Answer :

It is given that $\left(\dfrac{x}{3}+1, y-\dfrac{2}{3}\right)=\left(\dfrac{5}{3}, \dfrac{1}{3}\right)$

Since the ordered pairs are equal, the corresponding elements will also be equal.

Therefore, $\dfrac{x}{3}+1=\dfrac{5}{3}$ and $y-\dfrac{2}{3}=\dfrac{1}{3}$.

$\Rightarrow \ \dfrac{x}{3}+1=\dfrac{5}{3}$

$\Rightarrow \dfrac{x}{3}=\dfrac{5}{3}-1 $ $ \ \Rightarrow y-\dfrac{2}{3}=\dfrac{1}{3}$

$\Rightarrow \dfrac{x}{3}=\dfrac{2}{3} \quad \Rightarrow \ y=\dfrac{1}{3}+\dfrac{2}{3}$

$\Rightarrow x=2 \ \quad \Rightarrow \ y=1$

$\therefore \ \ x=2$ and $y=1$

Answer :

It is given that set $A$ has $3$ elements and the elements of set $B$ are $3, 4,$ and $5 .$

Number of elements in set $B=3$

Number of elements in $\left(A \times B\right)$ $=($ Number of elements in $A) \times($ Number of elements in $B)$

$\hspace{4.4cm}=3 \times 3=9$

Thus, the number of elements in $\left(A \times B\right)$ is $9 .$

Answer :

$G={7,8}$ and $H={5,4,2}$

We know that the Cartesian product $P \times Q$ of two non-empty sets $P$ and $Q$ is defined as

$\quad \ P \times Q={\left(p, q\right): p \in P, q \in Q}$

$\therefore \ G \times H={\left(7,5\right),\left(7,4\right),\left(7,2\right),\left(8,5\right),\left(8,4\right),\left(8,2\right)}$

$\quad H \times G={\left(5,7\right),\left(5,8\right),\left(4,7\right),\left(4,8\right),\left(2,7\right),\left(2,8\right)}$

Answer :

$(i)$ Given $P={m, n}$ and $Q={n, m}$ then

$ \qquad \mathrm{P} \times \mathrm{Q}={(\mathrm{m}, \mathrm{n}),(\mathrm{m}, \mathrm{~m}),(\mathrm{n}, \mathrm{n}),(\mathrm{n}, \mathrm{~m})} $

So, given value of $P \times Q$ is incorrect.

Hence, the given statement $(i)$ is false.

(ii) If $A$ and $B$ are non-empty sets, then $A \times B$ is a non-empty set of ordered pairs ( $x, y$ ) such that $x \in A$ and $y \in B$

Hence, the given statement $(ii)$ is true.

(iii) $A={1,2}$ and $B={3,4}$

$\qquad A \times(B \cap \phi)=A \times \phi=\phi$

So, $(iii)$ is true.

Answer :

It is known that for any non-empty set $A, A \times A \times A$ is defined as

$\begin{aligned} &A={-1,1}\\ &A \times A={-1,1} \times{-1,1}={(-1,-1),(-1,1),(1,-1),(1,1)}\\ &\Rightarrow A \times A \times A={-1,1} \times{(-1,-1),(-1,1),(1,-1),(1,1)} \end{aligned}$

$\therefore ~ A \times A \times A={\left(-1,-1,-1\right),\left(-1,-1,1\right),\left(-1,1,-1\right),\left(-1,1,1\right), \left(1,-1,-1\right),\left(1,-1,1\right),\left(1,1,-1\right),\left(1,1,1\right)}$

Answer :

It is given that $A \times B={\left(a, x\right),\left(a, y\right),\left(b, x\right),\left(b, y\right)}$

We know that the Cartesian product of two non-empty sets $P$ and $Q$ is defined as $P \times Q={\left(p, q\right): p \in P, q \in Q}$

$\therefore \ A$ is the set of all first elements and $B$ is the set of all second elements.

