discrete_mathematics_with_applications/chapter_8/notes.md

Page 512

Definition

Let R be a relation from A to B. Define the inverse relation R^{-1} from B to A as follows:

 R^{-1} = \{(y, x) \in B \times A | (x, y) \in R\} 

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Page 513

Definition

A relation on a set A is a relation from A to A.

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Page 514

Definition

Given sets A_1, A_2, \dots, A_n an $n$-ary relation R on A_1 \times A_2 \times \cdots \times A_n is a subset of A_1 \times A_2 \times \cdots \times A_n. The special cases of $2$-ary, $3$-ary, and $4$-ary relations are called binary, ternary, and quarternary relations, respectively.

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Page 518

Definition

Let R be a relation on a set A.

1. R is reflexive if, and only if, for every x \in A, x R x.

2. R is symmetric if, and only if, for every x, y \in A, \text{ if } x R y \text{ then } y R x.

3. R is transitive if, and only if, for every x, y, z \in A, \text{ if } x R y \text{ and } y R z \text{ then } x R z.

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Page 523

Proof of Reflexivity:

Suppose m is a particular but arbitrarily chosen integer. [We must show that m T m.] Now m - m = 0. But 3 | 0 since 0 = 3 \cdot 0. Hence 3 | (m - m). Thus, by definition of T, m T m [as was to be shown].

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Page 524

Proof of Symmetry:

Suppose m and n are particular but arbitrarily chosen integers that satisfy the condition m T n. [We must show that n T m.] By definition of T, since m T n then 3 | (m - n). By definition of "divides", this means that m - n = 3k, for some integer k. Multiplying both sides by -1 gives n - m = 3(-k). Since -k is an integer, this equation shows that 3 | (n - m). Hence, by definition of T, n T m [as was to be shown].

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Page 524

Proof of Transitivity:

Suppose m, n, and p are particular but arbitrarily chosen integers that satisfy the condition m T n and n T p. [We must show that m T p.] By definition of T, since m T n and n T p, then 3 | (m - n) and 3 | (n - p). By definition of "divides", this means that m - n = 3r and n - p = 3s, for some integers r and s. Adding the two equations gives (m - n) + (n - p) = 3r + 3s, and simplifying gives that m - p = 3(r + s). Since r + s is an integer, this equation shows that 3 | (m - p). Hence, by definition of T, m T p [as was to be shown].

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Page 525

Definition

Let A be a set and R a relation on A. The transitive closure of R is the relation R^t on A that satisfies the following three properties:

1. R^t is transitive.

2. R \subseteq R^t.

3. If S is any other transitive relation that contains R, then R^t \subseteq S.

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Page 529

Definition

Given a partition of a set A, the relation induced by the partition, R, is defined on A as follows: For every x, y \in A,

 x R y \Leftrightarrow \text{ there is a subset } A_i \text{ of the partition such that both } x \text{ and } y \text{ are in } A_i 

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Page 530

Theorem 8.3.1

Let A be a set with a partition and let R be the relation induced by the partition. Then R is reflexive, symmetric, and transitive.

Proof:

Suppose A is a set with a partition. In order to simplify notation, we assume that the partition consists of only a finite number of sets. The proof for an infinite partition is identical except for notation. Denote the partition subsets by

 A_1, A_2, \dots, A_n 

Then A_i \cap A_j = \emptyset whenever i \neq j, and A_1 \cup A_2 \cup A_3 \cdots \cup A_n = A. The relation R induced by the partition is defined as follows: For every x, y \in A,

 x R y \Leftrightarrow \text{ there is a set } A_i \text{ of the partition such that } x \in A_i \text{ and } y \in A_i 

[Idea for the proof of reflexivity: For R to be reflexive means that each element of a is related by R to itself. But by definition of R, for an element x to be related to itself means that x is in the same subset of the partition itself. Well, if x is in some subset of the partition, then it is certainly in the same subset as itself. And x is in some subset of the partition because the union of the subsets of the partition is all of A. This reasoning is formalized as follows.]

