mathematics-class-11-unit-01-chapter-03-set-theory-l-3-3-4qtlloo9png

Set theory lecture-3

Recall

$n(A)= \text { no. of elements in } A$

$n(A \cup B)= n(A)+n(B)-n(A \cap B)$

$n(A \cup B \cup C)= n(A)+n(B)+n(C) -n(A \cap B)-n(B \cap C) -n(A \cap C)+n(A \cap B \cap C)$

Also, $n(A)=n(A-B)+n(A \cap B)$

Set theory lecture-3

Problem 1

If $n(A)=6, n(B)=4$

What is the minimum & maximum value of $n(A-B)$ ?

Set theory lecture-3

Solution

$n(A)=n(A-B)+n(A \cap B)$

$\Rightarrow n(A-B)=n(A)-n(A \cap B)$

$= 6 - n(A \cap B)$

$\because A \cap B \subseteq B \Rightarrow n(A \cap B) \leq n(B) = 4$

$\therefore n(A-B) \geq 6-4=2 \leftarrow \text { Min. value }$

Also, if $A \cap B=\phi, n(A \cap B)=0$

$\therefore \operatorname{Max.} n(A-B)=6-0=6$

Set theory lecture-3

Problem 2

Suppose there are 60 people.
25 read newspaper $H$
26 read newspaper $T$
26 read newspaper $I$
9 read both $H$ and $I$
11 read both $H$ and $T$
8 read both $T$ and $I$

Also, 3 people read all three.

Find

Number of people who read at least one of the three.
Number of people who read exactly one newspaper.

Set theory lecture-3

Solution

(i) $n (H \cup T \cup I)$

$=25+10+5+12$ $=52$

or use $\quad n (H \cup T \cup I)$

$=n(H)+n(T)+n(I)-n(H \cap T)-n(T \cap I)$
$-n(H \cap I) + n(H \cap T \cap I)$

(ii) From the Venn diagram,

Number of people who read exactly one $=8+10+12=30$

alt text

Set theory lecture-3

Problem 3

Let $n(A \cap B)=x, \quad n(A-B)=6 x$

$n(B-A)=8+2 x, \quad n(A)=n(B)$

Find $x$ .

Set theory lecture-3

Solution

$n(A)=n(A-B)+n(A \cap B)=6 x+x=7 x$

$n(B)=n(B-A)+n(B \cap A)=8+2 x+x=8+3 x$

$\therefore \quad 7 x=8+3 x \Rightarrow x=2$

Set theory lecture-3

Problem 4

Suppose 70% Indians like apple, and $82$ % like mango. Let $x$ % like both.

Find the min. & max. possible $x$ .

Set theory lecture-3

Solution

Given: $n(A)=70, \quad n(M)=82$

$n(A \cap M)=x$

Since $\quad n(A \cup M) \leq 100,$

$n(A)+n(M)-n(A \cap M) \leq 100$

$\Rightarrow 70+82-x \leq 100$ $\Rightarrow x \geq 52$

Also, $A \cap M \subseteq A$

$\therefore n(A \cap M) \leq n(A)=70\qquad\quad$ So, $52 \leq x \leq 70$

Set theory lecture-3

Problem 5

Find $n P(P(P(\phi)))$ .

Set theory lecture-3

Solution

[Recall : $\quad P(A)=$ all subsets of $A$ .

$n(P(A))=2^{n(A)}$ ]

Since $n(\phi)=0, \quad n(P(\phi))=2^{0}=1$ .

$\Rightarrow n(P(P(\phi)))=2^1=2$

$\Rightarrow n(P(P(P(\phi))))=2^2=4 .$

Set theory lecture-3

Problem 6

If $n(A)=m, \quad n(B)=n$ ,

$n(P(A))-n(P(B))=112$ .

Find m and n.