Thus, $A={a, b}$ and $B={x, y}$

Answer :

(i) To verify: $A \times\left(B \cap C\right)=\left(A \times B\right) \cap\left(A \times C\right)$

We have $B \cap C={1,2,3,4} \cap{5,6}=\Phi$

$\therefore \ \ \text{L.H.S.} \ =A \times\left(B \cap C\right)=A \times \Phi=\Phi$

$A \times B={\left(1,1\right),\left(1,2\right),\left(1,3\right),\left(1,4\right),\left(2,1\right),\left(2,2\right),\left(2,3\right),\left(2,4\right)}$

$A \times C={\left(1,5\right),\left(1,6\right),\left(2,5\right),\left(2,6\right)}$

$\therefore \ \ \text{R.H.S.} \ \ =\left(A \times B\right) \cap\left(A \times C\right)=\Phi$

$\therefore \ \ \mathrm{L.H.S. = R.H.S.}$

Hence, $A \times\left(B \cap C\right)=\left(A \times B\right) \cap\left(A \times C\right)$

(ii): To verify: $A \times C$ is a subset of $B \times D$

$A \times C={\left(1,5\right),\left(1,6\right),\left(2,5\right),\left(2,6\right)}$

$B \times D={\left(1,5\right),\left(1,6\right),\left(1,7\right),\left(1,8\right),\left(2,5\right),\left(2,6\right),\left(2,7\right),\left(2,8\right),\left(3,5\right),\left(3,6\right),\left(3,7\right),\left(3,8\right),\left(4,5\right),\left(4,6\right),\left(4,7\right),\left(4,8\right)}$

We can observe that all the elements of set $A \times C$ are the elements of set $B \times D$

Therefore, $A \times C$ is a subset of $B \times D$.

Answer :

Given, $\mathrm{A}={1,2}$ and $\mathrm{B}={3,4}$

Now,

$ A \times B={(1,3),(1,4),(2,3),(2,4)} $

Since $A \times B$ contains 4 elements, so number of subsets of $A \times B$ is $2^4=16$.

Therefore, the set $A \times B$ has $2^{4}=16$ subsets. These are

$\Phi,{\left(1,3\right)},{\left(1,4\right)},{\left(2,3\right)},{\left(2,4\right)},{\left(1,3\right),\left(1,4\right)},{\left(1,3\right),\left(2,3\right)}$,

${\left(1,3\right),\left(2,4\right)},{\left(1,4\right),\left(2,3\right)},{\left(1,4\right),\left(2,4\right)},{\left(2,3\right),\left(2,4\right)}$,

${\left(1,3\right),\left(1,4\right),\left(2,3\right)},{\left(1,3\right),\left(1,4\right),\left(2,4\right)},{\left(1,3\right),\left(2,3\right),\left(2,4\right)}$,

${\left(1,4\right),\left(2,3\right),\left(2,4\right)},{\left(1,3\right),\left(1,4\right),\left(2,3\right),\left(2,4\right)}$

Answer :

It is given that $n(A)=3$ and $n(B)=2$ and ${(x, 1),(y, 2),(z, 1)}$ are in $A \times B$

We know that $A=$ Set of first elements of the ordered pair elements of $A \times B$

$B=$ Set of second elements of ordered pair elements of $A \times B$

$\therefore \ \ \mathrm{x}, \mathrm{y}$ and $z$ are the elements of $A$ and $1$ and $2$ are the elements of $B$

Since $n(A)=3$ and $n(B)=2$

it is clear that $A={x, y, z}$ and $B={1,2}$

Answer :

$ \mathrm{n}(\mathrm{~A} \times \mathrm{A})=9 $

Set $A={-1,0,1}$ Since $(-1,0)$ and $(0,1)$ are elements in $A \times A$ Remaining elements

$ \qquad \ \ ={(-1,-1),(-1,1),(0,-1),(0,0),(1,-1),(1,0),(1,1)} $