# How to Prove It - Solutions

## Chapter - 4, Relations

### Summary

• Relation: Suppose $A$ and $B$ are sets. Then a set $R \subseteq A \times B$ is called a relation from $A$ to $B$.
• Suppose $R$ is a relation from $A$ to $B$, then:
• Domain of relation $R$ is a set:
$dom(R) = \{a \in A \, \vert \, \exists b \in B((a, b) \in R)\}$.
• Range of relation $R$ is a set:
$Ran(R) = \{b \in B \, \vert \, \exists a \in A((a, b) \in R)\}$.
• The inverse of $R$ is the relation $R^{−1}$ from $B$ to $A$:
$R^{−1} = \{(b,a) \in B \times A \, \vert \, (a,b) \in R \}$.
• Composition: Suppose $R$ is a relation from $A$ to $B$ and $S$ is a relation from $B$ to $C$. Then the composition of $S$ and $R$ is the relation $S \circ R$ from $A$ to $C$ defined as follows:
$S \circ R = \{ (a, c) \in A \times C \, \vert \, \exists b \in B((a, b) \in R and (b, c) \in S) \}$.
• If $R$ is a relation from $A$ to $B$, $S$ is a relation from $B$ to $C$, and $T$ is a relation from $C$ to $D$. Then:
• $(R^{−1})^{−1} = R$.
• $Dom(R^{−1}) = Ran(R)$.
• $Ran(R^{−1}) = Dom(R)$.
• $T \circ (S \circ R) = (T \circ S) \circ R$.
• $(S \circ R)^{−1} = R^{−1} \circ S^{−1}$.

Soln1

(a) Domain is the set of living persons who are parents of some living people and Range is the set of all living persons who have a living parent.

(b) Domain is the set of all real numbers, $\mathbb R$. Range is the set of all positive numbers, $\mathbb R^+$.

Soln2

(a) Domain is the set of all living persons who are brother of some living person. Range is the set of all living persons who have a living brother.

(b) Domain : $% = 1 \} %]]>$. Note that if domain lies in $(-1, 1)$ then $1- \frac 2 {1 + x^2}$ will be negative. Thus there will be no $y \in \mathbb R$ such that $y^2$ is negative.
Range: $% $.

Soln3

(a) $L$ is a relation from $S$ to $R$. $L^{-1}$ is a relation from $R$ to $S$. Thus:
$L^{-1} \circ L = \{ (m,n) \in S \times S \, \vert \, \exists o \in R((m,o) \in L \land (o,n) \in L^{-1}) \}$.
$\quad = \{ (m,n) \in S \times S \, \vert \, \text{There exists a room o such that m lives in that room and also n lives in that room.} \}$.
$\quad = \{ (m,n) \in S \times S \, \vert \, \text{There exists a room such that m, and n both lives in that room} \}$.

(b) $L^{-1} \circ L$ is a relation from $S$ to $S$. $E$ is a relation from $S$ to $C$. Thus:
$E \circ (L^{−1} \circ L) = \{ (m,n) \in S \times C \, \vert \, \exists s \in S((m,s) \in L^{−1} \circ L \land (s,n) \in E) \}$.
$E \circ (L^{−1} \circ L) = \{ (m,n) \in S \times C \, \vert \, \text{There exists a student s such that m and s lives in same room and s is doing course n} \}$.
$E \circ (L^{−1} \circ L) = \{ (m,n) \in S \times C \, \vert \, \text{There exists a student s who lives in dorm which has a student doing course n} \}$.

Soln4

(a) $S \circ R = \{ \{1, 5\}, \{1, 6\}, \{1, 4\}, \{2, 4\}, \{3, 6\} \}$.

(b)

$S^{-1} = \{ \{5, 4\}, \{6, 4\}, \{4, 5\}, \{6, 6\}\}$.
$S \circ S^{-1} = \{ \{5, 5\}, \{5, 6\}, \{6, 5\}, \{6, 6\}, \{4, 4\} \}$.

Soln5

(a)

$S^{-1} = \{ \{7, 4\}, \{8, 4\}, \{6, 5\}\}$.
$S^{-1} \circ R = \{ \{1, 4\}, \{3, 5\}, \{3, 4\}\}$.

(b)

$R^{-1} = \{ \{7, 1\}, \{6, 3\}, \{7, 3\} \}$.
$R^{-1} \circ S = \{ \{4, 1\}, \{4, 3\}, \{5, 3\}\}$.

