Differentiation (Part-38) | S N De

 

14.  Let $S = \{a, b, c, d, e\}$ be the universal set and let $\,\,A = \{a, b, d\}\,\,$ and $\,\,B = \{b, d, e\}\,\,$ be two of its subsets. Find $\,\, (A \cap B)' \,;\, (A \cup B)' $

 Sol. $\,\,A \cap B=\{a,b,d\} \cap \{b,d,e\} =\{b,d\} \\ \therefore (A \cap B)'=S-(A \cap B)\\~~~~~~~~~~~~~~~~~~~=\{a,b,c,d,e\}-\{b,d\}\\~~~~~~~~~~~~~~~~~~~=\{a,c,e\} \\ A \cup B=\{a,b,d\} \cup \{b,d,e\} \\~~~~~~~~~~~=\{a,b,d,e\} \\ \therefore \,\,(A \cup B)'=S-(A \cup B)\\~~~~~~~~~~~~~~~~~~~~=\{a,b,c,d,e\}-\{a,b,d,e\}\\~~~~~~~~~~~~~~~~~~~~=\{c\}$

15. Let $\, S =\{1,2,3,4,5,6\}\,$ be the universal set . Let $\,\,A \cup B =\{2,3,4\}.\,\,$ Find $A^c \cap B^c,\,\,$ where  $\,A^c, B^c\,$ are complements of $\,A , B \,$ respectively. 

Sol. $\,\,A^c \cap B^c = (A \cup B)^c \\= S-(A \cup B)\\= \{1,2,3,4,5,6\}-\{2,3,4\} \\=\{1,5,6\}$

16. If $\,\, a \mathbb N=\{ax: x \in \mathbb N \},\,\,$ describe $\,\,3 \mathbb N \cap 7\mathbb N\,\,$ where $\,\, \mathbb N \,\,$ is the set of natural numbers.

Sol. $\,\, a \mathbb N=\{ax: x \in \mathbb N \}\\ \text{Let} x \in 3\mathbb N \cap 7 \mathbb N \,\,\text{be any element of the set.} \\ \implies x \in 3\mathbb N  \wedge x \in 7\mathbb N \\ \implies x =3k \wedge x=7p,\,\, \text{where}\,k,p \in \mathbb N  \\ \implies x \,\,\text{is divisible by both } 3 \,\,\text{and}\,\, 7 \\ \implies x\,\, \text{is divisible by the l.c.m. of }3 \,\,\text{and}\,\, 7 \\ \implies x =21 m \,\,, m \in  \mathbb N \\ \therefore 3\mathbb N  \cap 7\mathbb N  \subseteq 21 \mathbb N  \cdots (1)\\ \text{But,} \,21 \mathbb N  \subseteq 3\mathbb N  \wedge 21 \mathbb N \subseteq 7 \mathbb N \\ [\therefore 21k=3 \times (7k)=7 \times (3k) \\ \text{where,}\,\, 7k,3k \in \mathbb N ,\,\,\text{if}\, k \in \mathbb N ]\\ \therefore 21 \mathbb N  \subseteq 3 \mathbb N  \cap 7\mathbb N  \cdots (2)\\ \therefore \text{From (1) and (2), the result follows. } $  

Alternatively, $ 3 \mathbb N=\{3,6,9,12,15,18,21,24,......\} \\ 7\mathbb N=\{7,14,21,28,35,42,49,......\} \\ \therefore 3\mathbb N  \cap 7\mathbb N =\{21,42,....\}=21 \mathbb N $

17. For any 2 sets $\,\,A, B,\,\, \text{prove the followings:}\\ (i) (B-A) \cap A=\phi $ 

Sol. If possible let $(B-A) \cap A \neq \phi \\ \implies \exists \,\, x \in (B-A) \cap A \\ \implies x \in (B-A) \wedge x \in A \\ \implies \{x \in B \wedge x \not \in A\} \wedge x \in A \\ \implies x \in B \wedge \{x \not \in A   \wedge x \in A\}\cdots (1) \\ \text{From (1), we notice that } x \not \in A   \wedge x \in A\\  \text{ which can not happen simultaneously.}\\ \text{Hence the result follows.}$  

