Question paper, Set 4

$A \times B = \{(a, b) : a \in A, b \in B\}$

Pair each element of $A$ with every element of $B$.

$\therefore A \times B = \{(10, 3), (12, 3)\}$

Answer: $A \times B = \{(10, 3), (12, 3)\}$

$A \times B = \{(a, b) : a \in A, b \in B\}$

Pair each element of $A$ with every element of $B$.

$\therefore A \times B = \{(8, 3), (8, 7), (8, 12), (10, 3), (10, 7), (10, 12)\}$

Answer: $A \times B = \{(8, 3), (8, 7), (8, 12), (10, 3), (10, 7), (10, 12)\}$

$A \times B = \{(a, b) : a \in A, b \in B\}$

Pair each element of $A$ with every element of $B$.

$\therefore A \times B = \{(4, 9), (4, 10), (6, 9), (6, 10), (10, 9), (10, 10)\}$

$A \times A = \{(x, y) : x \in A, y \in A\}$

Pair each element of $A$ with every element of $A$.

$\therefore A \times A = \{(4, 4), (4, 6), (4, 10), (6, 4), (6, 6), (6, 10), (10, 4), (10, 6), (10, 10)\}$

Answer: $A \times B = \{(4, 9), (4, 10), (6, 9), (6, 10), (10, 9), (10, 10)\}$; $A \times A = \{(4, 4), (4, 6), (4, 10), (6, 4), (6, 6), (6, 10), (10, 4), (10, 6), (10, 10)\}$

$n(A \times B) = n(A) \times n(B) = mn$

$= 11 \times 12 = 132$

Answer: $n(A \times B) = 132$

$n(A \times B) = n(A) \times n(B) = 12 \times 2 = 24$

$n(B \times A) = n(B) \times n(A) = 2 \times 12 = 24$

Answer: $n(A \times B) = 24$, $n(B \times A) = 24$

$A \times B = \{(a, b) : a \in A, b \in B\}$

Any subset $R$ of $A \times B$ ($R \subseteq A \times B$) is called a relation from $A$ to $B$.

Answer: If $A$, $B$ are two nonempty sets, every subset of $A \times B$ is a relation from $A$ to $B$.

$A \times B = \{(a, b) : a \in A, b \in B\}$

Pair each element of $A$ with every element of $B$.

$\therefore A \times B = \{(a, 2), (a, 6), (c, 2), (c, 6)\}$

A relation from $A$ to $B$ is a subset of $A \times B$.

For example, $R_1 = \{(a, 2), (c, 6)\}$ and $R_2 = \{(a, 6)\}$ are subsets of $A \times B$.

Answer: $A \times B = \{(a, 2), (a, 6), (c, 2), (c, 6)\}$; for example $R_1 = \{(a, 2), (c, 6)\}$, $R_2 = \{(a, 6)\}$

A relation from $A$ to $B$ is a subset of $A \times B$.

A relation from $B$ to $A$ is a subset of $B \times A$.

First elements of $R_1$ are in $B$, second elements in $A$: $R_1 \subseteq B \times A$.

First elements of $R_2$ are in $A$, second elements in $B$: $R_2 \subseteq A \times B$.

Answer: $R_1$ is a relation from $B$ to $A$; $R_2$ is a relation from $A$ to $B$

Relations from $A$ to $B$ are the subsets of $A \times B$.

$\text{Number} = 2^{9} = 512$

Answer: $2^{9} = 512$

$\therefore A \times B = \{(5, q), (6, q), (10, q)\}$

$n(A \times B) = 3 \times 1 = 3$

Number of relations from $A$ to $B$ $=$ number of subsets of $A \times B$ $= 2^{mn}$.

$= 2^{3} = 8$

The relations are the subsets of $A \times B$:

$\phi$, $\{(5, q)\}$, $\{(6, q)\}$, $\{(10, q)\}$, $\{(5, q), (6, q)\}$, $\{(5, q), (10, q)\}$, $\{(6, q), (10, q)\}$, $\{(5, q), (6, q), (10, q)\}$

Answer: $A \times B = \{(5, q), (6, q), (10, q)\}$; 8 relations: $\phi$, $\{(5, q)\}$, $\{(6, q)\}$, $\{(10, q)\}$, $\{(5, q), (6, q)\}$, $\{(5, q), (10, q)\}$, $\{(6, q), (10, q)\}$, $\{(5, q), (6, q), (10, q)\}$

$12Ra$ means $(12, a) \in R$.

