Class 12: Relations – Miscellaneous Problems – ICSE / ISC / CBSE Mathematics Portal for K12 Students

Question 1: Three relations $R_1, R_2 \text{ and } R_3$ are defined on set $A = \{ a, b, c \}$ as follows:

(i) $R_1 = \{ (a, a), (a, b), (a, c), (b, b), (b, c), (c, a), (c, b), (c, c) \}$

(ii) $R_2= \{ (a, b), (b, a), (a, c), (c, a) \}$

(iii) $R_3= \{ ( a, b), (b, c), (c, a) \}$

Find whether each one of $R_1, R_2 \text{ and } R_3$ is reflexive, symmetric and transitive.

Answer:

(i) Reflexive: Clearly, $(a, a), (b, b) , (c, c) \in R_1$ . Hence, $R_1$ is reflexive on $A$ .

Symmetric: We observe that $( a, b) \in R_1 \text{ but } ( b, a) \notin R_1. R_1$ is not a symmetric relation on $A$ .

Transitive: We find that $(b, c) \in R_1 \text{ and } ( c, a) \in R_1 \text{ but } ( b, a) \notin R_1$ . Hence, $R1$ is not a transitive relation on $A$ .

(ii) Reflexive: Since $( a, a), ( b, b) \text{ and } ( c, c)$ are not in $R_2.$ Hence, it is not a reflexive relation on $A.$

Symmetric: We find that the ordered pairs obtained by interchanging the components of ordered pairs in $R_2$ are also in $R_2$ . Hence, $R_2$ is a symmetric relation on $A$ .

Transitive: Clearly, $(a, b) \in R_2 \text{ and } ( b, a) \in R_2 \text{ but } (a, a) \notin R_2.$

Hence, it is not a transitive relation on $R_2$ .

(iii) Reflexive: Since none of $(a, a), (b, b) \text{ and } (c, c)$ is an element of $R_3$ . Hence, $R_3$ is not reflexive on $A$ .

Symmetric: Clearly, $( b, c) \in R_3 \text{ but } ( c, b) \notin R_3$ . Hence, $R_3$ is not a symmetric relation on $A$ .

Transitive: Clearly, $(a, b) \in R_3 \text{ and } ( b, c) \in R_3 \text{ but } ( a, c) \notin R_3$ . Hence, $R_3$ is not a transitive relation on $A$ .

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Question 2: Show that the relation $R$ on the set $A = \{ 1, 2, 3 \}$ given by $R = \{ (1, 1), (2, 2), ( 3, 3) , (1, 2), (2, 3) \}$ is reflexive but neither symmetric nor transitive.

Answer:

Since $1,2, 3 \in A \text{ and } (1,1), (2,2), (3,3) \in R \text{ i.e. for each } a \in A,(a, a) \in R.$ Hence, $R$ is reflexive.

We observe that $(1, 2) \in R \text{ but } (2, 1) \in R.$ Hence, $R$ is not symmetric.

Also, $(1, 2) \in R \text{ and } (2, 3) \in R \text{ but } (1, 3) \notin R.$ Hence, $R$ is not transitive.

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Question 3: Show that the relation $R$ on the set $A = \{1 ,2, 3 \}$ given by $R = \{ (1.,2), (2,1) \}$ is symmetric but neither reflexive nor transitive.

Answer:

We observe that $(1, 1), (2,2) \text{ and } (3, 3)$ do not belong to $R.$ Hence, $R$ is not reflexive.

Clearly, $(1,2) \in R \text{ and } (2,1) \in R.$ Hence, $R$ is symmetric.

As $(1, 2) \in R \text{ and } (2,1) \in R \text{ but } (1, 1) \notin R.$ Hence, $R$ is not transitive

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Question 4: Check the following relations $R \text{ and } S$ for reflexivity, symmetry and transitivity:

$\text{(i) } aRb \text{ iff } \ b \text{ is divisible by } a, a, b \in N$

$\text{(ii) } l_1 \ S \ l_2 \text{ iff } \ l_1 \perp l_2, \text{ where } l_1 \text{ and } l_2 \text{ are straight lines in a plane. }$

Answer:

$\text{(i) We have, } aRb \Leftrightarrow a | b \text{ for all } a, b \in N.$

$\text{Reflexivity: For any } a \in N, \text{ we have } a | a \Leftrightarrow aRa.$

$\text{Thus, } aRa \text{ for all } a \in N. \text{ Hence, } R \text{ is reflexive on } N.$

Symmetry: $R$ is not symmetric because if $a|b,$ then b may not divide $a$ . For example, $2|6$ but $6|2.$

Transitivity: Let $a, b, c \in N$ such that $aRb$ and $bRc.$ Then,

$aRb \text{ and } bRc \Rightarrow a| b \text{ and } b|c \Rightarrow a|c \Rightarrow aRc$

Hence, $R$ is a transitive relation on $N.$

(ii) Let $L$ be the set of ail lines in a plane. We are given that

$l_1 \ S \ l_2 \Leftrightarrow l_1 \perp l_2 \text{ for all } l_1, l_2 \in L.$

Reflexivity: $S$ is not reflexive because a line cannot be perpendicular to itself i.e. $l \perp l$ is not true.

Symmetry: Let $l_1, l_2 \in L \text{ such that } l_1 \ S \ l_2.$ Then,

Hence, $S$ is symmetric on $L.$

Transitive: $S$ is not transitive, because $l_1 \perp l_2$ and $l_2 \perp l_3$ does not imply that $l_1 \perp l_3.$

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Question 5: Let a relation $R_1$ on the set $R$ of real numbers be defined as $(a,b) \in R_1 \Leftrightarrow 1 + ab >0$ for all $a, b \in R.$ Show that $R_1$ is reflexive and symmetric but not transitive.

Answer:

We observe the following properties:

Reflexivity: Let $a$ be an arbitrary element of $R.$ Then,

$a \in R$

$\Rightarrow 1 + a . a = 1 + a^2 > 0$

$\Rightarrow (a, a) \in R_1$

$\text{Thus, } (a, a) \in R_1 \text{ for all } a \in R. \text{ Hence, } R_1 \text{ is reflexive on } R.$

$\text{Symmetry: Let } (a,b) \in R. \text{ Then, }$

$(a, b) \in R_1$

$\Rightarrow 1 + ab > 0$

$\Rightarrow 1 + ba > 0$

$\Rightarrow (b, a) \in R_1$

$\text{Thus, } (a, b) \in R_1 \Rightarrow ( b, a) \in R_1 \text{ for all } a, b \in R. \text{ Hence, } R_1 \text{ is symmetric on } R.$

$\text{Transitivity: We observe that } ( 1, 1/2) \in R_1 \text{ and } ( 1/2, -1) \in R_1 \text{ but } \\ \\ ( 1, -1) \notin R_1 \text{ because } 1 + 1 \times (-1) = 0 \not> 0. \text{ Hence, } R_1 \text{ is not transitive on } R.$

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Question 6: Determine whether each of the following relations are reflexive, symmetric and transitive:

