Question 1. Let A = \{a, b, c, d \} , B = \{b, c, e \} and C = \{a, b, e \} . Find:

(i) A \cup B       (ii) B \cup C      (iii) A \cup C      (iv) A \cap B      (v) B \cap C      (vi) A \cap C  

Answer:

(i) A \cup B = \{a, b, c, d, e \}      (ii) B \cup C = \{a, b, c, e \}      (iii) A \cup C = \{a, b, c, d, e \}

(iv) A \cap B = \{b, c \}      (v) B \cap C = \{b, e \}      (vi) A \cap C = \{a, b \}

Note: The union of sets A and B , denoted by A \cup B , is the set of all those elements, each one of which is either in A or in B or in both A and B

If there is a set A = \{2, 3 \} and B = \{a, b \} , then A \cup B = \{2, 3, a, b \}

So if A \cup B = \{x | x \in A \ or \  x \in B \} then

x \in A \cup B which means x \in A or x \in B

And if x \notin A \cup B which means x \notin A or x \notin B

The intersection of sets A and B is denoted by A \cap B , and is a set of all elements that are common in sets A and B .

If A = \{1, 2, 3 \} and B = \{2, 4, 5 \} , then A \cap B = \{2 \} as 2 is the only common element.

Thus A \cap B = \{x: x \in A \ and \  x \in B \}

then x \in A \cap B i.e. x \in A and x \in B

And if x \notin A \cap B i.e. x \notin A and x \notin B

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Question 2. Let A = \{2, 3, 4, 6 \} , B = \{4, 6, 8, 9 \} and C = \{2, 7, 8, 9 \} . Find: (i) A \cup B     (ii) B \cup C      (iii) A \cup C      (iv) A \cap B      (v) B \cap C      (vi) A \cap C

Answer:

(i) A \cup B = \{2, 3, 4, 6, 8, 9 \}      (ii) B \cup C = \{2, 4, 6, 7, 8, 9 \}

(iii) A \cup C = \{2, 3, 4, 7, 8, 9 \}      (iv) A \cap B = \{4, 6 \}

(v) B \cap C = \{8, 9 \}      (vi) A \cap C = \{2 \}

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Question 3. Let A = \{1, 4, 7, 8 \} and B = \{4, 6, 8, 9 \} . Find i) A - B and ii) B - A

Answer:

(i) A - B = \{1, 4, 7, 8 \} - \{4, 6, 8, 9 \} = \{1, 7 \}

(ii) B - A = \{4, 6, 8, 9 \} - \{1, 4, 7, 8 \} = \{6, 9 \}

Note: For any two sets A and B , the difference A - B is a set of all those elements of A which are not in B .

i.e. if A = \{1, 2, 3, 4, 5 \} and B = \{4, 5, 6 \}

Then A - B = \{1, 2, 3 \} and B - A = \{6 \}

Therefore, A - B = \{x | x \in A \ and \  x \notin B \} , then

x \in A - B then x \in A but x \notin B

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Question 4. Let \xi = \{13, 14, 15, 16, 17, 18, 19, 20, 21 \} , A = \{13, 17, 19 \} and B = \{14, 16, 18, 20 \} . Find (i) A' and (ii) B'

Answer:

(i) A' = \{14, 15, 16, 18, 20, 21 \}

(ii) B' = \{13, 15, 17, 19, 21 \}

Note: Let x be the universal set and let A \subseteq x . Then the complement of A , denoted by A' is the set of all those elements of x which are not in A .

i.e. let \xi = \{1, 2, 3, 4, 5,6 ,7 ,8 \} and A = \{2, 3, 4 \} , then A' = \{1, 5, 6, 7, 8 \}

Thus A' = \{x | x \in \xi and x \notin A \} clearly x \in A' and x \notin A

Please note

  • \phi^{'} = \xi  and \phi^{'}= \xi 
  • A \cup A' = \xi  and A \cap A' = \phi

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Question 5. Let \xi = \{x | x \in Z, -4 \leq x \leq 4 \} , A = \{x | x \in W, x <4 \} and B = \{x | x \in N, 2 <x \leq 4 \} . Find (i) A' and (ii) B'

Answer:

Note: First find out the elements of the three given sets.

