Sets and Relations: Question paper, Set 8

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If $A = \{1, 8\}$ , find $A \times A$ .
1 mark

Answer and solution

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

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

$\therefore A \times A = \{(1, 1), (1, 8), (8, 1), (8, 8)\}$

Answer: $A \times A = \{(1, 1), (1, 8), (8, 1), (8, 8)\}$
Write the cartesian product $A \times B$ of $A = \{a, b, c\}$ and $B = \{c, y\}$ .
2 marks

Answer and solution

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

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

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

Answer: $A \times B = \{(a, c), (a, y), (b, c), (b, y), (c, c), (c, y)\}$
If $A = \{3, 7, 8\}$ and $B = \{c, t\}$ , find $A \times B$ and $A \times A$ .
5 marks

Answer and solution

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

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

$\therefore A \times B = \{(3, c), (3, t), (7, c), (7, t), (8, c), (8, t)\}$

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

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

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

Answer: $A \times B = \{(3, c), (3, t), (7, c), (7, t), (8, c), (8, t)\}$ ; $A \times A = \{(3, 3), (3, 7), (3, 8), (7, 3), (7, 7), (7, 8), (8, 3), (8, 7), (8, 8)\}$
If $A = \{2, 4, 6, 11, 12\}$ and $B = \{a, p, q, t, x\}$ , find $n(A \times B)$ without writing $A \times B$ .
1 mark

Answer and solution

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

$n(A \times B) = mn = 5 \times 5 = 25$

Answer: $n(A \times B) = 25$
If $A = \{p, q, t, x\}$ and $B = \{b, y\}$ , find $n(A \times B)$ and $n(A \times A)$ .
2 marks

Answer and solution

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

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

$n(A \times A) = 4 \times 4 = 16$

Answer: $n(A \times B) = 8$ , $n(A \times A) = 16$
If $A = \{1, 10, 11\}$ and $B = \{a, p\}$ , is $R = \{(1, a), (10, p), (11, q)\}$ a relation from $A$ to $B$ ? Give reason.
1 mark

Answer and solution

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

$q \notin B$ , so $(11, q) \notin A \times B$ .

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

Answer: No, $(11, q) \notin A \times B$ .
If $A = \{4, 10\}$ and $B = \{x, z\}$ , find $A \times B$ and write any two relations from $A$ to $B$ .
2 marks

Answer and solution

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

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

$\therefore A \times B = \{(4, x), (4, z), (10, x), (10, z)\}$

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

For example, $R_1 = \{(4, x), (10, z)\}$ and $R_2 = \{(4, z)\}$ are subsets of $A \times B$ .

Answer: $A \times B = \{(4, x), (4, z), (10, x), (10, z)\}$ ; for example $R_1 = \{(4, x), (10, z)\}$ , $R_2 = \{(4, z)\}$
If $A = \{3, 8\}$ and $B = \{a, b, p\}$ , find $A \times B$ . Write three relations from $A$ to $B$ having one, two and three ordered pairs.
3 marks

Answer and solution

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

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

$\therefore A \times B = \{(3, a), (3, b), (3, p), (8, a), (8, b), (8, p)\}$

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

For example $R_1 = \{(3, a)\}$ , $R_2 = \{(3, b), (3, p)\}$ , $R_3 = \{(3, a), (8, a), (8, p)\}$ .

Answer: $A \times B = \{(3, a), (3, b), (3, p), (8, a), (8, b), (8, p)\}$ ; for example $\{(3, a)\}$ , $\{(3, b), (3, p)\}$ , $\{(3, a), (8, a), (8, p)\}$
If $A \times B$ has 11 elements, how many relations are there from $A$ to $B$ ?
1 mark

Answer and solution

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

$\text{Number} = 2^{11} = 2048$

Answer: $2^{11} = 2048$
If $A = \{q, y\}$ and $B = \{1, 8, 9\}$ , find (i) $A \times B$ (ii) $n(A \times B)$ (iii) the number of relations from $A$ to $B$ (iv) any two relations from $A$ to $B$ .
5 marks

Answer and solution

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

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

$\therefore A \times B = \{(q, 1), (q, 8), (q, 9), (y, 1), (y, 8), (y, 9)\}$

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

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

$= 2^{6} = 64$

Any subsets of $A \times B$ , for example $\{(q, 1), (y, 9)\}$ and $\{(q, 8)\}$ .

