Class 12: Relations – Exercise 1.2 – ICSE / ISC / CBSE Mathematics Portal for K12 Students

Question 1: Show that the relation $R$ defined by $R = \{ (a, b): a - b \text{ is divisible by } 3; a, b \in Z \}$ is an equivalence relation.

Answer:

To prove that relation is equivalence, we need to prove that it is reflexive, symmetric and transitive.

Reflexivity:

$(a, a) \in R$

$\text{Let } a \in Z$

$\Rightarrow a - a = 0$

$\Rightarrow a - a \text{ is divisible by } 3$ $( \because 0 \text{ is divisible by } 3).$

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

Therefore R is reflexive

Symmetric:

$\text{If } (a, b) \in R, \text{ then } (b, a) \in R$

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

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

$\Rightarrow a - b = 3p \text{ (say) For some } p \in Z$

$\Rightarrow -( a - b) = -3p$

$\Rightarrow b - a = 3 \times (-p)$

$\Rightarrow b - a \in R$

Therefore R is symmetric

Transitive:

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

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

$\Rightarrow a - b = 3p \text{(say) and } b - c = 3q(say) \text{ For some } p, q \in Z$

$\Rightarrow a - c = 3 (p + q)$

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

Therefore R is transitive

Since, $R$ is reflexive, symmetric and transitive, $R$ is an equivalence relation.

$\\$

Question 2: Show that the relation $R$ on the set $Z$ of integers, given by $R = \{ (a, b) :2 \text{ divides } a -b \},$ is an equivalence relation.

Answer:

To prove equivalence relation, the given relation should be reflexive, symmetric and transitive.

Reflexivity:

Let $a$ be an arbitrary element of the set $Z$ .

$\text{Then, } a \in R$

$\Rightarrow a - a = 0 = 0 \times 2$

$\Rightarrow 2 \text{ divides } a - a$

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

Therefore, R is reflexive on Z.

Symmetry:

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

$\Rightarrow 2 \text{ divides } a - b$

$\displaystyle \Rightarrow \frac{(a - b)}{2} = p \text{ for some } p \in Z$

$\displaystyle \Rightarrow \frac{(b - a)}{2} = - p \text{ Here, } -p \in Z$

$\Rightarrow 2 \text{ divides } b - a$

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

Clearly, R is symmetric on Z

Transitivity:

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

$\Rightarrow 2 \text{ divides } a - b \text{ and } 2 \text{ divides } b - c$

$\displaystyle \Rightarrow \frac{(a - b)}{2} = p \text{ and } \frac{(b - c)}{2} = q \text{ for some } p, q \in Z$

On adding the above two equations, we get

$\displaystyle \frac{(a - b)}{2} + \frac{(b - c)}{2} = p + q$

$\displaystyle \Rightarrow \frac{(a - c)}{2} = p + q$

$\text{Here, } p + q \in Z$

$\Rightarrow 2 \text{ divides } a - c$

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

Thus, R is transitive on Z.

Therefore $R$ is reflexive, symmetric and transitive. Hence, $R$ is an equivalence relation on $Z$ .

$\\$

Question 3: Prove that the relation $R$ on $Z$ defined by $(a,b) \in R \Leftrightarrow a -b$ is divisible by 5 is an equivalence relation on $Z.$

Answer:

To prove equivalence relation, the relation should be reflexive, symmetric and transitive.

Reflexivity:

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

$\Rightarrow a - a = 0 = 0 \times 5$

$\Rightarrow a - a \text{ is divisible by } 5$

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

Therefore, R is reflexive on Z.

Symmetry:

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

$\Rightarrow a - b \text{ is divisible } \text{ by } 5$

$\Rightarrow a - b = 5p \text{ for some } p \in Z$

$\Rightarrow b - a = 5(-p)$

$\text{Here, } -p \in Z \text{ [Since }p \in Z]$

$\Rightarrow b - a \text{ is divisible by } 5$

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

Thus, R is symmetric on Z.

Transitivity:

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

$\Rightarrow a - b \text{ is divisible by } 5$

$\Rightarrow a - b = 5p \text{ for some } Z \text{ Also, } b - c \text{ is divisible by } 5$

$\Rightarrow b - c = 5q \text{ for some } Z$

On adding two equations above, we get

$a - b + b - c = 5p + 5q$

$\Rightarrow a - c = 5(p + q)$

$\Rightarrow a - c \text{ is divisible by } 5 \text{ Here, } p + q \in Z$

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

Therefore, R is transitive on Z.

