relation. Part(iv) It is reflexive, symmetric, and transitive, therefore it is an equivalence relation. Problem 2 Done in section. Problem 3 Proof: Part (a) In order to show that this relation defines an equivalence relation on A, we have to show that it is reflexive, symmetric, and transitive. Now, it is reflexive because given any (a, b) ∈ A, we have (a, b) ∼ (a, b), since ab = ba. It is also symmetric, since if (a, b) ∼ (c, d) then this means that ad = bc. From that we conclude that cb = da, in other words (c, d) ∼ (a, b). Now we have to show that it is transitive. Suppose that (a, b) ∼ (c, d), and (c, d) ∼ (e, f). This means that ad = bc and cf = de. We want to show that af = be. But adf = bcf = bde implies that d(af − be) = 0, and since d 6= 0, we conclude that af = be, thus this relation is transitive. Part (b) This should be simple. 1 2 Problem 5, page 183 Proof: We shall do this proof by induction on the cardinality of A. Base case: If |A| = 1 there is nothing to prove. Induction hypothesis: Suppose that it is true when |A| = k. Deduction: Let A be a set having k + 1 elements. Write it as A = {a1, . . . , ak+1} = B ∪ {ak+1}, where B = {a1, . . . , ak}. Since B has cardinality k, by the induction hypothesis, it has a minimal element, say b. If b < ak+1 then b is a minimal element of A. If b ≥ ak+1 then ak+1 is a minimal element of A. No matter what, we found a minimal element for A. This is what we wanted to prove. 2 Problem 11, page 184 Proof: (⇒) Suppose by contradiction that f is not surjective. This means that there exists a y ∈ Y such that f(x) 6= y for all x ∈ X. Since |X| = |Y |, we conclude by the Pigeonhole principle that there exist x1 and x2 in X such that x1 6= x2, but f(x1) = f(x2) and this is a contradiction with the fact that f is supposed to be injective. (⇐) The proof is similar, and you should be able to do it! 2 Problem 12, page 184 Proof: Suppose that X is finite, say of cardinality m. Let us restrict the function f to Nm+1, in this way we get an injective function between two finite sets: f : Nm+1 → X. By Corollary 11.1.1., we get that |Nm+1| ≤ |X|, but this is impossible since |Nm+1| = m + 1 and |X| = m. 2 Problem 14, page 184 Proof: Define the function f : A → N2n by the rule a 7→ f(a) = the greatest odd integer dividing a. There are exactly n odd integers in N2n, but the cardinality of A is n + 1. Therefore, by the Pigeonhole principle, there exist elements a and b in A such that a 6= b and f(a) = f(b). Moreover, without lost of generality we can suppose that a ≤ b. Since f(a) = f(b) we conclude that a and b differ only by a power of 2, and therefore a | b (do you clearly understand that?). 2