Class 11: Relations – Exercise 2.3 – ICSE / ISC / CBSE Mathematics Portal for K12 Students

Question 1: If $A = \{ 1,2,3 \}$ , $B = \{ 4,5,6 \}$ , which of the following are relations from $A$ to $B$ ? Give reasons in support of your answer: (i) $\{(1, 6), (3, 4), (5, 2) \}$ (ii) $\{ (1,5), (2, 6), (3,4), (3,6) \}$ (iii) $\{ (4, 2), (4,3), (5, 1) \}$ {iv} $A \times B$

Answer:

Given $A = \{ 1,2,3 \}$ , $B = \{ 4,5,6 \}$

$A \times B = \{ 1,2,3 \} \times \{ 4,5,6 \} \\ \hspace*{1.2cm} = \{ (1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6) \}$

i) $\{(1, 6), (3, 4), (5, 2) \}$ is not a relation from $A$ to $B$ as it is not a subset of $A \times B$

ii) $\{ (1,5), (2, 6), (3,4), (3,6) \}$ is a subset of $A \times B$ . So it is a relation from $A$ to $B$ .

iii) $\{ (4, 2), (4,3), (5, 1) \}$ is not a relation from $A$ to $B$ as it is not a subset of $A \times B$

iv) $A \times B$ is a relation from $A$ to $B$ .

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Question 2: A relation $R$ is defined from a set $A = \{ 2, 3, 4, 5 \}$ to a set $B = \{ 3, 6, 7, 10 \}$ as follows: $(x, y) \in R \Leftrightarrow x$ is relatively prime to $y$ . Express $R$ as a set of ordered pairs and determine its domain and range.

Answer:

Given $A = \{ 2, 3, 4, 5 \}$ and $B = \{ 3, 6, 7, 10 \}$

Also $(x, y) \in R \Leftrightarrow x$ is relatively prime to $y$

Therefore $2$ is a co-prime to $3$ and $7$

$3$ is a co-prime to $7$ and $10$

$4$ is a co-prime to $3$ and $7$

$5$ is a co-prime to $3, 6$ and $7$

$\therefore R = \{ (2, 3), (2, 7), (3, 7), (3, 10), (4, 3), (4, 7), (5, 3), (5, 6), (5, 7) \}$

Hence Domain of $R = \{ 2, 3, 4, 5 \}$

and Range of $R = \{ 3, 6,7, 10 \}$

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Question 3: Let $A$ be the set of first five natural numbers and let $R$ be a relation on $A$ defined as follows: $(x, y) \in R \Leftrightarrow x \leq y$ . Express $R$ and $R^{-1}$ as sets of ordered pairs. Determine also (i) the domain of $R^{-1}$ (ii) the range of $R$ .

Answer:

Given $A = \{ 1, 2, 3, 4, 5 \}$ $[ \because A$ is the set of first five natural numbers $]$

Also it is given that R is a relation on A defined as $(x, y) \in R \Leftrightarrow x \leq y$

$\therefore R = \{ (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (2, 2), (2, 3), (2, 4), (2, 5), (3, 3), (3, 4), \\ \hspace*{1.2cm} (3, 5), (4, 4), (4, 5) , (5,5) \}$

$R^{-1} = \{ (1, 1), (2, 1), (3, 1), (4, 1), (5, 1), (2, 2), (3, 2), (4, 2), (5, 2), (3, 3), (4, 3), (5, 3), \\ \hspace*{1.2cm} (4, 4), (5, 4), (5, 5) \}$

$\therefore$ i) Domain $(R^{-1}) = \{ 1, 2, 3, 4, 5 \}$

ii) Range $(R) = \{ 1, 2, 3, 4, 5 \}$

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Question 4: Find the inverse relation $R^{ - 1}$ in each of the following cases: (i) $R = \{ (1, 2), (1, 3), (2, 3), (3, 2), (5, 6) \}$ (ii) $R = \{ (x,y) : x , y \in N, x+2y =8 \}$ (iii) $R$ is a relation from $\{ 11, 12,13 \}$ to $\{8, 10, 12 \}$ defined by $y = x - 3$ .

