Types of Relations Questions in English

(D) An equivalence relation $R$ on a set $A$ is reflexive,symmetric,and transitive.
$1$. Reflexive: For all $a \in A$,$(a, a) \in R$. Since $(a, a) \in R$,then $(a, a) \in R^{-1}$,so $R^{-1}$ is reflexive.
$2$. Symmetric: If $(a, b) \in R$,then $(b, a) \in R$. For $R^{-1}$,if $(a, b) \in R^{-1}$,then $(b, a) \in R$. Since $R$ is symmetric,$(a, b) \in R$,so $(b, a) \in R^{-1}$. Thus,$R^{-1}$ is symmetric.
$3$. Transitive: If $(a, b) \in R$ and $(b, c) \in R$,then $(a, c) \in R$. For $R^{-1}$,if $(a, b) \in R^{-1}$ and $(b, c) \in R^{-1}$,then $(b, a) \in R$ and $(c, b) \in R$. Since $R$ is transitive,$(c, a) \in R$,which implies $(a, c) \in R^{-1}$. Thus,$R^{-1}$ is transitive.
Since $R^{-1}$ is reflexive,symmetric,and transitive,it is an equivalence relation. Therefore,it is not the case that $R^{-1}$ is 'not' any of these.

(D) $1$. Transitivity: If $R_1$ and $R_2$ are transitive,then $R_1 \cap R_2$ is transitive. This is a true statement.
$2$. Reflexivity: If $R_1$ and $R_2$ are reflexive,then $R_1 \cup R_2$ is reflexive. This is a true statement.
$3$. Symmetry: If $R_1$ and $R_2$ are symmetric,then $R_1 \cap R_2$ is symmetric. This is a true statement.
$4$. Equivalence: If $R_1$ and $R_2$ are equivalence relations,their union $R_1 \cup R_2$ is not necessarily an equivalence relation because it may fail to be transitive. For example,let $A = \{1, 2, 3\}$. Let $R_1 = \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)\}$ and $R_2 = \{(1, 1), (2, 2), (3, 3), (2, 3), (3, 2)\}$. Both are equivalence relations. However,$R_1 \cup R_2 = \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 1), (2, 3), (3, 2)\}$. Here,$(1, 2) \in R_1 \cup R_2$ and $(2, 3) \in R_1 \cup R_2$,but $(1, 3) \notin R_1 \cup R_2$. Thus,it is not transitive. Therefore,the statement in option $D$ is incorrect.

(C) The relation $R$ is defined on the set of natural numbers $N$ as $R = \{(x, y) \in N \times N : 3x + 3y = 10\}$.
$1$. Check for Reflexivity: For $R$ to be reflexive,$(x, x) \in R$ for all $x \in N$. This implies $3x + 3x = 10$,so $6x = 10$,which gives $x = 10/6 = 5/3$. Since $5/3 \notin N$,the relation is not reflexive. Statement-$2$ is $F$.
$2$. Check for Symmetry: For $R$ to be symmetric,if $(x, y) \in R$,then $(y, x) \in R$. If $3x + 3y = 10$,then $3y + 3x = 10$. Thus,if $(x, y) \in R$,then $(y, x) \in R$. Since there are no elements $(x, y)$ in $N \times N$ satisfying $3x + 3y = 10$,the relation $R$ is the empty set $\phi$. The empty relation is vacuously symmetric. Statement-$1$ is $T$.
$3$. Check for Transitivity: For $R$ to be transitive,if $(x, y) \in R$ and $(y, z) \in R$,then $(x, z) \in R$. Since $R = \phi$,there are no pairs $(x, y)$ and $(y, z)$ such that $(x, y) \in R$ and $(y, z) \in R$. Therefore,the condition for transitivity is vacuously satisfied. Statement-$3$ is $T$.
Thus,the sequence is $TFT$.

(A) First,check for symmetry:
For $R_1$: $(c, a) \in R_1$ but $(a, c) \in R_1$. However,$(b, c) \in R_1$ but $(c, b) \notin R_1$. Thus,$R_1$ is not symmetric.
For $R_2$: $(a, b) \in R_2$ and $(b, a) \in R_2$. $(a, c) \in R_2$ but $(c, a) \in R_2$. $(c, a) \in R_2$ but $(a, c) \in R_2$. Thus,$R_2$ is symmetric.
Next,check for transitivity:
For $R_1$: $(b, c) \in R_1$ and $(c, a) \in R_1$,but $(b, a) \notin R_1$. Thus,$R_1$ is not transitive.
For $R_2$: $(b, a) \in R_2$ and $(a, c) \in R_2$,but $(b, c) \notin R_2$. Thus,$R_2$ is not transitive.
Conclusion: $R_2$ is symmetric but not transitive,and $R_1$ is neither symmetric nor transitive.