Therefore R is reflexive, symmetric and transitive. Hence, R is an equivalence relation on Z.

$\\$

Question 4: Let $n$ be a fixed positive integer. Define a relation $R$ on $Z$ as follows: $(a, b) \in R \Leftrightarrow a -b$ is divisible by $n$ . Show that $R$ is an equivalence relation on $Z.$

Answer:

To prove equivalence relation, the relation should be reflexive, symmetric and transitive.

Reflexivity:

$\text{Let } a \in N$

$\text{Here, } a - a = 0 = 0 \times n$

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

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

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

Thus, R is reflexive on Z.

Symmetry:

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

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

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

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

$\Rightarrow b - a \text{ is divisible by } n \ \ \ \ \ [p \in Z \Rightarrow - p \in Z]$

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

Clearly, R is symmetric on Z.

Transitivity:

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

$\text{Here, } a - b \text{ is divisible by } n \text{ and } b - c \text{ is divisible by n. }$

$\Rightarrow a - b = n p \text{ for some } p \in Z$

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

$a - b + b - c = np + nq$

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

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

Thus, R is transitive on Z.

Therefore R is reflexive, symmetric and transitive. Hence, R is an equivalence relation on Z.

$\\$

Question 5: Let $Z$ be the set of integers. Show that the relation $R =\{(a, b) : a,b \in Z \text{ and } a+b \text{ is even } \}$ is an equivalence relation on $Z.$

Answer:

To prove equivalence relation the relation should be reflexive, symmetric and transitive.

Reflexivity:

$\text{Let } a \text{ be an arbitrary element of } Z.$

$\text{Then, } a \in R \text{ Clearly, } a + a = 2a \text{ is even for all } a \in Z.$

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

Thus, R is reflexive on Z.

Symmetry:

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

$\Rightarrow a + b \text{ is even }$

$\Rightarrow b + a \text{ is even }$

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

Therefore, R is symmetric on Z.

Transitivity:

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

$\Rightarrow a + b \text{ and } b + c \text{ are even }$

$\text{Now, let } a + b = 2x \text{ for some } x \in Z$

$\text{And } b + c = 2y \text{ for some } y \in Z$

Adding above two equations, we get

$A + 2b + c = 2x + 2y$

$\Rightarrow a + c = 2 (x + y - b), \text{ which is even for all } x, y, b \in Z$

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

Therefore, R is transitive on Z.

Therefore R is reflexive, symmetric and transitive. Hence, R is an equivalence relation on Z

$\\$

Question 6: $m$ is said to be related to $n$ if $m$ and $n$ are integers and $m - n$ is divisible by 13. Does this define an equivalence relation?

Answer:

To prove equivalence relation, the given relation should be reflexive, symmetric and transitive.

$\text{Let } R = {(m, n): m, n \in Z : m - n \text{is divisible by } 13}$

Reflexivity:

$\text{Let } m \text{ be an arbitrary element of } Z.$

$\text{Then, } m \in R \Rightarrow m - m = 0 = 0 \times 13$

$\Rightarrow m - m \text{ is divisible by } 13$

$\Rightarrow (m, m) \text{ is reflexive on } Z.$

Symmetry:

$\text{Let } (m, n) \in R.$

$\text{Then, } m - n \text{ is divisible by } 13$

$\Rightarrow m - n = 13p \text{ Here, } p \in Z$

$\Rightarrow n - m = 13(-p) \text{ Here, } -p \in Z$

$\Rightarrow n - m \text{ is divisible by } 13$

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

Therefore, R is symmetric on Z.

Transitivity:

$\text{Let } (m, n) \text{ and } (n, o) \in R$

$\Rightarrow m - n \text{ and } n - o \text{ are divisible by } 13$

$\Rightarrow m - n = 13p \text{ and } n - o = 13q \text{ for some } p, q \in Z$

Adding above two equations, we get

$m - n + n - o = 13p + 13q$

$\Rightarrow m - o = 13(p + q) \text{ Here, } p + q \in Z$

$\Rightarrow m - o \text{ is divisible by } 13$

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

So, R is transitive on Z.

