Question 1:  Find the smallest set A such that A \cup \{ 1, 2 \} = \{ 1, 2, 3, 5, 9 \}

Answer:

Smallest set A = \{ 1, 2, 3, 5, 9 \}  - \{ 1, 2 \} = \{  3, 5, 9 \}

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Question 2: Let A= \{ 1,2,4,5 \} \ B= \{ 2,3,5,6 \} \ C= \{ 4,5,6,7 \} . Verify the following identities:

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

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

(v) A -(B \cap C) =(A -B) \cup (A-C)    (vi) A \cap (B \triangle C) =(A \cap B) \triangle (A \cap C)

Answer:

(i)  Given A= \{ 1,2,4,5 \} \ B= \{ 2,3,5,6 \} \ C= \{ 4,5,6,7 \}

A \cup (B \cap C) = \{ 1,2,4,5 \} \cup (\{ 2,3,5,6 \} \cap \{ 4,5,6,7 \}) = \{ 1,2,4,5 \} \cup \{ 5,6 \}  = \{  1, 2, 4, 5, 6 \}

(A \cup B) \cap (A \cup C) = (\{ 1,2,4,5 \} \cup \{ 2,3,5,6 \}) \cap ( \{ 1,2,4,5 \} \cup \{ 4,5,6,7 \})  = \{ 1,2,3, 4,5,6 \} \cap \{ 1, 2, 4, 5, 6, 7 \} = \{ 1, 2, 4, 5, 6 \}

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

(ii)  Given A= \{ 1,2,4,5 \} \ B= \{ 2,3,5,6 \} \ C= \{ 4,5,6,7 \}

A \cap (B \cup C) = \{ 1,2,4,5 \} \cap (\{ 2,3,5,6 \} \cup \{ 4,5,6,7 \} )  = \{ 1,2,4,5 \} \cap \{ 2,3 ,4,5,6,7 \} = \{ 2, 4, 5 \} 

(A \cap B) \cup (A \cap C) = (\{ 1,2,4,5 \} \cap \{ 2,3,5,6 \}) \cup (\{ 1,2,4,5 \} \cap \{ 4,5,6,7 \})  = \{ 2, 5 \} \cup \{ 4, 5 \} = \{ 2, 4, 5 \}    

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

(iii) Given A= \{ 1,2,4,5 \} \ B= \{ 2,3,5,6 \} \ C= \{ 4,5,6,7 \}

B - C =  \{ 2,3,5,6 \} - \{ 4,5,6,7 \} = \{ 2, 3 \}

A \cap (B -C) = \{ 1,2,4,5 \} \cap \{ 2, 3 \} = \{ 2 \}

A \cap B = \{ 1,2,4,5 \} \cap \{ 2,3,5,6 \} = \{ 2, 5 \}

A \cap C = \{ 1,2,4,5 \} \cap \{ 4,5,6,7 \} = \{ 4, 5 \}

A \cap B - A \cap C = \{ 2 \}

Hence A \cap (B -C) =(A \cap B) -(A \cap C)

(iv) A= \{ 1,2,4,5 \} \ B= \{ 2,3,5,6 \} \ C= \{ 4,5,6,7 \}

B \cup C = \{ 2,3,5,6 \} \cup \{ 4,5,6,7 \} = \{2, 3, 4, 5, 6, 7 \}

A - (B \cup C) = \{ 1,2,4,5 \} - \{2, 3, 4, 5, 6, 7 \} = \{ 1 \}

A - B = \{ 1,2,4,5 \} - \{ 2,3,5,6 \} = \{1, 4 \}

A - C = \{ 1,2,4,5 \} - \{ 4,5,6,7 \} = \{ 1, 2 \}

(A -B) \cap (A-C) = \{1, 4 \} - \{ 1, 2 \} = \{1 \}

Hence A - (B \cup C) =(A -B) \cap (A-C)

(v)   Given A= \{ 1,2,4,5 \} \ B= \{ 2,3,5,6 \} \ C= \{ 4,5,6,7 \}

B \cap C = \{ 2,3,5,6 \} \cap \{ 4,5,6,7 \} = \{5, 6 \}

A -(B \cap C) = \{ 1,2,4,5 \} - \{5, 6 \} = \{1, 2, 4 \}

A-C = \{ 1,2,4,5 \} - \{ 4,5,6,7 \} = \{ 1, 2 \}

A - B = \{ 1,2,4,5 \} - \{ 2,3,5,6 \} = \{1, 4 \}

(A -B) \cup (A-C) = \{1, 4 \} - \{ 1, 2 \} = \{1, 2, 4 \}

Hence A -(B \cap C) =(A -B) \cup (A-C)

