Question paper, Set 3

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

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

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

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

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

$(12, t)$: $12 \in A$, $t \in B$, so it belongs.

$(10, t)$: $10 \in A$, $t \in B$, so it belongs.

$(t, 12)$: $t \notin A$, so it does not belong.

$(10, b)$: $b \notin B$, so it does not belong.

$\therefore$ $(12, t)$ and $(10, t)$ belong to $A \times B$.

Answer: $(12, t)$ and $(10, t)$

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

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

$\therefore A \times B = \{(x, b), (x, t), (x, y), (z, b), (z, t), (z, y)\}$

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

$\therefore B \times A = \{(b, x), (b, z), (t, x), (t, z), (y, x), (y, z)\}$

$(x, b) \in A \times B$ but $(x, b) \notin B \times A$.

$\therefore A \times B \ne B \times A$

Answer: $A \times B = \{(x, b), (x, t), (x, y), (z, b), (z, t), (z, y)\}$; $B \times A = \{(b, x), (b, z), (t, x), (t, z), (y, x), (y, z)\}$; No, $A \times B \ne B \times A$

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

$= 5 \times 5 = 25$

Answer: $n(A \times A) = 25$

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

$14 = 7 \times n(B)$

$n(B) = 14 \div 7 = 2$

Answer: $n(B) = 2$

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

$p \notin B$, so $(p, 5) \notin B \times A$.

So $R$ is not a relation from $B$ to $A$ (it is a relation from $A$ to $B$).

Answer: No, $(p, 5) \notin B \times A$.

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

Every pair of $R_2$ is in $A \times B$, so $R_2 \subseteq A \times B$.

$7 \notin B$, so $(c, 7) \notin A \times B$. So $R_1$ is not a relation.

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

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

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

$\therefore A \times B = \{(3, b), (3, c), (3, x), (4, b), (4, c), (4, x)\}$

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

All pairs of $R_1$ and $R_2$ are in $A \times B$.

$b \notin A$, so $(b, 3) \notin A \times B$. So $R_3$ is not a relation from $A$ to $B$.

Answer: $A \times B = \{(3, b), (3, c), (3, x), (4, b), (4, c), (4, x)\}$; $R_1$ and $R_2$ are relations, $R_3$ is not

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

Number of relations from $A$ to $A$ $= 2^{n(A \times A)}$

$= 2^{4} = 16$

Answer: $2^{4} = 16$

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

$\therefore A \times B = \{(6, b), (6, c), (6, t), (12, b), (12, c), (12, t)\}$

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

$\therefore B \times A = \{(b, 6), (b, 12), (c, 6), (c, 12), (t, 6), (t, 12)\}$

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

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

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

Answer: $A \times B = \{(6, b), (6, c), (6, t), (12, b), (12, c), (12, t)\}$; $B \times A = \{(b, 6), (b, 12), (c, 6), (c, 12), (t, 6), (t, 12)\}$; $64$ relations each way

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

$(4, a) \in R \Rightarrow 4Ra$

Answer: $4Ra$

$4Rb$ means $(4, b) \in R$.

Pairs of $R$ with first element $4$: $(4, a)$, $(4, z)$.

$\therefore b = a, z$

Answer: $a$ and $z$

$\phi \subseteq A \times B$, 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.

$A \times B = \{(b, a), (b, b), (b, t), (y, a), (y, b), (y, t)\}$

Answer: Null relation $= \phi$; universal relation $= \{(b, a), (b, b), (b, t), (y, a), (y, b), (y, t)\}$

$\phi \subseteq A \times B$, 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.

$A \times B = \{(3, c), (3, t), (3, z), (10, c), (10, t), (10, z)\}$

$n(A \times B) = 2 \times 3 = 6$

Answer: Null relation $\phi$ (0 pairs); universal relation $\{(3, c), (3, t), (3, z), (10, c), (10, t), (10, z)\}$ (6 pairs)

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

$\text{Domain of } R = \{2, 4, 9\}$

Answer: $\{2, 4, 9\}$

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

$A \times B = \{(7, c), (7, z), (8, c), (8, z), (12, c), (12, z)\}$

$\text{Domain of } A \times B = \{7, 8, 12\}$

$\text{Range of } A \times B = \{c, z\}$

Answer: Domain $= \{7, 8, 12\}$, Range $= \{c, z\}$

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 $R_1$.

$\text{Range of } R_1 = \{2, 3\}$

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

$\text{Range of } R_2 = \{2, 3, 5\}$

Answer: Domain of $R_1 = \{9, 11\}$, Range of $R_1 = \{2, 3\}$; Domain of $R_2 = \{9, 11\}$, Range of $R_2 = \{2, 3, 5\}$

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

$\text{Domain of } R_1 = \{12\}$

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

$\text{Range of } R_1 = \{c, p\}$

$\text{Domain of } R_2 = \{9, 12\}$

$\text{Range of } R_2 = \{c, p, t\}$

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

Answer: $R_1$: Domain $\{12\}$, Range $\{c, p\}$; $R_2$: Domain $\{9, 12\}$, Range $\{c, p, t\}$; $R_3$: Domain $\phi$, Range $\phi$

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

$R^{-1} = \{(x, 2), (p, 3), (x, 4), (p, 5)\}$

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

$= \{2, 3, 4, 5\}$

Answer: $R^{-1} = \{(x, 2), (p, 3), (x, 4), (p, 5)\}$; Range of $R^{-1} = \{2, 3, 4, 5\}$

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

$R^{-1} = \{(b, 1), (b, 8), (p, 8), (p, 11)\}$

$\text{Domain of } R = \{1, 8, 11\}$

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

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

$\text{Range of } R^{-1} = \{1, 8, 11\}$

Answer: (i) $R^{-1} = \{(b, 1), (b, 8), (p, 8), (p, 11)\}$ (ii) Domain $\{1, 8, 11\}$, Range $\{b, p\}$ (iii) Domain $\{b, p\}$, Range $\{1, 8, 11\}$