Therefore, R is reflexive, symmetric and transitive. Hence, R is an equivalence relation on Z.

$\\$

Question 7: Let $R$ be a relation on the set $A$ of ordered pairs of non-zero integers defined by $(x,y) R (u, v) \text{ iff } x v=yu.$ Show that $R$ is an equivalent relation.

Answer:

To prove that relation is equivalence, we need to prove that it is reflexive, symmetric and transitive.

Reflexivity :

$(a, a) \in R$

$\text{Since } xy = yu$

$\text{Therefore } (x, y) R (x, y)$

Therefore R is reflexive.

Symmetric :

$\text{If } (a, b) \in R, \text{ then } (b, a) \in R$

$\text{Let } (x, y) R (u, v)$

$\text{To prove that } (u, v) R (x, y)$

$\text{Given } xv = yu$

$\Rightarrow yu = xv$

$\Rightarrow uy = vx$

$\text{Therefore } (u, v) R (x, y)$

Therefore R is symmetric.

Transitive:

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

$\text{Let } (x, y) R (u, v) \text{ and } (u, v) R (p, q) \text{ ... ... ... ... ... i) }$

$\text{To prove that } (x, y) R (p, q)$

$\text{To prove that } xq = yp$

$\text{From i) } xv = yu \ \ \& \ \ \ uq = vp$

$xvuq = yuvp$

$xq = yp$

Therefore R is transitive.

Since R is reflexive, symmetric & transitive Therefore R is an equivalence relation.

$\\$

Question 8: Show that the relation $R$ on the set $A = \{ x \in Z; 0 \leq x \leq 12 \},$ given by $R =\{(a, b): a=b \},$ is an equivalence relation. Find the set of all elements related to 1.

Answer:

We have,

$A = \{x \in Z : 0 \leq x \leq 12\} \text{ be a set and }$

$R = \{(a, b) : a = b\} \text{ be a relation on } A$

Reflexivity:

$(a, a) \in R$

$\text{Let } a \in A$

$\Rightarrow a = a$

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

Therefore R is reflexive.

Symmetric :

$\text{If } (a, b) \in R, \text{ then } (b, a) \in R$

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

$\Rightarrow a = b$

$\Rightarrow b = a$

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

Therefore R is symmetric.

Transitive:

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

$\text{Let } a, b \ \& \ c \in A$

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

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

$\Rightarrow a = c$

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

Therefore R is transitive.

Since, R is being reflexive, symmetric and transitive, so R is an equivalence relation.

Also, we need to find the set of all elements related to 1.

Since the relation is given by, $R = \{(a, b) : a = b\}$ , and 1 is an element of A,

$\text{Therefore } R = \{(1, 1) : 1 = 1\}$ .

Thus, the set of all element related to 1 is 1.

$\\$

Question 9: Let $L$ be the set of all lines in XY-plane and $R$ be the relation in $L$ defined as $R = \{ (L_1, L_2) : L_1 \text{ is parallel to } L_2 \} .$ Show that $R$ is an equivalence relation. Find the set of all lines related to the line $y =2x + 4.$

Answer:

To prove that relation is equivalence, we need to prove that it is reflexive, symmetric and transitive.

Reflexivity:

$(a, a) \in R$ Since a line is always parallel to itself.

$\text{Therefore } (L_1, L_2) \in R$

Therefore R is reflexive.

Symmetric:

$\text{If } (a, b) \in R, \text{ then } (b, a) \in R$

$\text{Let } L_1, L_2 \in L\text{ and } (L_1, L_2) \in R$

$\Rightarrow L_1 \text{is parallel to } _L2$

$\Rightarrow L_2 \text{is parallel to } L_1$

$\Rightarrow (L_1, L_2) \in R$

Therefore R is symmetric.

Transitive:

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

$\text{Let } L_1, L_2 \text{ and } L_3 \in L \text{ such that } (L_1, L_2) \in R \text{ and } (L_2, L_3) \in R$

$\Rightarrow L_1\text{ is parallel to } L_2 \text{ and } L_2 \text{ is parallel to } L_3$

$\Rightarrow L_1\text{ is parallel to } L_3$

$\Rightarrow (L_1, L_3) \in R$

Therefore R is transitive.

Since, R is reflexive, symmetric and transitive, so R is an equivalence relation.