(vi) Given A= \{ 1,2,4,5 \} \ B= \{ 2,3,5,6 \} \ C= \{ 4,5,6,7 \}

A \cap C = \{ 1,2,4,5 \} \cap \{ 4 ,5,6,7 \} = \{ 4, 5 \} = X

A \cap B = \{ 1,2,4,5 \} \cap \{ 2,3,5,6 \} = \{ 2, 5 \} = Y

Y - X = \{ 2, 5 \} - \{ 4, 5 \} =\{ 2 \}

X - Y = \{ 4, 5 \} - \{ 2, 5 \} = \{ 4 \}

(A \cap B) \triangle (A \cap C) = Y \triangle X = (Y-X) \cup (X-Y) = \{ 2 \} \cup \{ 4 \} = \{ 2, 4 \}

B \triangle C = ( B - C ) \cup ( C - B ) = ( \{ 2,3,5,6 \} - \{ 4,5,6,7 \} ) \cup ( \{ 4,5,6,7 \} - \{ 2,3,5,6 \} ) = \{ 2,3 \} \cup \{ 4, 7 \} = \{ 2,3,4,7 \}

A \cap B \triangle C = \{ 1, 2, 4, 5 \} \cap \{ 2, 3, 4, 7 \} = \{ 2, 4 \}

Hence A \cap (B \triangle C) = (A \cap B) \triangle (A \cap C)

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Question 3: If U= \{ 2,3,5,7,9 \} is the universal set and A= \{3,7\} , B= \{2,5,7,9 \} , then prove that: (i) (A \cup B)'= A' \cap B'     (ii) (A \cap B)' = A' \cup B'

Answer:

(i)   Given U= \{ 2,3,5,7,9 \} ,A= \{3,7\} and B= \{2,5,7,9 \}

A \cup B = \{3,7\} \cup\{2,5,7,9 \} = \{ 2, 3, 5, 7, 9 \}

(A \cup B)' = \phi

A' = \{  2, 5, 9 \}

B' = \{  3 \}

A' \cap B' = \phi

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

(ii)  Given U= \{ 2,3,5,7,9 \} ,A= \{3,7\} and B= \{2,5,7,9 \}

A' = \{  2, 5, 9 \}

B' = \{  3 \}

A' \cup B' = \{  2, 3, 5, 9 \} 

A \cap B = \{3,7\} \cap\{2,5,7,9 \} = \{ 7 \}

(A \cap B)' = \{ 2, 3, 5, 9  \}

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

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Question 4: For any two sets A and B , prove that

(i) B \subset A \cup B (ii) A \cap B \subset A     (iii) A \subset B \Rightarrow  A \cap B=A

Answer:

(i)   B \subset A \cup B

For all x \in B

\Rightarrow x \in A or x \in B

\Rightarrow x \in A \cup B (Definition of union of sets)

\Rightarrow B \subset A \cup B

(ii) A \cap B \subset A

Fora all x \in A \cap B

\Rightarrow x \in A and x \in B (definition of intersection of sets)

\Rightarrow x \in A

(iii) A \subset B \Rightarrow  A \cap B=A

Let x \in A \subset B

\Rightarrow x \in B

Let x \in A \cap B

\Rightarrow x \in A and x \in B

\Rightarrow x \in A and x \in B (as A \subset B )

\therefore A \cap B = A

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Question 5: For any two sets A and B , show that the following statements are equivalent:

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

Answer:

In order to show that the four statements are equivalent, we need to show that i) \Rightarrow ii) ,   ii) \Rightarrow iii) iii) \Rightarrow iv) and  iv) \Rightarrow i)

i)    Let us assume A \subset B

Let x \in A , since A \subset B, x \in B

Now A - B  \Rightarrow  x \in B and x \notin B

Since A \subset B , each element of A is an element of B

\Rightarrow A - B = \phi

ii)   Let A - B = \phi

\Rightarrow x \in A, x \in B

Therefore every element of A is also an element of B .

