Answer :

$A=\{x: x \in R.$ and $x$ satisfies $.x^{2}-8 x+12=0\}$

2 and 6 are the only solutions of $x^{2}-8 x+12=0$.

$\therefore A=\{2,6\}$

$B=\{2,4,6\}, C=\{2,4,6,8 \ldots\}, D=\{6\}$

$\therefore D \subset A \subset B \subset C$

Hence, $A \subset B, A \subset C, B \subset C, D \subset A, D \subset B, D \subset C$

Answer :

(i) False

Let $A=\{1,2\}$ and $B=\{1,\{1,2\},\{3\}\}$

Now, $2 \in\{1,2\}$ and $\{1,2\} \in\{\{3\}, 1,\{1,2\}\}$

A $\in$ B

. However, $2 \notin\{\{3\}, 1,\{1,2\}\}$

(ii) False

Let $A=\{2\}, B=\{0,2\}$, and $C=\{1,\{0,2\}, 3\}$

As $A \subset B$

$B \in C$

However, $A \notin C$

(iii) True

Let $A \subset B$ and $B \subset C$.

Let $x \in A$

$\Rightarrow x \in B \quad[\because A \subset B]$

$\Rightarrow x \in C \quad[\because B \subset C]$

$\therefore A\subset C$

(iv) False

$ A=\{1,2\}, B=\{0,6,8\}, \text{ and } C=\{0,1,2,6,9\} $

Let Accordingly, $A \not \subset B$ and $B \not \subset C$.

However, $A \subset C$

(v) False

Let $A=\{3,5,7\}$ and $B=\{3,4,6\}$

Now, $ 5 \in A \text{ and } A \in B$

However, $5 \notin B$

(vi) True

Let $A \subset B$ and $x \notin B$.

To show: $x \notin A$

If possible, suppose $x \notin A$.

Then, $x \in B$, which is a contradiction as $x \notin B$

$\therefore x \notin A$

Answer :

Let, $A, B$ and $C$ be the sets such that $A \cup B=A \cup C$ and $A \cap B=A \cap C$.

To show: $B=C$

Let $x \in B$

$ \begin{matrix} \Rightarrow x \in A \cup B & {[B \subset A \cup B]} \\ \Rightarrow x \in A \cup C & {[A \cup B=A \cup C]} \\ \Rightarrow x \in A \text{ or } x \in C & \end{matrix} $

Case $I x$

$\in A$

Also, $x \in B$

$ \begin{aligned} & \therefore x \in A \cap B \\ & \Rightarrow x \in A \cap C \quad[\because A \cap B=A \cap C] \end{aligned} $

$\therefore x \in A$ and $x \in C$ :

$x \in C$

$\therefore B \in C $

Similarly, we can show that $ C \in B $

$\therefore B=C$

Answer :

First, we have to show that (i) $ \Leftrightarrow $ (ii).

Let $A \subset B$

To show: $A$ -$B = \phi $

If possible, suppose $A$ - $B \neq \Phi$

This means that there exists $x \in A, x \neq B$, which is not possible as $A \subset B$.

$\therefore A$ -$B=\Phi$

$ \therefore A \subset B \Rightarrow A - B = \phi $

(ii)$\Rightarrow$(i)

Let $A$ - $B=\Phi$

To show: $A \subset B$

Let $x \in A$

Clearly, $x \in B$ because if $x \notin B$, then $A$ - $B \neq \Phi$

$ \therefore A - B = 0 \Rightarrow A \subset B \neq \phi $

(ii)$\Rightarrow$(i)

Let $A \subset B$

To show: $A \cup B=B$

Clearly, $B \subset A \cup B$

Let $x \in A \cup B$

$\Rightarrow x \in A$ or $x \in B$

Case I: $x \in A$

$\Rightarrow x \in B \quad[\because A \subset B]$

$\therefore A \cup B \subset B$

Case II: $x \in B$

Then, $A \cup B=B$

Conversely, let $A \cup B=B$

Let $x \in A$

$ \begin{matrix} \Rightarrow x \in A \cup B & {[\because A \subset A \cup B]} \\ \Rightarrow x \in B & {[\because A \cup B=B]} \end{matrix} $

Now, we have to show that (i) $ \Rightarrow$ (iv).

$$ \begin{aligned} & \text{ Let } A \tilde{A} \hookrightarrow \mathring{A} B \\ & \text{ Clearly } A \cap B \subset A \\ & \text{ Let } x \in A \end{aligned} $$

We have to show that $x \in A \cap B$

As A $ \subset B, x \in B$

E

$x \in A \cap B$

$\therefore A \subset A \cap B$

$\therefore, A \subset A \cap B $

Hence, A = A $\cap $ B

Conversely, suppose $A \cap B =A$

Let $x \in A$

$\Rightarrow x \in A \cap B$

$\Rightarrow x \in A$ and $x \in B$

$\Rightarrow x \in B$

$\therefore A \subset B$

Hence, (i) $ \Rightarrow $ (iv).

