Question paper, Set 1

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

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

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

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

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

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

$\therefore B \times A = \{(7, 6), (7, 12), (9, 6), (9, 12), (11, 6), (11, 12)\}$

Answer: $B \times A = \{(7, 6), (7, 12), (9, 6), (9, 12), (11, 6), (11, 12)\}$

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

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

$\therefore A \times B = \{(2, 6), (2, 8), (2, 10), (12, 6), (12, 8), (12, 10)\}$

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

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

$\therefore B \times B = \{(6, 6), (6, 8), (6, 10), (8, 6), (8, 8), (8, 10), (10, 6), (10, 8), (10, 10)\}$

Answer: $A \times B = \{(2, 6), (2, 8), (2, 10), (12, 6), (12, 8), (12, 10)\}$; $B \times B = \{(6, 6), (6, 8), (6, 10), (8, 6), (8, 8), (8, 10), (10, 6), (10, 8), (10, 10)\}$

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

$= 8 \times 5 = 40$

Answer: $n(A \times B) = 40$

$n(A \times B) = n(A) \times n(B) = 4 \times 10 = 40$

$n(B \times A) = n(B) \times n(A) = 10 \times 4 = 40$

Answer: $n(A \times B) = 40$, $n(B \times A) = 40$

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

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

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

$a \notin A$, so $(a, 7) \notin A \times B$. So $R_2$ is not a relation.

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

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

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

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

$n(A \times B) = 5 \times 3 = 15$

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

$= 2^{15} = 32768$

Answer: $2^{15} = 32768$

$\therefore A \times B = \{(9, b), (11, b), (12, b)\}$

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

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 $A \times B$:

$\phi$, $\{(9, b)\}$, $\{(11, b)\}$, $\{(12, b)\}$, $\{(9, b), (11, b)\}$, $\{(9, b), (12, b)\}$, $\{(11, b), (12, b)\}$, $\{(9, b), (11, b), (12, b)\}$

Answer: $A \times B = \{(9, b), (11, b), (12, b)\}$; 8 relations: $\phi$, $\{(9, b)\}$, $\{(11, b)\}$, $\{(12, b)\}$, $\{(9, b), (11, b)\}$, $\{(9, b), (12, b)\}$, $\{(11, b), (12, b)\}$, $\{(9, b), (11, b), (12, b)\}$

$11Rc$ means $(11, c) \in R$.

Yes, since $(11, c) \in R$.

Answer: Yes, since $(11, c) \in R$.

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

$(y, 9) \in R \Rightarrow yR9$

$(y, 10) \in R \Rightarrow yR10$

$(z, 9) \in R \Rightarrow zR9$

$(z, 10) \in R \Rightarrow zR10$

Answer: $yR9$, $yR10$, $zR9$, $zR10$

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

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

$\therefore A \times B = \{(a, 6), (a, 9), (x, 6), (x, 9), (z, 6), (z, 9)\}$

Answer: $\{(a, 6), (a, 9), (x, 6), (x, 9), (z, 6), (z, 9)\}$

$A \times B = \{(2, b), (2, x), (2, y), (5, b), (5, x), (5, y)\}$

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

$R_1 = \phi$ is the null relation.

$R_2$ is a relation from $A$ to $B$, but neither null nor universal.

Answer: $R_3$ is universal, $R_1$ is null

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

$\text{Domain of } R = \{4, 12\}$

Answer: $\{4, 12\}$

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

$A \times B = \{(1, p), (1, q), (7, p), (7, q), (12, p), (12, q)\}$

$\text{Domain of } A \times B = \{1, 7, 12\}$

$\text{Range of } A \times B = \{p, q\}$

Answer: Domain $= \{1, 7, 12\}$, Range $= \{p, q\}$

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

$\text{Domain of } R_1 = \{1, 7\}$

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

$\text{Range of } R_1 = \{2, 10\}$

$\text{Domain of } R_2 = \{1, 7, 8\}$

$\text{Range of } R_2 = \{2, 10, 11\}$

Answer: Domain of $R_1 = \{1, 7\}$, Range of $R_1 = \{2, 10\}$; Domain of $R_2 = \{1, 7, 8\}$, Range of $R_2 = \{2, 10, 11\}$

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

$A \times B = \{(6, a), (6, b), (6, p), (11, a), (11, b), (11, p)\}$

$\text{Domain} = \{6, 11\}, \text{ Range} = \{a, b, p\}$

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

$\text{Domain of } R = \{6, 11\}$

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

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

Answer: $A \times B = \{(6, a), (6, b), (6, p), (11, a), (11, b), (11, p)\}$, Domain $\{6, 11\}$, Range $\{a, b, p\}$; $R$: Domain $\{6, 11\}$, Range $\{b, p\}$

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

$R^{-1} = \{(p, 5), (y, 7), (b, 8), (y, 8)\}$

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: $R^{-1} = \{(p, 5), (y, 7), (b, 8), (y, 8)\}$; 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.

$R = A \times B = \{(1, a), (1, p), (3, a), (3, p), (7, a), (7, p)\}$

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

$R^{-1} = \{(a, 1), (p, 1), (a, 3), (p, 3), (a, 7), (p, 7)\}$

$B \times A = \{(a, 1), (a, 3), (a, 7), (p, 1), (p, 3), (p, 7)\}$

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

Answer: $R = \{(1, a), (1, p), (3, a), (3, p), (7, a), (7, p)\}$; $R^{-1} = \{(a, 1), (p, 1), (a, 3), (p, 3), (a, 7), (p, 7)\} = B \times A$