Therefore A \cup B = B

iii)  Let us assume A \cup B = B

\Rightarrow x \in B and x \in B

Therefore A \cap B = A since every element of A is in B

iv)   Let A \cap B = A

\Rightarrow if x \in A, x \in B also

Therefore A \subset B . All elements of A are elements of B also.

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Question 6: For three sets A, B and C , show that

(i) A \cap B = A \cap C need not imply B=C     (ii) A \subset B \Rightarrow C - B \subset C-A

Answer:

(i)   A \cap B = A \cap C need not imply B=C

Let A = \{ 1, 2, 3 \}, B = \{ 2, 4, 6 \} and C = \{ 2, 5, 7 \}

Therefore A \cap B = \{ 2 \}

A \cap C = \{ 2 \}

Hence A \cap B = A \cap C

But we can clearly see that B \neq C

(ii)  A \subset B \Rightarrow C - B \subset C-A

Given A \subset B

Let x \in C - B

\Rightarrow x \in C but x \notin B (By definition  of C - B )

\Rightarrow x \in C and x \notin A (Since A \subset B )

Thus x \in C-B \Rightarrow x \in C - A . This is true for all x \in C - B

Therefore  C - B \subset C-A

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Question 7: For any two sets, prove that: (i) A \cup (A \cap B) = A     (ii) A \cap (A \cup B) = A

Answer:

(i)   A \cup (A \cap B) = A

A \cup ( A \cap B) = (A \cup A ) \cap (A \cup B) [Since union is distributive over intersection ]

= A \cap (A \cup B) [ Since A \cup A = A ]

= A [ Since A \subset (A \cup B) as union of two sets is biggest than each of the individual sets ]

Hence A \cup (A \cap B) = A . Hence Proved.

(ii)  A \cap (A \cup B) = A

A \cap (A \cup B) = (A \cap A ) \cup (A \cap B) [ Since intersection is distributive on union ]

= A \cup ( A \cap B) [Since A \cap A = A ]

= A

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Question 8: Find sets A, \ B and C such that A \cap B, A \cap C and B \cap C are non-empty sets and A \cap B \cap C = \phi

Answer:

Consider A = \{ 5, 6, 10 \}, B = \{ 6, 8, 9 \} and C = \{ 9, 10, 11 \}

Therefore A \cap B = \{ 6 \} \Rightarrow A \cap B \neq \phi

B \cap C = \{ 10 \} \Rightarrow B \cap C \neq \phi

Also A \cap B \cap C = \phi

Hence there are no common elements between all three sets.

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Question 9: For any two sets A and B , prove that A \cap B = \phi + A \subseteq B' .

Answer:

Given A \cap B  = \phi

\Rightarrow for x \in A, x \notin B

There are no common elements between A and B

\Rightarrow x \in B'

Thus x \in A and x \in B' \Rightarrow A \subseteq B'

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Question 10: If A and B are sets, then prove that A - B, A \cap B and B - A are pair wise disjoint.

Answer:

We need to show:

(A-B) \cap (A \cap B) = \phi

(A \cap B) \cap (B-A) = \phi

(A-B) \cap (B-A) = \phi

Let us first show (A-B) \cap (A \cap B) = \phi

Let x \in (A - B)  \Rightarrow x \in A but x \notin B

\Rightarrow x \notin A \cap B

This is true for all x \in (A - B)

Hence (A- B) \cap (A \cap B) = \phi

Similarly, let us prove (A \cap B) \cap (B-A) = \phi

x \in (B-A)  \Rightarrow x \in B but x \notin A

\Rightarrow x \notin A \cap B

Therefore (A \cap B ) \cap (B - A ) = \phi

Now prove (A-B) \cap (B-A) = \phi

If x \in (A - B) \Rightarrow x \in A but x \notin B

If y \in (B - A) \Rightarrow y \in B but y \notin A

We can see that there is no intersection of x and y . Hence (A-B) \cap (B-A) = \phi

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Question 11: Using properties of sets, show that for any two sets A and B , (A \cup B) \cap (A \cup B') = A .