$\text{(i) Relation } R \text{ on set } A = \{ 1, 2, 3, \ldots , 13, 14 \} \text{ defines as } \\ \\ R \{ (x, y) : 3x-y = 0 \}$

$\text{(ii) Relation } R \text{ on the set } N \text{ of all natural numbers defined as } \\ \\ R = \{ (x, y) : y = x + 5 \text{ and } x < 4 \}$

$\text{(iii) Relation } R \text{ on set } A = \{ 1, 2, 3, 4, 5, 6 \} \text{ defined as } \\ \\ R = \{ (x, y) : y \text{ is divisible by } \ x\}$

$\text{(iv) Relation } R \text{ on the set } Z \text{ of all integer defined as } \\ \\ R = \{ (x, y) : x-y \text{ is an integer} \}$

Answer:

$\text{(i) } R = \{ (x, y) : 3x - y = 0 \}, \text{ where } x, y \in A = \{ 1, 2, 3, \ldots , 13, 14 \}$

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

$\text{Reflexivity: Clearly, } (1, 1) \notin R. \text{ Hence, } R \text{ is not a reflexive relation on } A.$

$\text{Symmetry: We observe that } (1, 3) \in R \text{ but } (3, 1) \notin R. \\ \\ \text{ Hence, } R \text{ is not a symmetric relation } A.$

$\text{Transitivity: We observe that } (1, 3) \in R \text{ and } (3,9) \in R \text{ but } (1, 9) \notin R. \\ \\ \text{ Hence, } R \text{ is not a transitive relation } A.$

$\text{(ii) } R = \{ (x, y) : y = x + 5 \text{ and } x < 4 \}, \text{ where } x, y \in N$

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

Reflexivity: Clearly, $(1,1),(2,2)$ etc. are not in $R$ . Hence, $R$ is not reflexive.

Symmetry: We find that $(1, 6) \in R$ but $(6,7) \notin R.$ Hence, R is not symmetric.

Transitivity: Since $(1, 6) \in R$ and there is no order pair in $R$ which has $6$ as the first element. Same is the case for $(2,7)$ and $(3, 8).$ Hence, $R$ is transitive.

$\text{iii } R = \{ (x, y) : y \text{ is divisible by } x \} , \text{ where } x, y \in A=\{ 1,2,3,4,5,6 \} .$

Reflexivity: We know that $x$ is divisible by $x$ for all $x \in A$

$\therefore (x, x) \in R \text{ for all } x \in A$

$\Rightarrow R \text{ is reflexive on set } A$

Symmetry: We observe that 6 is divisible by 2 but 2 is not divisible by 6. This means that $(2, 6) \in R \text{ but } (6, 2) \notin R.$

Hence, $R$ is not symmetric on set $A.$

Transitivity: Let $(x, y) \in R \text{ and } (y, z) \in R. \text{ Then, } (x, y) \in R \text{ and } (y, z) \in R.$

$\Rightarrow y$ is divisible by $x$ and $z$ is divisible by $y$

$\Rightarrow z$ is divisible by $x$

$\Rightarrow (x, z) \in R$

Hence, $R$ is transitive relation on $A.$

$\text{(iv) } R = \{ (x, y): x - y \text{ is an integer } \} , \text{ where } x, y \in Z$

Reflexivity: We have, $x-x = 0,$ which is an integer for all $x \in Z$

$\Rightarrow (x, x) \in R \text{ for all } x \in Z$

$\Rightarrow R \text{ is reflexive on } Z$

$\text{Symmetry: Let } (x, y) \in R. \text{ Then, } ( x, y) \in R$

$\Rightarrow x - y \text{ is an integer, say } \lambda$

$\Rightarrow y - x = -\lambda$

$\Rightarrow y-x \text{ is an integer }$

$\Rightarrow (y, x) \in R$

$\text{Therefore, } (x, y) \in R \Rightarrow (y, x) \in R \text{ for all } x, y \in Z$

$\text{Hence, } R \text{ is symmetric on } Z$

$\text{Transitivity: Let } (x, y) \in R \text{ and } (y, z) \in R. \text{ Then, } (x, y) \in R \text{ and } ( y, z) \in R$

$\Rightarrow x-y \text{ and } y - z \text{ are integers }$

$\Rightarrow (x-y) + (y-z) \text{ is an integer }$

$\Rightarrow x-z \text{ is an integer }$

$\Rightarrow (x, z) \in R$

$\text{Hence, } R \text{ is transitive on } Z$

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Question 7: Show that the relation $R \text{ on } R$ defined as $R = \{ (a, b) : a \leq b \}$ , is reflexive and transitive but not symmetric.

Answer:

$\text{ We have, } R = \{ (a,b): a \leq b \}, \text{ where } a, b \in R$

$\text{ Reflexivity: For any } a \in R. \ \ \ a \leq a$

$\Rightarrow (a, a) \in R \text{ for all } a \in R$

$\Rightarrow R \text{ is reflexive.}$

$\text{ Symmetry: We observer that} (2, 3) \in R \text{ but } (3, 2) \in R. \\ \\ \text{ Hence, } R \text{ is not symmetric.}$

$\text{ Transitivity: Let } (a, b) \in R \text{ and } (b, c) \in R. \text{ Then, } (a, b) \in R \text{ and } (b, c) \in R$

$\Rightarrow a \leq b \text{ and } b \leq c$

$\Rightarrow a \leq c$

$\Rightarrow (a, c) \in R$

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Question 8: Let $S$ be the set of all points in a plane and $R$ be a relation on $S$ defined as

$R = \{ (P, Q): \text{ Distance between } P \text{ and } Q \text{ is less than 2 units } \}.$

Show that $R$ is reflexive and symmetric but not transitive.

Answer:

We observe the following properties of relation $R:$

Reflexivity: For any point $P$ in set $S$ , we find that

Distance between $P$ and itself is 0 which is less than 2 units.

$\Rightarrow (P,P) \in R$

$\text{Thus, } (P, P) \in R \text{ for all } P \in S$

Hence, $R$ is reflexive on $S$

Symmetry: Let $P$ and $Q$ be two points in $S$ such that $(P, Q) \in R.$

Then, $(P,Q) \in R$

$\Rightarrow$ Distance between $P$ and $Q$ is less than 2 units.

