Binary Relations and Functions (Due on Wednesday, September 30Th, 2015)

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 respectively 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 = {a, b}.

5. (5 points) An integer q ∈ N divides an integer p ∈ N if there is an integer r ∈ N such that p = rq. We write q|p 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 ∈ 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) ⇐⇒ ∃k ∈ Z | 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:

∀x, y ∈ 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.