Proof that R is reflexive:

Suppose x \in A. Since A_1, A_2, \dots A_n is a partition of A, it follows that x \in A_i, for for some i, and so the statement

there is a set A_i of the partition such that x \in A_i and x \in A_i

is true. Thus by definition of R, x R x.

[Idea for the proof of symmetry: For R to be symmetric means that any time one element is related to a second, then the second is related to the first. Now for one element x to be related to a second element y means that x and y are in the same subset of the partition. But if this is the case, then y is in the same subset of the partition as x, so y is related to x by definition of R. This reasoning is formalized as follows.]

Proof that R is symmetric:

Suppose x and y are elements of A such that x R y. Then there is a subset A_i of the partition such that x \in A_i and y \in A_i by definition of R. It follows that the statement

there is a subset A_i of the partition such that y \in A_i and x \in A_i

is also true. Hence, by definition of R, y R x.

[Idea for the proof of transitivity: For R to be transitive means that any time one element of A is related by R to a second and that second is related to a third, then the first element is related to the third. But for one element to be related to another means that there is a subset of the partition that contains both. So suppose x, y, and z are elements such that x is in the same subset as y and y is in the same subset as z. Must x be in the same subset as z? Yes, because the subsets 9f the partition are mutually disjoint. Since the subset that contains x and y has an element in common with the subset that contains y and z (namely, y), the two subsets are equal. But this means that x, y, and z are all in the same subset, and so, in particular, x and z are in the same subset. Hence x is related by R to z. This reasoning is formalized as follows.]

Proof that R is transitive:

Suppose x, y, and z are in A and x R y and y R z. By definition of R, there are subsets A_i and A_j of the partition such that

 x \text{ and } y  \text{ are in } A_i \quad \text{ and } \quad y \text{ and } z \text{ are in } A_j 

Suppose A_i \neq A_j. [We will deduce a contradiction.] Then A_i \cap A_j = \emptyset since \{A_1, A_2, A_3, \dots, A_n\} is a partition of A. But y is in A_i and y is in A_j also. Hence A_i \cap A_j \neq \emptyset. [This contradicts the statement that A_i \cap A_j = \emptyset.] Thus A_i = A_j. It follows that x, y, and z are all in A_i, and so, in particular,

 x \text{ and } z \text{ are in } A_i 

Thus x R z by definition of R.

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Page 531

Definition

Let A be a set and R a relation on A. R is an equivalence relation if, and only if, R is reflexive, symmetric, and transitive.

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Page 533

Definition

Suppose A is a set and R is an equivalence relation on A. For each element a in A, the equivalence class of $a$, denoted [a] and called the class of $a$ for short, is the set of all elements x in A such that x is related to a by R.

In symbols:

 [a] = \{x \in A | x R a\} 

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Page 536

Lemma 8.3.2

Suppose A is a set, R is an equivalence relation on A, and a and b are elements of A. If a R b, then [a] = [b].

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Page 536

Proof of Lemma 8.3.2

Let A be a set, let R be an equivalence relation on A, and suppose

 a \text{ and } b \text{ are elements of } A \text{ such that } a R b 

[We must show that [a] = [b].]

Proof that [a] \subseteq [b]:

Let x \in [a]. [We must show that x \in [b].]

Since

 x \in [a] 

then

 x R a 

by definition of class. But

 a R b 

by hypothesis. Thus, by transitivity of R,

 x R b 

Hence

 x \in [b] 

by definition of class. [This is what was to be shown.]

**Proof that [b] \subseteq [a].

Let x \in [b]. [We must show that x \in [a].]

Since

 x \in [b] 

then

 x R b 

by definition of class. Now

 a R b 

by hypothesis. Thus, since R is symmetric,

 b R a 

also. Then, since R is transitive and x R b and b R a,

 x R a 

Hence,

 x \in [a] 

by definition of class. [This is what was to be shown.]

Since [a] \subseteq [b] and [b] \subseteq [a], it follows that [a] = [b] by definition of set equality.