$\Rightarrow$ Distance between $Q$ and $P$ is less than 2 units

$\Rightarrow (Q, P) \in R$

Hence, $R$ is symmetric on $S.$

Transitivity: Consider points $P, Q \text{ and } R$ having coordinates $(0,0), (1.5,0) \text{ and } (3.2,0).$ We observe that the distance between $P \text{ and } Q$ is 1.5 units which is less than 2 units and the distance between $Q \text{ and } R$ is 1.7 units which is also less than 2 units. But, the distance between $P \text{ and } R$ is 3.2 which is not less than 2 units. This means that $(P, Q) \in R \text{ and } (Q, R) \in R \text{ but } (P, R) \notin R.$ Hence, $R$ is not transitive on $S.$

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$\displaystyle \text{Question 9: Let } X = \{ 1, 2, 3, 4, 5, 6, 7, 8, 9 \} . \text{ Let } R_1 \text{ be a relation on } \\ \\ X \text{ given by } R_1 = \{ (x, y) : x - y \text{ is divisible by } 3 \} \text{ and } R_2 \text{ be another relation on } \\ \\ X \text{ given by } R_2 = \{ (x, y) : \{x, y \} \subset \{1, 4, 7 \} \text{ or } \\ \\ \{x, y \} \subset \{2, 5, 8 \} \text{ or } \{x , y \} \subset \{ 3, 6, 9 \}. \text{ Show that } R_1 = R_2$

Answer:

Clearly, $R_1 \text{ and } R_2$ are subsets of $X \times X.$ In order to prove that $R_1 = R_2,$ it is sufficient to show that $R_1 \subset R_2 \text{ and } R_2 \subset R_1.$

We observe that the difference between any two elements of each of the sets $\{1,4,7 \}, \{2,5,8\} \text{ and } \{3, 6, 9\}$ is a multiple of 3.

Let $(x, y)$ be an arbitrary element of $R_1.$ Then, $(x, y) \in R_1$

$\Rightarrow x-y \text{ is divisible by } 3$

$\Rightarrow x-y \text{ is a multiple of } 3$

$\Rightarrow \{x, y \} \subset \{1, 4, 7\} \text{ or } \{x, y \} \subset \{2, 5, 8\} \text{ or } \{x, y \} \subset \{3, 6, 9\}$

$\Rightarrow (x, y) \in R_2$

$\text{ Therefore } (x, y) \in R_1 \Rightarrow (x, y) \in R_2$

$\text{ Hence, } R_1 \subset R_2 \text{ ... ... ... ... ... i)}$

Now, let $(a, b)$ be an arbitrary element of $R_2.$ Then, $(a, b) \in R_2$

$\Rightarrow \{a, b \} \subset \{1, 4, 7\} \text{ or } \{a, b \} \subset \{2, 5, 8\} \text{ or } \{a, b \} \subset \{3, 6, 9\}$

$\Rightarrow a-b \text{ is divisible by } 3$

$\Rightarrow (a, b) \in R_1$

$\text{ Therefore } (a, b) \in R_2 \Rightarrow (a,b) \in R_1$

$\text{ Hence, } R_2 \subset R_1 \text{ ... ... ... ... ... ii)}$

Therefore for (i) and (ii) we get $R_1 = R_2$

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Question 10: Show that the relation R on the set R of all real numbers, defined as $R = \{ (a, b) : a \leq b^2 \}$ is neither reflexive nor symmetric nor transitive.

Answer:

$\text{ We have, } R = \{ (a, b) : a \leq b^2 \}, \text{ where } a, b \in R.$

$\displaystyle \text{ Reflexivity: We observe that } \frac{1}{2} \leq \Big( \frac{1}{2} \Big)^2 \text{ is not true. Therefore, } \Big( \frac{1}{2} , \frac{1}{2} \Big) \notin R.$

Therefore, $R$ is not reflexive.

$\displaystyle\text{ Symmetry: We observe that } \\ \\ -1 \leq 3^2 \text{ but } 3 \nleq (-1)^2 \text{ i.e. } (-1, 3) \in R \text{ but } ( -1, 3) \in R \text{ but } ( 3, -1) \notin R.$

Hence, $R$ is not symmetric.

Transitivity: We observe that

$2 \leq (-3)^2 \text{ and } -3 \leq 1^2 \text{ but } 2 \nleq 1^2 \text{ i.e. } (2, -3) \in R \text{ and } (-3, -1) \in R \text{ but } ( 2, 1 ) \notin R.$

Hence, $R$ is not transitive.

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Question 11: Let $A : \{ 1, 2, 3 \}.$ Then, show that the number of relations containing $(1,2)$ and $(2,3)$ which are reflexive and transitive but not symmetric is three.

Answer:

The smallest reflexive relation on set $A$ containing $(1, 2) \text{ and } (2, 3)$ is $R = \{ (1, 1), (2,2), (3, 3), (1,2), (2, 3) \}$

Since $(1,2) \in R \text{ and } (2, 3) \in R \text{ but } (1,3) \notin R.$ Hence, $R$ is not transitive.

To make it transitive we have to include $(1,3)$ in $R.$ Including $(1,3)$ in $R,$ we get $R_1 = \{ (1, 1), (2,2), (3, 3), (1.,2), (2, 3), (1, 3)\}$

This is reflexive and transitive but not symmetric as $(1, 3) \in R_1$ but $(3, 1) \notin R_1.$

Now, if we add the pair $(2, 1)$ to $R_1$ to get $R_2 = \{ (1,1), (2,2), (3,3), (1,2), (2,3), (1,3), (2,1) \} .$

The relation $R_2$ is reflexive and transitive but not symmetric. Similarly, by adding $(3,2)$ and $(3, 1)$ respectively to $R_1$ we get

$R_3 = \{ (1, 1), (2,2), (3, 3), (1,2), (2, 3), (1, 3),(3,2) \}$

$R_3 = \{ (1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3), (3, 1), (3, 2) \}$

These relations are reflexive and transitive but not symmetric.

We observe that out of ordered pairs $(2,1), (3,2)$ and $(3, 1)$ at a time if we add any two ordered pairs at a time to $R_1,$ then to maintain the transitivity we will be forced to add the remaining third pair and in this process the relation will become symmetric also which is not required. Hence, the total number of reflexive, transitive but not symmetric relations containing $(1,2)$ and $(2,3)$ is three.

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Question 12: Let $R$ be a relation on the set of all lines in a plane defined by $(l_1, l_2) \in R \Leftrightarrow \text{ line } l_1 \text{ is parallel to line } l_2$ . Show that $R$ is an equivalence relation.

Answer:

Let $L$ be the given set of lines in a plane. Then, we observe the following properties.

Reflexive: For each line $l \in L$ , we have

$l \parallel l \Rightarrow (l, l) \in R \text{ for all } l \in L$

$\Rightarrow R$ is reflexive

$\text{ Symmetric: Let } l_1, l_2 \in L \text{ such that } l_1, l_2 \in R. \text{ Then, }$

$(l_1, l_2) \in R \Rightarrow l_1 \parallel l_2 \Rightarrow l_2 \parallel l_1 \Rightarrow (l_2, l_1) \in R.$

$\text{ Hence, } R \text{ is symmetric on } L.$

$\text{ Transitive: Let } l_1, l_2, l_3 \in L \text{ such that } (l_1, l_2) \in R \text{ and } (l_2, l_3) \in R. \text{ Then, }$

$(l_1, l_2) \in R \text{ and } (l_2, l_3) \in R \Rightarrow l_1 \parallel l_2 \text{ and } l_2 \parallel l_3 \Rightarrow l_1 \parallel l_3 \in R .$

Hence, R is transitive on L.

Hence, R being Reflexive, symmetric and transitive is an equivalence relation on L.

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Question 13: Show that the relation is congruent to on the set of all triangles in a plane is an equivalence relation.

