Question paper, Set 9

$(a, b) \in A \times B$ only when $a \in A$ and $b \in B$.

Here $7 \in A$ and $a \in B$.

$\therefore (7, a) \in A \times B$

Answer: Yes, since $7 \in A$ and $a \in B$.

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

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

$\therefore B \times A = \{(1, a), (1, b), (5, a), (5, b), (6, a), (6, b)\}$

Answer: $B \times A = \{(1, a), (1, b), (5, a), (5, b), (6, a), (6, b)\}$

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

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

$\therefore A \times B = \{(8, a), (8, q), (8, x), (11, a), (11, q), (11, x)\}$

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

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

$\therefore B \times B = \{(a, a), (a, q), (a, x), (q, a), (q, q), (q, x), (x, a), (x, q), (x, x)\}$

Answer: $A \times B = \{(8, a), (8, q), (8, x), (11, a), (11, q), (11, x)\}$; $B \times B = \{(a, a), (a, q), (a, x), (q, a), (q, q), (q, x), (x, a), (x, q), (x, x)\}$

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

$= 6 \times 6 = 36$

Answer: $n(A \times A) = 36$

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

$99 = 9 \times n(B)$

$n(B) = 99 \div 9 = 11$

Answer: $n(B) = 11$

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

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

$\therefore R \subseteq B \times A$

Answer: Yes, $R \subseteq B \times A$.

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

Every pair of $R_1$ is in $A \times B$, so $R_1 \subseteq A \times B$.

$5 \notin A$, so $(5, c) \notin A \times B$. So $R_2$ is not a relation.

Answer: $R_1$ is a relation from $A$ to $B$; $R_2$ is not.

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

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

$\therefore A \times B = \{(c, 7), (c, 9), (c, 10), (t, 7), (t, 9), (t, 10)\}$

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

All pairs of $R_1$ and $R_3$ are in $A \times B$.

$10 \notin A$, so $(10, t) \notin A \times B$. So $R_2$ is not a relation from $A$ to $B$.

Answer: $A \times B = \{(c, 7), (c, 9), (c, 10), (t, 7), (t, 9), (t, 10)\}$; $R_1$ and $R_3$ are relations, $R_2$ is not

$n(A \times B) = 1 \times 8 = 8$

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

$= 2^{8} = 256$

Answer: $2^{8} = 256$

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

$\therefore A \times B = \{(c, 3), (c, 9), (x, 3), (x, 9), (y, 3), (y, 9)\}$

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

$\therefore B \times A = \{(3, c), (3, x), (3, y), (9, c), (9, x), (9, y)\}$

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

$\text{From } A \text{ to } B: 2^{3 \times 2} = 2^{6} = 64$

$\text{From } B \text{ to } A: 2^{2 \times 3} = 2^{6} = 64$

Answer: $A \times B = \{(c, 3), (c, 9), (x, 3), (x, 9), (y, 3), (y, 9)\}$; $B \times A = \{(3, c), (3, x), (3, y), (9, c), (9, x), (9, y)\}$; $64$ relations each way

If $(a, b) \in R$ then it can be written $aRb$.

$(1, x) \in R \Rightarrow 1Rx$

Answer: $1Rx$

If $(a, b) \in R$ then it can be written $aRb$.

$(4, p) \in R \Rightarrow 4Rp$

$(4, x) \in R \Rightarrow 4Rx$

$(7, p) \in R \Rightarrow 7Rp$

$(12, p) \in R \Rightarrow 12Rp$

Answer: $4Rp$, $4Rx$, $7Rp$, $12Rp$

$\phi \subseteq A \times B$, so $\phi$ is a relation from $A$ to $B$: the null (void) relation.

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

$A \times B = \{(5, 1), (5, 5), (5, 12), (8, 1), (8, 5), (8, 12)\}$

Answer: Null relation $= \phi$; universal relation $= \{(5, 1), (5, 5), (5, 12), (8, 1), (8, 5), (8, 12)\}$

$\phi \subseteq A \times B$, so $\phi$ is a relation from $A$ to $B$: the null (void) relation.

$\phi$ has no ordered pairs.

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

$A \times B = \{(9, a), (9, c), (11, a), (11, c)\}$

$n(A \times B) = 2 \times 2 = 4$

Answer: Null relation $\phi$ (0 pairs); universal relation $\{(9, a), (9, c), (11, a), (11, c)\}$ (4 pairs)

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

$\text{Domain of } R = \{7, 10, 11\}$

Answer: $\{7, 10, 11\}$

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

$A \times B = \{(b, 9), (b, 12), (z, 9), (z, 12)\}$

$\text{Domain of } A \times B = \{b, z\}$

$\text{Range of } A \times B = \{9, 12\}$

Answer: Domain $= \{b, z\}$, Range $= \{9, 12\}$

Domain of $R_1$ $=$ set of first elements of the pairs in $R_1$.

$\text{Domain of } R_1 = \{9, 11\}$

Range of $R_1$ $=$ set of second elements of the pairs in $R_1$.

$\text{Range of } R_1 = \{q, t\}$

$\text{Domain of } R_2 = \{2, 9, 11\}$

$\text{Range of } R_2 = \{a, q, t\}$

Answer: Domain of $R_1 = \{9, 11\}$, Range of $R_1 = \{q, t\}$; Domain of $R_2 = \{2, 9, 11\}$, Range of $R_2 = \{a, q, t\}$

Domain of $R_1$ $=$ set of first elements of the pairs in $R_1$.

$\text{Domain of } R_1 = \{t\}$

Range of $R_1$ $=$ set of second elements of the pairs in $R_1$.

$\text{Range of } R_1 = \{9, 11\}$

$\text{Domain of } R_2 = \{c, y\}$

$\text{Range of } R_2 = \{3, 9, 11\}$

$R_3 = \phi$ has no pairs, so its domain and range are both $\phi$.

Answer: $R_1$: Domain $\{t\}$, Range $\{9, 11\}$; $R_2$: Domain $\{c, y\}$, Range $\{3, 9, 11\}$; $R_3$: Domain $\phi$, Range $\phi$

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

$R^{-1} = \{(x, 9), (b, 11), (x, 11)\}$

Range of $R^{-1}$ $=$ second elements of its pairs.

$= \{9, 11\}$

Answer: $R^{-1} = \{(x, 9), (b, 11), (x, 11)\}$; Range of $R^{-1} = \{9, 11\}$

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

$R^{-1} = \{(y, 1), (y, 6), (b, 7), (y, 10)\}$

$\text{Domain of } R = \{1, 6, 7, 10\}$

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

$\text{Domain of } R^{-1} = \{b, y\}$

$\text{Range of } R^{-1} = \{1, 6, 7, 10\}$

Answer: (i) $R^{-1} = \{(y, 1), (y, 6), (b, 7), (y, 10)\}$ (ii) Domain $\{1, 6, 7, 10\}$, Range $\{b, y\}$ (iii) Domain $\{b, y\}$, Range $\{1, 6, 7, 10\}$