Question 1: i) If $\displaystyle \Big( \frac{a}{3} + 1 , b - \frac{2}{3} \Big) = \Big( \frac{5}{3} , \frac{1}{3} \Big) \text{ find the values } a \text{ and } b$ (ii) if $\displaystyle (x+1,1): (3,y -2) \text{, find the values } x \text{ and } y$.

$\displaystyle \text{i) } \Big( \frac{a}{3} + 1 , b - \frac{2}{3} \Big) = \Big( \frac{5}{3} , \frac{1}{3} \Big)$

By definition of equality of ordered pairs, we have

$\displaystyle \Rightarrow \frac{a}{3} +1 = \frac{5}{3} \text{ and } b - \frac{2}{3} = \frac{1}{3}$

$\displaystyle \Rightarrow \frac{a}{3} = \frac{2}{3} \text{ and } b = 1$

$\displaystyle \Rightarrow a = 2 \text{ and } b= 1$

$\displaystyle \text{ii) } (x+1,1): (3,y -2)$

By definition of equality of ordered pairs, we have

$\displaystyle \Rightarrow x + 1 = 3 \text{ and } y -2 = 1$

$\displaystyle \Rightarrow x = 2 \text{ and } y = 3$

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Question 2: If the ordered pairs $\displaystyle (x,-1) \text{ and } (5,y) \text{ belong to the set } \{ (a,b):b=2a-3 \} \text{, find the values } x \text{ and } y$.

The ordered pairs $\displaystyle (x,-1) \text{ and } (5,y) \text{ belong to the set } \{ (a,b):b=2a-3 \}$

Thus we have

$\displaystyle x = a \text{ and } -1 = b \text{ such that } b = 2a - 3$

$\displaystyle \therefore -1 = 2x - 3$

$\displaystyle \Rightarrow 2x = 2 \Rightarrow x = 1$

$\displaystyle \text{Also } 5-a \text{ and } y - b \text{ such that } b - 2a = 3$

$\displaystyle \Rightarrow y = 2(5) - 3 = 10 - 3 = 7$

Thus we get $\displaystyle x = 1 \text{ and } y = 7$

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Question 3: If $\displaystyle a \in \{ 1,2,3,4,5\} \text{ and } b \in \{ 0,3,6 \}$, write the set of all ordered pairs $\displaystyle (a,b) \text{ such that } a+b=5$.

$\displaystyle \text{Given } a \in \{ 1,2,3,4,5\} \text{ and } b \in \{ 0,3,6 \}$

Since $\displaystyle a + b = 5$, we know that the possible ordered pairs are $\displaystyle \{ (-1, 6), (2, 3), (5, 0) \}$

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Question 4: If $\displaystyle a \in \{ 2, 4,6,9 \} \text{ and } b \in \{4,6,18,27 \}$, then form the set of all ordered pairs $\displaystyle (a,b)$ such that a divides $\displaystyle b \text{ and } a .

$\displaystyle \text{Given } a \in \{ 2, 4,6,9 \} \text{ and } b \in \{4,6,18,27 \}$. Here

$\displaystyle 2 \text{ divides } 4, 6, 18 \text{ and } 2$ is less than all of them

$\displaystyle 6 \text{ divides } 18 \text{ and } 6$ is less than $\displaystyle 18$

$\displaystyle 9 \text{ divides } 18, 27 \text{ and } 9$ is less than $\displaystyle 18 \text{ and } 27$

Therefore set of all ordered pairs $\displaystyle (a, b) \text{ such that } a \text{ divides } b \text{ and } a < b$ Hence $\displaystyle R = \{ (2, 4), (2, 6), (2, 18), (6, 18), (9, 18), (9, 27) \}$

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Question 5: If $\displaystyle A= \{ 1, 2 \} B=\{ 1,3 \} \text{, find } A \times B \text{ and } B \times A$.

$\displaystyle \text{Given } A= \{ 1, 2 \} B=\{ 1,3 \}$

$\displaystyle \therefore A \times B = \{ 1, 2 \} \times \{ 1,3 \} = \{ (1, 1), (1, 3), (2, 1), (2, 2) \}$

$\displaystyle B \times A = \{ 1,3 \} \times \{ 1, 2 \} = \{ (1, 1), (1, 2), (3, 1), (3, 2) \}$

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Question 6: Let $\displaystyle A = \{1, 2, 3 \} \text{ and } B = \{ 3, 4 \}$. Find $\displaystyle A \times B$ and show it graphically.

$\displaystyle \text{Given } A = \{1, 2, 3 \} \text{ and } B = \{ 3, 4 \}$

$\displaystyle A \times B = \{1, 2, 3 \} \times \{ 3, 4 \} = \{ (1, 3), (1, 4), (2, 3), (2, 4), (3, 3), (3, 4) \}$

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Question 7: If $\displaystyle A = \{1, 2, 3\} \text{ and } B = \{2,4\} \text{, what are } A \times B, B \times A, A \times A, B \times B$, and $\displaystyle (A \times B) \cap (B \times A)$ ?