Answer:

Let $S$ be the set of all triangles in a plane and let $R$ be the relation on $S$ defined by

$(\triangle_1, \triangle_2) \in R \Leftrightarrow \text{ triangle } \triangle_1 \text{ is congruent to triangle } \triangle_2.$

We observe the following properties of relation $R:$

$\triangle \cong \triangle \Rightarrow ( \triangle, \triangle) \in R \text{ for all } \triangle \in R \Rightarrow R \text{ is reflexive on } S$

$\text{Symmetry: Let } \triangle_1, \triangle_2 \in S \text{ such that } ( \triangle_1, \triangle_2) \in R. \text{ Then, }$

$(\triangle_1, \triangle_2) \in R \Rightarrow \triangle_1 \cong \triangle_2 \Rightarrow \triangle_2 \cong \triangle_1 \Rightarrow ( \triangle_2, \triangle_1) \in R.$

Hence, R is symmetric on S

$\text{Transitivity: Let } \triangle_1, \triangle_2, \triangle_3 \in S \text{ such that } (\triangle_1, \triangle_2) \in R \text{ and } (\triangle_2, \triangle_3) \in R. \text{ Then, }$

$(\triangle_1, \triangle_2) \in R \text{ and } (\triangle_2, \triangle_3) \in R \Rightarrow \triangle_1 \cong \triangle_2 \text{ and } \triangle_2 \cong \triangle_3 \\ \\ \Rightarrow \triangle_1 \cong \triangle_3 \Rightarrow ( \triangle_1, \triangle_3) \in R$

Hence, R is transitive on S. Hence, R being reflexive, symmetric and transitive, is an equivalence relation on S

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Question 14: Show that the relation $R$ defined on the set $A$ of all triangles in a plane as $R : \{ (T_1,T_2) :T_1 \text{ is similar to } T_2 \}$ is an equivalence relation. Consider three right angle triangles $T_1$ with sides $3,4,5;\ T_2$ with sides $5,12,13$ and $T_3$ with sides $6,8,10$ . Which triangles among $T_1, T_2 \text{ and } T_3$ are relate?

Answer:

We observe the following properties of relation R.

Reflexivity: We know that every triangle is similar to itself.

$\therefore (T,T) \in R \text{ for all } T \in A \Rightarrow R \text{ is reflexive. }$

$\text{Symmetry: Let } (T_1, T_2)\in R. \text{ Then, } (T_1, T_2) \in R$

$\Rightarrow T_1 \text{ is similar to } T_2$

$\Rightarrow T_2 \text{ is similar to } T_1$

$\Rightarrow (T_2, T_1) \in R$

Hence, R is symmetric.

$\text{Transitivity: Let } T_1,T_2,T_3 \in A \text{ such that } (T_1, T_2) \in R \text{ and } (T_2, T_3) \in R. \text{ Then, }$

$(T_1, T_2) \in \text{ and } (T_2, T_3) \in R$

$\Rightarrow T_1 \text{ is similar to } T_2 \text{ and } T_2 \text{ is similar to } T_3$

$\Rightarrow T_1 \text{ is similar to } T_3$

$\Rightarrow (T_1, T_3) \in R$

Hence, R is transitive.

Hence, R is an equivalence relation on set A.

In triangles $T_1$ and $T_3$ , we observe that the corresponding angles are equal and the corresponding sides are proportional

$\displaystyle \text{i.e. } \frac{3}{6} = \frac{4}{8}= \frac{5}{10}. \text{Hence, } T_1 \text{ and } T_3 \text{ are related. }$

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Question 15: Let n be a positive integer. Prove that the relation R on the set Z of all integers numbers defined by $(x, y) \in R \Leftrightarrow x - y$ is divisible by $n$ , is an equivalence relation on Z.

Answer:

We observe the following properties of relation R.

Reflexivity: For any $a \in N$

$a-a = 0 = 0 \times n$

$\Rightarrow a-a \text{ is divisible by } n$

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

$\text{Thus, } (a, a) \in R \text{ for all } a \in Z. \text{Hence, } R \text{ is reflexive on } Z$

$\text{Symmetry: Let } (a,b) \in R. \text{ Then, } (a,b) \in R$

$\Rightarrow (a -b) \text{ is divisible by } n$

$\Rightarrow (a -b) = np \text{ for some } p \in Z$

$\Rightarrow b-a = n(-p)$

$\Rightarrow b - a \text{ is divisible by } n$

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

$\text{ Thus, } (a,b) \in R \Rightarrow (b,a) \in R \text{ for all } a,b \in Z. \text{ Hence, } R \text{ is symmetric on } Z.$

$\text{ Transitivity: Let } a,b, c \in Z \text{ such that } (a,b) \in R \text{ and } (b, c) \in R. \\ \\ \text{ Then, } (a,b) \in R$

$\Rightarrow (a -b) \text{ is divisible by } n$

$\Rightarrow a -b =np \text{ for some } p \in Z \text{ and, } (b, c) \in R$

$\Rightarrow (b - c) \text{ is divisible by } n$

$\Rightarrow b-c= nq \text{ for some } q \in Z$

$\therefore (a,b) \in R \text{ and } (b, c) \in R$

$\Rightarrow a-b = np \text{ and } b -c = nq$

$\Rightarrow (a-b)+(b-c) =np+nq$

$\Rightarrow a-c =n(p+q)$

$\Rightarrow a - c \text{ is divisible by } n$

$\Rightarrow (a,c) \in R$

$\text{Thus, } (a,b) \in R \text{ and } (b, c) \in R= (a,c) \in R \text{ for all } a,b,c \in Z.$

Hence, R is transitive relation on Z.

Thus, R being reflexive, symmetric and transitive, is an equivalence relation on Z.

$\\$

Question 16: Show that the relation R on the set A of all the books in a library of a college given by
$R : \{ (x, y) : x \text{ and } y \text{ have the same number of pages. } \}$ is an equivalence relation.

Answer:

We observe the following properties of relation R

Reflexivity: For any book $x$ in set $A$ , we observe that $x$ and $x$ have the same number of pages.

$\Rightarrow (x, x) \in R$

$\text{ Thus, } (x, x) \in R \text{ for all } x \in A.$

Hence, $R$ is reflexive.

$\text{ Symmetry: Let } (x, y) \in R. \text{ Then, } (x, y) \in R$

$\Rightarrow x$ and $y$ have the same number of pages

$\Rightarrow y$ and $x$ have the same number of pages

$\Rightarrow (y, x) \in R$

$\text{ Thus, } (x, y) \in R \Rightarrow (y, x) \in R$ Hence, $R$ is symmetric.

Transitivity: Let $(x, y) \in R \text{ and } (y, z) \in R$ . Then, $(x,y) \in R \text{ and } (y,z) \in R$

$\Rightarrow (x \text{ and } y$ have the same number of pages) and ( $y \text{ and } z$ have the same number of pages)

$\Rightarrow x \text{ and } z$ have the same number of pages.