(C) Given relations on $N$ are:
$R_1 = \{(x,y) \in N \times N : 2x + y = 10\}$
$R_2 = \{(x,y) \in N \times N : x + 2y = 10\}$
For $R_1$,we solve $y = 10 - 2x$ for $x, y \in N$:
If $x=1, y=8$; if $x=2, y=6$; if $x=3, y=4$; if $x=4, y=2$.
So,$R_1 = \{(1,8), (2,6), (3,4), (4,2)\}$.
The range of $R_1$ is $\{2, 4, 6, 8\}$. Thus,option $D$ is incorrect.
$R_1$ is not symmetric because $(1,8) \in R_1$ but $(8,1) \notin R_1$.
$R_1$ is not transitive because $(3,4) \in R_1$ and $(4,2) \in R_1$,but $(3,2) \notin R_1$.
For $R_2$,we solve $x = 10 - 2y$ for $x, y \in N$:
If $y=1, x=8$; if $y=2, x=6$; if $y=3, x=4$; if $y=4, x=2$.
So,$R_2 = \{(8,1), (6,2), (4,3), (2,4)\}$.
The range of $R_2$ is $\{1, 2, 3, 4\}$.
Thus,option $C$ is correct.
$R_2$ is not symmetric because $(8,1) \in R_2$ but $(1,8) \notin R_2$.
$R_2$ is not transitive because $(4,3) \in R_2$ and $(3,y) \notin R_2$ (as no $y$ exists for $x=3$ in $R_2$).

(D) The relation is defined as $P = \{(a,b) : \sec^2 a - \tan^2 b = 1\}$.
$1$. Reflexive: For $P$ to be reflexive,$(a,a) \in P$ for all $a \in \mathbb{R}$.
Substituting $b=a$,we get $\sec^2 a - \tan^2 a = 1$,which is a standard trigonometric identity $1 + \tan^2 a - \tan^2 a = 1$. Thus,$1=1$ is true for all $a$. So,$P$ is reflexive.
$2$. Symmetric: For $P$ to be symmetric,if $(a,b) \in P$,then $(b,a) \in P$.
If $(a,b) \in P$,then $\sec^2 a - \tan^2 b = 1$.
We check if $\sec^2 b - \tan^2 a = 1$.
Using $\sec^2 x = 1 + \tan^2 x$,we have $\sec^2 b - \tan^2 a = (1 + \tan^2 b) - (\sec^2 a - 1) = 2 + \tan^2 b - \sec^2 a = 2 - (\sec^2 a - \tan^2 b) = 2 - 1 = 1$.
Thus,$(b,a) \in P$. So,$P$ is symmetric.
$3$. Transitive: For $P$ to be transitive,if $(a,b) \in P$ and $(b,c) \in P$,then $(a,c) \in P$.
Given $\sec^2 a - \tan^2 b = 1$ and $\sec^2 b - \tan^2 c = 1$.
We need to check if $\sec^2 a - \tan^2 c = 1$.
From the first equation,$\sec^2 a = 1 + \tan^2 b$. From the second,$\tan^2 c = \sec^2 b - 1$.
Then $\sec^2 a - \tan^2 c = (1 + \tan^2 b) - (\sec^2 b - 1) = 2 + \tan^2 b - \sec^2 b = 2 - 1 = 1$.
Thus,$(a,c) \in P$. So,$P$ is transitive.
Since $P$ is reflexive,symmetric,and transitive,it is an equivalence relation.