$\displaystyle \text{Given } A = \{1, 2, 3\} \text{ and } B = \{2,4\}$

$\displaystyle \therefore A \times B = \{1, 2, 3\} \times \{2,4\} = \{ (1, 2), (1, 4), (2, 2), (2, 4), (3, 2), (3, 4) \}$

$\displaystyle B \times A = \{2,4\} \times \{1, 2, 3\} = \{ (2, 1), (2, 2), (2, 3), (4, 1), (4, 2), (4, 3) \}$

$\displaystyle A \times A = \{1, 2, 3\} \times \{1, 2, 3\} \\ \hspace*{1.2cm} = \{ (1, 1), (1,2), (1, 3), (2, 1), (2,2), (2, 3),(3, 1), (3,2), (3, 3) \}$

$\displaystyle \therefore (A \times B) \cap (B \times A) = \{ (2, 2) \}$

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Question 8: If $\displaystyle A \text{ and } B$ are two sets haying $\displaystyle 3$ elements in common. It $\displaystyle n(A)=5, n(B)=4 \text{, find } n(A \times B) \text{ and } n[ (A \times B) \cap (B \times A)]$.

$\displaystyle \text{Given } n(A) = 5 \text{ and } n(B ) = 4$

Therefore $\displaystyle n(A \times B) = 5 \times 4 = 20$

$\displaystyle A \text{ and } B$ are two sets having $\displaystyle 3$ elements in common

Let $\displaystyle A = \{ a, a, a, b, c \} and B = \{ a, a, a, d \}$

$\displaystyle \therefore A \times B = \{ (a, a), (a, a), (a, a), (a, d), (a, a), (a, a), (a, a), \\ \hspace*{2.4cm} (a, d), (a, a), (a, a), (a, a), (a, d), (b, a), (b, a), \\ \hspace*{2.4cm} (b, a), (b, d), (c, a), (c, a), (c, a), (c, d) \}$

$\displaystyle B \times A = \{ (a, a), (a, a), (a, a), (a, b), (a, c), (a, a), (a, a), (a, a), \\ \hspace*{2.1cm} (a, b), (a, c), (a, a), (a, a), (a, a), (a, b), (a, c), \\ \hspace*{2.1cm} (d, a), (d, a), (d, a), (d, b), (d, c) \}$

$\displaystyle (A \times B) \cap (B \times A) = \{ (a, a), (a, a), (a, a), (a, a), (a, a), (a, a), (a, a), (a, a), (a, a) \}$

$\displaystyle \therefore n [A \times B \cap B \times A] = 9$

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Question 9: Let $\displaystyle A \text{ and } B$ be two sets. Show that the sets $\displaystyle A \times B \text{ and } B \times A$ have an element in common iff the sets $\displaystyle A \text{ and } B$ have an element in common.

Case 1: Let $\displaystyle A = \{ a, b, c \} \text{ and } B = \{ e, f \}$

$\displaystyle \therefore A \times B = \{ a, b, c \} \times \{ e, f \} = \{ (a, e), (a, f), (b, e), (b, f), (c, e), (c, f) \}$

$\displaystyle B \times A = \{ e, f \} \times \{ a, b, c \} = \{ (e, a), (e, b), (e, c), (f, a), (f, b), (f, c) \}$

Therefore they have no common elements.

Case 2: Let $\displaystyle A = \{ a, b, c \} \text{ and } B = \{ a, f \}$

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

$\displaystyle B \times A = \{ a, f \} \times \{ a, b, c \} = \{ (a, a), (a, b), (a, c), (f, a), (f, b), (f, c) \}$

Hence $\displaystyle (A \times B) \cap (B \times A) = \{(a, a), (a, a) \}$

Therefore $\displaystyle A \times B \text{ and } B \times A$ will have common elements if and only if sets A and set B have elements in common.

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Question 10: Let $\displaystyle A \text{ and } B$ be two sets such that $\displaystyle n(A) = 3 \text{ and } n(B) = 2$. If $\displaystyle (x, 1), (y, 2), (z, 1)$ are in $\displaystyle A \times B$ find $\displaystyle A \text{ and } B$ where $\displaystyle x, y$, and $\displaystyle z$ are distinct elements.

Since $\displaystyle (x, 1), (y, 2), (z, 1)$ are elements of $\displaystyle A \times B$, therefore $\displaystyle x, y, z, \in A \text{ and } 1, 2, \in B$

It is given that $\displaystyle n(A) = 3 \text{ and } n(B) = 3$

$\displaystyle \Rightarrow A = \{ x, y, z \}$

Similarly, $\displaystyle 1, 2 \in B \text{ and } n(B) = 2$

$\displaystyle \Rightarrow B = \{1, 2 \}$

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Question 11: Let $\displaystyle A = \{1, 2, 3, 4 \} \text{ and } R = \{(a, b) : a \in A, b \in A, a \text{ divides } b \}$. Write $\displaystyle R$ explicitly.