No, since $(12, a) \notin R$.

Answer: No, since $(12, a) \notin R$.

$aRb$ means $(a, b) \in R$.

$cR5 \Rightarrow (c, 5) \in R$

$xR2 \Rightarrow (x, 2) \in R$

$xR5 \Rightarrow (x, 5) \in R$

$\therefore R = \{(c, 5), (x, 2), (x, 5)\}$

Answer: $R = \{(c, 5), (x, 2), (x, 5)\}$

$A \times B \subseteq A \times B$, so $A \times B$ is a relation from $A$ to $B$: the universal relation.

Pair each element of $A$ with every element of $B$.

$\therefore A \times B = \{(p, 9), (p, 10), (p, 12), (q, 9), (q, 10), (q, 12)\}$

Answer: $\{(p, 9), (p, 10), (p, 12), (q, 9), (q, 10), (q, 12)\}$

$A \times B = \{(a, a), (a, x), (a, z), (x, a), (x, x), (x, z)\}$

$A \times B \subseteq A \times B$, so $A \times B$ is a relation from $A$ to $B$: the universal relation.

$R_3$ has all pairs of $A \times B$, so $R_3$ is universal.

$R_2 = \phi$ is the null relation.

$R_1$ is a relation from $A$ to $B$, but neither null nor universal.

Answer: $R_3$ is universal, $R_2$ is null

Range of $R$ $=$ set of second elements of the pairs in $R$.

$\text{Range of } R = \{4, 7\}$

Answer: $\{4, 7\}$

Domain of $R$ $=$ set of first elements of the pairs in $R$.

$\text{Domain of } R = \{3, 5, 12\}$

Range of $R$ $=$ set of second elements of the pairs in $R$.

$\text{Range of } R = \{c, y\}$

Answer: Domain $= \{3, 5, 12\}$, Range $= \{c, y\}$

Every pair of $R$ has first element in $A$ and second element in $B$.

$\therefore R \subseteq A \times B$

So $R$ is a relation from $A$ to $B$.

$\text{Domain of } R = \{4, 8\}$

$\text{Range of } R = \{b, c\}$

Answer: $R \subseteq A \times B$; Domain $= \{4, 8\}$, Range $= \{b, c\}$

$A \times B \subseteq A \times B$, so $A \times B$ is a relation from $A$ to $B$: the universal relation.

$A \times B = \{(p, 2), (p, 4), (p, 6), (x, 2), (x, 4), (x, 6)\}$

$\text{Domain} = \{p, x\}, \text{ Range} = \{2, 4, 6\}$

Domain of $R$ $=$ set of first elements of the pairs in $R$.

$\text{Domain of } R = \{p, x\}$

Range of $R$ $=$ set of second elements of the pairs in $R$.

$\text{Range of } R = \{2, 4, 6\}$

Answer: $A \times B = \{(p, 2), (p, 4), (p, 6), (x, 2), (x, 4), (x, 6)\}$, Domain $\{p, x\}$, Range $\{2, 4, 6\}$; $R$: Domain $\{p, x\}$, Range $\{2, 4, 6\}$

$R^{-1}$: interchange the two elements of every pair of $R$.

$R^{-1} = \{(c, 2), (c, 5), (t, 5), (t, 8)\}$

Every pair of $R^{-1}$ has first element in $B$ and second element in $A$.

$\therefore R^{-1} \subseteq B \times A$

So $R^{-1}$ is a relation from $B$ to $A$.

Answer: $R^{-1} = \{(c, 2), (c, 5), (t, 5), (t, 8)\}$; yes, $R^{-1} \subseteq B \times A$

$A \times B \subseteq A \times B$, so $A \times B$ is a relation from $A$ to $B$: the universal relation.

$R = A \times B = \{(b, 1), (b, 3), (b, 12), (c, 1), (c, 3), (c, 12)\}$

$R^{-1}$: interchange the two elements of every pair of $R$.

$R^{-1} = \{(1, b), (3, b), (12, b), (1, c), (3, c), (12, c)\}$

$B \times A = \{(1, b), (1, c), (3, b), (3, c), (12, b), (12, c)\}$

Both have the same pairs, so $R^{-1} = B \times A$.

Answer: $R = \{(b, 1), (b, 3), (b, 12), (c, 1), (c, 3), (c, 12)\}$; $R^{-1} = \{(1, b), (3, b), (12, b), (1, c), (3, c), (12, c)\} = B \times A$