$\Rightarrow (x,z) \in R \text{ Hence, R is transitive. }$

$\text{Thus, R is reflexive, symmetric and transitive. }$

$\text{Hence, R is an equivalence relation. }$

$\\$

Question 17: Show that the relation $R$ on the set $A = \{1,, 2, 3, 4,5|\}$ given by $R = \{ (a,b): | a -b | \text{ is even } \} , \text{ is an equivalence relation. }$ Show that all the elements of $\{1,3,5 \}$ are related to each other and all the elements of $\{2 , 4\}$ are related to each other. But, no element of $\{1,3,5 \}$ is related to any element of $\{ 2, 4 \}.$

Answer:

We have,
$R = \{ (a,b): | a -b | \text{ is even } \}, \text{ where } a, b \in A = \{ 1, 2, 3, 4, 5 \}.$

We observe the following properties of relation R.

$\text{Reflexivity: For any } a \in A, \text{ we have } |a-a| = 0 , \text{ which is even }$

$\therefore (a, a) \in R \text{ for all } a \in A$

Hence, R is reflexive.

$\text{Symmetry: Let } (a, b) \in R. \text{ Then, } (a,b) \in R$

$\Rightarrow |a - b | \text{ is even }$

$\Rightarrow |b - a | \text{ is even }$

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

$\text{Thus, } (a,b) \in R \Rightarrow (b,a) \in R$

Hence, R is symmetric.

$\text{Transitivity: Let } (a, b) \in R \text{ and } (b, c) \in R. \text{ Then, } (a,b) \in R \text{ and } (b, c) \in R$

$\Rightarrow |a-b | \text{ is even and } |b-c| \text{ is even }$

$\Rightarrow (a \text{ and } b \text{ both are even or both are odd } ) \\ \\ \text{ and } (b \text{ and } c \text{ both are even or both are odd } )$

Now two cases arise:

Case 1:

When $b$ is even in this case,

$(a,b) \in R \text{ and } (b, c) \in R$

$\Rightarrow | a -b | \text{ is even and } | b - c | \text{ is even }$

$\Rightarrow a \text{ is even and } c \text{ is even }$

$\Rightarrow | a - c | \text{ is even }$

$\Rightarrow (a,c) \in R$

Case 2:

When $b$ is odd in this case,

$(a,b) \in R \text{ and } (b, c) \in R$

$\Rightarrow | a -b | \text{ is even and } | b - c | \text{ is even }$

$\Rightarrow a \text{ is odd and } c \text{ is odd }$

$\Rightarrow | a - c | \text{ is even }$

$\Rightarrow (a,c) \in R$

Therefore, $(a,b) \in R \text{ and } (b, c) \in R \Rightarrow (a, c) \in R$

Hence, R is an equivalence relation.

We know that the difference of any two odd (even) natural numbers is always an even natural number. Therefore, all the elements of set $\{1, 3, 5\}$ are related to each other and all the elements of $\{ 2, 4 \}$ are related to each other.

We know that the difference of an even natural number and an odd natural number is an odd natural number. Therefore, no element of $\{ 1,3,5 \}$ is related to any element of $\{ 2,4 \}$ .

$\\$

Question 18: Show that the relation R on the set $A = \{ x \in Z " 0 \leq x \leq 12 \}, \text{ given by } R= \{ (a, b) : | a-b | \text{ is a multiple of 4 } \}$ is an equivalence relation. Find the set of all elements related to 1 i.e. equivalence class [1].

Answer:

We have,

$R= \{ (a, b) : | a-b | \text{ is a multiple of 4 } \} , \text{ where } \\ \\ a, b \in A = \{ x \in Z " 0 \leq x \leq 12 \} = \{ 0, 1, 2, \ldots , 12 \}.$

We observe the following properties of relation R.

Reflexivity: For any $a \in A$ , we have

$|a-a|=0 , \text{ which is a multiple of 4. }$

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

$\text{ Therefore, } (a, a) \in R \text{ for all } a \in A$

Hence, R is reflexive.

$\text{ Symmetric: Let } (a, b) \in R. \text{ Then, } (a, b) \in R$

$\Rightarrow |a- b| \text{ is a multiple of 4 }$

$\Rightarrow |a-b| = 4 \lambda \text{ for some } \lambda \in N$

$\Rightarrow | b-a | = 4 \lambda \text{ for some } \lambda \in N$

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

Hence, R is symmetric.

$\text{Transitivity: Let } (a,b) \in R \text{ and } (b, c) \in R. \text{ Then, } (a,b) \in R \text{ and } (b, c) \in R$

$\Rightarrow |a - b | \text{ is a multiple of 4 } \text{ and } |b-c | \text{ is a multiple of 4}$

$\Rightarrow |a - b | = 4\lambda \text{ and } |b-c| = 4\mu \text{ for some } \lambda, \mu \in N$

$\Rightarrow a-b = \pm 4 \lambda \text{ and } b-c = \pm 4 \mu$

$\Rightarrow a-c = \pm 4\lambda \pm 4 \mu$

$\Rightarrow a-c \text{ is a multiple of 4 }$

$\Rightarrow |a - c| \text{ is a multiple of 4 }$

$\Rightarrow (a, c) \in R$

$\text{Thus, } (a, b) \in R \text{ and } (b, c) \in R \Rightarrow (a,c) \in R$

Hence, R is transitive.

Hence, R is an equivalence relation.

Let $x$ be an element of $A$ such that $(x, 1) \in R. \text{ Then, }$

$|x-1| \text{ is a multiple of } 4$

$\Rightarrow |x-1| = 0, 4, 8, 12$

$\Rightarrow x-1 = 0, 4, 8, 12$

$\Rightarrow x = 1, 5, 9$

Hence, the set of all elements of A which are related to 1 is $\{1, 5, 9 \} \text{ i.e. } [1] = \{ 1, 5, 9 \}.$

$\\$

Question 19: Show that the relation R on the set A of points in a plane, given by

$R = \{ (P, Q): \text{Distance of the point P from the origin is same as the distance} \\ \\ \text{ of the point Q from the origin, } \}$ is an equivalence relation. Further shaw that the set of alt points related to a point $P \neq (0, 0)$ is the circle passing through P with origin as center.

Answer:

Let $O$ denote the origin in the given plane. Then,

$R = \{ (P, Q) : OP =OQ \}$

We observe the following properties of relation $R.$

Reflexivity: For any point $P$ in set $A$ , we have

$OP =OP$

$\Rightarrow (P,P) \in R$

$\text{Thus, } (P, P) \in R \text{ for all } P \in A$

Hence, R is reflexive.

Symmetry: Let $P$ and $Q$ be two points in set $A$ such that

$(P,Q) \in R$

$\Rightarrow OP =OQ$

$\Rightarrow OQ=OP$

$\Rightarrow (Q, P) \in R$

$\text{Thus, } (P,Q) \in R \Rightarrow (Q, P) \in R \text{ for all } P, Q \in A$

Hence, R is symmetric.

Transitivity: Let P , Q and S be three points in set A such that

$(P,Q) \in R \text{ and } (Q, S) \in R$

$\Rightarrow OP =OQ \text{ and } OQ =OS$

$\Rightarrow OP=OS$

$\Rightarrow (P,S) \in R$

$\text{Thus, } (P,Q) \in R \text{ and } (Q, S) \in R \Rightarrow (P, S) \in R \text{ for all } P,Q,S \in A$

Hence, R is transitive.

