If $A=\{-1,2,3\}$ and $B=\{1,3\}$, then determine
(i) $A \times B$
(ii) $B \times A$
(iii) $B \times B$
(iv) $A \times A$
$$A=\{-1,2,3\} \text { and } B=\{1,3\}$$
(i) $A \times B=\{(-1,1),(-1,3),(2,1),(2,3),(3,1),(3,3)\}$
(ii) $B \times A=\{(1,-1),(1,2),(1,3),(3,-1),(3,2),(3,3)\}$
(iii) $B \times B=\{(1,1),(1,3),(3,1),(3,3)\}$
(iv) $A \times A=\{(-1,-1),(-1,2),(-1,3),(2,-1),(2,2),(2,3),(3,-1),(3,2),(3,3)\}$
If $P=\{x: x<3, x \in N\}, \quad Q=\{x: x \leq 2, x \in W\}, \quad$ then find $(P \cup Q) \times(P \cap Q)$, where $W$ is the set of whole numbers.
We have,$$P=\{x: x<3, x \in N\}=\{1,2\}$$
and $$Q=\{x: x \leq 2, x \in W\}=\{0,1,2\}$$
$\therefore \quad P \cup Q=\{0,1,2\}$ and $P \cap Q=\{1,2\}$
$\begin{aligned}(P \cup Q) \times(P \cap Q) & =\{0,1,2\} \times\{1,2\} \\ & =\{(0,1),(0,2),(1,1),(1,2),(2,1),(2,2)\}\end{aligned}$
If $A=\{x: x \in W, x< 2\}, B=\{x: x \in N, 1< x< 5\}$ and $C=\{3,5\}$, then find
(i) $A \times(B \cap C)$
(ii) $A \times(B \cup C)$
$$\begin{aligned} \text{We have,}\quad A & =\{x: x \in W, x<2\}=\{0,1\} \\ \text{and}\quad B & =\{x: x \in N, 1< x<5\} \\ & =\{2,3,4\} \text { and } C=\{3,5\} \end{aligned}$$
$\begin{aligned} \text { (i) } & \because & B \cap C=\{3\} \\ & \therefore & A \times(B \cap C)=\{0,1\} \times\{3\}=\{(0,3),(1,3)\}\end{aligned}$
$$\begin{aligned} \text { (ii) }\quad \because(B \cup C)=\{2,3,4,5\} & \\ \therefore \quad A \times(B \cup C) & =\{0,1\} \times\{2,3,4,5\} \\ & =\{(0,2),(0,3),(0,4),(0,5),(1,2),(1,3),(1,4),(1,5)\} \end{aligned}$$
In each of the following cases, find $a$ and $b$.
(i) $(2 a+b, a-b)=(8,3)$
(ii) $\left(\frac{a}{4}, a-2 b\right)=(0,6+b)$
(i) We have, $(2 a+b, a-b)=(8,3)$
$$\Rightarrow \quad 2 a+b=8 \text { and } a-b=3$$
[since, two ordered pairs are equal, if their corresponding first and second elements are equal]
On substituting, $b=a-3$ in $2 a+b=8$, we get
$$\begin{aligned} & 2 a+a-3=8 \Rightarrow 3 a-3=8 \\ & \Rightarrow \quad 3 a=11 \Rightarrow a=\frac{11}{3} \end{aligned}$$
Again, $\quad$ substituting $a=\frac{11}{3}$ in $b=a-3$, we get
$\begin{array}{ll} & b=\frac{11}{3}-3=\frac{11-9}{3}=\frac{2}{3} \\ \therefore \quad & a=\frac{11}{3} \text { and } b=\frac{2}{3}\end{array}$
$\begin{array}{rlrl} & \text { (ii) We have, } & \left(\frac{a}{4}, a-2 b\right) =(0,6+b) \\ \Rightarrow & \frac{a}{4} =0 \Rightarrow a=0\end{array}$
$$\begin{aligned} \text{and}\quad a-2 b & =6+b \\ 0-2 b & =6+b \\ -3 b & =6 \end{aligned}$$
$$\begin{aligned} & \therefore \quad b=-2 \\ & \therefore \quad a=0, b=-2 \end{aligned}$$
$A=\{1,2,3,4,5\}, S=\{(x, y): x \in A, y \in A\}$, then find the ordered which satisfy the conditions given below.
(i) $x+y=5$
(ii) $x+y<5$
(iii) $x+y>8$
We have, $A=\{1,2,3,4,5\}$ and $S=\{(x, y): x \in A, y \in A\}$
(i) The set of ordered pairs satisfying $x+y=5$ is,
$$\{(1,4),(2,3),(3,2),(4,1)\}$$
(ii) The set of ordered pairs satisfying $x+y<5$ is $\{(1,1),(1,2),(1,3),(2,1),(2,2),(3,1)$.
(iii) The set of ordered pairs satisfying $x+y>8$ is $\{(4,5),(5,4),(5,5)\}$.