Page 512 **Definition** Let $R$ be a relation from $A$ to $B$. Define the inverse relation $R^{-1}$ from $B$ to $A$ as follows: $$ R^{-1} = \{(y, x) \in B \times A | (x, y) \in R\} $$ --- Page 513 **Definition** A **relation on a set** A is a relation from $A$ to $A$. --- 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. --- 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 $x, y, z \in A, \text{ if } x R y \text{ and } y R z \text{ then } x R z$. --- 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]_. --- 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]_. --- 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]_. --- 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 $R^t \subseteq S$. --- 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 $x, y \in A$, $$ x R y \Leftrightarrow \text{ there is a subset } A_i \text{ of the partition such that both } x \text{ and } y \text{ are in } A_i $$ --- 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 $R$. --- 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. --- 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: $$ [a] = \{x \in A | x R a\} $$ --- 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 $[a] = [b]$. --- 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. --- 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 $$ \text{either } [a] \cap [b] = \emptyset \quad \text{ or } \quad [a] = [b] $$ --- 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]._ --- 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. --- 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]._ --- 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 $[a] = S$. -- 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: $$ m \equiv n(\mod d) \Leftrightarrow d | (m - n) $$ --- Page 542 **Example 8.3.12** _Rational Numbers are Really Equivalence Classes Let $A$ be the set of all ordered pairs of integers for which the second element of the pair is nonzero. Symbolically: $$ A = \mathbb{Z} \times (\mathbb{Z} - \{0\}) $$ Define a relation $R$ on $A$ as follows: For all pairs $(a, b)$ and $(c, d)$ in $A$, $$ (a, b) R (c, d) \Leftrightarrow ad = bc $$ The fact is that $R$ is an equivalence relation.