17. For any 2 sets $\,\,A, B,\,\, \text{prove the followings:}\\ (ii) A^c-B^c=B-A $ 

Sol. Let $x \in  A^c-B^c \,\,\text{be arbitrary element of the set.}\\ \implies x \in A^c \wedge x \not \in B^c \\ \implies x \not \in A \wedge x \in B \\ \implies   x \in B  \wedge x \not \in A \\ \implies x \in B-A \\ \therefore (A^c-B^c) \subseteq (B-A) \cdots (1) \\ \text{Let}\,\,y \in (B-A) \,\,\text{be arbitrary  element}\\  \text{ of the set.} \\ \implies y \in B \wedge y \not \in A \\ \implies y \not \in B^c \wedge y \in A^c\\ \implies y \in A^c \wedge y \not \in B^c \\ \implies y \in (A^c-B^c) \\ \therefore (B-A) \subseteq (A^c-B^c) \cdots (2) \\ \therefore \,\,\text{From (1), (2) the result follows.}$

17. For any 2 sets $\,\,A, B,\,\, \text{prove the followings:}\\ (iii) A-B=A-(A\cap B)$ 

Sol. Let $\,\,x \in (A-B)\,\,\text{be arbitrary element of the set.}\\ \implies x \in A \wedge x \not \in B \\ \implies x \in A \wedge x \not \in (A\cap B) \\ \implies x \in A-(A \cap B) \\ \therefore (A-B) \subseteq A-(A \cap B) \cdots (1) \\ \text{Let}\,\, \,\,y \in A-(A\cap B)\,\,\text{be arbitrary element of the set.} \\ \implies y \in A \wedge y \not \in (A \cap B)\\ \implies y \in A \wedge \{y \not \in A \vee y \not \in B\} \\ \implies \{y \in A \wedge y \not \in A \} \vee \{y \in A \wedge y \not \in B\} \\ \implies \{y \in A \wedge y  \in A^c \} \vee \{y \in A \wedge y \not \in B\} \\ \implies \{y \in (A \cap A^c)\} \vee \{y \in (A-B) \} \\ \implies y \in  (A \cap A^c) \cup (A -B)\\ \implies y \in \phi \cup (A-B)\\ \implies y \in (A-B) \\ \therefore A-(A \cap B) \subseteq (A-B) \cdots (2) \\ \therefore \text{From (1) and (2),  }    A-B=A-(A\cap B) $

17. For any 2 sets $\,\,A, B,\,\, \text{prove the followings:}\\ (iv)\, A-B=A \cap B'$ 

Sol. Let $\,\,x \in (A-B)\,\,\text{be arbitrary element of the set.}\\ \implies x \in A \wedge x \not \in B \\ \implies x \in A \wedge x  \in B' \\ \implies x \in A \cap B' \\ \therefore (A-B) \subseteq A \cap B' \cdots (1) \\ \text{Let}\,\, \,\,y \in A \cap B' \,\,\text{be arbitrary element of the set.} \\ \implies y \in A \wedge y  \in B' \\ \implies y \in A \wedge y \not \in B \\ \implies y \in A-B  \\ \therefore A \cap B'  \subseteq (A-B)  \cdots (2) \\ \therefore \text{From (1)  and  (2),}\,\,A-B=A \cap B'  $

17(v). For any 2 sets $\,\,A, B,\,\, \text{prove the followings:}\\ (v)\, B-A^c=A \cap B$ 

Sol. Let $\,\,x \in (B-A^c)\,\,\text{be arbitrary element of the set.}\\ \implies x \in B \wedge x \not \in A^c \\ \implies x \in B \wedge x  \in A \\ \implies x \in A \cap B \\ \therefore (B-A^c) \subseteq A \cap B \cdots (1) \\ \text{Let}\,\, \,\,y \in A \cap B \,\,\text{be arbitrary element of the set.} \\ \implies y \in A \wedge y  \in B \\ \implies y \in B \wedge y \not \in A^c \\ \implies y \in B-A^c  \\ \therefore A \cap B  \subseteq (B-A^c)  \cdots (2) \\ \therefore \text{From (1)  and  (2),}\,\,B-A^c=A \cap B  $

17(vi) For any 2 sets $\,A,\,B\,\,$ prove that: $B \subseteq (A-B)^c$

Sol. For any 2 sets $\,\,A \,\,\text{and}\,\,B,\, B \subseteq A^c \cup B \\ \implies B \subseteq (A \cap B^c)^c \,\,\,[\text{De Morgan's Law}] \\ \implies B \subseteq (A-B)^c \,\,[\text{Since,}\,\,A-B=A \cap B^c]$ 

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