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Page 537

Lemma 8.3.3

If A is a set, R is an equivalence relation on A, and a and b are elements of A, then

 \text{either } [a] \cap [b] = \emptyset \quad \text{ or } \quad [a] = [b] 

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Page 537

Proof of Lemma 8.3.3

Suppose A is a set, R is an equivalence relation on A, a and b are elements of A, and

 [a] \cap [b] \neq \emptyset 

[We must show that [a] = [b].]

Since [a] \cap [b] \neq \emptyset, there exists an element x in A such that x \in [a] \cap [b]. By definition of intersection,

 x \in [a] \quad \text{ and } \quad x \in [b]

and so

 x R a \quad \text{ and } \quad x R b 

by definition of class. Since R is symmetric [being an equivalence relation] and x R a, then a R x. But R is also transitive [since it is an equivalence relation], and so, since a R x and x R b,

 a R b 

Now A and b satisfy the hypothesis of Lemma 8.3.2. Hence, by that lemma,

 [a] = [b] 

[as was to be shown].

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Page 537

Theorem 8.3.4 The Partition Induced by an Equivalence Relation

If A is a set and R is an equivalence relation on A, then the distinct equivalence classes of R form a partition of A; that is, the union of the equivalence classes is all of A, and the intersection of any two distinct classes is empty.

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Page 538

Proof of Theorem 8.3.4

Suppose A is a set and R is an equivalence relation on A. For notational simplicity, we assume that R has only a finite number of distinct equivalence classes, which we denote

 A_1, A_2, \dots, A_n 

where n is a positive integer. (When the number of classes is infinite, the proof is identical except for notation.)

Proof that A = A_1 \cup A_2 \cup \cdots \cup A_n:

[We must show that A \subseteq A_1 \cup A_2 \cup \cdots \cup A_n and that A_1 \cup A_2 \cup \cdots \cup A_n \subseteq A.]

To show that A \subseteq A_1 \cup A_2 \cup \cdots \cup A_n, suppose x is any element of A. [We must show that x \in A_1 \cup A_2 \cup \cdots A_n.] By reflexivity of R, x R x. And this implies that x \in [x] by definition of class. Since x is in some equivalence class, it must be in one of the distinct equivalence classes A_1, A_2, \dots, or A_n. Thus x \in A_i for some index i, and hence x \in A_1 \cup A_2 \cup \cdots \cup A_n by definition of union [as was to be shown].

To show that A_1 \cup A_2 \cup \cdots \cup A_n \subseteq A, suppose x \in A_1 \cup A_2 \cup \cdots \cup A_n. [We must show that x \in A.] Then x \in A_i for some i = 1, 2, \dots, n, by definition of union. Now each A_i is an equivalence class of R, and equivalence classes are subsets of A. Hence A_i \subseteq A and so x \in A [as was to be shown].

Since A \subseteq A_1 \cup A_2 \cup \cdots A_n and A_1 \cup A_2 \cup \cdots \cup A_n \subseteq A, then by definition of set equality, A = A_1 \cup A_2 \cup \cdots \cup A_n.

Proof that the distinct classes of R are mutually disjoint:

Suppose that A_i and A_j are any two distinct equivalence classes of R. [We must show that A_i and A_j are disjoint.] Since A_i and A_j are distinct, then A_i \neq A_j. And since A_i and A_j are equivalence classes of R, there must exist elements a and b in A such that A_i = [a] and A_j = [b].

By Lemma 8.3.3,

 \text{either } [a] \cap [b] = \emptyset \quad \text{ or } \quad [a] = [b]

Now [a] \neq [b] because A_i \neq A_j, and hence [a] \cap [b] = \emptyset. Thus A_i \cap A_j = \emptyset, and so A_i and A_j are disjoint [as was to be shown].

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Page 540

Definition

Suppose R is an equivalence relation on a set A and S is an equivalence class of R. A representative of the class S is any element a such that [a] = S.

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Page 541

Definition

Let m and n be integers and let d be a positive integer. We say that m is congruent to n modulo $d$ and write

 m = n (\mod d) 

if, and only if,

 d | (m - n) 

Symbolically:

 m \equiv n(\mod d) \Leftrightarrow d | (m - n)