RD Sharma Class 12 Chapter 1 Exercise 1.2 Relation Solutions Maths - Download PDF Free Online

Relation Exercise1.2 Question 2

Answer: R is an equivalence relation on Z.
Hint: To prove equivalence relation it is necessary that the given relation should be reflexive, symmetric and transitive.
$Given: R=\left \{\left ( a, b \right ):2 \text{ divides } a-b \right \}is\: a\, relation \: def\! ined \: on \: Z.$
Explanation:
Let us check these properties on R.
Reflexivity:
Let a be an arbitrary element of the set Z
$Then, a-\! a=0=0\! \times \! 2$
$\Rightarrow 2\: divides\: a-a$
$\Rightarrow (a, a) \: \epsilon \: R\: f\! or \: all\: a \: \epsilon \: Z.$
So, R is reflexive on Z.
Symmetry:
$Let (a, b)\: \epsilon \: R$
$\Rightarrow 2 \: divides\: a-b$
$\Rightarrow \frac{a-b}{2}=\! p\: f\! or\: some\: p \: \epsilon \: Z$
$\Rightarrow \frac{b-a}{2}=-p$
$Here, -p \: \epsilon \: Z$
$\Rightarrow 2\: divides\: b-a$
$(b, a) \: \epsilon \: R \: f\! or \: all\: a,b \: \epsilon \: Z$
So, R is symmetric on Z
Transitivity:
$Let (a, b) and (b, c) \: \epsilon \: R$
$\Rightarrow 2\: divides\: a-b\: and\: 2 \: divides\: b-c$
$\Rightarrow \frac{a-b}{2}=p$ $...(i)$
$\Rightarrow\frac{b-c}{2}=q$ $...(ii)$
$f\! or\: some\: p, q\: \epsilon \: Z$
Adding eq. (i) and (ii)
$\frac{a-b}{2}+\frac{b-c}{2}=p+q$
$\frac{a-c}{2}=p+q$
$Here, p+q \: \epsilon \: Z$
$\Rightarrow 2\: divided\: a-c$
$(a,c) \: \epsilon \: R\: f\! or\: all\: a, c \: \epsilon \: Z$
So, R is transitive on Z
Therefore, R is reflexive, symmetric and transitive.
Hence, R is an equivalence relation on Z.

Relation Exercise 1 Question 3

Answer:R is an equivalence relation on Z.
Hint: To prove equivalence relation it is necessary that the given relation should be reflexive,symmetric and transitive.$Given:\! R=\left \{\left ( a, b \right ):\left ( a-b\right ) \: is\: divisible \: by\: 5\right \}\: is\: a\: relation\: de\! f\! ined\: on\: Z$.
Explanation:
Let us check these properties on R.
Reflexivity:
Let a be an arbitrary element of the set Z
$\Rightarrow a-a=0=0\times 5$
$\Rightarrow a-a\: is\: divisible\: by \: 5$
$(a, a)\: \epsilon \: R\: f\! or \: all\: a \: \epsilon \: Z.$
So, R is reflexive on Z.
Symmetry:
$Let (a, b)\: \epsilon \: R$
$\Rightarrow a-b\: is\: divisible\: by\: 5$
$a-b=5p\: f\! or\: some\: p\: \epsilon \: Z$
$then\: b-a=5(-p)$
$Here, -p\: \epsilon \: Z \: \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; [\because p \: \epsilon \: Z]$
$b-a\: is\: divisible\: by\: 5$
$(b, a)\: \epsilon \: R \: f\! or\: all \: a, b \: \epsilon \: Z$
So, R is symmetric on Z
Transitivity:
$\text{Let (a, b) and (b, c)}\: \epsilon R$
$\Rightarrow a-b\: is\: divisible\: by\: 5$
$a-b=5p \: f\! or\: some\: p \: \epsilon \: Z$ $...(i)$
$Also,b-c \: is\: divisible\: by\: 5$$...(ii)$
$b-c=5q\: f\! or\: some\: q \: \epsilon \: Z$
Adding eq. (i) and (ii)
$a-b+b-c=5p+5q$
$a-c=5\left (p+q \right )$
$a-c\: is\: divisible\: by\: 5$
$Here, p+q \: \epsilon \: Z$
($a, c) \: \epsilon \: R\: is\: transitive\: on\: Z$
Therefore R is reflexive, symmetric and transitive.
Hence, R is an equivalence relation on Z.