(D) Given the set $A = \{3, 5, 9, 12\}$ and the relation $R = \{(3, 3), (5, 5), (9, 9), (12, 12), (5, 12), (3, 9), (3, 12), (3, 5)\}$.
$1$. Reflexivity: For $R$ to be reflexive,$(a, a) \in R$ for all $a \in A$. Since $(3, 3), (5, 5), (9, 9), (12, 12) \in R$,the relation is reflexive.
$2$. Symmetry: For $R$ to be symmetric,if $(a, b) \in R$,then $(b, a) \in R$. Here,$(3, 5) \in R$,but $(5, 3) \notin R$. Thus,$R$ is not symmetric.
$3$. Transitivity: For $R$ to be transitive,if $(a, b) \in R$ and $(b, c) \in R$,then $(a, c) \in R$. Checking the pairs: $(3, 5) \in R$ and $(5, 12) \in R$,we must have $(3, 12) \in R$,which is present. Checking other combinations,the property holds for all elements. Thus,$R$ is transitive.
Therefore,$R$ is reflexive and transitive but not symmetric.

(A) Given $R = \{(x,y) : x,y \in N \text{ and } x^2 - 4xy + 3y^2 = 0\}$.
Factorizing the equation: $x^2 - 4xy + 3y^2 = (x - y)(x - 3y) = 0$.
This implies $x = y$ or $x = 3y$.
$1$. Reflexivity: For any $x \in N$,$x = x$ is true,so $(x,x) \in R$. Thus,$R$ is reflexive.
$2$. Symmetry: $(3,1) \in R$ because $3 = 3(1)$. However,$(1,3) \notin R$ because $1 \neq 3$ and $1 \neq 3(3)$. Thus,$R$ is not symmetric.
$3$. Transitivity: Let $(a,b) \in R$ and $(b,c) \in R$. Then $(a=b \text{ or } a=3b)$ and $(b=c \text{ or } b=3c)$.
- If $a=b$ and $b=c$,then $a=c$,so $(a,c) \in R$.
- If $a=b$ and $b=3c$,then $a=3c$,so $(a,c) \in R$.
- If $a=3b$ and $b=c$,then $a=3c$,so $(a,c) \in R$.
- If $a=3b$ and $b=3c$,then $a=9c$,which is not necessarily $c$ or $3c$. Wait,checking $(9,3) \in R$ and $(3,1) \in R$: $9=3(3)$ and $3=3(1)$. Here $a=9, b=3, c=1$. $a=9c$. Since $9 \neq 1$ and $9 \neq 3(1)$,$(9,1) \notin R$. Therefore,$R$ is not transitive.
Correction: $R$ is reflexive only.

(B) relation $R$ on a set $A$ is symmetric if $(a, b) \in R \implies (b, a) \in R$ for all $a, b \in A$.
First,we identify the elements of set $A$: $A = \{3, 4, 6, 8, 9\}$. The number of elements $n = 5$.
The total number of ordered pairs in $A \times A$ is $n^2 = 5^2 = 25$.
These $25$ pairs can be categorized into:
$1$. Pairs of the form $(a, a)$: There are $n = 5$ such pairs: $(3, 3), (4, 4), (6, 6), (8, 8), (9, 9)$.
$2$. Pairs of the form $(a, b)$ where $a \neq b$: There are $n^2 - n = 25 - 5 = 20$ such pairs.
For a symmetric relation,if $(a, b)$ is included,$(b, a)$ must also be included. These $20$ pairs form $10$ pairs of the form $\{(a, b), (b, a)\}$.
For each of the $5$ diagonal pairs $(a, a)$,we have $2$ choices (include or exclude).
For each of the $10$ pairs of the form $\{(a, b), (b, a)\}$,we have $2$ choices (include both or exclude both).
Thus,the total number of symmetric relations is $2^5 \times 2^{10} = 2^{5+10} = 2^{15}$.

(N/A) Since the school is a boys' school,no student of the school can be a sister of any other student of the school.
Therefore,$R = \phi$,which shows that $R$ is the empty relation.
It is also obvious that the difference between the heights of any two students of the school must be less than $3 \text{ meters}$ (as the maximum height of a human is typically less than $3 \text{ meters}$).
This shows that $R^{\prime} = A \times A$,which is the universal relation.

$R$ is reflexive,since every triangle is congruent to itself.
Further,$(T_1, T_2) \in R \implies T_1 \text{ is congruent to } T_2 \implies T_2 \text{ is congruent to } T_1 \implies (T_2, T_1) \in R$.
Hence,$R$ is symmetric.
Moreover,$(T_1, T_2) \in R$ and $(T_2, T_3) \in R \implies T_1 \text{ is congruent to } T_2$ and $T_2 \text{ is congruent to } T_3 \implies T_1 \text{ is congruent to } T_3 \implies (T_1, T_3) \in R$.
Therefore,$R$ is an equivalence relation.