\xi = \{-4, -3, -2, -1, 0, 1, 2, 3, 4 \}, A = \{0, 1, 2, 3 \} and B = \{3, 4 \}

(i) A' = \{-4, -3, -2, -1, 4 \}      (ii) B' = \{-4, -3, -2, -1, 0, 1, 2 \}

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Question 6. Let \xi = \{x | x \in N, x \ is \ a \ factor \ of 144 \} , A = \{x | x \in N, x \ is \ a \ factor \ of 24 \} , B = \{x | x \in N, x \ is \ a \ factor \ of 36 \} and C = \{x | x \in N, \ x \ is \ a \ factor \ of \ 48 \} . Find:

(i) A'      (ii) B'      (iii) C'      (iv) A \cup B       (v) B \cup C      (vi) A \cup C       (vii) A \cap B'      (viii) B \cap C'       (ix) C - A       (x)  A - (B \cap C)

Answer:

Note: First calculate what the elements of each of these sets are. Therefore, we have the following:

\xi = \{1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 36, 48, 72, 144 \} . This is the universal set.

A = \{1, 2, 3, 4, 6, 8, 12, 24 \}      B = \{1, 2, 3, 4, 6, 9, 12, 18, 36 \}

C = \{1, 2, 3, 4, 6, 8, 12, 16, 24, 48 \}

Now we can easily calculate.

(i) A' = \{9, 16, 18, 36, 48, 72, 144 \}      (ii) B' = \{8, 16, 24, 48, 72, 144 \}

(iii) C' = \{9, 18, 36, 72, 144 \}      (iv) A \cup B = \{1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36 \}

(v) B \cup C = \{1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 26, 48 \}

(vi) A \cap C = \{1, 2, 3, 4, 6, 8, 12, 16, 24, 48 \}      (vii) A \cap B' = \{8, 24 \}

(viii) B \cap C' = \{9, 18, 36 \}      (ix) C - A = \{16, 48 \}      (x) A - (B \cap C)

First calculate (B \cap C) = \{1, 2, 3, 4, 6, 12 \}

Now Calculate A - (B \cap C) = \{8, 24 \}

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Question 7. Considering the sets given in Q.6 state whether the following statements are True or False.

(i) A \cap (B \cup C) = A (ii) A \subset C (iii) B \subseteq C (iv) A \cap C' = \phi

Answer:

Note: First calculate what the elements of each of these sets are. Therefore, we have the following:

\xi = \{1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 36, 48, 72, 144 \} . This is the universal set.

A = \{1, 2, 3, 4, 6, 8, 12, 24 \}

B = \{1, 2, 3, 4, 6, 9, 12, 18, 36 \}

C = \{1, 2, 3, 4, 6, 8, 12, 16, 24, 48 \}

(i) A \cap (B \cup C) = A

A = \{1, 2, 3, 4, 6, 8, 12, 24 \}

B \cup C = \{1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 36, 48 \}

Therefore, A \cap (B \cup C) = \{1, 2, 3, 4, 6, 8, 12, 24 \} = A

Hence Proved. TRUE

(ii) A \subset C

A = \{1, 2, 3, 4, 6, 8, 12, 24 \}

C = \{1, 2, 3, 4, 6, 8, 12, 16, 24, 48 \} Every element of A is in C and there are two elements (16 and 18) which are in C but not in A . A is a proper subset of B .

Hence, TRUE.

(iii) B \subseteq C

B = \{1, 2, 3, 4, 6, 9, 12, 18, 36 \}

C = \{1, 2, 3, 4, 6, 8, 12, 16, 24, 48 \}

Here every element of B is not in C . Hence it is not a subset of C . Hence FALSE.

(iv) A \cap C' = \phi 

\xi = \{1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 36, 48, 72, 144 \} . This is the universal set.

