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 .

Answer:

\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 .

Answer:

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 .

Answer:

\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 <b .

Answer:

\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 .

Answer:

\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.

Answer:

\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) ?

Answer:

\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)] .

Answer:

\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.

Answer:

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.

Answer:

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.

Answer:

\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 .

Answer:

\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

Answer:

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 .

Answer:

\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

Answer:

\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) \}