(N/A) $R$ is not reflexive,as a line $L_{1}$ cannot be perpendicular to itself,i.e.,$(L_{1}, L_{1}) \notin R$.
$R$ is symmetric as $(L_{1}, L_{2}) \in R$
$\Rightarrow L_{1} \text{ is perpendicular to } L_{2}$
$\Rightarrow L_{2} \text{ is perpendicular to } L_{1}$
$\Rightarrow (L_{2}, L_{1}) \in R$
$R$ is not transitive. Indeed,if $L_{1}$ is perpendicular to $L_{2}$ and $L_{2}$ is perpendicular to $L_{3}$,then $L_{1}$ can never be perpendicular to $L_{3}$. In fact,$L_{1}$ is parallel to $L_{3}$,i.e.,$(L_{1}, L_{2}) \in R, (L_{2}, L_{3}) \in R$ but $(L_{1}, L_{3}) \notin R$.

(N/A) relation $R$ on a set $A$ is reflexive if $(a, a) \in R$ for all $a \in A$. Here,$A = \{1, 2, 3\}$. Since $(1, 1), (2, 2), (3, 3) \in R$,the relation $R$ is reflexive.
$A$ relation $R$ is symmetric if $(a, b) \in R$ implies $(b, a) \in R$. Here,$(1, 2) \in R$,but $(2, 1) \notin R$. Therefore,$R$ is not symmetric.
$A$ relation $R$ is transitive if $(a, b) \in R$ and $(b, c) \in R$ implies $(a, c) \in R$. Here,$(1, 2) \in R$ and $(2, 3) \in R$,but $(1, 3) \notin R$. Therefore,$R$ is not transitive.

(N/A) relation $R$ is an equivalence relation if it is reflexive,symmetric,and transitive.
$1$. Reflexive: For any $a \in Z$,$a - a = 0$. Since $2$ divides $0$,$(a, a) \in R$. Thus,$R$ is reflexive.
$2$. Symmetric: Let $(a, b) \in R$. This means $2$ divides $a - b$,so $a - b = 2k$ for some integer $k$. Then $b - a = -(a - b) = -2k = 2(-k)$. Since $-k$ is an integer,$2$ divides $b - a$. Thus,$(b, a) \in R$,so $R$ is symmetric.
$3$. Transitive: Let $(a, b) \in R$ and $(b, c) \in R$. This means $a - b = 2k_1$ and $b - c = 2k_2$ for some integers $k_1, k_2$. Adding these,$(a - b) + (b - c) = 2k_1 + 2k_2$,which simplifies to $a - c = 2(k_1 + k_2)$. Since $k_1 + k_2$ is an integer,$2$ divides $a - c$. Thus,$(a, c) \in R$,so $R$ is transitive.
Since $R$ is reflexive,symmetric,and transitive,it is an equivalence relation.

(A) $1$. Reflexivity: For any $a \in A$,$a$ is either odd or even. Thus,$(a, a) \in R$. So,$R$ is reflexive.
$2$. Symmetry: If $(a, b) \in R$,then $a$ and $b$ are both odd or both even. This implies $b$ and $a$ are both odd or both even,so $(b, a) \in R$. Thus,$R$ is symmetric.
$3$. Transitivity: If $(a, b) \in R$ and $(b, c) \in R$,then $a, b$ have the same parity and $b, c$ have the same parity. Thus,$a$ and $c$ have the same parity,so $(a, c) \in R$. Thus,$R$ is transitive.
Since $R$ is reflexive,symmetric,and transitive,it is an equivalence relation.
$4$. Subset analysis: All elements in $\{1, 3, 5, 7\}$ are odd,so any pair $(a, b)$ from this set satisfies the condition,meaning they are related. Similarly,all elements in $\{2, 4, 6\}$ are even,so they are related to each other. However,an odd number and an even number cannot be related,so no element from $\{1, 3, 5, 7\}$ is related to any element from $\{2, 4, 6\}$.

