Class 11, Exercise 1.7, Sets (Class 11)

Class 11: Sets – Exercise 1.7

Date: April 29, 2019Author: ICSE CBSE ISC Board Mathematics Portal for Students 0 Comments

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.

$\\$

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

$\\$

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 $C - B \subset C - A$

$\\$

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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