Chapter 2 Relations And Functions EXERCISE 2.1
EXERCISE 2.1
1. If $(\dfrac{x}{3}+1, y-\dfrac{2}{3})=(\dfrac{5}{3}, \dfrac{1}{3})$, find the values of $x$ and $y$.
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Answer :
It is given that $(\dfrac{x}{3}+1, y-\dfrac{2}{3})=(\dfrac{5}{3}, \dfrac{1}{3})$
Since the ordered pairs are equal, the corresponding elements will also be equal.
Therefore, $\dfrac{x}{3}+1=\dfrac{5}{3}$ and $y-\dfrac{2}{3}=\dfrac{1}{3}$.
$\dfrac{x}{3}+1=\dfrac{5}{3}$
$\Rightarrow \dfrac{x}{3}=\dfrac{5}{3}-1 \quad y-\dfrac{2}{3}=\dfrac{1}{3}$
$\Rightarrow \dfrac{x}{3}=\dfrac{2}{3} \Rightarrow y=\dfrac{1}{3}+\dfrac{2}{3}$
$\Rightarrow x=2 \quad \Rightarrow y=1$
$\therefore x=2$ and $y=1$
2. If the set $A$ has 3 elements and the set $B=\{3,4,5\}$, then find the number of elements in $(A \times B)$.
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Answer :
It is given that set A has 3 elements and the elements of set B are 3, 4, and 5 .
$\Rightarrow$ Number of elements in set $B=3$
Number of elements in $(A \times B)$
$=($ Number of elements in $A) \times($ Number of elements in $B)$
$=3 \times 3=9$
Thus, the number of elements in $(A \times B)$ is 9 .
3. If $G=\{7,8\}$ and $H=\{5,4,2\}$, find $G \times H$ and $H \times G$.
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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
$P \times Q=\{(p, q): p \in P, q \in Q\}$
$\therefore G \times H=\{(7,5),(7,4),(7,2),(8,5),(8,4),(8,2)\}$
$H \times G=\{(5,7),(5,8),(4,7),(4,8),(2,7),(2,8)\}$
4. State whether each of the following statements are true or false. If the statement is false, rewrite the given statement correctly.
(i) If $P=\{m, n\}$ and $Q=\{n, m\}$, then $P \times Q=\{(m, n),(n, m)\}$.
(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 $y \in B$.
(iii) If $A=\{1,2\}, B=\{3,4\}$, then $A \times(B \cap \phi)=\phi$.
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Answer :
(i) False
If $P=\{m, n\}$ and $Q=\{n, m\}$, then
$P \times Q=\{(m, m),(m, n),(n, m),(n, n)\}$
(ii) True
(iii) True
5. If $A=\{-1,1\}$, find $A \times A \times A$.
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Answer :
It is known that for any non-empty set $A, A \times A \times A$ is defined as
$A \times A \times A=\{(a, b, c): a, b, c \in A\}$
It is given that $A=\{-1,1\}$
$\therefore A \times A \times A=\{(-1,-1,-1),(-1,-1,1),(-1,1,-1),(-1,1,1)$,
$(1,-1,-1),(1,-1,1),(1,1,-1),(1,1,1)\}$
6. If $A \times B=\{(a, x),(a, y),(b, x),(b, y)\}$. Find $A$ and $B$.
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Answer :
It is given that $A \times B=\{(a, x),(a, y),(b, x),(b, y)\}$
We know that the Cartesian product of two non-empty sets $P$ and $Q$ is defined as $P \times Q=\{(p, q): p \in P, q \in Q\}$
$\therefore A$ is the set of all first elements and $B$ is the set of all second elements.
Thus, $A=\{a, b\}$ and $B=\{x, y\}$
7. Let $A=\{1,2\}, B=\{1,2,3,4\}, C=\{5,6\}$ and $D=\{5,6,7,8\}$. Verify that (i) $A \times(B \cap C)=(A \times B) \cap(A \times C)$. (ii) $A \times C$ is a subset of $B \times D$.
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Answer :
(i) To verify: $A \times(B \cap C)=(A \times B) \cap(A \times C)$
We have $B \cap C=\{1,2,3,4\} \cap\{5,6\}=\Phi$
$\therefore$ L.H.S. $=A \times(B \cap C)=A \times \Phi=\Phi$
$A \times B=\{(1,1),(1,2),(1,3),(1,4),(2,1),(2,2),(2,3),(2,4)\}$
$A \times C=\{(1,5),(1,6),(2,5),(2,6)\}$
$\therefore$ R.H.S. $=(A \times B) \cap(A \times C)=\Phi$
$\therefore$ L.H.S. $=$ R.H.S
Hence, $A \times(B \cap C)=(A \times B) \cap(A \times C)$
(ii) To verify: $A \times C$ is a subset of $B \times D$
$A \times C=\{(1,5),(1,6),(2,5),(2,6)\}$
$B \times D=\{(1,5),(1,6),(1,7),(1,8),(2,5),(2,6),(2,7),(2,8),(3,5),(3,6),(3,7),(3,8),(4,5),(4,6),(4,7),(4,8)\}$
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 $B \times D$.
8. Let $A=\{1,2\}$ and $B=\{3,4\}$. Write $A \times B$. How many subsets will $A \times B$ have? List them.
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Answer :
$A=\{1,2\}$ and $B=\{3,4\}$
$\therefore A \times B=\{(1,3),(1,4),(2,3),(2,4)\}$
$\Rightarrow n(A \times B)=4$
We know that if $C$ is a set with $n(C)=m$, then $n[P(C)]=2^{m}$.
Therefore, the set $A \times B$ has $2^{4}=16$ subsets. These are
$\Phi,\{(1,3)\},\{(1,4)\},\{(2,3)\},\{(2,4)\},\{(1,3),(1,4)\},\{(1,3),(2,3)\}$,
$\{(1,3),(2,4)\},\{(1,4),(2,3)\},\{(1,4),(2,4)\},\{(2,3),(2,4)\}$,
$\{(1,3),(1,4),(2,3)\},\{(1,3),(1,4),(2,4)\},\{(1,3),(2,3),(2,4)\}$,
$\{(1,4),(2,3),(2,4)\},\{(1,3),(1,4),(2,3),(2,4)\}$
9. Let $A$ and $B$ be two sets such that $n(A)=3$ and $n(B)=2$. If $(x, 1),(y, 2),(z, 1)$ are in $A \times B$, find $A$ and $B$, where $x, y$ and $z$ are distinct elements.
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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 the ordered pair elements of $A \times B$.
$\therefore x, y$, and zare 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 $B=\{1,2\}$.
10. The Cartesian product $A \times A$ has 9 elements among which are found $(-1,0)$ and $(0,1)$. Find the set $A$ and the remaining elements of $A \times A$.
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Answer :
We know that if $n(A)=p$ and $n(B)=q$, then $n(A \times B)=p q$.
$\therefore n(A \times A)=n(A) \times n(A)$
It is given that $n(A \times A)=9$
$\therefore n(A) \times n(A)=9$
$\Rightarrow n(A)=3$
The ordered pairs $(-1,0)$ and $(0,1)$ are two of the nine elements of $A \times A$.
We know that $A \times A=\{(a, a): a \in A\}$. Therefore, $-1,0$, and 1 are elements of $A$.
Since $n(A)=3$, it is clear that $A=\{-1,0,1\}$.
The remaining elements of set $A \times A$ are $(-1,-1),(-1,1),(0,-1),(0,0)$,
$(1,-1),(1,0)$, and $(1,1)$