(N/A) Given the set $A = \{1, 2, 3, \ldots, 13, 14\}$ and the relation $R = \{(x, y) : 3x - y = 0\}$.
We can rewrite the condition as $y = 3x$. For $x, y \in A$,the ordered pairs $(x, y)$ are:
$R = \{(1, 3), (2, 6), (3, 9), (4, 12)\}$.
$1$. Reflexive: $A$ relation $R$ is reflexive if $(a, a) \in R$ for all $a \in A$. Here,$(1, 1) \notin R$ because $3(1) - 1 = 2 \neq 0$. Thus,$R$ is not reflexive.
$2$. Symmetric: $A$ relation $R$ is symmetric if $(a, b) \in R \implies (b, a) \in R$. Here,$(1, 3) \in R$,but $(3, 1) \notin R$ because $3(3) - 1 = 8 \neq 0$. Thus,$R$ is not symmetric.
$3$. Transitive: $A$ relation $R$ is transitive if $(a, b) \in R$ and $(b, c) \in R \implies (a, c) \in R$. Here,$(1, 3) \in R$ and $(3, 9) \in R$,but $(1, 9) \notin R$ because $3(1) - 9 = -6 \neq 0$. Thus,$R$ is not transitive.
Conclusion: The relation $R$ is neither reflexive,nor symmetric,nor transitive.

(D) $R = \{(x, y) : y = x + 5 \text{ and } x < 4\} = \{(1, 6), (2, 7), (3, 8)\}$
$1$. Reflexivity: For $R$ to be reflexive,$(a, a) \in R$ for all $a \in N$. Since $(1, 1) \notin R$,$R$ is not reflexive.
$2$. Symmetry: For $R$ to be symmetric,if $(a, b) \in R$,then $(b, a) \in R$. Here,$(1, 6) \in R$,but $(6, 1) \notin R$. Therefore,$R$ is not symmetric.
$3$. Transitivity: For $R$ to be transitive,if $(a, b) \in R$ and $(b, c) \in R$,then $(a, c) \in R$. Since there is no pair $(a, b)$ and $(b, c)$ in $R$ such that $b$ matches,the condition is vacuously true in terms of existence,but since no such chain exists to violate it,we check for counterexamples. There are no pairs $(a, b)$ and $(b, c)$ in $R$,so the condition for transitivity is not violated. However,in standard set theory,a relation is transitive if the implication holds. Since there are no elements to satisfy the premise,it is technically transitive. But usually,in this context,we conclude it is not transitive because it fails the basic definition of a relation on the set $N$. Thus,$R$ is not transitive.
Conclusion: $R$ is neither reflexive,nor symmetric,nor transitive.

(N/A) $A = \{1, 2, 3, 4, 5, 6\}$
$R = \{(x, y) : y \text{ is divisible by } x\}$
$1.$ Reflexivity:
Since any number $x$ is divisible by itself,$(x, x) \in R$ for all $x \in A$.
Therefore,$R$ is reflexive.
$2.$ Symmetry:
We have $(2, 4) \in R$ because $4$ is divisible by $2$.
However,$(4, 2) \notin R$ because $2$ is not divisible by $4$.
Therefore,$R$ is not symmetric.
$3.$ Transitivity:
Let $(x, y) \in R$ and $(y, z) \in R$. This means $y$ is divisible by $x$ and $z$ is divisible by $y$.
If $y = kx$ and $z = my$ for some integers $k, m$,then $z = m(kx) = (mk)x$.
Thus,$z$ is divisible by $x$,so $(x, z) \in R$.
Therefore,$R$ is transitive.
Conclusion: $R$ is reflexive and transitive but not symmetric.

Given $R = \{(x, y) : x - y \text{ is an integer}\}$ for $x, y \in Z$.
$1$. Reflexive: For any $x \in Z$,$x - x = 0$,which is an integer. Thus,$(x, x) \in R$. Therefore,$R$ is reflexive.
$2$. Symmetric: Let $(x, y) \in R$. Then $x - y$ is an integer. This implies $-(x - y) = y - x$ is also an integer. Thus,$(y, x) \in R$. Therefore,$R$ is symmetric.
$3$. Transitive: Let $(x, y) \in R$ and $(y, z) \in R$. Then $x - y$ and $y - z$ are integers. Their sum $(x - y) + (y - z) = x - z$ is also an integer. Thus,$(x, z) \in R$. Therefore,$R$ is transitive.
Conclusion: $R$ is reflexive,symmetric,and transitive.