And, the set of lines parallel to the line $y = 2x + 4 is y = 2x + c$ For all $c \in R$ where R is the set of real numbers.

$\\$

Question 10: Show that the relation $R,$ defined on the set $A$ of all polygons as $R = \{ (P_1, P_2): P_1 \text{ and } P_2 \text{ have same number of sides } \},$ is an equivalence relation. What is the set of all elements in A related to the right angle triangle T with sides 3, 4 and 5?

Answer:

$R = \{(P_1, P_2): P_1 \text{ and } P_2 \text{ have same the number of sides } \}$

Reflexivity:

$(a, a) \in R$

R is reflexive since $(P_1, P_1) \in R$ as the same polygon has the same number of sides with itself.

Symmetric:

$\text{If } (a, b) \in R, \text{ then } (b, a) \in R$

$\text{Let } (P_1, P_2) \in R.$

$\Rightarrow P_1 \text{ and } P_2 \text{ have the same number of sides. }$

$\Rightarrow P_2 \text{ and } P_1 \text{ have the same number of sides. }$

$\Rightarrow (P_2, P_1) \in R$

Therefore R is symmetric.

Transitive:

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

$\text{Now, } (P_1, P_2), (P_2, P_3) \in R$

$\Rightarrow P_1 \text{ and } P_2 \text{ have the same number of sides. }$

$\text{Also, } P_2 \text{ and } P_3 \text{ have the same number of sides. }$

$\Rightarrow P_1 \text{ and } P_3 \text{ have the same number of sides. }$

$\Rightarrow (P_1, P_3) \in R$

Therefore R is transitive.

Hence, R is an equivalence relation. And, now the elements in A related to the right-angled triangle (T) with sides 3, 4 and 5 are those polygons which have three sides (since T is a polygon with three sides).

Hence, the set of all elements in A related to triangle T is the set of all triangles.

$\\$

Question 11: Let $O$ be the origin. we define a relation between two points $P$ and $Q$ in a plane if $OP=OQ .$ Show that the relation, so defined is an equivalence relation.

Answer:

Let A be set of points on the plane. Let $R = \{(P, Q) : OP = OQ\}$ be a relation on A where O is the origin.

Reflexivity:

$(a, a) \in R$

$\text{Let } p \in A$

$\text{Since } OP = OP \Rightarrow (P, P) \in R$

Therefore R is reflexive.

Symmetric:

$\text{If } (a, b) \in R, \text{ then } (b, a) \in R$

$\text{Let } (P, Q) \in R \text{ for } P, Q \in R$

$\text{Then } OP = OQ$

$\Rightarrow OP = OP$

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

Therefore R is symmetric.

Transitive:

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

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

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

$\Rightarrow OP = OS$

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

Therefore R is transitive.

Since, R is reflexive, symmetric and transitive, so R is an equivalence relation on A.

$\\$

Question 12: Let $R$ be the relation defined on the set $A = \{1,2,3,4,5,6,7 \}$ by $R = \{ (a,b) : \text{ both } a \text{ and } b \text{ are either odd or even } \}.$ Show that $R$ is an equivalence relation. Further, show that all the elements of the subset $\{ 1, 3 , 5 ,7 \}$ are related to each other and all the elements of the subset $\{2, 4, 6 \}$ are related to each other, but no element of the subset $\{1, 3,5 ,7 \}$ is related to any element of the subset $\{ 2, 4, 6 \}.$

Answer:

$\text{Given } A = \{1, 2, 3, 4, 5, 6, 7\} \text{ and } \\ \\ R = \{ (a, b) : \text{ both } a \text{ and } b \text{ are either odd or even number } \}$

Therefore, $R = \{(1, 1), (1, 3), (1, 5), (1, 7), (3, 3), (3, 5), (3, 7), (5, 5), (5, 7), (7, 7), \\ \\ (7, 5), (7, 3), (5, 3), (6, 1), (5, 1), (3, 1), (2, 2), (2, 4), (2, 6), (4, 4), (4, 6), \\ \\ (6, 6), (6, 4), (6, 2), (4, 2)\}$

Reflexivity :

$(a, a) \in R$

$\text{Here } (1,1), (2,2), (3,3), (4,4), (5,5), (6,6), (7,7) \text{ all } \in R$

From the relation R it is seen that R is reflexive.