Answer: (i) $A \times B = \{(q, 1), (q, 8), (q, 9), (y, 1), (y, 8), (y, 9)\}$ (ii) $6$ (iii) $2^{6} = 64$ (iv) for example $\{(q, 1), (y, 9)\}$ , $\{(q, 8)\}$
If $11Rq$ for a relation $R$ , which ordered pair belongs to $R$ ?
1 mark

Answer and solution

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

$11Rq \Rightarrow (11, q) \in R$

Answer: $(11, q) \in R$
A relation $R$ from $A = \{1, 5, 8\}$ to $B = \{p, t\}$ has only $1Rt$ , $5Rp$ and $8Rt$ . Write $R$ in roster form.
2 marks

Answer and solution

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

$1Rt \Rightarrow (1, t) \in R$

$5Rp \Rightarrow (5, p) \in R$

$8Rt \Rightarrow (8, t) \in R$

$\therefore R = \{(1, t), (5, p), (8, t)\}$

Answer: $R = \{(1, t), (5, p), (8, t)\}$
If $A = \{c, t, x\}$ and $B = \{c, q\}$ , is $R = \{(x, q), (t, q), (x, c)\}$ the universal relation from $A$ to $B$ ? Give reason.
2 marks

Answer and solution

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

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

$(c, c) \in A \times B$ but $(c, c) \notin R$ , so $R \ne A \times B$ : no.

Answer: No, $(c, c) \notin R$ , so $R \ne A \times B$ .
If $A = \{t, x\}$ and $B = \{5, 8\}$ , write the universal relation from $A$ to $B$ and the universal relation from $B$ to $A$ . Are they equal?
3 marks

Answer and solution

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

$A \times B = \{(t, 5), (t, 8), (x, 5), (x, 8)\}$

$B \times A = \{(5, t), (5, x), (8, t), (8, x)\}$

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

Answer: $A \times B = \{(t, 5), (t, 8), (x, 5), (x, 8)\}$ ; $B \times A = \{(5, t), (5, x), (8, t), (8, x)\}$ ; not equal
Find the range of the relation $R = \{(4, c), (4, z), (10, c), (12, z)\}$ .
1 mark

Answer and solution

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

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

Answer: $\{c, z\}$
$R = \{(3, 7), (3, 9), (3, 10), (12, 10)\}$ is a relation from $A = \{3, 4, 7, 12\}$ to $B = \{7, 9, 10\}$ . Find the domain and range of $R$ .
2 marks

Answer and solution

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

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

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

$\text{Range of } R = \{7, 9, 10\}$

Answer: Domain $= \{3, 12\}$ , Range $= \{7, 9, 10\}$
If $A = \{1, 10, 12\}$ and $B = \{6, 8, 11\}$ , show that $R = \{(1, 11), (10, 6), (10, 11)\}$ is a relation from $A$ to $B$ and find its domain and range.
3 marks

Answer and solution

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 $B$ .

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

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

Answer: $R \subseteq A \times B$ ; Domain $= \{1, 10\}$ , Range $= \{6, 11\}$
Let $A = \{a, p, x\}$ and $B = \{9, 11\}$ . (i) Find $A \times B$ . (ii) Show that $R = \{(a, 9), (p, 9), (p, 11)\}$ is a relation from $A$ to $B$ . (iii) Find the domain and range of $R$ .
5 marks

Answer and solution

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

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

$\therefore A \times B = \{(a, 9), (a, 11), (p, 9), (p, 11), (x, 9), (x, 11)\}$

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 $R$ .

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

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

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

Answer: (i) $A \times B = \{(a, 9), (a, 11), (p, 9), (p, 11), (x, 9), (x, 11)\}$ (ii) $R \subseteq A \times B$ (iii) Domain $= \{a, p\}$ , Range $= \{9, 11\}$
If $R = \{(8, a), (9, a), (11, a)\}$ , find $R^{-1}$ and the domain of $R^{-1}$ .
3 marks

Answer and solution

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

$R^{-1} = \{(a, 8), (a, 9), (a, 11)\}$

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

$= \{a\}$

Answer: $R^{-1} = \{(a, 8), (a, 9), (a, 11)\}$ ; Domain of $R^{-1} = \{a\}$
$R_1 = \{(b, 12), (x, 1), (x, 12)\}$ and $R_2 = \{(a, 2), (b, 2), (x, 2)\}$ are relations from $A$ to $B$ . Find $R_1^{-1}$ and $R_2^{-1}$ , and the domain and range of each inverse.
5 marks

Answer and solution

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

$R_1^{-1} = \{(12, b), (1, x), (12, x)\}$

$\text{Domain} = \{1, 12\}, \text{ Range} = \{b, x\}$

$R_2^{-1} = \{(2, a), (2, b), (2, x)\}$

$\text{Domain} = \{2\}, \text{ Range} = \{a, b, x\}$

Answer: $R_1^{-1} = \{(12, b), (1, x), (12, x)\}$ (Domain $\{1, 12\}$ , Range $\{b, x\}$ ); $R_2^{-1} = \{(2, a), (2, b), (2, x)\}$ (Domain $\{2\}$ , Range $\{a, b, x\}$ )

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Question paper, Set 8