A = \{1, 2, 3, 4, 6, 8, 12, 24 \}

C = \{1, 2, 3, 4, 6, 8, 12, 16, 24, 48 \} Therefore C' = \{9, 18, 36, 72, 144 \}

If we look at the two sets A and C' , we see that there are no common elements and hence

A \cap C' = \phi or null set. Hence TRUE

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Question 8. Let A = \{a, b, c, d, e \} , B = \{a, c, e, g \} and C = \{b, e, f, g \} . Then verify the following identities:

(i) B \cup C = C \cup B (ii) B \cap C = C \cap B (iii) A \cup (B \cup C) = (A \cup B) \cup C

(iv) A \cap (B \cap C) = (A \cap B) \cap C (v) A \cup (B \cap C) = (A \cup B) \cap (A \cup C)

(vi) A \cap (B \cup C) = (A \cap B) \cup (A \cap C)

Answer:

(i) B \cup C = C \cup B

B = \{a, c, e, g \} & C = \{b, e, f, g \} . Therefore, B \cup C = \{a, b, c, e, f, g \} .

Similarly, C \cup B = \{a, b, c, e, f, g \} .

Hence B \cup C = C \cup B

(ii) B \cap C = C \cap B

B = \{a, c, e, g \} & C = \{b, e, f, g \} . Therefore, B \cap C = \{a, b, c, e, f, g \} .

Similarly, C \cup B = \{a, b, c, e, f, g \} .

(iii) A \cup (B \cup C) = (A \cup B) \cup C

A = \{a, b, c, d, e \} , B = \{a, c, e, g \} , C = \{b, e, f, g \}

Therefore:

LHS: A \cup (B \cup C) = \{a, b, c, d, e \} \cup \{a, b, c, e, f, g \} = \{a, b, c, d, e, f, g \}

RHS: (A \cup B) \cup C = \{a, b, c, d, e, g \} \cup \{b, e, f, g \} = \{a, b, c, d, e, f, g \}

Therefore, LHS = RHS. Hence Proved.

(iv) A \cap (B \cap C) = (A \cap B) \cap C

A = \{a, b, c, d, e \} , B = \{a, c, e, g \} , C = \{b, e, f, g \}

Therefore:

LHS: A \cap (B \cap C) = \{a, b, c, d, e \} \cap \{e, g \} = \{e \}

RHS: (A \cap B) \cap C = \{a, c, e \} \cap \{b, e, f, g \} = \{e \}

Therefore, LHS = RHS. Hence Proved.

(v) A \cup (B \cap C) = (A \cup B) \cap (A \cup C)

A = \{a, b, c, d, e \} , B = \{a, c, e, g \} , C = \{b, e, f, g \} Therefore:

LHS: A \cup (B \cap C) = \{a, b, c, d, e \} \cup \{e, g \} = \{a, b, c, d, e, g \}

RHS: (A \cup B) \cap (A \cup C) = \{a, b, c, d, e, g \} \cap \{a, b, c, d, e, f, g \} = \{a, b, c, d, e, g \}

Therefore, LHS = RHS. Hence Proved.

(vi) A \cap (B \cup C) = (A \cap B) \cup (A \cap C)

A = \{a, b, c, d, e \} , B = \{a, c, e, g \} , C = \{b, e, f, g \} Therefore:

LHS: A \cap (B \cup C) = \{a, b, c, d, e \} \cap \{a, b, c, e, f, g \} = \{a, b, c, e \}

RHS: (A \cap B) \cup (A \cap C) = \{a, c, e \} \cap \{b, e \} = \{a, b, c, e \}

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Question 9. Let A = \{b, c, d, e \} and B = \{d, e, f, g \} be the two sub sets of the universal set \xi = \{b, c, d, e, f, g \} . Then verify the following:

(i) (A \cup B)' = (A' \cap B')     (ii)  (A \cap B)' = (A' \cup B')

Answer:

(i) (A \cup B)' = (A' \cap B')

LHS: A \cup B = \{b, c, d, e \} \cup \{d, e, f, g \} = \{b, c, d, e, f, g \}

Therefore, (A \cup B)' = \phi

RHS: A' = \{f, g \} and B' = \{b, c \}

Therefore, A' \cap B' = \phi 

Therefore, LHS = RHS. Hence proved.