Symmetric:

$\text{If } (a, b) \in R, \text{ then } (b, a) \in R$

From the relation R, it is seen that R is symmetric.

Transitive:

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

$[ \text{If } (a, b) \text{ are odd and } (b, c) \text{ are odd then } (a, c) \text{ are also odd numbers }]$

From the relation R, it is seen that R is transitive too.

Also, from the relation R, it is seen that $\{1, 3, 5, 7\}$ are related with each other only and $\{2, 4, 6\}$ are related with each other.

$\\$

Question 13: Let $S$ be a relation on the set $R$ of all real numbers defined by $S = \{ (a, b) \in R \times R : a^2 + b^2 = 1 \}$ . Prove that $S$ is not an equivalence relation on $R$ .

Answer:

$S = \{ (a, b) \in R \times R : a^2 + b^2 = 1 \}$

$(a, a) \in R$

$\text{Let } a = 1 \Rightarrow 1^2 + 1^2 = 2 \neq 1$

$\Rightarrow ( 1, 1 ) \notin R$

As this relation is not reflexive so it can’t be an equivalence relation.

$\\$

Question 14: Let $Z$ be the set of all integers and $Z_0$ be the set of all non-zero integers. Let a relation $R$ on $Z \times Z_0$ be defined as follows:

$(a, b) \ R \ ( c, d) \Leftrightarrow ad = bc \text{ for all } (a, b) , ( c, d) \in Z \times Z_0$

Prove that $R$ is an equivalence relation on $Z \times Z_0.$

Answer:

We have, $Z$ be set of integers and $Z_0$ be the set of non-zero integers.

$R = \{(a, b) (c, d) : ad = bc\} \text{ be a relation on } Z \text{ and } Z_0$ .

Reflexivity:

$(a, a) \in R$

$(a, b) \in Z \times Z_0$

$\Rightarrow ab = ba$

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

Therefore R is reflexive.

Symmetric:

$\text{If } (a, b) \in R, then (b, a) \in R$

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

$\Rightarrow ad = bc \Rightarrow cd = da$

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

Therefore R is symmetric.

Transitive:

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

$\text{Let } (a, b), (c, d) \in R \text{ and } (c, d), (e, f) \in R$

$\Rightarrow ad = bc \text{ and } cf = de$

$\displaystyle \Rightarrow \frac{a}{d} = \frac{c}{d} \text{ and } \frac{c}{d} = \frac{e}{f}$

$\displaystyle \Rightarrow \frac{a}{b} = \frac{e}{f}$

$\Rightarrow af = be$

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

Therefore R is transitive.

Hence, R is an equivalence relation on $Z \times Z_0$

$\\$

Question 15: If $R$ and $S$ are relations on a set $A$ , then prove the following:

(i) $R$ and $S$ are symmetric $\Rightarrow R \cap S$ and $R \cup S$ are symmetric

(ii) $R$ is reflexive and $S$ is any relation $\Rightarrow R \cup S$ is reflexive.

Answer:

R and S are two symmetric relations on set A

(i) To prove: $R \cap S$ is symmetric

Symmetric:

$\text{If } (a, b) \in R, \text{ then } (b, a) \in R$

$\text{Let } (a, b) \in R \cap S$

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

$\Rightarrow (b, a) \in R \text{ and } (b, a) \in S [ \text{Therefore R and S are symmetric }]$

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

$\Rightarrow R \cap S \text{ is symmetric }$

To prove: $R \cup S \text{ is symmetric }$

Symmetric:

$\text{If } (a, b) \in R, \text{ then } (b, a) \in R$

$\text{ Let } (a, b) \in R \cup S$

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

$\Rightarrow (b, a) \in R \text{ or } (b, a) \in S [ \text{Therefore R and S are symmetric }]$

$\Rightarrow (b, a) \in R \cup S \Rightarrow R \cup S \text{ is symmetric }$

(ii) R and S are two relations on a such that R is reflexive.

To prove : $R \cup S$ is reflexive

Reflexivity:

$(a, a) \in R \text{ Suppose } R \cup S \text{ is not reflexive. }$

$\text{This means that there is } a \in R \cup S \text{ such that } (a, a) \notin R \cup S$

$\text{Since } a \in R \cup S,$

$\text{Therefore } a \in R \text{ or } a \in S$

$\text{ If } a \in R, \text{ then } (a, a) \in R [ \text{ Since R is reflexive }]$

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

Hence, $R \cup S$ is reflexive.

