Question paper, Set 2

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

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

$\therefore A \times A = \{(t, t), (t, x), (x, t), (x, x)\}$

Answer: $A \times A = \{(t, t), (t, x), (x, t), (x, x)\}$

$A$ is the set of first elements of the pairs.

$A = \{2, 4, 12\}$

$B$ is the set of second elements of the pairs.

$B = \{3, 8\}$

Answer: $A = \{2, 4, 12\}$, $B = \{3, 8\}$

$A$ is the set of first elements of the pairs.

$A = \{3, 5, 6\}$

$B$ is the set of second elements of the pairs.

$B = \{q, x\}$

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

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

$\therefore B \times A = \{(q, 3), (q, 5), (q, 6), (x, 3), (x, 5), (x, 6)\}$

Answer: $A = \{3, 5, 6\}$, $B = \{q, x\}$; $B \times A = \{(q, 3), (q, 5), (q, 6), (x, 3), (x, 5), (x, 6)\}$

$n(A) = 3, n(B) = 5$

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

Answer: $n(A \times B) = 15$

$n(A) = 2, n(B) = 5$

$n(A \times B) = 2 \times 5 = 10$

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

Answer: $n(A \times B) = 10$, $n(A \times A) = 4$

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

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

$\therefore R \subseteq A \times B$

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

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

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

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

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

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

For example, $R_1 = \{(10, c), (12, q)\}$ and $R_2 = \{(10, q)\}$ are subsets of $A \times B$.

Answer: $A \times B = \{(10, c), (10, q), (12, c), (12, q)\}$; for example $R_1 = \{(10, c), (12, q)\}$, $R_2 = \{(10, q)\}$

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

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

$\therefore A \times B = \{(5, b), (5, q), (11, b), (11, q), (12, b), (12, q)\}$

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

For example $R_1 = \{(5, b)\}$, $R_2 = \{(5, q), (11, b)\}$, $R_3 = \{(5, b), (11, q), (12, q)\}$.

Answer: $A \times B = \{(5, b), (5, q), (11, b), (11, q), (12, b), (12, q)\}$; for example $\{(5, b)\}$, $\{(5, q), (11, b)\}$, $\{(5, b), (11, q), (12, q)\}$

$n(A) = 4, n(B) = 2$

$n(A \times B) = 4 \times 2 = 8$

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

$= 2^{8} = 256$

Answer: $2^{8} = 256$

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

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

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

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

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

$= 2^{4} = 16$

Any subsets of $A \times B$, for example $\{(3, a), (12, z)\}$ and $\{(3, z)\}$.

Answer: (i) $A \times B = \{(3, a), (3, z), (12, a), (12, z)\}$ (ii) $4$ (iii) $2^{4} = 16$ (iv) for example $\{(3, a), (12, z)\}$, $\{(3, z)\}$

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

$pR3 \Rightarrow (p, 3) \in R$

Answer: $(p, 3) \in R$

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

$(6, b) \notin R$, so $6Rb$ is false.

$(6, t) \in R$, so $6Rt$ is true.

$(10, b) \in R$, so $10Rb$ is true.

Answer: $6Rt$ and $10Rb$ are true; $6Rb$ is false.

The universal relation from $A$ to $B$ is $A \times B$.

$A \times B = \{(5, a), (5, c), (5, p), (8, a), (8, c), (8, p)\}$

$R$ has every pair of $A \times B$, so $R = A \times B$: yes.

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

Universal relation from $A$ to $B$ is $A \times B$; from $B$ to $A$ it is $B \times A$.

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

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

$(5, q) \in A \times B$ but $(5, q) \notin B \times A$, so they are not equal.

Answer: $A \times B = \{(5, q), (5, y), (10, q), (10, y)\}$; $B \times A = \{(q, 5), (q, 10), (y, 5), (y, 10)\}$; not equal

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

$\text{Range of } R = \{10, 11\}$

Answer: $\{10, 11\}$

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

$\text{Domain of } R = \{c, t, x\}$

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

$\text{Range of } R = \{a, b, t\}$

Answer: Domain $= \{c, t, x\}$, Range $= \{a, b, t\}$

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 $B$.

$\text{Domain of } R = \{b, q, t, y\}$

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

Answer: $R \subseteq A \times B$; Domain $= \{b, q, t, y\}$, Range $= \{y, z\}$

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

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

$\therefore A \times B = \{(4, p), (4, q), (8, p), (8, q), (9, p), (9, q)\}$

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

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

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

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

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

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

Answer: (i) $A \times B = \{(4, p), (4, q), (8, p), (8, q), (9, p), (9, q)\}$ (ii) $R \subseteq A \times B$ (iii) Domain $= \{4, 9\}$, Range $= \{p, q\}$

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

$R^{-1} = \{(a, 4), (t, 9), (z, 10)\}$

Domain of $R^{-1}$ $=$ first elements of its pairs.

$= \{a, t, z\}$

Answer: $R^{-1} = \{(a, 4), (t, 9), (z, 10)\}$; Domain of $R^{-1} = \{a, t, z\}$

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

$R_1^{-1} = \{(t, 3), (t, 11), (x, 12)\}$

$\text{Domain} = \{t, x\}, \text{ Range} = \{3, 11, 12\}$

$R_2^{-1} = \{(t, 3), (x, 11), (x, 12)\}$

$\text{Domain} = \{t, x\}, \text{ Range} = \{3, 11, 12\}$

Answer: $R_1^{-1} = \{(t, 3), (t, 11), (x, 12)\}$ (Domain $\{t, x\}$, Range $\{3, 11, 12\}$); $R_2^{-1} = \{(t, 3), (x, 11), (x, 12)\}$ (Domain $\{t, x\}$, Range $\{3, 11, 12\}$)