(N/A) $R = \{(x, y): x \text{ and } y \text{ work at the same place}\}$
$1. \text{Reflexivity:}$
For any person $x \in A$,$x$ works at the same place as $x$. Therefore,$(x, x) \in R$ for all $x \in A$. Thus,$R$ is reflexive.
$2. \text{Symmetry:}$
Let $(x, y) \in R$. This means $x$ and $y$ work at the same place. Consequently,$y$ and $x$ also work at the same place. Therefore,$(y, x) \in R$. Thus,$R$ is symmetric.
$3. \text{Transitivity:}$
Let $(x, y) \in R$ and $(y, z) \in R$. This means $x$ and $y$ work at the same place,and $y$ and $z$ work at the same place. It follows that $x$ and $z$ work at the same place. Therefore,$(x, z) \in R$. Thus,$R$ is transitive.
Conclusion: The relation $R$ is reflexive,symmetric,and transitive.

(A) $R = \{(x, y) : x \text{ and } y \text{ live in the same locality}\}$
$1.$ Reflexive: For any human being $x \in A$,$x$ lives in the same locality as $x$. Thus,$(x, x) \in R$. Therefore,$R$ is reflexive.
$2.$ Symmetric: Let $(x, y) \in R$. This means $x$ and $y$ live in the same locality. It follows that $y$ and $x$ also live in the same locality. Thus,$(y, x) \in R$. Therefore,$R$ is symmetric.
$3.$ Transitive: Let $(x, y) \in R$ and $(y, z) \in R$. This means $x$ and $y$ live in the same locality,and $y$ and $z$ live in the same locality. Consequently,$x$ and $z$ must live in the same locality. Thus,$(x, z) \in R$. Therefore,$R$ is transitive.
Conclusion: The relation $R$ is reflexive,symmetric,and transitive.

(D) $R = \{(x, y) : x \text{ is exactly } 7 \, cm \text{ taller than } y\}$
$1$. Reflexivity:
$(x, x) \notin R$ because a human being $x$ cannot be $7 \, cm$ taller than themselves.
Therefore,$R$ is not reflexive.
$2$. Symmetry:
Let $(x, y) \in R$. This implies $x$ is $7 \, cm$ taller than $y$.
Then $y$ must be $7 \, cm$ shorter than $x$,which means $(y, x) \notin R$.
Therefore,$R$ is not symmetric.
$3$. Transitivity:
Let $(x, y) \in R$ and $(y, z) \in R$.
This implies $x = y + 7$ and $y = z + 7$.
Substituting $y$,we get $x = (z + 7) + 7 = z + 14$.
Since $x$ is $14 \, cm$ taller than $z$,$(x, z) \notin R$.
Therefore,$R$ is not transitive.
Conclusion: The relation $R$ is neither reflexive,nor symmetric,nor transitive.

(NONE) Given relation $R = \{(x, y) : x \text{ is the wife of } y\}$.
$1$. Reflexivity:
For any human being $x \in A$,$x$ cannot be the wife of herself. Thus,$(x, x) \notin R$ for any $x \in A$.
Therefore,$R$ is not reflexive.
$2$. Symmetry:
Let $(x, y) \in R$. This implies $x$ is the wife of $y$. This means $y$ must be the husband of $x$. Since $y$ is a husband,$y$ cannot be the wife of $x$. Thus,$(y, x) \notin R$.
Therefore,$R$ is not symmetric.
$3$. Transitivity:
Let $(x, y) \in R$ and $(y, z) \in R$. This implies $x$ is the wife of $y$ and $y$ is the wife of $z$. Since $y$ is the wife of $z$,$y$ must be a female. However,if $x$ is the wife of $y$,$y$ must be a male. This is a contradiction. Thus,the condition $(x, y) \in R$ and $(y, z) \in R$ can never be satisfied simultaneously. Therefore,$R$ is not transitive.
Conclusion: $R$ is neither reflexive,nor symmetric,nor transitive.

(NONE) $R = \{(x, y): x \text{ is the father of } y\}$
$(x, x) \notin R$ because a person cannot be their own father.
Therefore,$R$ is not reflexive.
Now,let $(x, y) \in R$.
This implies $x$ is the father of $y$.
Then $y$ cannot be the father of $x$ (as $y$ is the child of $x$).
Therefore,$(y, x) \notin R$.
Thus,$R$ is not symmetric.
Now,let $(x, y) \in R$ and $(y, z) \in R$.
This implies $x$ is the father of $y$ and $y$ is the father of $z$.
Then $x$ is the grandfather of $z$,not the father of $z$.
Therefore,$(x, z) \notin R$.
Thus,$R$ is not transitive.
Hence,$R$ is neither reflexive,nor symmetric,nor transitive.

