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 = 3 (p + q)

\Rightarrow (a, c) \in R

Therefore R is transitive

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

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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 .

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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.

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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.

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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

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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.

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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.

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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.

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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.

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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.

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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.

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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.

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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.

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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

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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.

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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

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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 }

\displaystyle \text{Since } z^2 = - 1

\displaystyle \Rightarrow \text{ Real part } = 0

\displaystyle \Rightarrow 2x_2y_1 - 2x_1 y_2 = 0

\displaystyle \Rightarrow x_2 y_1 = x_1 y_2

\displaystyle \Rightarrow \frac{x_1}{x_2} = \frac{y_1}{y_2}  \text{   ... ... ... ... ... i) }

\displaystyle \text{Similarly } (z_2, z_3) \in R \Rightarrow \frac{z_2 - z_3}{z_2 + z_ 3}\text{ is a real number }

\displaystyle \text{Then } \frac{x_2}{x_3} = \frac{y_2}{y_3}  \text{   ... ... ... ... ... ii) }

From equation i) and ii) we get

\displaystyle \frac{x_1}{x_2} \times \frac{x_2}{x_3} = \frac{y_1}{y_2}  \times \frac{y_2}{y_3}

\displaystyle \Rightarrow \frac{x_1}{x_3} = \frac{y_1}{y_3}

\displaystyle \Rightarrow \frac{z_1 - z_3}{z_1 + z_3} \text{ is a real number }

\displaystyle \Rightarrow (z_1, z_3) \in R

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

\displaystyle \Rightarrow ( z_1, z_3) \in R \ \forall \ z_1, z_2, z_3 \in C_0

Therefore R is transitive relation