Answer:

i) Given $R = \{ (1, 2), (1, 3), (2, 3), (3, 2), (5, 6) \}$

$\Rightarrow R^{-1} = \{ (2, 1), (3, 1), (3, 2), (2, 3), (6, 5) \}$

ii) Given $R = \{ (x,y) : x , y \in N, x+2y =8 \}$

Now $x + 2y = 8 \Rightarrow x = 8 - 2y$

Putting $y = 1, 2, 3$ we get $x = 6, 4, 2$ respectively

For $y = 4$ , we get $x = 0 \notin N$ and for $y > 4 , x \notin N$

$\therefore R = \{ (6, 1), (4, 2), (2, 3) \}$

Hence $R^{-1} = \{ (1, 6), (2, 4), (3, 2) \}$

iii) Given $R$ is a relation from $\{ 11, 12,13 \}$ to $\{8, 10, 12 \}$ defined by $y = x - 3$

Now, $y = x - 3$

Putting $x = 11, 12, 13$ we get $y = 8, 9, 10$ respectively.

$\Rightarrow (11, 8) \in R, (12, 9) \notin R \ and \ (13, 10) \in R$

$\therefore R = \{ (11, 8), (13, 10) \}$

and $R^{-1} = \{ (8, 11), (10, 13) \}$

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Question 5: Write the following relations as the sets of ordered pairs:
(i) A relation $R$ from the set $\{ 2, 3, 4 , 5, 6 \}$ to the set $\{1, 2, 3\}$ defined by $x =2y$ .
(ii) A relation $R$ on the set $\{1,2,3,4,5,6,7\}$ defined by $(x,y) \in R \Leftrightarrow x$ is relatively prime to $y$ .
(iii) A relation $R$ on the set $\{ 0, 1, 2, ...,10 \}$ defined by $2x + 3y =12$ .
(iv) A relation $R$ from a set $A = \{ 5,6,7,8 \}$ to the set $B= \{ 10,12,15,16,18 \}$ defined by $(x,y) \in R \Leftrightarrow x$ divides $y$ .

Answer:

i) Given $x = 2y$ ,

Putting $y = 1, 2, 3$ we get $x = 2, 4, 6$ respectively

$\therefore R = \{ (2, 1), (4, 2), (6, 1) \}$

ii) Given relation $R$ on the set $\{1,2,3,4,5,6,7\}$ defined by $(x,y) \in R \Leftrightarrow x$ is relatively prime to $y$

Therefore $2$ is relatively prime to $3, 5$ and $7$

$3$ is relatively prime to $2, 4, 5$ and $7$

$4$ is relatively prime to $3, 5$ and $7$

$5$ is relatively prime to $2, 3, 4, 6$ and $7$

$6$ is relatively prime to $5$ and $7$

$\therefore R = \{ (2, 3), (2, 5), (2, 7), (3, 2), (3, 4), (3, 5), (3, 7), (4, 3), \\ \hspace{2.0cm} (4, 5), (4, 7), (5, 2), (5, 3),(5, 4), (5, 6), (5, 7), (6, 5), (6, 7) \}$

iii) Given $2x+ 3y = 12 \Rightarrow 2x = 12 - 3y \Rightarrow x = \frac{1}{2} (12-3y) \}$

Putting $y = 0, 2, 4$ we get $x = 6, 3, 0$ respectively

$\therefore R = \{ (6, 0), (3, 2), (0, 4) \}$

iv) Given a relation $R$ from a set $A = \{ 5,6,7,8 \}$ to the set $B= \{ 10,12,15,16,18 \}$ defined by $(x,y) \in R \Leftrightarrow x$ divides $y$

Here $5$ divides $10$ and $15$

$6$ divides $12$ and $18$

$8$ divides $16$

$\therefore R = \{ (5, 10), (5,15), (6, 12), (6, 18), (8, 16) \}$

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Question 6: Let $R$ be a relation in $N$ defined by $(x,y) \in R \Leftrightarrow x+2y=8$ . Express $R$ and $R^{-1}$ assets of ordered pairs.