Hence, R is an equivalence relation.

Lei P be a fixed point in set A and Q be any point in set A such that $(P, Q) \in R \text{. Then, } (P,Q) \in R$

$\Rightarrow OP =OQ$

$\Rightarrow Q$ moves in the plane in such a way that its distance from the origin $O (0, 0)$ is always same and is equal to OP.

$\Rightarrow$ Locus of Q is a circle with center at the origin and radius OP.

Hence, the set of all points related to P is the circle passing through F with origin O as center.

$\\$

Question 20: Prove that the relation R on the set $N \times N$ defined by

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

is an equivalence relation.

Answer:

We observe the following properties of relation R.

Reflexivity: Let $(a, b)$ be an arbitrary element of $N \times N$ . Then,

$(a,b) \in N \times N$

$\Rightarrow a,b \in N$

$\Rightarrow a+b =b+a$

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

$\text{Therefore } (a,b) \ R \ (a,b) \text{ for all } (a, b) \in N \times N. \\ \\ \text{Hence, R is reflexive on } N \times N.$

$\text{Symmetry: Let } (a,b), (c, d) \in N \times N \text{ be such that } (a,b) \ R\ (c, d). \text{ Then, }$

$(a,b) \ R \ (a,b)$

$\Rightarrow a+d =b+c$

$\Rightarrow c+b =d+a$

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

$\therefore (a,b) \ R \ (c,d) \Rightarrow (c, d) \ R \ (a, b) ) \text{ for all } (a,b), (c, d) \in N \times N.$

Hence, R is symmetric on $N \times N.$

$\text{Transitivity: Let } (a,b), (c,d),(e,f) \in N \times N \text{ such that } (a,b) R (c,d) \text{ and } (c,d) R (e,f). \text{ Then, }$

$\displaystyle \begin{array}{ccc} (a,b) \ R \ (c,d) \Rightarrow a+d = b+c \\ \\ (c, d) \ R \ (e, f) \Rightarrow c+f = d+e \\ \end{array} \Bigg\} \Rightarrow (a+d) + ( c+f) = ( b+c) + ( d+e) \\ { \hspace{6.3cm} \Rightarrow a+f = b+e \Rightarrow (a,b) \ R \ (e, f) }$

$\text{Therefore, } (a,b) \ R \ (c, d) \text{ and } (c, d) \ R \ (e, f) \Rightarrow (a,b) \ R \ (e, f) \\ \\ \text{ for all } (a, b), (c, d), (e, f) \in N \times N.$

Hence, R is transitive on $N \times N$

Hence, R being reflexive, symmetric and transitive, is an equivalence relation on $N \times N$ .

$[(2, 3)] = \{ (x, y) \in N \times N : (x, y) \ R \ (2,3) \}$

$\Rightarrow [(2, 3)] = \{ (x, y) \in N \times N : x+3 = y+2 \} = \{(x, y) \in N \times N : x - y = 1\}$

$= \{ (x, y) \in N \times N : y = x+1 \}$

$= \{ ( x, x+1) : x \in N \}$

$= \{ (1, 2), (2, 3), (3, 4), (4, 5), \ldots \}$

$[(7 , 3)] = \{ (x, y) \in N \times N :(x, y) \ R \ (7 , 3) \}$

$= \{ (x, y) \in N \times N : x+3 = y+7 \}$

$= \{ (x,y) \in N \times N : y = x-4 \}$

$= \{ (x,x-4) \in N \times N : x \in N \}$

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

$\\$

Question 21: Let $A= \{ 1,2,3, \ldots ,9 \}$ and R be the relation on $A \times A$ defined by $(a,b) \ R \ (c,d) if a+ d=b + c$ for all $(a, b) ,( c, d) \in A \times A$ . Prove that R is an equivalence relation and also obtain the equivalence class $[(2, 5)]$ .

Answer:

We observe the following properties of relation R.

Reflexivity: Let $(a, b)$ be an arbitrary element of $A \times A$ . Then,

$(a,b) \in A \times A$

$\Rightarrow a, b \in A$

$\Rightarrow a+b =b+a$

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

Thus, $(a, b) \ R \ (a,b)$ for all $(a,b) \in A \times A$ . Hence, R is reflexive on $A \times A.$

Symmetry: Let $(a,b), (c, d) \in A \times A$ be such that $(a, b) \ R \ (c, d)$ . Then,

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

$\Rightarrow a+d=b+c$

$\Rightarrow c+b =d+a$

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

$\text{Therefore } (a,b) \ R \ (c, d) \Rightarrow (c, d) \ R \ (a,b) \text{ for all } (a, b), (c, d) \in A \times A.$

Hence, R is symmetric on $A \times A$

$\text{Transitivity : Let } (a, b), (c, d), (e, f) \in A \times A \text{ such that } \\ \\ (a, b) \ R \ (c, d) \text{ and } (c, d) \ R \ (e, f) . \text{ Then, }$

$\text{Thus, } (a,b) \ R \ (c,d) \text{ and } (c,d) \ R \ (e,f) \\ \\ \Rightarrow (a,b) \ R \ (e,f) \text{ for all } (a,b),(c,d),(e,f) \in A \times A.$

Hence, R is a transitive relation on $A \times A.$

Hence, R is an equivalence relation on $A \times A.$

Now, $[(2,5)] = \{ (x, y) \in A \times A : (x, y) \ R \ (2,5) \}$

$= \{ (x, y) \in A \times A : x+5 =y +2 \} = \{ (x, y) \in A \times A : y = x+3 \}$

$= \{ (x, x + 3) :x \in A \text{ and } x + 3 \in A \} = \{ (1, 4), (2, 5), (3, 6), (4, 7), (5, 8), (6, 9) \}$

$\\$

Question 22: Let N be the set of all natural numbers and let R be a relation on $N \times N,$ defined by

$(a,b) \ R \ (c,d) \Leftrightarrow ad = bc \text{ for all } (a,b), (c, d) \in N\times N.$

Shorn that R is an equivalence relation on $N\times N$ . Also, find the equivalence class $[(2, 6)].$

Answer:

We observe the following properties of relation R.

