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)\} $