# How to Prove It - Solutions

## Chapter - 4, Relations

### Summary

• Ordered Pair:
• Lets take an example. Suppose $P(x,y)$ means “x is capital city of country y”. The pair $(x,y)$ is considered as an ordered pair since $(x,y) \ne (y,x)$.
• In an ordered pair $(a,b)$, $a$ is called first coordinate and $b$ is called second coordinate.
• Cartesian Product:
Suppose $A$ and $B$ are sets. Then the Cartesian product of $A$ and $B$, denoted $A \times B$, is the set of all ordered pairs in which the first coordinate is an element of A and the second is an element of B. In other words,
$A \times B = \{(a, b) \vert a \in A \land b \in B \}$.
• Suppose $A, B, C$ and $D$ are sets, then:
• $A \times (B \cap C) = (A \times B) \cap (A \times C)$.
• $A \times (B \cup C) = (A \times B) \cup (A \times C)$.
• $(A \times B) \cap (C \times D) = (A \cap C) \times (B \cap D)$.
• $(A \times B) \cup (C \times D) \subseteq (A \cup C) \times (B \cup D)$.
• $A \times \phi = \phi \times A = \phi$.
• Suppose $A$ and $B$ are sets. Then $A \times B$ = $B \times A$ iff either $A = ∅, B = ∅,$ or $A = B$.
• Truth set of a statement:
Suppose $P(x, y)$ is a statement with two free variables in which $x$ ranges over a set $A$ and $y$ ranges over another set $B$. Then $A \times B$ is the set of all assignments to $x$ and $y$ that make sense in the statement $P(x, y)$. The truth set of $P(x, y)$ is the subset of $A \times B$ consisting of those assignments that make the statement come out true. Thus,
truth set of $P(x, y) = \{(a, b) \in A \times B \vert P(a, b) \}$

Soln1

(a) $\{(x,y) \; \vert P \; \times P \; \vert \; \text{x is a parent of y} \}$

(b) $\{(x,y) \; \vert \; C \times U \; \vert \; \text{Someone lives in x and attends y} \}$

Soln2

(a) $\{(x,y) \vert P \times C \; \vert \; \text{x lives in y} \}$.

(b) $\{(x,y) \; \vert \; C \times \mathbb N \; \vert \; \text{x is the population of y} \}$.

Soln3

It requires plotting points in the plane. So leaving it :)

Soln4

$A = \{1,2,3 \}$, $B = \{1,4\}$, $C = \{3,4\}$, and $D = \{5\}$

$B \cap C = \{ 4 \}$.
$B \cup C = \{ 1, 3, 4 \}$.

$A \cap C = \{ 3 \}$.
$A \cup C = \{ 1, 2, 3, 4 \}$.

$B \cap D = \phi$.
$B \cup D = \{ 1, 4, 5 \}$.

$A \times B = \{ \{1,1\}, \{1,4\}, \{2,1\}, \{2,4\}, \{3,1\}, \{3,4\} \}$.
$A \times C = \{ \{1,3\}, \{1,4\}, \{2,3\}, \{2,4\}, \{3,3\}, \{3,4\} \}$.
$C \times D = \{ \{3,5\}, \{4,5\}\}$.

(a)

$A \times (B \cap C) = \{ \{1, 4\}, \{2, 4\}, \{3, 4\} \}$
$(A \times B) \cap (A \times C) = \{ \{1, 4\}, \{2, 4\}, \{3, 4\} \}$

Thus, $A \times (B \cap C) = (A \times B) \cap (A \times C)$.

(b)

$A \times (B \cup C) = \{ \{1, 1\}, \{1, 3\}, \{1, 4\}, \{2, 1\}, \{2, 3\}, \{2, 4\}, \{3, 1\}, \{3, 3\}, \{3, 4\} \}$
$(A \times B) \cup (A \times C) = \{ \{1, 1\}, \{1, 3\}, \{1, 4\}, \{2, 1\}, \{2, 3\}, \{2, 4\}, \{3, 1\}, \{3, 3\}, \{3, 4\} \}$.

Thus, $A \times (B \cup C) = (A \times B) \cup (A \times C)$

(c)

$(A \times B) \cap (C \times D) = \phi$
$(A \cap C) \times (B \cap D) = (A \cap C) \times \phi = \phi$

Thus, $(A \times B) \cap (C \times D) = (A \cap C) \times (B \cap D)$.