Set theory lecture-3

Solution

Since, $2^m-2^n=112(\Rightarrow m>n)$

$\Rightarrow 2^n\left(2^{m-n}-1\right)=112=16 \times 7=2^4 \times 7$

$\Rightarrow n=4$ & $2^{m-n}-1=7$

$\Rightarrow n=4$ & $2^{m-n}=8=2^3$

$\Rightarrow n=4$ & $m-n=3$

$\Rightarrow n=4$ & $m=7$

alt text

Set theory lecture-3

Problem 7

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

Find the no. of sets $C$ such that

$A \cap B \subseteq C \subseteq A \cup B .$

Set theory lecture-3

Solution

$A \cap B=\{2,4\}$

$A \cup B=\{1,2,3,4,6\}$

So, { $2,4$ } $\subseteq C$ $\subseteq$ { $1,2,3,4,6$ } $\Rightarrow \quad C$ ={2,4} $\cup C^{\prime}$

$\text { where }C^{\prime} \subseteq$ {1,3,6}

$\therefore$ Number of such C= Number of $C^{\prime}$ $\subseteq$ {1,3,6}

$=n(P($ { $1,3,6$ } $))$ $=2^3=8 .$

Set theory lecture-3

Problem 8

Let $X=\{4^n-3 n-1: n \in \mathbb{N}\}$

$Y=\{9(n-1): n \in \mathbb{N}\}$

Then $X \cup Y$ equals

(a) $\mathbb{N}$ $\quad$ (b) $Y-X$

(c) $X$ $\quad$ (d) $Y$ .

Set theory lecture-3

Solution

$X=\{0,9,54, \cdots\}$

$Y=\{0,9,18,27, \cdots\}$

Clearly, $1 \notin X \cup Y$

So, (a) is FALSE.

Also, $Y \nsubseteq X(\because 18 \in Y, 18 \notin X)$

Set theory lecture-3

Solution

$\therefore X \cup Y \neq X$ So, (c) is False.

$Y-X \neq X \cup Y$ because $0 \in X \cup Y$ but $0 \notin Y-X$ .

By elimination (d) must be true if it is given that one of the staments is true.

Let’s try to prove this.

Claim: $\quad X \subseteq Y \quad(\Rightarrow X \cup Y=Y)$

ie. $4^n-3 n-1$ is divisible by 9 for all $n \in \mathbb{N}$ .

Set theory lecture-3

Solution

Note that for $n=1: 4^n-3 n-1=0$ which is divisible by $9$

Suppose $4^k-3 k-1$ is divisible by $9$ .

We’ ll show that $4^{k+1}-3(k+1)-1$ is also divisible by $9$ .

$4^{k+1}-3(k+1)-1$

$= 4\left(4^k-3 k-1\right)+9 k$ ,which is divisible by 9.

Set theory lecture-3

Problem 9

$A=\{n^3+(n+1)^3+(n+2)^3: n \in \mathbb{N}\}$

$B=\{9 n: n \in \mathbb{N}\}$

Then which of the following is/are true?

(a) $A \subseteq B$

(b) $B \subseteq A$

(c) $A=B$

(d) $A \subsetneq B$ .

Set theory lecture-3

Solution

$A=\{36,2^3+3^3+4^3, \cdots\}$

Smallest no. in $A$ is 36 .

So, $A \neq B$ and $B \nsubseteq A$ .

(b) & (c) are FALSE.

Set theory lecture-3

Solution

Claim: $n^3+(n+1)^3+(n+2)^3$ is a multiple of $9$ for every $n \in \mathbb{N}$ .

$n^3+(n+1)^3+(n+2)^3$

$= n^3+\left(n^3+3 n^2+3 n+1\right)+\left(n^3+6 n^2+12 n+8\right)$

$= 3 n^3+9 n^2+15 n+9$ $= 9\left(n^2+1\right)+3 n\left(n^2+5\right)$

Claim: $n\left(n^2+5\right)$ is a multiple of 3.

If $n=3 k$ , then O.K, $\qquad$ If $n=3 k+1 ; \quad n^2+5=(3 k+1)^2+5$

Set theory lecture-3

Solution

$(3 k+1)^2+5=9 k^2+6 k+6$

which is a multiple of 3 .

Also, $(3 k+2)^2+5 =9 k^2+12 k+4+5$

$=3\left(3 k^2+4 k+3\right)$ multiple of $3$ .

Hence, $n^3+(n+1)^3+(n+2)^3$ is a multiple of $9$ for all $n \in \mathbb{N}$ .

So, $A \subsetneq B$ .Hence (a) & (d) are correct. .

Thank you