$\displaystyle \text{Given } A = \{1, 2, 3, 4 \}$

$\displaystyle R = \{(a, b) : a \in A, b \in A, a \text{ divides } b \}$

We know: $\displaystyle 1 \text{ divides } 1, 2, 3, \text{ and } 4$

$\displaystyle 2 \text{ divides } 2 \text{ and } 4$

$\displaystyle 3 \text{ divides } 3$

$\displaystyle 4 \text{ divides } 4$

$\displaystyle \therefore R = \{ (1, 1,), (1, 2), (1, 3), (1,4), (2,2), (2,4), (3,3), (4,4) \}$

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Question 12: If $\displaystyle A= \{-1,1 \}$ , find $\displaystyle A \times A \times A$.

$\displaystyle \text{Given } A= \{-1,1 \}$

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

$\displaystyle \therefore A \times A \times A = \{ (-1, -1, -1), (-1, -1, 1), (-1, 1, -1), (-1, 1, 1) , \\ \hspace*{3.0cm} (1, -1, -1), (1, -1, 1), (1, 1, -1), (1, 1, 1) \}$

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Question 13: State whether each of the following statements are true or false. If the statement is false, re-write the given statement correctly:

(i) If $\displaystyle P = \{m, n\} \text{ and } Q = \{n, m\}$ , then $\displaystyle P \times Q = \{ (m, n), (n, m) \}$

(ii) If $\displaystyle A \text{ and } B$ are non-empty sets, then $\displaystyle A \times B$ is a non-empty set of ordered pairs $\displaystyle (x,y) \text{ such that } x \in B \text{ and } y \in A$.

(iii) If $\displaystyle A = \{1, 2 \} , B = {3,4}$, then $\displaystyle A \times (B \cap \phi ) = \phi$

i) False

If $\displaystyle P = \{m, n\} \text{ and } Q = \{n, m\}$, then

$\displaystyle P \times Q = \{ (m, n), (m, m), (n,n) , (n, m) \}$

ii) False

If $\displaystyle A \text{ and } B$ are non-empty sets then $\displaystyle A \times B$ is a non-empty set of ordered pairs $\displaystyle (x, y) \text{ such that } x \in A \text{ and } y \in B$.

iii) True

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Question 14: If $\displaystyle A = \{1, 2\}$, form the set $\displaystyle A \times A \times A$.

$\displaystyle \text{Given } A = \{1, 2\}$

$\displaystyle A \times A = \{1, 2\} \times \{1, 2\} = \{ (1, 1), (1, 2), (2, 1), (2, 2) \}$

$\displaystyle \therefore A \times A \times A = \{1, 2\} \times \{ (1, 1), (1, 2), (2, 1), (2, 2) \} \\ \hspace*{2.0cm} = \{ (1, 1, 1), (1, 1, 2), (1, 2, 1), (1, 2, 2) , (2, 1, 1), (2, 1, 2), (2, 2, 1), (2, 2, 2) \}$

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Question 15: If $\displaystyle A = \{1, 2, 4 \} \text{ and } B = \{1, 2, 3\}$, represent following sets graphically: $\displaystyle \text{(i) } A \times B \hspace{1.0cm} \text{(ii) } B \times A \hspace{1.0cm} \text{(iii) } A \times A \hspace{1.0cm} \text{(iv) } B \times B$

$\displaystyle \text{Given } A = \{1, 2, 4 \} \text{ and } B = \{1, 2, 3\}$
$\displaystyle \text{i) } A \times B = \{1, 2, 4 \} \times \{1, 2, 3\} \\ \hspace*{1.2cm} = \{ (1, 1), (1, 2), (1,3), (2, 1), (2, 2), (2,3), (4, 1), (4, 2), (4,3) \}$
$\displaystyle \text{ii) } B \times A = \{1, 2, 3\} \times \{1, 2, 4 \} \\ \hspace*{1.2cm} = \{ (1, 1), (1, 2), (1, 4), (2, 1), (2, 2), (2, 4), (3, 1), (3, 2), (3, 4) \}$
$\displaystyle \text{iii) } A \times A = \{1, 2, 4 \} \times \{1, 2, 4 \} \\ \hspace*{1.2cm} = \{ (1,1), (1,2), (1, 4), (2,1), (2,2), (2, 4), (4,1), (4,2), (4, 4) \}$
$\displaystyle \text{iv) } B \times B = \{1, 2, 3\} \times \{1, 2, 3\} \\ \hspace*{1.2cm} = \{ (1,1), (1,2), (1,3), (2,1), (2,2), (2,3), (3,1), (3,2), (3,3) \}$