Relation Exercise1.2 Question 4

Answer: R is an equivalence relation on Z.
Hint: To prove equivalence relation it is necessary that the given relation should be reflexive, symmetric and transitive.$Given\! : \! \! R=\left \{\left ( a, b \right ) :a-b \: is\: divisible\: by\: n \right \}\: is\: a \: relation \: de\! f\! ined\: on\: Z$
Explanation:
Let us check these properties on R
Reflexivity:
$Let \: a \: \epsilon \: N$
$Here, a-a=0\times n$
$a-a\: is\: divisible\: by\: n$
$(a, a) \: \epsilon \: R\: f\! or\: all\: a \: \epsilon \: Z$
So, R is reflexive on Z
Symmetry:
$Let (a, b) \: \epsilon \: R$
$Here, a-b\: is \: divisible\: by\: n$
$a-b=np\: f\! or \: some \: p \: \epsilon \: Z$
$b-a=n(-p)$
$b-a\: is\: divisible\: by\: n \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; [i\! f p \: \epsilon \: Z\:, then -p \: \epsilon \: Z]$
$(b, a) \: \epsilon \: R$
So, R is symmetric on Z.
Transitivity:
$Let (a, b) and (b, c) \: \epsilon \: R$
$Here, a-b\: is\: divisible\: by\: n \: and\: b-c\: is\: divisible\: by \: n$
$a-b=np \: f\! or\: some\: p \: \epsilon \: Z \: \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; ...(i)$
$b-c=nq\: f\! or\: some\: q \: \epsilon \: Z \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; ....(ii)$
Adding eq. (i) and (ii)
$a-b+b-c=np+nq$
$a-c=np+q \; \; \; \; \; \; \; \; \; \; \; \; \; [Here, p+q \: \epsilon \: Z]$
$(a, c) \: \epsilon \: R \: f\! or\: all \: a, c \: \epsilon \: Z$
So, R is transitive on Z.
Therefore, R is reflexive, symmetric and transitive.
Hence, R is an equivalence relation on Z.

Relation Exercise1.2 Question 5

Answer: R is an equivalence relation on Z
Hint: To prove equivalence relation it is necessary that the given relation should be reflexive, symmetry and transitive.
Explanation:
$Given:\! Z\: be\: set\: o\! f\: integers\: and\: R=\left \{\left ( a,b \right ):a,b \: \epsilon \: Z\: and\: a+b\: is\: even\right \}$
Let us check these properties on R.
Reflexivity:
Let a be an arbitrary element of Z.
$Then, a+a=2a\: is\: even\: f\! or\: all\: a \: \epsilon \: Z.$
$(a, a) \: \epsilon \: R\: f\! or\: all\: a \: \epsilon \: Z.$
So, R is reflexive on Z.
Symmetry:
$Let (a, \! b) \: \epsilon \: R$
$then, a+b\: is\: even$
$\Rightarrow b+a\: is\: even$
$(b, a) \: \epsilon \: R\: f\! or\: all\: a, b \: \epsilon \: Z$
So, R is symmetric on Z.
Transitivity:
$Let (a, b) and (b, c) \: \epsilon \: R$
$then, a+b\: and\: b+c\: are\: even$
$Now, let\: a+b=2x\: f\! or\: some\: x \: \epsilon \: Z \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: ...(i)$
$b+c=2y\: f\! or\: some\: y \: \epsilon \: Z \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; ...(ii)$
Adding (i) and (ii)
$\Rightarrow a+b+b+c=2x+2y$
$\Rightarrow a+2b+c=2x+2y$
$\Rightarrow a+c=2x+2y-2b$
$\Rightarrow a+c=2(x+y-b), which\: is\: even\: f\! or\: all\: x, y, b \: \epsilon \: Z$
$Thus, (a, c) \: \epsilon \: R$
So, R is transitive on Z
Therefore, R is reflexive, symmetry and transitive.
Hence, R is an equivalence relation on Z.