Answer:

Given $(x,y) : x , y \in N, x+2y =8$

Now $x + 2y = 8 \Rightarrow x = 8 - 2y$

Putting $y = 1, 2, 3$ we get $x = 6, 4, 2$ respectively

For $y = 4$ , we get $x = 0 \notin N$ and for $y > 4 , x \notin N$

$\therefore R = \{ (6, 1), (4, 2), (2, 3) \}$

Hence $R^{-1} = \{ (1, 6), (2, 4), (3, 2) \}$

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Question 7: Let $A= \{ 3,5 \}$ and $B = \{ 7,11 \}$ . Let $R = \{a,b \}: a \in A , b \in B, a-b \ is \ odd \}$ . Show that $R$ is an empty relation from $A$ into $B$ .

Answer:

Given $A= \{ 3,5 \}$ and $B = \{ 7,11 \}$

Also $R = \{a,b \}: a \in A , b \in B, a-b \ is \ odd \}$

For elements of $A$ and $B, 3-7 = -4, 3-11 = -8, 5-7=-2, 5-11=-6$

They are all even

Therefore R is an empty relation from $A$ to $B$ .

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Question 8: Let $A = \{1, 2\}$ and $B = \{3, 4 \}$ . Find the total number of relations from $A$ into $B$ .

Answer:

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

$\Rightarrow n(A) = 2 \ \ \ \ \ \& \ \ \ \ \ n(B) = 2$

$\Rightarrow n(A) \times n(B) = 2 \times 2 = 4$

$\Rightarrow n(A \times B) = 4$

$[ \because n(A \times B) = n(A) \times n(B) ]$

Therefore there are $2^4 = 16$ relations from $A$ to $B$

$[ \because n(x) = a , n(y) = b \Rightarrow$ total number of relations $= 2^{ab} ]$

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Question 9: Determine the domain and range of the relation R defined by:

(i) $R = \{ (x,x +5) : x \in \{0, 1,2, 3, 4,5 \} \}$

(ii) $R = \{ ( x, x^3 ) : x \ is \ a \ prime \ number \ less \ than \ 10 \}$

Answer:

i) Given $R = \{ (x,x +5) : x \in \{0, 1,2, 3, 4,5 \} \}$

Therefore for elements given we get

$R = \{ (0, 5), (1, 6), (2, 7), (3, 8), (4, 9), (5, 10) \}$

Therefore Domain $(R) = \{ 0, 1, 2, 3, 4, 5 \}$

Range $(R) = \{5, 6, 7, 8, 9, 10 \}$

ii) Given $R = \{ ( x, x^3 ) : x \ is \ a \ prime \ number \ less \ than \ 10 \}$

$\therefore x = 2, 3, 5, 7$

$\Rightarrow R = \{ (2, 8), (3, 27), (5, 125), (7, 343) \}$

Therefore Domain $(R) = \{ 2, 3, 5, 7 \}$

Range $(R) = \{ 8, 27, 125, 343 \}$

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Question 10: Determine the domain and range of the following relations

(i) $R = \{ (a, b): a \in N, a < 5, b = 4 \}$

ii) $S = \{ (a, b) : b = |a-1|, a \in Z \ and \ |a| \leq 3 \}$

Answer:

i) Given $R = \{ (a, b): a \in N, a < 5, b = 4 \}$

$\Rightarrow a = 1, 2,3 ,4$ $\& \ \ \ b = 4$

$\therefore R = \{ (1, 4), (2, 4), (3 ,4), (4,4) \}$

Therefore Domain $(R) = \{ 1, 2, 3, 4 \}$

Range $(R) = \{ 4 \}$

ii) Given $S = \{ (a, b) : b = |a-1|, a \in Z \ and \ |a| \leq 3 \}$

$\Rightarrow a = -3, -2, -1, 0, 1, 2, 3 \therefore b = 4, 3, 2, 1, 0, 1, 2$ respectively

$\therefore S = \{(-3, 4), (-2, 3), (-1, 2), ), (0, 1), (1, 0), (2, 1), (3, 2) \}$

Therefore Domain $(S) = \{ -3, -2, -1, 0, 1, 2, 3 \}$

Range $(S) = \{ 0, 1, 2, 3, 4 \}$

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Question 11: Let $A = \{ a,b \}$ . List all relations on $A$ and find their number.