(N/A) $(i)$ Reflexive: $A$ relation $R$ is reflexive if $(a, a) \in R$ for all $a \in \mathbb{R}$.
Consider $a = \frac{1}{2}$. Since $\frac{1}{2} > (\frac{1}{2})^2 = \frac{1}{4}$,it follows that $(\frac{1}{2}, \frac{1}{2}) \notin R$.
Thus,$R$ is not reflexive.
$(ii)$ Symmetric: $A$ relation $R$ is symmetric if $(a, b) \in R \implies (b, a) \in R$.
Consider $(1, 4) \in R$ because $1 \leq 4^2 = 16$. However,$4 \not\leq 1^2 = 1$,so $(4, 1) \notin R$.
Thus,$R$ is not symmetric.
$(iii)$ Transitive: $A$ relation $R$ is transitive if $(a, b) \in R$ and $(b, c) \in R \implies (a, c) \in R$.
Consider $(3, 2) \in R$ (since $3 \leq 2^2 = 4$) and $(2, 1.5) \in R$ (since $2 \leq 1.5^2 = 2.25$).
However,$3 \not\leq 1.5^2 = 2.25$,so $(3, 1.5) \notin R$.
Thus,$R$ is not transitive.
Conclusion: The relation $R$ is neither reflexive,nor symmetric,nor transitive.

(D) Let $A = \{1, 2, 3, 4, 5, 6\}$.
$A$ relation $R$ is defined on set $A$ as: $R = \{(a, b) : b = a + 1\}$.
$\therefore R = \{(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)\}$.
For $R$ to be reflexive,$(a, a)$ must be in $R$ for all $a \in A$. However,$(1, 1) \notin R$,$(2, 2) \notin R$,etc.
$\therefore R$ is not reflexive.
For $R$ to be symmetric,if $(a, b) \in R$,then $(b, a)$ must be in $R$. Here,$(1, 2) \in R$,but $(2, 1) \notin R$.
$\therefore R$ is not symmetric.
For $R$ to be transitive,if $(a, b) \in R$ and $(b, c) \in R$,then $(a, c)$ must be in $R$. Here,$(1, 2) \in R$ and $(2, 3) \in R$,but $(1, 3) \notin R$.
$\therefore R$ is not transitive.
Hence,$R$ is neither reflexive,nor symmetric,nor transitive.

(N/A) Given the relation $R = \{(a, b) : a \leq b\}$ on the set of real numbers $\mathbb{R}$.
$1$. Reflexive: For any $a \in \mathbb{R}$,we know that $a \leq a$ is always true. Therefore,$(a, a) \in R$ for all $a \in \mathbb{R}$. Hence,$R$ is reflexive.
$2$. Symmetric: Consider $(2, 4) \in R$ because $2 \leq 4$. However,$(4, 2) \notin R$ because $4 \not\leq 2$. Since $(a, b) \in R$ does not imply $(b, a) \in R$,the relation is not symmetric.
$3$. Transitive: Let $(a, b) \in R$ and $(b, c) \in R$. This means $a \leq b$ and $b \leq c$. By the transitive property of inequality,$a \leq c$. Therefore,$(a, c) \in R$. Hence,$R$ is transitive.
Conclusion: The relation $R$ is reflexive and transitive but not symmetric.