Relation Exercise 1.2 Question 6

Answer: R is an equivalence relation on Z
Hint: To prove equivalence relation it is necessary that the given relation should be reflexive, symmetry and transitive.
$Given: \! R=\left \{\left (m, n \right ) \! :\! m-n\: is\: divisible\: by\: 13 \right \} be\: a \: relation\: on\: Z$
Explanation:
Let us check these properties on R.
Reflexivity:
Let m be an arbitrary element of Z.
$Then, m-m=0=0\times 13$
$\Rightarrow m-m \: is \: divisible\: by\: 13.$
$\Rightarrow (m, m) \: \epsilon \: R$
Hence, R is reflexive on Z.
Symmetry:
$Let (m, n) \: \epsilon \: R$
$Then, m-n\: is\: divisible\: by \: 13$
$m-n=13p$
$Here, p \: \epsilon \: Z$
$n-m=13(-p)$
$Here, -p \: \epsilon \: Z$
$n-m\: is\: divisible \: by\: 13$
$(n, m) \: \epsilon \: R \: f\! or\: all\: m, n \: \epsilon \: Z$
So, R is symmetric on Z.
Transitivity:
$Let (m, n) and (n, o) \: \epsilon \: R$
$m-n\: and\: n-o \: are\: divisible\: by \: 13$
$m-n=13p\: f\! or\: some\: p \: \epsilon \: Z \: \: \: \: \: \: \: \: \: ...(i)$
$n-o=13q \: f\! or\: some\: q \: \epsilon \: Z \: \: \: \: \: \: \: ...(ii)$
Adding (i) and (ii)
$m-n+n-o=13p+13q$
$m-o=13(p+q)$
$Here, p+q \: \epsilon \: Z$
$m-o\: is \: divisible\: by \: 13$
$(m, o) \: \epsilon \: R\: f\! or\: all\: m, o \: \epsilon \: Z$
So, R is transitive on Z
Therefore, R is reflexive, symmetric and transitive.
Hence, R is an equivalence relation on Z.

Relation Exercise 1.2 Question 7

Answer: R is an equivalence relation.
Hint: To prove equivalence relation it is necessary that the given relation should be reflexive, symmetric and transitive.
Explanation:
Given: R be a relation on A
$Set\: A\: o\! f\: order\: pair\: o\! f \: integers\: de\! f\! ined\: by\: (x, y) R(u, v)\: if\: xv=yu$
Reflexivity:
$Let (a, b) be\: an\: arbitrary\: element\: o\! f\: the\: set\: A.$
$Then, (a, b) \: \epsilon \: A$
$ab=ba$
$\left (a, b \right )\: R\: (a, b)$
Thus, R is reflexive on A.
Symmetry:
$\text{Let (x, y) and (u, v) }\: \epsilon \: A \: \text{ such that (x, y) R(u, v), Then}$
$xv=yu$
$\Rightarrow vx=uy$
$\Rightarrow uy=vx$
$\left ( u, v \right )R\: (x, y)$
So, R is symmetric on A.
Transitivity:
$Let\left ( x, y\right ), (u, v) and (p, q) \: \epsilon \: R\: such\: that\left ( x, y \right )R (u, v) and\left ( u, v \right ) R (p, q)$
$xv=yu \: \: \: \: \: \: \: \: ...(i)$
$uq=vp \: \: \: \: \: \: ...(ii)$
Multiply eq. (i) and (ii)
$xv\! \times \! uq=yu\! \times\! vp$
$xvuq = yuvp$
cancelling out vu it is common on both sides
$xq=yp$
$\left ( x, y\right ) R (p,q)$
So, R is transitive on A.
Therefore, R is reflexive, symmetric and transitive.
Hence, R is an equivalence relation on A.