(d)

$(A \times B) \cup (C \times D) = \{ \{1,1\}, \{1,4\}, \{2,1\}, \{2,4\}, \{3,1\}, \{3,4\}, \{3,5\}, \{4,5\} \}$
$(A \cup C) \times (B \cup D) = \{ \{1, 1\}, \{1, 4\}, \{1, 5\}, \{2, 1\}, \{2, 4\}, \{2, 5\}, \{3, 1\}, \{3, 4\}, \{3, 5\}, \{4, 1\}, \{4, 4\}, \{4, 5\} \}$

Thus, $(A \times B) \cup (C \times D) \subseteq (A \cup C) \times (B \cup D)$

(e) $A \times \phi = \phi$ and $\phi \times A = \phi$, thus $A \times \phi = \phi \times A = \phi$.

Soln5

Proof for: $A \times (B \cup C) \, = \, (A \times B) \cup (A \times C)$

($\to$)Let $p$ be an arbitrary element of $A \times (B \cap C)$. Thus $p = (x,y)$ is an ordered pair with in first coordinate $x \in A$ and second coordinate in $y \in (B \cap C)$. Thus $y$ can be either in $B$ or in $C$. Thus we have following cases:

• Case 1: $x \in A \land y \in B$:
Thus we can say that $p = (x,y) \in (A \times B)$. It follows that $p \in (A \times B) \cup (A \times C)$.

• Case 2: $x \in A \land y \in C$:
Thus we can say that $p = (x,y) \in (A \times C)$. It follows that $p \in (A \times B) \cup (A \times C)$.

Thus from both cases, if $p \in A \times (B \cup C)$, then $p \in (A \times B) \cup (A \times C)$.

($\leftarrow$)Suppose $p = (x,y) \in (A \times B) \cup (A \times C)$. Thus we have two cases:

• Case 1: $p \in (A \times C)$:
Thus $x \in A$ and $y \in C$. Since $y \in C$, it follows that $y \in B \cup C$. Now since $x \in A$ and $y \in B \cup C$, it follows that $p \in A \times (B \cup C)$.

• Case 2: $p \in (A \times B)$:
Thus $x \in A$ and $y \in B$. Since $y \in B$, it follows that $y \in B \cup C$. Now since $x \in A$ and $y \in B \cup C$, it follows that $p \in A \times (B \cup C)$.

Thus from both cases, if $p \in (A \times B) \cup (A \times C)$, then $p \in A \times (B \cup C)$.

Thus from both directions $A \times (B \cup C) \; = \; (A \times B) \cup (A \times C)$.

Proof for: $(A \times B) \cap (C \times D) = (A \cap C) \times (B \cap D)$

($\to$)Let $p = (x,y)$ be an arbitrary element of $(A \times B) \cap (C \times D)$. Thus $(x,y) \in (A \times B)$ and $(x,y) \in (C \times D)$. It follows that $x \in A \land x \in C$, or $x \in A \cap C$. Similarly, since $y \in B \land y \in D$, it follows $y \in B \cap D$. Thus $(x,y) \in (A \cap C) \times (B \cap D)$.

($\leftarrow$)Let $p = (x,y)$ be an arbitrary element of $(A \cap C) \times (B \cap D)$. Thus $x \in A \land y \in B$, it follows that $(x,y) \in A \times B$. Similarly, since $x \in C \land y \in D$, it follows that $(x,y) \in (C \times D)$. Thus we have $p = (x,y) \in (A \times B) \cap (C \times D)$.

Thus from both directions we have $(A \times B) \cap (C \times D) = (A \cap C) \times (B \cap D)$.

Soln6

It does not cover all the possible cases. Following cases are missed:

• Case 1: $x \in A \land y \in D$.
• Case 2: $x \in C \land y \in B$.

Soln7 $m \times n$.

Soln8

Theorem is correct.

A×(B\C) = (A×B)(A×C)

($\to$)Suppose $p=(x,y) \in A \times (B \setminus C)$. Thus $x \in A$ and $y \in B \land y \notin C$. It follows that $(x,y) \in A \times B$, but $(x,y) \notin A \times C$. Thus $(x,y) \in (A \times B) \setminus (A \times C)$.

($\leftarrow$)Suppose $p=(x,y) \in (A \times B) \setminus (A \times C)$. Thus $(x,y) \in A \times B$ and $(x,y) \notin A \times C$. Since $x \in A$ and $(x,y) \notin A \times C$, it follows that $y \notin C$. Thus $x \in A$ and $y \in B \land y notin C$ , or $y \in B \setminus C$. Thus we have $(x,y) \in A \times (B \times C)$.

Since $p=(x,y)$ is arbitrary, we can conclude that $A \times (B \setminus C) = (A \times B) \setminus (A \times C)$.

Soln9

($\to$)Suppose $p=(x,y) \in (A \times B) \setminus (C \times D)$. Thus $(x,y) \in (A \times B)$ and $(x,y) \notin (C \times D)$. Since $(x,y) \in (A \times B)$, it follows that $x \in A$ and $y \in B$. Since $(x,y) \notin (C \times D)$, we can have following possible cases:

• Case 1: $x \notin C$:
Since $x \in A$ and $y \in B$, it follows that $(x,y) \in (A \setminus C) \times B$. Thus $(x,y) \in [((A \setminus C) \times B) \cup (A \times (B \setminus D))]$.