Reflexivity: Let $(a, b)$ be an arbitrary element of $N \times N$ . Then

$(a,b) \in N \times N$

$\Rightarrow a,b \in N$

$\Rightarrow ab = ba$

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

$\text{Thus, }(a,b) \ R \ (a,b) \text{ for all } (a, b) \in N \times N$

Hence, R is reflexive on $N \times N$

$\text{Symmetry: Let } (a, b), (c, d) \in N \times N \text{ be such that } (a,b) \ R \ (c, d). \text{ Then, }$

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

$\Rightarrow ad =bc$

$\Rightarrow cb =da$

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

$\text{Thus, } (a,b) \ R \ (c, d) \Rightarrow (c, d) \ R \ (a,b) \text{ for all } (a,b), (c, d) \in N \times N.$

Hence, R is symmetric on $N \times N.$

$\text{Transitivity: Let } (a,b),(c,d),(e,f) \in N \times N \\ \\ \text{ such that } (a,b) \ R \ (c,d) \text{ and } (c,d) \ R \ (e,f). \text{ Then, }$

$\displaystyle \begin{array}{ccc} (a, b) \ R \ (c,d) \Rightarrow ad = bc \\ \\ (c, d) \ R \ (e, f) \Rightarrow cf = de \\ \end{array} \Bigg\} \Rightarrow (ad) ( cf) = ( bc) ( de) \\ { \hspace{5.3cm} \Rightarrow af = be \Rightarrow (a,b) \ R \ (e, f) }$

$\text{Thus, } (a,b) \ R \ (c,d) \text{ and } (c,d) \ R \ (e,f) \Rightarrow (a,b) \ R \ (e,f) \\ \\ \text{ for all } (a,b), (c,d),(e, f) \in N \times N.$

Hence, R is transitive on $N \times N.$

Hence, R being reflexive, symmetric and transitive, is an equivalence relation on $N \times N.$

$[(2, 6)] = \{ ( x, y) \in N \times N : ( x, y) \ R \ (2, 6) \}$

$= \{ ( x, y) \in N \times N : 3x=y \}$

$= \{ ( x, y) x \in N \} = \{ (1, 3), (2, 6), (3, 9), (4, 12), \ldots \}$

$\\$

Question 23: Let N denote the set of all natural numbers and R be the relation on $N \times N$ defined by $(a, b) \ R \ (c, d) \Leftrightarrow ad (b+c) = bc ( a+d) .$ Check whether R is an equivalence relation on $N \times N.$

Answer:

We observe the following properties of relation R.

Reflexivity: Let $(a, b)$ be an arbitrary element of $N \times N$ . Then,

$(a,b) \in N \times N$

$\Rightarrow a, b \in N$

$\Rightarrow ab (b + a) =ba(a +b)$

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

$\text{Thus, } (a, b) \ R \ (a, b) \text{ for all } (a, b) \in N \times N.$

Hence, R is reflexive on $N \times N$ .

$\text{Symmetry: Let } (a, b), (c, d) \in N \times N \text{ be such that } (a, b) \ R \ (c, d). \text{ Then, }$

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

$\Rightarrow ad(b+c) = bc( a+d)$

$\Rightarrow cd ( d+a) = da ( c+ b)$

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

$\text{Thus, } ( a, b) \ R \ (c, d) \Rightarrow ( c, d) \ R \ ( a, b) \text{ for all } (a, b), ( c, d) \in N \times N$

Hence, R is symmetric on $N \times N$ .

$\text{Transitivity: Let } (a, b), ( c, d), (e, f) \in N \times N \text{ such that } \\ \\ (a,b) \ R \ (c,d) \text{ and } (c,d) \ R \ (e,f). \text{ Then, }$

$\displaystyle ( a, b) \ R \ (c, d) \Rightarrow ad(b+c) = bc(a+d) \\ \\ \Rightarrow \frac{b+c}{bc} = \frac{a+d}{ad} \Rightarrow \frac{1}{b} + \frac{1}{c} = \frac{1}{a} + \frac{1}{d} \text{ ... ... ... ... ... i) }$

and,

$\displaystyle ( c, d) \ R \ (e, f) \Rightarrow cf(d+e) = de(c+f) \\ \\ \Rightarrow \frac{d+e}{de} = \frac{c+f}{cf} \Rightarrow \frac{1}{d} + \frac{1}{e} = \frac{1}{c} + \frac{1}{f} \text{ ... ... ... ... ... ii) }$

Adding i) and ii) we get

$\displaystyle \Big( \frac{1}{b} + \frac{1}{c} \Big) + \Big( \frac{1}{d} + \frac{1}{e} \Big) = \Big( \frac{1}{a} + \frac{1}{d} \Big) + \Big( \frac{1}{c} + \frac{1}{f} \Big)$

$\displaystyle \frac{1}{b} + \frac{1}{e} = \frac{1}{a} + \frac{1}{f} \Rightarrow \frac{b+e}{be} = \frac{a+f}{af} \Rightarrow af(b+e) = be ( a+f) \Rightarrow (a, b) \ R \ (e, f)$

$\text{Therefore, } (a, b) \ R \ ( c, d) \text{ and } (c, d) \ R \ ( e, f) \Rightarrow (a, b) \ R \ (e, f) \\ \\ \text{ for all } (a, b), ( c, d), (e, f) \in N \times N$

Hence, R is transitive on $N \times N$

$\\$

Question 24: Prove that the relation ‘congruence modulo $m$ ‘ on the set Z of all integers is an equivalence relation.

Answer:

We observe the following properties of the given relation.

Let a be an arbitrary integer. Then,

$a-a = 0 = 0 \times m \Rightarrow a-a \text{ is divisible by } m \Rightarrow a = a (mod m)$

$\text{Thus, } a \equiv a ( \text{ mod } m) \text{ for all } a \in Z.$

So, “congruence modulo m” is reflexive.

Symmetry: Let a,b in Z such that a =b (mod m). Then,

$a \equiv b ( \text{ mod } \ m )$

$\Rightarrow a - b \text{ is divisible by } m$

$\Rightarrow a-b = \lambda m \text{ for } \lambda \in Z$

$\Rightarrow b - a = ( - \lambda) m$

$\Rightarrow b-a \text{ is divisible by } m$

$\Rightarrow b \equiv a ( \text{ mod } \ m )$

So, “congruence modulo m” is symmetric on Z.

$\text{ Transitivity: Let } a, b, c \in Z \text{ such that } a \equiv b (\text{ mod } m) \text{ and } b \equiv c ( \text{ mod } m). \text{ Then, }$

$a \equiv b (\text{ mod } m)\Rightarrow a -b \text{ is divisible by } m \Rightarrow a -b =\lambda_1m \text{ for some } \lambda_1 \in Z$

$b \equiv c ( \text{ mod } m) \Rightarrow b - c \text{ is divisible by } m \Rightarrow b - c = \lambda_2m \text{ for some } \lambda_2 \in Z$

$(a-b)+(b -c) =\lambda_1m + \lambda_2m = (\lambda_1 + \lambda_2) m$

$\Rightarrow a - c = \lambda_ 3 m , \text{ where } \lambda_3 = \lambda_1 + \lambda_2 \in Z$

$\Rightarrow a \equiv c ( \text{ mod } m )$

$\text{Thus, } a \equiv b (\text{ mod } m) \text{ and } b \equiv c (\text{ mod } m)= a \equiv c (\text{ mod } m).$

$\text{So, "congruence modulo } m \text{" is transitive on Z. }$

$\text{Hence, "congruence modulo } m\text{" is an equivalence relation on Z }$

$\\$

Question 25: Show that the number of equivalence relations on the set $\{ 1, 2, 3 \}$ containing $(1, 2) and$ latex (2, 1) is two.