Relation Exercise 1.2 Question 8

Answer:R is an equivalence relation and the set of all elements related to 1 is 1.
Hint: To prove equivalence relation it is necessary that the given relation should be reflexive, symmetric and transitive.
$Given: \! Set A=\left \{ x \: \epsilon \: Z ;0 \leq x \leq 12 \right \}$
$Also\: given\: that\: relation\: R=\left \{ \left ( a, b \right )\! :\! a=b \right \} is \: defined\: on \: set\: A$
Explanation:
Reflexivity:
Let a be an arbitrary element of A.
Then,
$a=a \; \; \; \; \; \; [Since, every\: element \: is\: equal\: to\: itsel\! f]$
$(a, a) \: \epsilon \: R\: f\! or\: all\: a \: \epsilon \: A$
So, R is reflexive on A
Symmetry:
$Let (a, b)\: \epsilon \: R$
$a=b$
$b=a$
$(b, a) \: \epsilon \: R\: f\! or\: all\: a,b \: \epsilon \: A$
So, R is symmetric on A.
Transitivity:
$Let\: a, b\: and\: (b, c) \: \epsilon \: R$
$a=b ...(i)\: and \: b=c ...(ii)$
multiplying eqn (i) and (ii), we get
$ab=bc$
$a=c$
$there\! f\! ore, (a, c) \: \epsilon \: R$
So, R is transitive on A.
Therefore, R is reflexive, symmetric and transitive.
Hence, R is an equivalence relation on A.
$R=\left \{ \left ( a, b \right ):a=b\: \right \} and\: 1 \: is\: on\: element\: o\! f \: A.$
$R=\! \left \{\left ( 1, 1 \right ):1=1 \right \}$
Thus, the set of all elements related to 1 is 1.

Relation Exercise 1.2 Question 9

Answer: R is an equivalence relation.
$T\! he\: set\: o\! f \: line\: parallel\: to\: the \: line \: y=2x+4$
$y=2x+C\: f\! or \: all\: C \: \epsilon \: R \: where\: R\: is \: the \: set\: o\! f \: real \: numbers.$
Hint:To prove equivalence relation it is necessary that the given relation should be reflexive, symmetric and transitive.
Given: We have, L is the set of lines
$R=\! \left \{\left (L1, L2 \right ) :L1\: is\: parallel\: to \: L2 \right \} be\: a \: relation\: on\: L.$
Explanation:
Now,
Reflexivity:
$Let \: L1 \: \epsilon \: L$
Since, one line is always parallel to itself.
$(L1, L1) \: \epsilon \: R$
R is reflexive.
Symmetric:
$Let,L1, L2 \: \epsilon \: L \: and (L1,L2) \: \epsilon \: R$
$L1\: is \: parallel\: to\: L2$
$L2\: is\: parallel\: to\: L1$
$(L2, L1) \: \epsilon \: R$
R is symmetric
Transitive:
$Let,L1, L2 \: and\: L3 \: \epsilon \: L\: such\: that (L1, L2) \: \epsilon \: R\: and (L2, L3) \: \epsilon \: R$
$L1\: is\: parallel\: to\: L3\: and\: L2\: is\: parallel\: to\: L3$
$L1\: is \: parallel\: to\: L3$
$(L1, L3) \: \epsilon \: R$
R is transitive
Therefore, R is reflexive, symmetric and transitive.
Hence, R is an equivalence relation.
The set of lines parallel to the line $y=2x+4$ is
$y=2x+C f\! or\: all\: C \: \epsilon \: R$
where R is the set of real numbers.

Relation Exercise 1.2 Question 10

Answer: R is an equivalence relation.
The set of all elements in A related to triangle T is the set of all triangles.
Hint: To prove equivalence relation it is necessary that the given relation should be reflexive, symmetric and transitive.
$Given: \! R= \left \{ \left ( P_{1}, P_{2}\right ):P_{1}\, and\, P_{2} \: have\: same\: number\: o\! f\: sides \right \}$
Explanation:
Reflexive:
$R\: is\: re\! f\! lexive\: \text{ since } (P_{1}, P_{1})\epsilon \: R \: \text{ is }\: the\: same \: polygon\: as\: itsel\! f.$
Symmetric:
$Let (P_{1}, P_{2}) \epsilon \: R$
$P_{1}\, \text{ and }\, P_{2}\: have \: the \: same\: numbers\: o\! f\: sides.$
$P_{2}\, \text{ and }\, P_{1} \: \text{ have }\: the\: same \: number \: o\! f \: side.$
$(P_{2}, P_{1}) \epsilon \: R$
R is symmetric
Transitive:
$Let\:\left ( P_{1}, P_{2} \right ) , (P_{2}, P_{3})\: \epsilon \: R$
$P_{1} \text{ and } P_{2}\: have \: the\: same\: number \: o\! f \: sides$
$Also\: P_{2}\, and\, P_{3}\: have\: the\: same\: number\: o\! f\: sides.$
$P_{1}\, and\, P_{3} \: have \: the\: same\: number\: o\! f \: sides$.
$(P_{1}, P_{3}) \: \epsilon \: R$
R is transitive.
Therefore, R is reflexive, symmetric and transitive.
Hence, R is an equivalence relation.
The elements in A related to the right-angle triangle (T) with sides 3, 4, and 5 are those polygons that have 3 sides (since T is a polygon with 3 sides).
Hence, the set of all elements in A related to triangle T is the set of all triangles.