Answer:

Given $A = \{ a,b \}$

We know that number of relations $= 2^{mn} = 2^4 = 16$

Ordered pairs $= (a, a), (a, b), (b, a), (b, b)$

$\therefore R = \Big\{ \{ (a, a), (a, a) \}, \{ (a, a), (a, b) \}, \{ (a, a), (b, a) \}, \{ (a, a), (b, b) \}, \{ (a, b), (a, a) \}, \\ \hspace*{1.5cm} \{ (a, b), (a, b) \}, \{ (a, b), (b, a) \}, \{ (a, b), (b, b) \}, \{ (b, a), (a, a) \}, \{ (b, a), (a, b) \}, \\ \hspace*{1.5cm} \{ (b, a), (b, a) \}, \{ (b, a), (b, b) \}, \{ (b, b), (a, a) \}, \{ (b, b), (a, b) \}, \{ (b, b), (b, a) \}, \\ \hspace*{1.5cm} \{ (b, b), (b, b) \} \Big\}$

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Question 12: Let $A = \{ x, y, z \}$ and $B = \{ a, b \}$ . Find the total number of relations from $A$ into $B$ .

Answer:

Given $A = \{ x, y, z \}$ and $B = \{ a, b \}$

$\Rightarrow n(A) = 3$ and $n(B) = 2$

$\Rightarrow n(A) \times n(B) = 3 \times 2 = 6$

$\Rightarrow n(A \times B) = 6$

Therefore there are $2^6 = 64$ relations from $A$ to $B$

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Question 13: Let $R$ be a relation from $N$ to $N$ defined by $R = \{ (a,b):a,b \in N$ and $a=b^2$ . Are the following statements true?

(i) $(a,a) \in R$ for all $a \in N$

(ii) $(a,b) \in R \Rightarrow (b,a) \in R$

(iii) $(a,b) \in R$ and $(b,c) \in R \Rightarrow (a,c) \in R$

Answer:

Given $R = \{ (a,b):a,b \in N$ and $a=b^2 \}$

i) $(a,a) \in R$ for all $a \in N$

False: Statement is not true because $(2, 2) \notin R$

ii) $(a, b) \in R \Rightarrow (b, a) \in R$

False: The statement is not true because $(4, 2) \in R$ but $(2, 4) \notin R$

iii) $(a, b) \in R$ and $(b, c) \in R \Rightarrow (a, c) \in R$

False: Statement is not true because $(36, 6) \in R$ and $(25, 5) \in R$ but $(36, 5) \notin R$

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Question 14: Let $A= \{ 1,2,3,...,14 \}$ . Define a relation on a set $A$ by $R = \{ (x,y):3x-y=0, \ where \ x,y \in A \}$ . Depict this relationship using an arrow diagram. Write down its domain, co-domain and range.

Answer:

Given $3x-y = 0 \Rightarrow y = 3x$

Putting $x = 1, 2, 3, 4$ then $y = 3, 6, 9, 12$ respectively

For $x > 4, y > 14$ which does not belong to $A$

$\therefore R = \{ (1, 3), (2, 6), (3, 9), (4, 12) \}$

Therefore Domain $(R) = \{ 1, 2, 3, 4 \}$

Range $(R) = \{ 3, 6, 9, 12\}$

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Question 15: Define a relation $R$ on the set $N$ of natural numbers by $R=\{(x,y):y=x+5, \ x \ is \ a \ natural \ number \ less \ than \ 4,x,y \in N \}$ Depict this relationship using (i) roster form (ii) an arrow diagram. Write down the domain and range or R.

Answer:

Given $R=\{(x,y):y=x+5, \ x \ is \ a \ natural \ number \ less \ than \ 4,x,y \in N \}$

i) Putting $x = 1, 2,3$ we get $y = 6, 7, 8$ respectively.