• Case 2: $y \notin D$:
Since $x \in A$ and $y \in B$, it follows that $(x,y) \in A \times (B \setminus D)$. Thus $(x,y) \in [((A \setminus C) \times B) \cup (A \times (B \setminus D))]$.

Thus from both cases, $p=(x,y) \in [((A \setminus C) \times B) \cup (A \times (B \setminus D))]$.

($\leftarrow$)Suppose $p=(x,y) \in [((A \setminus C) \times B) \cup (A \times (B \setminus D))]$. Thus we have following cases:

• Case 1: $(x,y) \in ((A \setminus C) \times B)$.
Here since $x \in A$ and $y \in B$, it follows that $(x,y) \in A \times B$. Also since $x \notin C$, it follows that $(x,y) \notin C \times D$. Thus we have, $(x,y) \in A \times B$ and $(x,y) \notin C \times D$. It follows that $(x,y) \in (A \times B) \setminus (C \times D)$.

• Case 2: $(x,y) \in (A \times (B \setminus D)$.
Here since $x \in A$ and $y \in B$, it follows that $(x,y) \in A \times B$. Also since $y \notin D$, it follows that $(x,y) \notin C \times D$. Thus we have, $(x,y) \in A \times B$ and $(x,y) \notin C \times D$. It follows that $(x,y) \in (A \times B) \setminus (C \times D)$.

Thus from both cases, $p=(x,y) \in (A \times B) \setminus (C \times D)$.

Thus from both directions, we have $(A \times B) \setminus (C \times D) = [((A \setminus C) \times B) \cup (A \times (B \setminus D))]$.

Soln10

Suppose $A \times B \cap C \times D = \phi$. Suppose $(x,y) \in A \times B$. It follows that $(x,y) \notin C \times D$. Thus there are following cases:

• Case 1: $x \notin C$:
Since $x \in A$, it follows that $A \cap C = \phi$.

• Case 2: $y \notin D$: Since $y \in B$, it follows that $B \cap D = \phi$.

Thus either $A \cap C = \phi$, or $B \cap D = \phi$.

Soln11

(a)

Suppose $(x,y) \in \cup_{i \in I}(A_i \times B_i)$. Thus there atleast exists an $p \in I$ such that $(x,y) \in (A_p \times B_p)$. Thus $x \in A_p$ and $y \in B_p$. Since $x \in A_p$, it follows $x \in \cup_{i \in I}A_i$. Similarly, since $y \in B_p$, it follows $y \in \cup_{i \in I}B_i$. Thus $(x,y) \in \cup_{i \in I}A_i \times \cup_{i \in I}B_i$. Since $(x,y)$ is arbitrary, it follows that $\cup_{i \in I}(A_i \times B_i) \subseteq \cup_{i \in I}A_i \times \cup_{i \in I}B_i$.

(b)

∪p∈PCp =(∪i∈IAi)×(∪i∈IBi)

($\to$)Suppose $(x,y) \in \cup_{p \in P}C_p$. Thus $(x,y)$ exist in atleast one of the set, say $C_k$ such that $k \in P$. Thus $x \in A_k$ and $y \in B_k$. It follows that $x \in \cup_{i \in I}A_i$ and $y \in \cup_{i \in I}B_i$. Thus we have $(x,y) \in \cup_{i \in I}A_i \times \cup_{i \in I}B_i$. Since $(x,y)$ is arbitrary, it follows that $\cup_{p \in P}C_p \subseteq \cup_{i \in I}A_i \times \cup_{i \in I}B_i$.

($\leftarrow$)Suppose $(x,y) \in \cup_{i \in I}A_i \times \cup_{i \in I}B_i$. Thus $x \in A_m$ such that $m \in I$ and $y \in B_n$ such that $n \in I$. Thus $(x,y) \in A_m \times B_n$, or $(x,y) \in C_t$ where $C_t=(A_m \times B_n)$. Since $m \in I$ and $n \in I$, it follows that $t = (m,n) \in I \times I$, or $t \in P$. Now, since $(x,y) \in C_t$ and $t \in P$, it follows that $(x,y) \in \cup_{p \in P}C_p$. Since $(x,y)$ is arbitrary, it follows that $\cup_{i \in I}A_i \times \cup_{i \in I}B_i \subseteq \cup_{p \in P}C_p$.

Thus, from both directions we have: $\cup_{p \in P}C_p = \cup_{i \in I}A_i \times \cup_{i \in I}B_i$.

Soln12

The theorem and proof both are not correct. It does not consider if some of the sets $A, B, C,$ or $D$ are empty.