Relation Exercise 1.2 Question 11

Answer: R is an equivalence relation on A
Hint: To prove equivalence relation it is necessary that the given relation should be reflexive, symmetric and transitive.
$Given: O\; be\; the\; origin\; and\; R=\left \{ \left (P, Q \right ):OP=OQ \right \} be\; a\; relation \; on\; A \; where\; O\; is \; the\; origin$.
Explanation:
Let A be a set of points on a plane.
Reflexivity:
$Let\: P \: \epsilon \: A$
$Since, OP=OP=(P, P) \: \epsilon \: R$
R is reflexive.
Symmetric:
$Let (P, Q) \: \epsilon \: R \: f\! or\: P, Q \: \epsilon \: A$
$Then \: OP=OQ$
$OQ=OP$
$(Q, P) \: \epsilon \: R$
R is symmetric
Transitive:
$Let (P, Q) \: \epsilon \: R\: and\: (Q, S) \: \epsilon \: R$
$OP=OQ \: \: ...(i)\: and \: \: OQ=OS \: \:....(ii)$
Putting (ii) in (i), we get
$OP=OS$
$(P, S)\: \epsilon \: R$
R is transitive
Therefore, R is reflexive, symmetric and transitive.
Thus, R is an equivalence relation on A.

Relation Exercise 1.2 Question 12

Answer: R is an equivalence relation on A.
No number of the subset {1, 3, 5, 7} is related to any number of the subset {2, 4, 6}.
Hint: To prove an equivalence relation it is necessary that the given relation should be reflexive, symmetric and transitive.
$Given: \! \left \{ A=1, 2, 3, 4, 5, 6, 7 \right \}$
$R=\left \{\left ( a, b \right ) : both\: a\: and \: b\: are\: either \: odd \: or\: even\: number \right \}$
Explanation:
$R=\left \{ \left ( 1,1 \right ),\left ( 1 ,3\right ),\left (1, 5 \right ) , (1,7), \left ( 3, 1\right ),\left ( 3, 3 \right ), \left ( 3, 5 \right ),\left ( 3, 7 \right ), \left (5, 1 \right ),\left ( 5, 3 \right ),\left (5, 5 \right ),\left ( 5, 7 \right ),\left ( 7, 1 \right ) ,\left ( 7, 3 \right ),\left ( 7, 5 \right )\left (7,7 \right ), \left ( 2, 2\right ), \left (2, 4 \right ),\left ( 2, 6 \right ),\left (4, 2 \right ),\left ( 4, 4 \right ) ,\left (4, 6 \right ),(6, 2)\left ( 6, 4 \right ),\left ( 6, 6 \right ) \right \}$We observe the following properties of R on A.
Reflexivity:
$Clearly \: \left (1,1 \right ) ,\left ( 2,2 \right ), \left ( 3, 3 \right ),\left ( 4, 4 \right ),\left (5, 5 \right ) ,\left ( 6,6 \right ),\left ( 7,7 \right ) \: \epsilon \: R$.
So, R is a reflexive relation in A
Symmetric:
$Let \: a, b \: \epsilon \: A\: \text{ be such that (a,b) }\: \epsilon \: R.$
$Then \: (a, b)\: \epsilon \: R$
Both a and b are either odd or even.
Both b and a are either odd or even.
$(b, a) \epsilon \: R$
$Thus, (a, b) \epsilon \: R$
$\Rightarrow (b, a) \epsilon \: R\: f\! or\: al\! l\: a, b \: \epsilon \: A$
So, R is a symmetric relation on A
Transitivity:
$Let \: a, b, c \: \epsilon \: A\: be\: such \: that (a, b) \: \epsilon \: R, (b, c) \: \epsilon \: R$
$Then (a, b) \epsilon \: R$
Both a and b are either odd or even.
$(b, c) \epsilon \: R$
⇒ Both b and c are odd or even.
$\\\text{If both a and b are even, then (b, c) }\epsilon \: R\: \text{ both b and c are even}$$\\\text{If both a and b are odd, then (b, c) }\epsilon \: R\: \text{ both b and c are odd}$
⇒ Both a and c are even or odd.
$\therefore (a, c) \epsilon \: R$
$So, (a, b) \epsilon \: R \: \text{ and (b, c) }\epsilon \: R$
$(a, c) \epsilon \: R$
R is a transitive relation on A
Thus, R is reflexive, symmetric and transitive.
Hence, R is an equivalence relation on A.
We observe that two numbers in A are related if both are odd or both are even.
Since, {1, 3, 5, 7} has all odd numbers of A
So, all the numbers of {1, 3, 5, 7} are related to each other
Similarly, all the numbers of {2, 4, 6} are related to each other as it contains all even numbers of set A.
An even, odd number in A is related to an even, odd number in A respectively.
So, no number of the subset {1, 3, 5, 7} is related to any number of the subset {2, 4, 6}