Soln6

(a)

$Ran(R^{-1})$ and $Dom(R)$ are both subsets of $A$. Let $a$ is an arbitrary element of $A$. Then,

$a \in Ran(R^{−1})$ iff $\exists b \in B((b,a) \in R^{−1})$.
$iff \exists a ∈ A((a,b) ∈ R)$ iff $b ∈ Ran(R)$.

Since $a$ is arbitrary, $Ran(R^{-1}) = Dom(R)$.

(b)

Suppose $S = R^{-1}$, then $S^{-1} = (R^{-1})^{-1} = R$. Since $Dom(R^{−1}) = Ran(R)$, it follows that $Dom(S) = Ran(S^{-1})$, or $Ran(S^{-1} = Dom(S)$.

(c)

Clearly $(T \circ S) \circ R$ and $T \circ (S \circ R)$ are both relations from $A$ to $D$. Let $(a, d)$ be an arbitrary element of $A \times D$.

Suppose $(a,d) \in (T \circ S) \circ R$. Thus there must exist an element $b$ such that $(a, b) \in R$ and $(b,d) \in T \circ S$. Since $(b,d) \in T \circ S$, there must exist a n element $c$ such that $(b,c) \in S$ and $(c,d) \in T$. Now since $(a,b) \in R$ and $(b,c) \in S$, it follows that $(a,c) \in S \circ R$. Thus we have $(a,c) \in S \circ R$ and $(c,d) \in T$, thus it follows that $(a,d) \in T \circ (S \circ R)$. Since $(a,d)$ is arbitrary, it follows that $(T \circ S) \circ R \subseteq T \circ (S \circ R)$.

(d)

Clearly $S \circ R$ is a relation from $A$ to $C$. Thus $(S \circ R)^{-1}$ is a relation from $C$ to $A$. Also, it can be noted that $R^{−1} \circ S^{−1}$ is also a relation from $C$ to $A$. Suppose $(c,a)$ is an arbitrary element of $C$ to $A$.

($\to$)Suppose $(c,a) \in (S \circ R)^{-1}$. It follows that $(a,c) \in (S \circ R)$. Thus there exist an element $b$ such that $(a,b) \in R$ and $(b,c) \in S$. Since $(a,b) \in R$, it follows that $(b,a) \in R^{-1}$. Similarly, since $(b,c) \in S$, it follows that $(c,b) \in S^{-1}$. Now, we have $(c,b) \in S^{-1}$ and $(b,a) \in R^{-1}$, it follows $(c,a) \in R^{-1} \circ S^{-1}$. Now since $(c,a)$ is arbitrary, it follows that $(S \circ R)^{-1} \subseteq R^{-1} \circ S^{-1}$.

($\leftarrow$)Similarly, suppose $(c,a) \in R^{-1} \circ S^{-1}$. Thus there exists an element $b \in B$ such that $(c,b) \in S^{-1}$ and $(b,a) \in R^{-1}$. Thus it follows that $(a,b) \in R$ and $(b,c) \in S$. Now since $(a,b) \in R$ and $(b,c) \in S$, it follows that $(a,c) \in S \circ R$. And since $(a,c) \in S \circ R$, it follows that $(c,a) \in (S \circ R )^{-1}$. Since $(c,a)$ is arbitrary, we can conclude that $R^{-1} \circ S^{-1} \subseteq (S \circ R)^{-1}$.

Thus from both directions, we can conclude that $(S \circ R)^{-1} = R^{-1} \circ S^{-1}$.

Soln7

Suppose $p$ is the enemy of $q$ and $q$ is the enemy of $r$, then it is given that $p$ is the friend of $r$. Thus we have if $(p,q) \in E$ and $(q,r) \in E$, then $(p,r) \in F$. Now since $(p,q) \in E$ and $(q,r) \in E$, it follows $(p,r) \in E \circ E$. Thus the given statement says: if $(p,r) \in E \circ E$, then $(p,r) \in F$, or Enemy of an enemy is a friend. Since $(p,r)$ is arbitrary, it follows that $E \circ E \subseteq F$.

Soln8

(a)

Suppose $a \in Dom(S \circ R)$. Thus there exist a pair $(b,c) \in S$ such that $(a,b) \in R$. Since $(a,b) \in R$, it follows that $a \in Dom(R)$. Thus if $a \in Dom(S \circ R)$, then $a \in Dom(R)$. Since $a$ is arbitrary, we can conclude that $Dom(S \circ R) \subseteq Dom(R)$.

