# RELATION AND MAPPING : Part -1

In  this chapter, we will discuss about Relations and Mapping (Function) related mathematical problems and their solutions as a part of our S.N.dey Math solution series (For class XI). We hope that this discussions will be helpful for numerous students across the country.

## Hence, from (1) and (2), the result follows.

6. If $\,A=(0,1),\,\,$ find $\,(i)\, A \times A \quad (ii) A \times A \times A$

Sol. (i) $\,\,A \times A \\=(0,1) \times (0,1)\\=\{(0,0),(0,1),(1,0),(1,1)\} \\ (ii)\,\,A \times A \times A\\=(0,1)\times(0,1) \times(0,1)\\=(0,1)\times \{(0,1)\times(0,1)\}\\= (0,1)\times \{(0,0),(0,1),(1,0),(1,1)\} \\=\{(0,0,0),(0,0,1),(0,1,0),(0,1,1),\\(1,0,0),(1,0,1),(1,1,0),(1,1,1)\}$

## Short Answer Type Questions :  [Exercise: 2A]

1. If $\,\, A \times B=\{(1,2),(3,4),(5,2),(1,4),\\(3,2),(5,4)\},\,\,\,$
find $\,\,B \times A.$

Sol. $\,\, B \times A=\{(2,1),(4,3),(2,5), \\(4,1),(2,3),(4,5)\}.\,\,\,$

2. If $\,\, P \times Q= \{(2,-1),(3,0),(2,1),(3,1),\\(2,0),(3,-1),\}\,\,$
find $\,\, P, \,\,Q.$

Sol. $\,\,P=\{x: (x,y) \in P \times Q\} \\ \therefore P=\{2,3\}$
and $\,\, Q=\{y: (x,y) \in P \times Q\} \\ \therefore Q=\{-1,0,1\}$

3. If $\, A=\{x : x \in \mathbb{N} \wedge 1 <x \leq 3\}$ and
$B= \{x : x \in \mathbb{Z} \wedge -2 <x <2\},$ find $\,\,B \times A.$

Sol. Since $\, A=\{x : x \in \mathbb{N} \wedge 1 <x \leq 3\} \\ \implies A=\{2,3\}$

Again since by question we have $B= \{x : x \in \mathbb{Z} \wedge -2 <x <2\}, \\ \implies B=\{-1,0,1\}$

Now, $\, B \times A =\{(x,y): x \in B \wedge y \in A\} \\ \implies B \times A \\=\{(-1,2),(-1,3),(0,2),(0,3),(1,2),(1,3)\}$

4. Let $\,\,A=\{x: x \in \mathbb Z \wedge -1 < x \leq 1\}, \\B=\{x: x \in \mathbb N \wedge 1 <x<5\}, \\ ​C=\{x:x \,\, \text{is an odd positive integer x and}\\ 1<x \leq6\}\,$
​then show that,  (i)$\,\,A \times (B \cap C)=(A \times B) \cap (A \times C) \\ (ii) (A \times B) \cup (A \times C)= A \times (B \cup C)$

Sol. $\,\,A=\{x: x \in \mathbb Z \wedge -1 < x \leq 1\} \\ \implies A=\{0,1\} \\ B=\{x: x \in \mathbb N \wedge 1 <x<5\} \\ \implies B=\{2,3,4\} \\ ​C=\{x:x \,\, \text{is an odd positive integer x and}\\1<x \leq6\} \\ \implies C=\{3,5\}$

Now, $\, A \times B \cap C=\{3\} \\ (B \cap C)=\{(x,y) : X \in A, y \in B \cap C\} \\ =\{0,1\} \times \{3\}\\=\{(0,3),(1,3)\} \cdots (1)$

Again, $A \times B=\{(x,y): x \in A \wedge y \in B\} \\ A \times B=\{(0,2),(0,3),(0,4),\\(1,2),(1,3),(1,4)\} \\ A \times C=\{(x,y): x \in A \wedge y \in C \}\\=\{(0,3),(0,5),(1,3),(1,5)\} \\ (A \times B)\cap (A \times C) =\{(0,3),(1,3)\} \cdots (2)$

Hence, from (1) and (2), it follows that $\,\,A \times (B \cap C)=(A \times B) \cap (A \times C)$

(ii)$A \times B=\{(0,2),(0,3),(0,4), \\(1,2),(1,3),(1,4)\} \\ C=\{(0,3),(0,5),(1,3),(1,5)\} \\ \therefore (A \times B) \cup (A \times C)\\=\{(0,2),(0,3),(0,4),\\(0,5),(1,2),(1,3),(1,4),(1,5)\} \cdots (3) \\ B \cup C =\{2,3,4,5\} \\ A \times (B \cup C)\\=\{(x,y): x \in A \wedge y \in (B \cup C)\}\\=\{(0,2),(0,3),(0,4),(0,5),(1,2),(1,3),\\(1,4),(1,5)\} \cdots (4)$

Hence, from (3) and (4), the result follows.