Relation Exercise 1.2 Question 13

Answer: Hence prove, S is not an equivalence relation on R.
Hint: To prove an equivalence relation it is necessary that the given relation should be reflexive, symmetric and transitive.
Explanation:
$Given: S\left \{ \left ( a, b \right ):a^{2}+b^{2}=1 \right \}$
Now,
Reflexivity:
$Let \: a=\frac{1}{2} \, \epsilon \, R$
$Then, a^{2}+a^{2}=\frac{1}{4}+\frac{1}{4}=\frac{1}{2}\neq 1$
$(a, a) \notin S$
S is not reflexive.
When one the three properties are not valid, relation not an equivalence.
Hence, S is not an equivalence relation on R.

Relation Exercise 1.2 Question 14

Answer: R is an equivalence relation on$Z\times Z_{0}$
Hint: To prove equivalence relation it is necessary that the given relation should be reflexive, symmetric and transitive.
$Given: Z\: be\: set\: o\! f \: integers\: and\: Z_{0}\: be\: the\: set\: o\! f \: non-zero \: integers.$
$R=\left \{ ((a, b),( c, d)):ad=bc \right \}\: be\: a\: relation\: on\: Z\times Z_{0}$
Explanation:
Reflexivity:
$(a, b) \: \epsilon \: Z\times Z_{0}$
$ab=ba$
$(a, b), (a, b)) \: \epsilon \: R$
R is reflexive
Symmetric:
$Let\left ( (a, b), (c, d)\right ) \: \epsilon \: R$
$ad=bc$
$cb=da$
$\left ((c,d), (a,b) \right )\epsilon \: R$
R is symmetric
Transitive:
$Let ((a, b), (c, d)) \epsilon \: R\: and\: (c, d), (e, f)) \epsilon \: R$
$ad=bc\: and\: cf=de$
$\frac{a}{b}=\frac{c}{d} \: \:\: \: \: \: ...(i)\: and \: \frac{c}{d} =\frac{e}{f}\: \: \: \: \: ....(ii)$
Using (ii) in (i), we get
$\frac{a}{b}=\frac{e}{f}$
$a\! f=bc$
$((a, b),(e, f)) \epsilon \: R$
R is transitive
Therefore, R is reflexive, symmetric and transitive.
Hence, R is an equivalence relation on$Z\times Z_{0}$

Relation Exercise 1.2 Question 15 (i)

Given:$Hence\: proved, R\cap S \: and \: R\cup S\: are \: symmetric.$
Hint: For symmetric relation.
$\text{a=b is true then b=a is also true.}$
Given: R and S are relations on a set A.
R and S are two symmetric relations on set A.
$To\: Prove: R\cap S\: is \: symmetric$
Explanation:
$Let\left ( a, b \right ) \: \epsilon \: R\cap S$
$(a, b) \: \epsilon \: R \: and\: (a, b) \: \epsilon \: S$
$(b, a) \: \epsilon \: R\: and\: (b, a) \: \epsilon \: S$
$(b, a) \, \epsilon \, R\cap S \; \; \; \; \because [R\: and\: S\: are\: symmetric]$
$(b, a) \: \epsilon \: R\cap S$
$R\cap S\: is \: symmetric$
$To\: prove: \! R\cup S \: is \: symmetric$
$Let (a, b) \: \epsilon \: R\cup S$
$(a, b) \: \epsilon \: R \: or (a, b) \: \epsilon \: S$
$(b, a) \: \epsilon \: R \: or\: (b, a) \: \epsilon \: S$
$(b, a) \: \epsilon \: R\cup S \; \; \; \; \; \; \; \; \because [R\: and\: S\: are\: symmetric]$
$R\cup S\: is\: symmetric$
$Hence \: proved, R\cap S\: and\: R\cup S\: are\: symmetric.$