(b)

Suppose $Ran(R) \subseteq Dom(S)$. Suppose $a \in Dom(R)$. Thus there must exist an element $b \in B$ such that $(a,b) \in R$. Thus $b \in Ran(R)$. Since $Ran(R) \subseteq Dom(S)$, it follows that $b \in Dom(S)$. Thus there must exist an element $c \in C$ such that $(b,c) \in S$. Now since $(a,b) \in R$ and $(b,c) \in S$, it follows that $(a,c) \in (S \circ R)$. Thus $a \in Dom(S \circ R)$. Thus we have, if $a \in Dom(R)$, then $a \in Dom(S \circ R)$. Since $a$ is arbitrary, we can conclude that $Dom(R) \subseteq Dom(S \circ R)$.

Thus from part (a) and this proof we can conclude that $Dom(S ◦ R) = Dom(R)$.

(c)

• $Ran(S \circ R) \subseteq Ran(S)$.
Suppose $c \in (S \circ R)$. Thus there must exist a pair $(a,c) \in S \circ R$. Since $(a,c) \in S \circ R$, there must exist an element $b \in B$ such that $(a,b) \in R$ and $(b,c) \in S$. Now since $(b,c) \in S$, it follows that $c \in Ran(S)$. Since $c$ is arbitrary, we can conclude that $Ran(S \circ R) \subseteq Ran(S)$.

• $Ran(S \circ R) = Ran(S)$.
Suppose $Dom(S) \subseteq Ran(R)$. Suppose $c \in Ran(S)$. Thus there must exist an element $b \in B$ such that $(b,c) \in Ran(S)$. Thus $b \in Dom(S)$. Now since $Dom(S) \subseteq Ran(R)$, it follows that $b \in Ran(R)$. Thus there must exist an element $a \in A$ such that $(a,b) \in R$. Thus we have $(a,b) \in R$ and $(b,c) \in S$, it follows that $(a,c) \in S \circ R$. Thus $c \in Ran(S \circ R)$. Thus we have, if $c \in Ran(S)$, then $c \in Ran(S \circ R)$. Since $c$ is arbitrary, we can conclude that $Ran(S) \subseteq Ran(S \circ R)$.

Since $Ran(S \circ R) \subseteq Ran(S)$ and $Ran(S) \subseteq Ran(S \circ R)$, we can conclude that $Ran(S \circ R) = Ran(S)$.

Soln9

(a). True.

Suppose $(a,b) \in R$. Then $a \in Dom(R)$ and $b \in Ran(R)$. Thus $(a,b) \in Dom(R) \times Ran(R)$. Thus we can say that, if $(a,b) \in R$, then $(a,b) \in Dom(R) \times Ran(R)$. Since $(a,b)$ is arbitrary, we can conclude that $R \subseteq Dom(R) \times Ran(R)$.

(b) True.

Suppose $R \subseteq S$. Suppose $(b,a) \in R^{-1}$. Thus $(a,b) \in R$. Since $R \subseteq S$, it follows $(a,b) \in S$. Now since $(a,b) \in S$, it follows $(b,a) \in S^{-1}$. Thus if $(b,a) \in R^{-1}$, then $((b,a) \in S^{-1}$. Since $(b,a)$ is arbitrary, we can conclude that $R^{-1} \subseteq S^{-1}$.

(c) True.

Suppose $(a,b) \in (R \cup S)^{−1}$.
iff $(b,a) \in (R \cup S)$.
iff $((b,a) \in R) \lor ((b,a) \in S)$.
iff $((a,b) \in R^{-1}) \lor ((a,b) \in S^{-1})$.
iff $(a,b) \in (R^{-1} \cup \in S^{-1})$.

Since $(a,b)$ is arbitrary, we can conclude that $(R \cup S)^{-1} = R^{-1} \cup S^{-1}$.

Soln10

We will prove it by contra-positive.

($\to$)Suppose $Ran(R) \cap Dom(S) \ne \phi$. Thus there must exist some element $b \in Ran(R)$ and $b \in Dom(S)$. Since $b \in Ran(R)$, there must exist an element $a \in A$ such that $(a,b) \in R$. Similarly, since $b \in Dom(S)$, there must exist an element $c \in C$ such that $(b,c) \in S$. Thus we have $(a,b) \in R$ and $(b,c) \in S$, it follows that $(a,c) \in S \circ R$. Thus $S \circ R \ne \phi$. Hence by contrapositive we can conclude that if $S \circ R = \phi$, then $Ran(R) \cap Dom(S) = \phi$.