$\therefore R = \{ (1, 6), (2, 7), (3, 8) \}$

ii) Domain $(R) = \{ 1, 2, 3 \}$

Range $(R) = \{ 6, 7, 8 \}$

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Question 16: $A = \{1,2,3,5 \}$ and $B= \{4,6,9 \}$ . Define a relation $R$ from $A$ to $B$ by $R=\{(x,y): \ the \ difference \ between x \ and \ y \ is \ odd, \ x \in A, \ y \in B \}$ . Write $R$ in Roster form.

Answer:

Given $A = \{1,2,3,5 \}$ and $B= \{4,6,9 \}$

Also $R=\{(x,y): \ the \ difference \ between x \ and \ y \ is \ odd, \ x \in A, \ y \in B \}$

For $x = 1, y = 4, 6$

For $x =2, y = 9$

For $x = 3, y = 4, 3$

For $x = 5, y = 4, 6$

$\therefore R = \{ (1, 4), (1, 6), (2, 9), (3, 4), (3, 3), (5, 4), (5, 6) \}$

$\\$

Question 17: Write the relation $R=\{(x,x^3):x \ is \ a \ prime \ number \ less \ than 10 \}$ in roster form.

Answer:

Given $R=\{(x,x^3):x \ is \ a \ prime \ number \ less \ than 10 \}$

$\therefore x = 2, 3, 5, 7$

Hence $R = \{ (2, 8), (3, 27), (5, 125), (7, 343) \}$

$\\$

Question 18: Let $A = \{ 1,2, 3, 4,5, 6 \}$ . Let $R$ be a relation on $A$ defined by $\{(a,b):a,b \in A,b \ is \ exactly \ divisible \ by a \}$ (i) Write $R$ in roster form (ii) Find the domain of $R$ (iii) Find the range of $R$

Answer:

Given $A = \{ 1,2, 3, 4,5, 6 \}$ and $R$ be a relation on $A$ defined by $\{(a,b):a,b \in A, b \ is \ exactly \ divisible \ by \ a \}$

For $a = 1, b = 1, 2, 3, 4, 5, 6$

For $a = 2, b = 2, 4, 6$

For $a = 3, b = 3, 6$

For $a = 4, b = 4$

For $a = 5, b = 5$

For $a = 6, b = 6$

i) $R = \{ (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 2), (2, 4), (2, 6), (3, 3), \\ \hspace*{1.0cm} (3, 6), (4, 4), (5, 5) , (6, 6) \}$

ii) Domain $(R) = \{ 1, 2, 3, 4, 5, 6 \}$

iii) Range $(R) = \{ 1, 2, 3, 4, 5, 6 \}$

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Question 19: Figure below shows a relationship between the sets $P$ and $Q$ . Write this relation in (i) set builder form (ii) roster form. What is its domain and range?

Answer:

Given $R = \{ (x, y) : y = x-2, x \in P , y \in Q \}$

$P = \{ 5, 6, 7 \}$ and $Q = \{ 3, 4, 5 \}$

When $x = 5 \Rightarrow y = 3$

$x = 6 \Rightarrow y = 4$

$x = 7 \Rightarrow y = 5$

$\therefore R = \{ (5, 3), (6, 4), (7, 5) \}$

Therefore Domain $(R) = \{ 5, 6, 7 \}$

and Range $(R) = \{ 3, 4, 5 \}$

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Question 20: Let $R$ be the relation on $Z$ defined by $R= \{(a,b): a,b \in Z, a-b \ is \ an \ integer \}$ . Find the domain and range of $R$ .

Answer:

Given $R= \{(a,b): a,b \in Z, a-b \ is \ an \ integer \}$

Difference between two integers would also be an integer.

Therefore Domain $(R) = Z$

And Range $(R) = Z$

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Question 21: For the relation $R_1$ defined on $R$ by the rule $(a,b) \in R_1 \Leftrightarrow 1 + ab>0$ . Prove that $(a,b) \in R_1$ and $(b , c) \in R_1 \Rightarrow (a, c) \in R_1$ is not true for all $a, b, c \in R$ .

Answer:

To prove: $(a,b) \in R_1$ and $(b , c) \in R_1 \Rightarrow (a, c) \in R_1$ is not true for all $a, b, c \in R$ .