$\\$

Question 16: If R and S are transitive relations on a set A, then prove that $R \cup S$ may not be a transitive relation on A.

Answer:

We will prove this using an example.

Let $A =\{a, b, c\}$ be a set and

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

$S = \{(a, a) (b, b) (c, c) (b, c) (c, d)\}$ are two relations on A

Clearly R and S are transitive relation on A

Now, $R \cup S = \{ (a, a) (b, b) (c, c) (a, b) (b, a) (b, c) (c, b) \}$

Here, $(a, b) \in R \cup S \text{ and } (b, c) \in R \cup S \text{ but } (a, c) \notin R \cup S$

Therefore $R \cup S$ is not transitive

$\\$

Question 17: Let $C$ be the set of all complex numbers and $C_0$ be the set of all non-zero complex numbers. Let a relation $R$ on $C_0$ be defined as

$\displaystyle z_1 \ R \ z_2 \Leftrightarrow \frac{z_1 - z_2}{z_1 + z_2} \text{ is real for all } z_1, z_2 \in C_0$

Show that $R$ is an equivalence relation.

Answer:

Given: Set $C_0 =$ set of the non-zero complex number and a relation R in $C_0$ is defined as:

$\displaystyle z_1 \ R \ z_2 \Leftrightarrow \frac{z_1 - z_2}{z_1 + z_2} \text{ is real for all } z_1, z_2 \in C_0$

To prove that R is equivalence relation, we have to prove that R is reflexive, symmetric and transitive.

(i) Reflexivity:

$\text{Let } z \in C_0$

$\displaystyle \therefore z \in X_o \Rightarrow \frac{z-z}{z+z} = 0 \text{ is a real number }$

$\Rightarrow zRz$

$\Rightarrow (z, z) \in R \ \forall \ z \in C_0$

Therefore R is reflexive.

(ii) Symmetricity:

$\displaystyle \text{Let } z_1, z_2 \in C_0$

$\displaystyle (z_1, z_2) \in R \Rightarrow \frac{z_1 - z_2}{z_1 + z_2} \text{ is a real number }$

$\displaystyle \Rightarrow \frac{z_2 - z_1}{z_2 + z_1} = - \Big( \frac{z_1 - z_2}{z_1 + z_2} \Big)$

$\displaystyle \Rightarrow ( z_2, z_1) \in R$

$\displaystyle \text{Therefore } (z_1, z_2) \in R \Rightarrow ( z_2, z_1) \in R$

Therefore R is a symmetric relation.

(iii) Transitivity:

$\displaystyle \text{Let } z_1, z_2 \text{ and } z_3 \in C_0$

$\displaystyle (z_1,z_2) \in R \text{ and } (z_2, z_3) \in R$

$\displaystyle \text{Let } z_1 = x_1 + iy_1$

$\displaystyle z_2 = x_2 + iy_2$

$\displaystyle z_3 = x_3 + iy_3$

$\displaystyle \text{Let } x_1, y_1, x_2, y_2 \text{ and } x_3, y_3 \text{ are real numbers. }$

$\displaystyle \Rightarrow \frac{z_1 - z_2}{z_1 + z_2} \text{ is a real number }$

$\displaystyle \Rightarrow \frac{x_1 + iy_1 - x_2 - iy_2}{x_1 + iy_1 + x_2 + iy_2}\text{ is a real number }$

$\displaystyle \Rightarrow \frac{(x_1 -x_2)+ i(y_1-y_2)}{(x_1 +x_2)+ i(y_1+y_2)} \text{ is a real number }$

$\displaystyle \Rightarrow \frac{[(x_1 -x_2)+ i(y_1-y_2)][(x_1 +x_2)- i(y_1+y_2)]}{[(x_1 +x_2)+ i(y_1+y_2)][(x_1 +x_2)- i(y_1+y_2)]} \text{ is a real number }$

$\displaystyle \Rightarrow \frac{[({x_1}^2 - {x_2}^2)+({y_1}^2 - {y_2}^2) ] +i[ 2x_2y_1 - 2 x_1 y_2 ] }{(x_1+x_2)^2+(y_1+y_2)^2} \text{ is a real number }$