(ii) (A \cap B)' = (A' \cup B')

x = \{b, c, d, e, f, g \}

LHS: A \cap B = \{b, c, d, e \} \cap \{d, e, f, g \} = \{d, e \}

Therefore, (A \cap B)' = \{b, c, f, g \}

RHS: A' = \{ f, g \} and B' = \{ b, c \}

Therefore, A' \cup B' = \{b, c, f, g \}

Therefore, LHS = RHS. Hence proved.

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Question 10. Fill in the blanks

(i) A \cup A = \ldots      (ii) A \cap A = \ldots      (iii) A \cup \phi = \ldots

(iv) A \cap \phi = \ldots (v) (A \cup B)' = \ldots       (vi) (A \cap B)' = \ldots

Answer:

(i) A \cup A = A      (ii) A \cap A = A      (iii) A \cup \phi= A      (iv) A \cap \phi= A

(v) (A \cup B)' = (A' \cap B')       (vi) (A \cap B)' = (A' \cup B')

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Question 11. Let \xi = \{x | x \in N 4 \leq x < 18 \} and A, B , and C are subsets of \xi . Given A = \{x | x \ is \ a \ multiple \ of \ 2 \} , B = \{x | x \ is \ a \ multiple \ of \ 3 \} , C = \{x | x \in N x < 11 \} . Then verify, the following:

(i) (A \cup B)' = (A' \cap B')       (ii) (A \cap B)' = (A' \cup B')

(iii) A - B = A \cap B'       (iv) A \cup (B \cap C) = (A \cup B) \cap (A \cup C)

Answer:

Note: First identify the elements of all the 4 sets in this question. That would make the solution easier.

\xi = \{4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17 \}

A = \{4, 6, 8, 10, 12, 14, 16 \}

B = \{6, 9, 12, 15 \}

C = \{4, 5, 6, 7, 8, 9, 10 \}

(i) (A \cup B)' = (A' \cap B')

LHS: (A \cup B) = \{4, 6, 8, 9, 10, 12, 14, 15, 16 \}

(A \cup B)' = \{5, 7, 11, 13, 17 \}

RHS: (A' \cap B') = \{5, 7, 9, 11, 13, 15, 17 \} \cap \{4, 5, 7, 8, 10, 11, 13, 14, 16, 17 \}

(A' \cap B') = \{5, 7, 11, 13, 17 \}

Therefore, LHS = RHS. Hence proved.

(ii) (A \cap B)' = (A' \cup B')

LHS: (A \cap B)' = \{4, 5, 7, 8, 9, 10, 11, 13, 14, 15, 16, 17 \}

RHS: (A' \cup B') = \{5, 7, 9, 11, 13, 15, 17 \} \cup \{4, 5, 7, 8, 10, 11, 13, 14, 16, 17 \}

= \{4, 5, 7, 8, 9, 10, 11, 13, 14, 15, 16, 17 \}

Therefore, LHS = RHS. Hence proved.

(iii) A - B = A \cap B'

A - B = \{4, 6, 8, 10, 12, 14, 16 \} - \{6, 9, 12, 15 \} = \{4, 8, 10, 14, 16 \}

A \cap B' = \{4, 6, 8, 10, 12, 14, 16 \} \cap \{4, 5, 7, 8, 10, 11, 13, 14, 16, 17 \} = \{4, 8, 10, 14, 16 \}

Therefore, LHS = RHS. Hence proved.

(iv) A \cup (B \cap C) = (A \cup B) \cap (A \cup C)

A = \{4, 6, 8, 10, 12, 14, 16 \}

B = \{6, 9, 12, 15 \}

C = \{4, 5, 6, 7, 8, 9, 10 \}

A \cup (B \cap C) = \{4, 6, 8, 9, 10, 12, 14, 16 \}

(A \cup B) \cap (A \cup C) = \{4, 6, 8, 9, 10, 12, 14, 15, 16 \} \cap \{4, 5, 6, 7, 8, 9, 10, 12, 14, 16 \} = \{4, 6, 8, 9, 10, 12, 14, 16 \}

Therefore, LHS = RHS. Hence proved.