Answer:

The smallest equivalence relation $R_1$ containing $(1,2)$ and $(2, 1)$ is $R_1 = \{ (1, 1), (2,2), (3, 3), (1,2), (2,1) \}$

Now, we are left with four ordered pairs namely $(2,3 )(3,2), (1,3)$ and $( 3, 1)$ . If we add any one, say $(2,3)$ to $R_1$ , then for symmetry we must add $(3, 2)$ and then for transitivity we are forced to add $(1,3)$ and $(3, 1)$ . Thus, the only equivalence relation other than $R_1$ is the universal relation. Hence, the total number of equivalence relations containing $(1,2)$ and $(2, 1)$ is two.

$\\$

Question 26: Given a non-empty set $X$ , consider $P (X)$ which is the set of all subsets of $X$ . Define a relation in $P (X)$ as follows:
$\text{For subsets } A,B \text{ in } P(X), A \ R \ B \text{ if } A \subset B .$ Is $R$ an equivalence relation on $P(X)?$ Justify your answer.

Answer:

$\text{ It is given that for any } A, B \text{ in } P (X): A \ R \ B \Leftrightarrow A \subset B$

$\text{ We observe the following properties of } R.$

$\text{ Reflexivity: For any } A \text{ in } P (X), \text{ we have }$

$A \subset A \Rightarrow A \ R \ A$

$\text{ So, R is reflexive on } P (X).$

$\text{ Symmetry: Let } A, B \text{ in } P (X) \text{ be such that } A \ R \ B. \text{ Then, }$

$A \ R \ B \Rightarrow A \subset B$

$\text{ This need not imply that } B \subset A. \text{ In fact it is possible only when } A = B.$

$\text{ Also, we know that } \{ 1,2 \} \subset \{ 1,2, 3\}, \text{ but } \{ 1, 2, 3 \} \not\subset \{1,2 \}.$

$\text{ So, R is not a symmetric relation on } P (X)$

$\text{ Transitivity: Let } A, B,C \text{ be in } P (X) \text{ such that }$

$A \ R \ B \text{ and } B\ R \ C \Rightarrow A \subset B \text{ and } B \subset C \Rightarrow A \subset C \Rightarrow A \ R \ C$

$\text{ So, R is a transitive relation on } P (X).$

$\text{ Thus, } R \text{ is reflexive and transitive relation on } P (X) \text{ but it is not symmetric. }$

$\text{ Hence, } R \text{ is not an equivalence relation on } P (X).$

$\\$

Question 27: Let $R$ be the equivalence relation in the set $A =\{ 0,1,2, 3, 4,5 \}$ given by $R = \{ (a ,b) : 2$ divides $(a -b).$ Write the equivalence class $[0].$

Answer:

Clearly, the equivalence class $[0]$ is the set of those elements in $A$ which are related to $0$ under the relation $R. \text{ i.e. } [0] =\{ (a,0) \in R: a \in A \}.$

$\text{ Now, } (a, 0) \in R$

$\displaystyle \Rightarrow a-0 \text{ is divisible by } 2 \text{ and } a \in A$

$\displaystyle \Rightarrow a \in A \text{ such that } 2 \text{ divides a }$

$\Rightarrow a=0,2,4$

$\displaystyle \text{ Thus, } [0] = \{ 0, 2, 4 \}$

$\\$

Question 28: On the set N of all natural numbers, a relation R is defined as follows:
$n \ R \ m \Leftrightarrow$ Each of the natural numbers $n$ and $m$ leaves the same remainder less than 5 when divided by 5.

Show that R is an equivalence relation. Also, obtain the pairwise disjoint subsets determined by R.

Answer:

We observe the following properties of relation R.

Reflexivity: Let a be an arbitrary element of N. Then, either $a$ is less than 5 and if $a \geq 5$ , then on dividing $a$ by 5 we obtain a remainder as one of the numbers $0, 1, 2,3,4$ .

Thus, $a R a$ for all $a \in N$ . So, R is reflexive on N.

Symmetry: Let $a,b \in N$ such that $a R b$ . Then,

$a R b \Rightarrow$ Each of $a$ and $b$ leaves the same remainder less than 5 when divided by 5

$\Rightarrow$ Each of $b$ and $a$ leave the same remainder less than 5 when divided by 5

$\Rightarrow b R a$

Thus, $a R b \Rightarrow b R a$ for all $a,b \in N$ . So, R is symmetric.

Transitivity : Let $a, b, c \in N$ be such that $aRb$ and $bRc.$ Then,

$aRb \Rightarrow$ Each of $a$ and $b$ leaves the same remainder less than 5 when divided by 5

$bRc \Rightarrow$ Each of $b$ and $c$ leaves the same remainder less than 5 when divided by 5

Therefore Each of $a$ and $c$ leaves the same remainder less than 5 when divided by 5

$\Rightarrow aRc$

Thus, $aRb$ and $bRc \Rightarrow aRc$ for all $a, b , c \in N$

So, R is a transitive relation on N.

Hence, R is an equivalence relation on N.

Let us now find the equivalence classes.

$[1] = \{ x \in N : x \ R \ 1\ \}$

$= \{ x \in N : x \text{ and } 1 \text{ leave the remainder less than 5 when divided by 5} \}$

$= \{ x \in N : x \text{ leaves the remainder 1 when divided by 5} \}$

$=\{ 1,6,11,16,21, \ldots \}$

$[2] = \{ x \in N : x \ R \ 2\ \}$

$= \{ x \in N : x \text{ and } 2 \text{ leave the remainder less than 5 when divided by 5} \}$

$= \{ x \in N : x \text{ leaves the remainder 2 when divided by 5} \}$

$=\{ 2, 7, 12, 17, 22, \ldots \}$

$[3] = \{ x \in N : x \ R \ 3\ \}$

$= \{ x \in N : x \text{ and } 3 \text{ leave the remainder less than 5 when divided by 5} \}$

$= \{ x \in N : x \text{ leaves the remainder 3 when divided by 5} \}$

$=\{ 3, 8, 13, 18, 23, \ldots \}$

$[4] = \{ x \in N : x \ R \ 4\ \}$

$= \{ x \in N : x \text{ and } 4 \text{ leave the remainder less than 5 when divided by 5} \}$

$= \{ x \in N : x \text{ leaves the remainder 4 when divided by 5} \}$

$=\{ 4, 9, 14, 19, 24, \ldots \}$

$[5] = \{ x \in N : x \ R \ 5\ \}$

$= \{ x \in N : x \text{ and } 5 \text{ leave the remainder less than 5 when divided by 5} \}$

$= \{ x \in N : x \text{ leaves the remainder 0 when divided by 5} \}$

$=\{ 5, 10, 15, 20, 25, \ldots \}$

Proceeding in this manner we find that

$[1]=[6]=[11] \ldots$

$[2]=[7]=[12] \ldots$

$[3]=[8]=[13] \ldots$

$[4]=[9]=[14] \ldots$

and $[5]=[10]=[15] \ldots$

Thus, we obtain the following disjoint equivalence classes:

$[1], [2], [3], [4], [5] \text{ such that } N = [1] \cup [2] \cup [3] \cup [4] \cup [5]$

ICSE / ISC / CBSE Mathematics Portal for K12 Students

Mathematics Made Easy for School Students

Class 12: Relations – Miscellaneous Problems

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