Relation Exercise 1.2 Question 16

Answer:$R\cup S\: is\: not\: transitive$
Hint: For transitive relation
$x=y \: and\: y=z, then\: also\: x=z.$
Given:$\\\text{R and S are transitive relation on a set A}$
To Prove:$R\cup S\: \text{ may not be a transitive relation on A}$
Explanation:
We will prove this by means of an example.
$Let \: A=\left \{ a, b, c \right \} be\: a\: set$
$R= \left \{ (a,a),(b, b),(c, c),(a,b),(b, a)\right \} and$
$S=\left \{ (a, a),(b, b),(c,c),(b, c),(c, b) \right \} are\: two\: relations\: on\: A$
$\\\text{Clearly, R and S are transitive relation on A.}$
$Now, R\cup S=\left \{ (a, a),(b, b),(c, c),(a, b),(b, a),(b, c),(c, b) \right \}$
$Here, (a, b) \: \epsilon \: R\cup S\: and \: (b, c)\: \epsilon\: R\cup S$
$but \: (a,c) \: \notin\: R\cup\: S$
$R\cup S\: is\: not\: transitive$

Relation Exercise 1.2 Question 17

Answer: Hence prove, R is an equivalence relation.
Hint: To prove an equivalence relation it is necessary that the given relation should be reflexive, symmetric and transitive.
Explanation:
Given: $\text{C\: be\: the\: set\: o\! f\: all\: complex \: numbers.}$
$C_{0}\: \text{be\: the\: set\: o\! f\: all\: non-zero\: complex\: numbers.}$
$\text{Relation R on }C_{0} \text{ be defined as}$
$z_{1}\: R\: z_{2}\Leftrightarrow\frac{ z_{1}-z_{2}}{z_{1}+z_{2}} \text{ is real for all } z_{1}, z_{2}\: \epsilon \: C_{0}$
(i) Test for reflexivity
$Since, \frac{ z_{1}-z_{1}}{z_{1}+z_{1}}=0, which\: is\: a \: real \: number$
$So, (z_{1}, z_{1})\: \epsilon \: R$
Hence, R is a reflexive relation.
(ii) Test for symmetric
$\frac{z_{1}-z_{2}}{z_{1}+z_{2}}=x , where\: x\: is\: real$
$-\frac{ z_{1}-z_{2}}{z_{1}+z_{2}}= -x$
$\frac{z_{2}-z_{1}}{z_{2}+z_{1}}= -x \: is\: also\: a\: real\: number$
$So, (z_{2},z_{1}) \: \epsilon \: R$
Hence, R is a symmetric relation.
(iii) Test for transitivity
$Let (z_{1}, z_{2}) \: \epsilon \: R \: and\: (z_{2}, z_{3} ) \: \epsilon \: R$
$Then\: \frac{z_{1}-z_{2}}{z_{1}+z_{2}}= x$
$z_{1}-z_{2}=xz_{1}+xz_{2}$
$\frac{z_{1}}{z_{2}}=\frac{1+x}{1-x }$ $...(i)$
$Also, \frac{ z_{2}-z_{3}}{z_{2}+z_{3}}=y$
$z_{2}-z_{3}=yz_{2}+yz_{3}$
$z_{2}\left ( 1-y \right )=z_{3}(1+y)$
$\frac{z_{2}}{z_{3}}=\frac{1+y}{1-y }$ $...(ii)$
Dividing (i) and (ii), we get
$\frac{z_{1}}{z_{3}}=\frac{1+x}{1-x}\times \frac{1+y}{1-y}$
$=z\: where\: z\: is\: a\: real \: number.$
$\frac{z_{1}-z_{3}}{z_{1}+z_{3}}=\frac{z-1}{z+1}, which \: is\: real$
$(z_{1}, z_{3})\: \epsilon \: R$ $...(iii)$
From (i), (ii) & (iii)
$\text{Hence, R\: is\: transitivity\: relation.}$
$\text{Hence\: proved, R \: is\: an\: equivalence \: relation.}$