($\leftarrow$)Suppose $S \circ R \ne \phi$. Thus there exist atleast one pair $(a,c) \in S \circ R$. Thus there exist an element $b \in B$ such that $b \in Ran(R)$ and $b \in Dom(S)$. Thus it follows $b \in Ran(R) \cap Dom(S)$, or $Ran(R) \cap Dom(S) \ne \phi$. Hence by contrapositive, we can conclude that if $Ran(R) \cap Dom(S) = \phi$, then $S \circ R = \phi$.

Soln11

(a)

Clearly $S \circ R$, $T \circ R$, and $(S \setminus T) \circ R$ are relations from $A$ \to $C$. Suppose $(a,c)$ is an arbitrary element in $(S \circ R)\setminus(T \circ R)$. Thus $(a,c) \in S \circ R$ and $(a,c) \notin T \circ R$. It follows that $a \in Dom(R)$ and $c \in Ran(S)$ and $c \notin Ran(T)$. Thus $c \in Ran(S) \setminus Ran(T)$, which is equivalent to saying that $c \in Ran(S \setminus T)$. Now since $a \in Dom(R)$ and $c \in Ran(S \setminus T)$, there exist an element $b \in B$ such that $(a,b) \in R$ and $(b,c) \in S \setminus T$. Thus it follows that $(a,c) \in (S \setminus T) \circ R$. Sicne $(a,c)$ is arbitrary, we can conclude that $(S \circ R)\setminus(T \circ R) \, \subseteq \, (S \setminus T) \circ R$.

(b)

The theorem and proof both not correct. In the proof, incorrect statement is:

Since $(a,b) \in R$ and $(b,c) \notin T, (a,c) \notin T \circ R$.

The problem is, there might exist some other element $b'$ such that $(a,b') \in R$ and $(b',c) \in T$. Thus even if $(b,c) \notin T$, it might be a case that $(a,c)$ may still is in $T \circ R$.

(c)

As noted in the above part b, we can easily find such element:

$R = \{ \{ 1, 2\}, \{1, 1\} \}$.
$S = \{ \{ 2, 7\} \}$.
$T = \{ \{ 1, 7\} \}$.

$(S \setminus T) \circ R = \{ 1, 7 \}$.

$S \circ R = \{ 1, 7 \}$.
$T \circ R = \{ 1, 7 \}$.
$(S \circ R) \setminus (T \circ R) = \phi$.

Soln12

(a) True.

Suppose $S \subseteq T$. Suppose $(a,c) \in S \circ R$. Thus there exists an element $b \in B$ such that $(a,b) \in R$ and $(b,c) \in S$. Since $S \subseteq T$, it follows that $(b,c) \in T$. Now since $(a,b) \in R$ and $(b,c) \in T$, it follows $(a,c) \in T \circ R$. Thus if $(a,c) \in S \circ R$, then $(a,c) \in T \circ R$. Since $(a,c)$ is arbitrary, we can conclude $S \circ R \subseteq T \circ R$.

(b) True.

Suppose $(a,c) \in (S \cap T) \circ R$. Thus there exists an element $b \in B$ such that $(a,b) \in R$ and $(b,c) \in S \cap T$. It follows that $(b,c) \in S$ and $(b,c) \in T$. Since $(a,b) \in R$ and $(b,c) \in S$, $(a,c) \in S \circ R$. Similarly, since $(a,b) \in R$ and $(b,c) \in T$, $(a,c) \in T \circ R$. Thus $(a,c) \in (S \circ R) \cap (T \circ R)$.

(c) False.

$R = \{ \{ 1, 2\}, \{1, 1\} \}$.
$S = \{ \{ 2, 7\} \}$.
$T = \{ \{ 1, 7\} \}$.

$S \circ R = \{ 1, 7 \}$.
$T \circ R = \{ 1, 7 \}$.
$(S \circ R) \cap (T \circ R) = \{ 1, 7 \}$.

$S \cap T = \phi$
$(S \cap T) \circ R = \phi$

Thus, $(S \circ R) \cap (T \circ R) \ne (S \cap T) \circ R$.

(d)

Suppose $(a,c) \in (S \cup T) \circ R$. Thus,

iff, $\exists b \in B,$ such that, $(a,b) \in R \land ((b,c) \in S \lor (b,c) \in T)$.
iff, $((a,b) \in R \land (b,c) \in S) \lor ((a,b) \in R \land (b,c) \in T)$.
iff, $((a,c) \in S \circ R) \lor ((a,c)\in T \circ R)$.
iff, $(a,c) \in (S \circ R) \cup (T \circ R)$.

Since, $(a,c)$ is arbitrary, we can conclude that $(S \cup T) \circ R \, = \, (S \circ R) \cup (T \circ R)$.