Given $R_1 = \{ (a, b): 1 + ab > 0 \}$

Let $a = 1, b = -0.5, c = -4$

Here, $(1, -0.5) \in R_1 \ \ \ [ \because 1+(1 \times -0.5) = 0.5 > 0 ]$

And, $(-0.5, -4) \in R_1 \ \ \ [ \because 1+(-0.5 \times -4) = 3 > 0]$

But, $(1, -4) \notin R_1 \ \ \ [ \because 1+(1\times -4) = -3 < 0]$

$\therefore (a, b) \in R_1$ and $(b,c) \in R_1 \Rightarrow (a, c) \in R_1$ is not true for all $a, b, c \in R$

Hence Proved.

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Question 22: Let $R$ be a relation on $N \times N$ defined by $(a,b) R (c,d) \Leftrightarrow a+d=b+c$ for all $(a,b),(c,d) \in N \times N$ . Show that:

(i) $(a,b) \ R \ (a,b)$ for all $(a, b) \in N \times N$

(ii) $(a, b) \ R \ (c, d) \Rightarrow (c, d) \ R \ (a, b)$ for all $(a, b), (c, d) \in N \times N$

(iii) $(a,b) \ R \ (c,d)$ and $(c,d) \ R \ ( e,f) \Rightarrow (a,b) \ R \ (e,f)$ for all $(a,b),(c,d),(e,f) \in N \times N$

Answer:

Given $(a,b) R (c,d) \Leftrightarrow a+d=b+c$ for all $(a,b),(c,d) \in N \times N$

i) $(a,b) \ R \ (a,b)$ for all $(a, b) \in N \times N$

$\because (a+b) = (b + a)$ for all $a, b \in N$

$\therefore (a, b) \ R \ (a, b)$ for all $a, b \in N$

ii) $(a, b) \ R \ (c, d) \Rightarrow (c, d) \ R \ (a, b)$ for all $(a, b), (c, d) \in N \times N$

$(a, b) R (c, d) \Rightarrow a + d = b + c$

$\Rightarrow c + d = d + a$

$\Rightarrow (c, d) \ R \ (a, b)$

iii) $(a,b) \ R \ (c,d)$ and $(c,d) \ R \ ( e,f) \Rightarrow (a,b) \ R \ (e,f)$ for all $(a,b),(c,d),(e,f) \in N \times N$

$\therefore (a+b) \ R \ (c, d)$ and $(c, d) \ R \ ( e, f)$

$\Rightarrow a + d = b + c$ and $c + f = d + e$

$\Rightarrow a + d + c + f = b + c + d + e$

$\Rightarrow a + f = b + e$

$\Rightarrow (a, b) \ R \ (e, f)$

$\\$

Question 23: Find the linear relation between the components of the ordered pairs of the relation $R$ , where (i) $R = \{ (0, 2), (- 1,5), (2, -4), \ldots \}$ (ii) $R = \{ (- 1, -1), (0,2), (1, 5), \ldots \}$

Answer:

i) Given $R = \{ (0, 2), (- 1,5), (2, -4), \ldots \}$

$x$ and $y$ are related to each other by linear equation

Let the relation be defined as $y = mx + c$

For $(0, 2)$ we get $c = 2$

For $(-1, 5)$ we get $5 = -m + 2 \Rightarrow m = -3$

Therefore $y = -3x + 2$ or $y = 2 - 3x$

ii) Given $R = \{ (- 1, -1), (0,2), (1, 5), \ldots \}$

$x$ and $y$ are related to each other by linear equation

Let the relation be defined as $y = mx + c$

For $( 0, 2)$ we get $2 = 0 + c \Rightarrow c = 2$

For $(-1, -1)$ we get $-1 = -m + 2 \Rightarrow m = 3$

Therefore $y = 3x + 2$

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Question 24: If $A = \{ 1, 3,5, 6 \}$ and $B = \{ 3,4,5 \}$ , write the relation $R$ as a set of ordered pairs, if
(i) $R = \{(x,y) : (x, y) \in A \times B, x+y \ is \ even \}$