Answer:

To prove: (A \cup B) \cap (A \cup B') = A

Now (A \cup B) \cap (A \cap B') = [ (A \cup B )\cap A ] \cap B'   (using associative property)

= \Big( (A \cap A) \cup (B \cap A ) \Big) \cap B' [ Since A \cap A = A and B \cap A = A \cap B ]

= A \cap B' [Since A \cup (A \cap B ) = A ]

= A

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Question 12: For any two sets of A and B , prove that:   (i) A' \cup B = U \Rightarrow A \subset B     (ii) B' \subset A' \Rightarrow A \subset B

Answer:

(i) To prove: A' \cup B = U \Rightarrow A \subset B

Let x \in A  \Rightarrow x \notin A' [ Since A \cap A' = \phi ]

Therefore x \in A and A \subset U

\Rightarrow x \in U

\Rightarrow x \in (A' \cup B) [Since U = A' \cup B ]

\Rightarrow x \in A' or x \in B

But x \notin A' \Rightarrow x \in B

Thus x \in  A \Rightarrow x \in B

This is true for all x \in A

Therefore A \subset B

(ii)  To prove B' \subset A' \Rightarrow A \subset B

Let x \in A \Rightarrow x \notin A' [ Since A \cap A' = \phi ]

\Rightarrow x \notin B' [Since B' \subset A' ]

\Rightarrow x \in B [Since B \cap B' = \phi ]

Thus x \in A \Rightarrow x \in B

This is true for all x \in A . Therefore A \subset B

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Question 13: Is it true that for any sets A and B , P (A) \cup P (B) = P (A \cup B) Justify your answer.

Answer:

False

Let X \in P(A) \cup P(B)

\Rightarrow X \in P(A) or X  \in P(B)

\Rightarrow X \subset A or X \subset B

\Rightarrow X \subset A \cup B

\Rightarrow X \in P(A \cap B)

Therefore P(A) \cup P(B) \subset P(A \cup B)

Again let X \in P(A \cup B) but X \notin P(A) or X \notin P(B)

Therefore X \notin P(A) \cup P(B)

Thus P(A \cup B) is not necessarily a subset of P(A) \cup P(B)

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Question 14: Show, that for any sets A and B ,   (i) A=(A \cap B) \cap (A-B)      (ii) A \cup (B-A) =A \cup B

Answer:

(i)   We know that (A \cap B) \subset A and (A - B) \subset A

\Rightarrow (A \cap B) \cap (A - B) … … … … … i)

Let x \in (A \cap B) \cap (A - B)

\Rightarrow x \in (A \cap B) \Rightarrow x \in A and x \in B

and x \in (A - B) \Rightarrow x \in A but x \notin B

\Rightarrow x \in A [ Since x \in B and x \notin B are not possible simultaneously]

Therefore (A \cap B) \cap (A - B) \subset A     … … … … … ii)

From i) and ii) we get

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

(ii)  Let x \in A \cup (B-A)

\Rightarrow x \in A or x \in (B-A) \Rightarrow x \in B or x \notin A

\Rightarrow x \in A or x \in B

\Rightarrow x \in ( A \cup B)

Therefore A \cup (B - A) \subset (A \cup B)    … … … … … i)

Let x \in (A \cup B) \Rightarrow x \in A or x \in B

\Rightarrow x \in A or x \in B and x \notin A

\Rightarrow x \in A or x \in (B-A)

\Rightarrow x \in A \cup ( B - A)

Therefore (A \cup B) \subset A \cup (B - A)    … … … … … ii)

From i) and ii) we get A \cup (B-A) =A \cup B

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Question 15: Each set X , contains 5 elements and each set Y , contains 2 elements and \underset{r=1}{\overset{20}{U}} X_r = S = \underset{r=1}{\overset{n}{U}} Y_r . If each element of S belongs to exactly 10 of the {X'}_r^s and to exactly 4 of {Y'}_r^s , find the value of n .

Answer:

It is given that each set X contains 5 elements \underset{r=1}{\overset{20}{U}} X_r = S

Therefore n(S) = 20 \times 5 = 100

But it is given that each element of S belongs to exactly 10 of that X_r's

Number of distinct elements in S = \frac{100}{10} = 10   … … … … … i)

It is also given that each set Y contains 2 elements and \underset{r=1}{\overset{n}{U}} Y_r = S

Therefore n(S) = n \times 2 = 2n

Also each element of S belongs to exactly 4 of Y_r's = \frac{1}{4} (2n) = \frac{n}{2}   … … … … … ii)

Therefore from i) and ii) we get

10 = \frac{n}{2} \Rightarrow n = 20

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