Answer :

Let $A \subset B$

To show: C - B C C - A

Let $x \in C-B \Rightarrow x \in C$ and

$x \notin B \Rightarrow x \in C$ and $x \notin A[A$

$\subset B] \Rightarrow x \in C-A$

$\therefore C-B \subset C-A$

Answer :

To show:

$ \begin{aligned} & (A \cap B) \cup(A-B) \\ & =(A \cap B) \cup\left(A \cap B^{\prime}\right) \\ & =A \cap\left(B \cup B^{\prime}\right) \text { (By distributive law) } \\ & =A \cap U=A \end{aligned} $

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

Also $A \cup(B-A)$

$ \begin{aligned} & =A \cup\left(B \cap A^{\prime}\right) \\ & =(A \cup B) \cap\left(A \cup A^{\prime}\right) \text { (By distributive law) } \\ & =(A \cup B) \cap U \\ & =A \cup B \end{aligned} $

Hence $A \cup(B-A)=A \cup B$.

Answer :

(i)

From the distributive property:

$ \begin{aligned} & A \cup(B \cap C)=(A \cup B) \cap(A \cup C) \\ & A \cup(A \cap B)=(A \cup A) \cap(A \cup B)=A \cap(A \cup B)=A \end{aligned} $

Hence,

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

(ii)

From the distributive property:

$ \begin{aligned} & A \cap(B \cup C)=(A \cap B) \cup(A \cap C) \\ & A \cap(A \cup B)=(A \cap A) \cup(A \cap B)=A \cup(A \cap B) \end{aligned} $

From part (i),

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

Hence,

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

Answer :

$ \begin{aligned} & \text { Let } \mathrm{A}=\{0,1\} \ & \mathrm{B}=\{\mathrm{O}, 2,3\} \ & \mathrm{C}=\{0,4,5\} \end{aligned} $

So,

$A \cap B=\{0\}$

and $\mathrm{A} \cap \mathrm{C}=\{\mathrm{O}\}$

Here, $\mathrm{A} \cap \mathrm{B}=\mathrm{A} \cap \mathrm{C}=\{\mathrm{O}\}$

However, $\mathrm{B} \neq \mathrm{C}$ as $2 \in \mathrm{B}$ and $2 \notin \mathrm{C}$

Answer :

Let $\mathrm{A}$ and $\mathrm{B}$ be two sets such that

$\mathrm{A} \cap \mathrm{X}=\mathrm{B} \cap \mathrm{X}=\phi$ and $\mathrm{A} \cup \mathrm{X}=\mathrm{B} \cup \mathrm{X}$ for some set $\mathrm{X}$.

To show: $\mathrm{A}=\mathrm{B}$

It can be seen that

$ \begin{aligned} & A=A \cap(A \cup X) \\ & =A \cap(B \cup X)(A \cup X=B \cup X) \\ & =(A \cap B) \cup(A \cap X) \text { (Distributive law )} \\ & =(A \cap B) \cup \phi(\because A \cap X=\phi) \\ & =A \cap B . . . .(1) \end{aligned} $

Now, $\mathrm{B}=\mathrm{B} \cap(\mathrm{B} \cup \mathrm{X})$

$ \begin{aligned} & =\mathrm{B} \cap(\mathrm{A} \cup \mathrm{X})(\because \mathrm{A} \cup \mathrm{X}=\mathrm{B} \cup \mathrm{X}) \\ & =(\mathrm{B} \cap \mathrm{A}) \cup(\mathrm{B} \cap \mathrm{X}) \text { (Distributive law) } \\ & =(\mathrm{B} \cap \mathrm{A}) \cup \phi(\because \mathrm{B} \cap \mathrm{X}=\phi) \\ & =\mathrm{B} \cap \mathrm{A} \\ & =\mathrm{A} \cap \mathrm{B} . . . . .(2) \end{aligned} $

Hence, from (1) and (2), we get $ \mathrm{A}=\mathrm{B} $

Answer :

Let $A=\{0,1\}, B=\{1,2\}$, and $C=\{2,0\}$.

Accordingly,

$A \cap B= \{1\}$,

$B \cap C=\{2\}$, and $A \cap C=\{0\}$.

$\therefore A \cap B, B \cap \odot C$, and $A \cap \odot C$ are non-empty.

However, $A \cap \odot B \cap \odot C=\Phi$