Question 1: For any two sets A and B , prove that : A' - B' = B - A

Answer:

To show: A' -B' = B - A

We show that A' -B' \subseteq B - A and vice versa

Let x \in A' - B' \Rightarrow x \in A' and x \notin B'

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

\Rightarrow x \in B and x \notin A

\Rightarrow x \in B - A

This is true for all x \in A' - B' . Hence A' - B' \subseteq B - A

Conversely, let x \in B - A

\Rightarrow x \in B and x \notin A

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

\Rightarrow x \in A' - B'

This is true for all x \in B - A

Hence B - A \subseteq A' - B' .

Therefore A' - B' = B - A . Hence proved.

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Question 2:  For any two sets A and B , prove the following :

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

Answer:

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

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

= (A \cap A') \cup ( A \cap B) [ Since Intersection distributes over union ]

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

= A \cap B [ Since \phi \cup X = X for any set X ]

= RHS. Hence proved.

(ii)  To prove: A-(A-B)=A \cap B

For any set A and B , we have De-morgan’s Laws.

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

LHS = A - ( A - B)

= A \cap (A - B)'

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

= A \cap ( A' \cup ( B')') [ By De-morgans law]

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

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

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

= A \cap B [ Since \phi \cup X = X for any set X ]

= RHS. Hence proved.

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

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

= A \cap (A' \cap B') [ By De-morgans law]

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

= \phi \cap B'

= \phi =  RHS.

(iv) To prove: A-B = A \triangle (A \cap B)

RHS = A \triangle ( A \cap B)

= ( A - ( A \cap B )) \cup ( A \cap B - A) [ Since E \triangle F = ( E - F)  \cup ( F - E) ]

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

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

[ By De-morgans law and associative law ]

(A \cap A') \cup (A \cap B') \cup (\phi \cap B)

[ Since intersection distributes over union and A \cap A' = \phi ]

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

= A \cap B'   [Since A \cap B' = A - B ]

= A - B = LHS

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Question 3: If A, \ B, \ C are three sets such that A \subset B , then prove that C -B \subset C -A

Answer:

We have sets A, B and C

To show C - B \subset C - A

Let x \in C -B and x \in C and x \notin B

\Rightarrow x \in C and x \notin A [ Since A \subset B ]

\Rightarrow x \in C - A

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

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

Answer:

i)    To prove: (A \cup B)-B=A-B

( A \cup B ) - B = ( A - B ) \cup ( B - B)

= ( A - B ) \cup \phi

= A - B

ii)   To prove: A-(A \cap B) = A - B

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

= \phi \cap ( A - B )

= A - B

iii)   To prove: A-(A-B)=A \cap B 

Let x \in A - ( A - B )

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

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

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

\Rightarrow x \in (A \cap B)

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

iv)   To prove: A \cup (B-A)=A \cup B

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

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

\Rightarrow x \in B

\Rightarrow x \in ( A \cup B )   [ Since B \subset ( A \cup B ) ]

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

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

Conversely,

Let x \in ( A \cup B )

\Rightarrow x \in A and x \in B

\Rightarrow x \in A or x \in ( B - A ) [ Since B \subset ( 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 )

v)   To prove: (A -B) \cup (A \cap B)= A

Let x \in A

The either x \in ( A - B ) or x \in ( A \cap B )

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

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

Conversely,

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

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

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

\Rightarrow x \in A

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

From i) and ii) we get

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

 

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