Harvard University, Math 101, Fall 2015 Problem Set 3: Binary relations and functions (Due on Wednesday, September 30th, 2015) 1. (12 points) In this problem we think of a relation as a set of ordered pairs. We denote respec- tively by Dom(T ) and Ran(T ) the domain and the range of a relation T . Prove or disprove 3 othe following propositions where R and S are two relations: (a) Dom(R [ S) = Dom(R) [ Dom(S) (b) Ran(R [ S) = Ran(R) [ Ran(S) (c) Dom(R \ S) = Dom(R) \ Dom(S) (d) Ran(R \ S) = Ran(R) \ Rng(S) (e) Dom(R − S) = Dom(R) − Dom(S) (f) Ran(R − S) = Ran(R) − Ran(S) 2. (14 points) Prove or give a counter-example for the following propositions1 (a) g ◦ f injective =) g injective (b) g ◦ f injective =) f injective (c) g ◦ f surjective =) g surjective (d) g ◦ f surjective =) f surjective (e) g ◦ f bijective =) g bijective (f) g ◦ f bijective =) f bijective (g) g and f injective =) g ◦ f bijective 3. (20 points) In this problem A and B are two finite sets with respectively n and m elements. Answer (with formal proofs) the following questions (a) How many binary relations can we define on A? (b) How many functions can we have from A to B? (c) How many injective maps can we have from A to B? 4. (8 points) Find all possible binary relations on a pair A = fa; bg. 5. (5 points) An integer q 2 N divides an integer p 2 N if there is an integer r 2 N such that p = rq. We write qjp to indicate that q divides p (or equivalently p is divisible by q). Prove that divisibility in N is a partial order2. Show that 0 is a maximal element while 1 is a minimal element of this partial order. 1A bijection is a function that is both injective and surjective. 2A binary relation is said to be a relation of order if it is reflexive, transitive, and antisymmetric. A relation of order R is said to be partial if there exists at least two elements x and y that are not in relation to each other (they are not comparable): you cannot have xRy nor yRx. 1 6. (6 points) Given an integer n 2 Z, we define a binary relation called congruence modulo n. When two integers x; y are congruent modulo n, we write x ≡ y (mod n). By definition: x ≡ y (mod n) () 9k 2 Z j x − y = kn: (a) Show that congruence modulo n in Z is a relation of equivalence3. (b) What goes wrong if in the definition, we restrict k to be an element of N? (c) Present a definition of congruence modulo n that only involves N but still give a relation of equivalence. 7. (6 points) Given an arbitrary function f : A ! B (a) Shows that the binary relation ≡f defined below is a relation of equivalence: 8x; y 2 A : x ≡f y () f(x) = f(y): (b) What are the classes of equivalence of this relation? (c) What happens to the classes of equivalence when f is injective? 3A binary relation is said to be a relation of equivalence if it is reflexive, transitive, and symmetric. 2.