5. If $\,\,A = \{1, 2, 3\}$ and  $\,B=(6, 7)\,$ find the number of subsets of the set $\,A \times B.$

Sol. $\,\,A \times B=\{(x,y): x \in A \wedge y \in B\} \\ A \times B=\{(1,6),(1,7),(2,6),(2,7),\\(3,6),(3,7)\}$

So, $n(A \times B)=$ the no. of elements of the set $\,\,(A \times B)=6$

Hence, the total number of subsets of $(A \times B)=2^6=64$

6. The cartesian product $\,P \times P$ has $\,9\,$ elements; if its two
elements are $\,(-3,-2) , (-2,-1)\,$,
find the remaining terms of  $\,P \times P$

Sol. $\,\,n(P \times P)=9 \\ \therefore n(P)=3$
Since $\,(-3,-2) , (-2,-1) \in P \times P \\ \therefore \{-1,-2,-3\} \subseteq P$
But, since  $\,\,n(P)=3,\,\,\{-1,-2,-3\}=P \\ P \times P=\{(x,y): x \in P \wedge y \in P\}\\=\{(-1,-1),(-1,-2),(-1,-3),(-2,-1),\\(-2,-2),(-2,-3),(-3,-1),(-3,-2),\\(-3,-3)\}$

So, the remaining terms of  $\,P \times P =\{(-1,-1),(-1,-2),(-1,-3),\\(-2,-2),(-2,-3),(-3,-1),(-3,-3)\}$

7. Let $A = \{x:x \in \mathbb N \wedge \, \text{ x is a prime} \in [10, 19]\}$
and $\,B=\{2, 3\};\,$ find  $\,A \times B.$

Sol. Since,  $A = \{x:x \in \mathbb N \, \text{and x is a prime } \in [10, 19]\} \\=\{11,13,17,19\} \\ A \times B=\{(x,y): x \in A \wedge y \in B\}\\=\{11,13,17,19\} \times \{2,3\}\\=\{(11,2),(13,2),(17,2),(19,2),\\(11,3),(13,3),(17,3),(19,3)\}$

8. Two sets  $\,A\,$ and $\,B\,$ have $\,4\,$ common elements. If  $\,n(A) = 6\,$ and
$\,n(B)=7\,$ then find the values of $\,n(A \times B),n[(A \times B)\cap (B \times A)].$

Sol. By problem, $n(A \times B)=n(A)n(B)= 6 \times 7=42$.
Since two sets  $\,A\,$ and $\,B\,$ have $\,4\,$ common elements,  so the sets
$n(A \times B),\,n(B \times A)$ have $\,4 \times 4=16$ common elements.
Hence,$\,n(A \times B),n[(A \times B)\cap (B \times A)]=16.$

9. In each of the following cases state whether the given
statement is true or false :
(i) If $\,A = \{1, 2, 3\}, B = \{4, 5\}\,$, then, $\,A \times (B \cup \phi)=\phi \,$ where is $\,\phi\,$ is the null set.

Sol. $B \cup \phi = \{4, 5\} \cup \phi \\=\{4,5\}\,$
and so  $\,A \times (B \cup \phi)\\=\{1,2,3\} \times \{4,5\}\\ \neq \phi$

So, the given statement is false.

(ii) If $\,X=\{a,b,c\}, Y=\{c,a,b\},\,\,$ then $X \times Y=Y \times X$

Sol. $\,X \times Y=\{(x,y): x \in X \wedge y \in Y\}\\=\{(a,c),(a,a),(a,b),(b,c),(b,a)\\,(b,b),(c,c),(c,a),(c,b)\} \cdots (1)\\ Y \times X=\{(c,a),(c,b),(c,c),(a,a),(a,b),\\(a,c),(b,a),(b,b),(b,c)\} \cdots (2)$

So, from (1) and (2), it follows that the given statement is true.

(iii) If $\,\, A=\{3,4,5\} ,\,B=\{1,2\},\,$ then $\,\, A \times (B \cap \phi)=\phi,\,$ where $\,\,\phi\,$ is a null set.

Sol. $\,B \cap \phi \\=\{1,2\} \cap \phi \\=\phi$
Hence, $\,A \times (B \cap \phi)\\=A \times \phi\\=\phi$
So, the given statement is true.

(iv) If $\,A=\{1,0,-1\},$ then the value of $\,n(A \times A \times A)=\phi,\,\, \phi,$ being a null set.

Sol. $\,A=\{1,0,-1\},\\\ \implies n(A)=3 \\ n(A \times A \times A)\\=n(A) \times n(A) \times n(A)\\=3 \times 3 \times 3\\=27 \neq \phi$

So, the given statement is false.

10. For any three sets $\,\,A,B,C\,\,$ prove that $\,\, A \times (B-C) =(A \times B)-(A \times C)$

Sol. Let $\,\,(x,y) \in A \times (B-C) \\ \implies x \in A \wedge y \in (B-C) \\ \implies x \in A \wedge (y \in B \wedge y \not \in C) \\ \implies (x \in A \wedge y \in B)\wedge (x \in A \wedge y \not\in C) \\ \implies (x,y)\in A \times B \wedge (x,y) \not \in A \times C \\ \implies (x,y) \in (A \times B) - (A \times C) \\ \therefore A \times (B-C) \subseteq (A \times B)- (A \times C) \cdots (1)$

Again, let us suppose
$\,\, (a,b) \in (A \times B)- (A \times C) \\ \implies (a,b) \in A \times B \wedge (a,b) \not \in A \times C \\ \implies (a \in A \wedge b \in B)\wedge (a \in A \wedge b \not \in C) \\ \implies a \in A \wedge (b \in B \wedge b \not \in C) \\ \implies a \in A \wedge b \in (B-C) \\ (a,b) \in A \times(B-C) \\ \implies (A \times B)- (A \times C) \subseteq A \times (B-C) \cdots (2)$

From (1) and (2), we can conclude that  $\,\, A \times (B-C) =(A \times B)-(A \times C).$ (proved)

11. If $\,\,n(A \times B \times C)=60,\,\,n(B)=4,\,\,n(C)=3,\,\,$ find the value of $\,n(A).$

Sol. $\,\,n(A \times B \times C)=60 \\ \implies n(A).n(B).n(C)=60 \\ \implies n(A) \times 4 \times 3=60 \\ \implies n(A)= \frac{60}{4 \times 3}=5$

To be continued....