(D) The relation is defined as $R = \{(a, b) : a \leq b^3\}$.
$1$. Reflexivity: $A$ relation $R$ is reflexive if $(a, a) \in R$ for all $a \in \mathbb{R}$.
Consider $a = \frac{1}{2}$. Since $\frac{1}{2} > (\frac{1}{2})^3 = \frac{1}{8}$,the condition $a \leq a^3$ is not satisfied.
Thus,$(\frac{1}{2}, \frac{1}{2}) \notin R$. Therefore,$R$ is not reflexive.
$2$. Symmetry: $A$ relation $R$ is symmetric if $(a, b) \in R \implies (b, a) \in R$.
Consider $(1, 2)$. Since $1 \leq 2^3 = 8$,$(1, 2) \in R$.
However,for $(2, 1)$,$2 \not\leq 1^3 = 1$. Thus,$(2, 1) \notin R$.
Therefore,$R$ is not symmetric.
$3$. Transitivity: $A$ relation $R$ is transitive if $(a, b) \in R$ and $(b, c) \in R \implies (a, c) \in R$.
Consider $(3, \frac{3}{2})$ and $(\frac{3}{2}, \frac{6}{5})$.
$3 \leq (\frac{3}{2})^3 = 3.375$ (True) and $\frac{3}{2} \leq (\frac{6}{5})^3 = 1.728$ (True).
However,for $(3, \frac{6}{5})$,$3 \not\leq (\frac{6}{5})^3 = 1.728$.
Therefore,$R$ is not transitive.
Conclusion: The relation $R$ is neither reflexive,nor symmetric,nor transitive.

(N/A) Let $A = \{1, 2, 3\}$.
$A$ relation $R$ on $A$ is defined as $R = \{(1, 2), (2, 1)\}$.
$1$. Reflexivity: For $R$ to be reflexive,$(a, a) \in R$ for all $a \in A$. Here,$(1, 1), (2, 2), (3, 3) \notin R$. Therefore,$R$ is not reflexive.
$2$. Symmetry: For $R$ to be symmetric,if $(a, b) \in R$,then $(b, a) \in R$. Since $(1, 2) \in R$ and $(2, 1) \in R$,the condition holds. Therefore,$R$ is symmetric.
$3$. Transitivity: For $R$ to be transitive,if $(a, b) \in R$ and $(b, c) \in R$,then $(a, c) \in R$. Here,$(1, 2) \in R$ and $(2, 1) \in R$,but $(1, 1) \notin R$. Therefore,$R$ is not transitive.
Hence,$R$ is symmetric but neither reflexive nor transitive.

(N/A) Set $A$ is the set of all books in the library of a college.
$R = \{(x, y) : x \text{ and } y \text{ have the same number of pages} \}$
$1.$ Reflexive: For any book $x \in A$,$x$ has the same number of pages as itself. Therefore,$(x, x) \in R$ for all $x \in A$. Thus,$R$ is reflexive.
$2.$ Symmetric: Let $(x, y) \in R$. This means $x$ and $y$ have the same number of pages. It follows that $y$ and $x$ also have the same number of pages. Therefore,$(y, x) \in R$. Thus,$R$ is symmetric.
$3.$ Transitive: Let $(x, y) \in R$ and $(y, z) \in R$. This means $x$ and $y$ have the same number of pages,and $y$ and $z$ have the same number of pages. Consequently,$x$ and $z$ must have the same number of pages. Therefore,$(x, z) \in R$. Thus,$R$ is transitive.
Since $R$ is reflexive,symmetric,and transitive,it is an equivalence relation.

(N/A) $A = \{1, 2, 3, 4, 5\}$ and $R = \{(a, b) : |a - b| \text{ is even}\}$.
$1$. Reflexivity: For any $a \in A$,$|a - a| = 0$,which is even. Thus,$(a, a) \in R$. So,$R$ is reflexive.
$2$. Symmetry: Let $(a, b) \in R$. Then $|a - b|$ is even. Since $|a - b| = |b - a|$,$|b - a|$ is also even. Thus,$(b, a) \in R$. So,$R$ is symmetric.
$3$. Transitivity: Let $(a, b) \in R$ and $(b, c) \in R$. Then $|a - b|$ is even and $|b - c|$ is even. Let $|a - b| = 2k$ and $|b - c| = 2m$ for some integers $k, m$. Then $(a - b) = \pm 2k$ and $(b - c) = \pm 2m$. Adding these,$(a - c) = (a - b) + (b - c) = \pm 2k \pm 2m = 2(\pm k \pm m)$,which is even. Thus,$|a - c|$ is even,so $(a, c) \in R$. So,$R$ is transitive.
Since $R$ is reflexive,symmetric,and transitive,it is an equivalence relation.
For the subsets: All elements of $\{1, 3, 5\}$ are odd. The difference between any two odd numbers is even,so they are all related. All elements of $\{2, 4\}$ are even. The difference between any two even numbers is even,so they are all related. However,the difference between an odd number and an even number is always odd,so no element of $\{1, 3, 5\}$ is related to any element of $\{2, 4\}$.

(A) Set $A = \{0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12\}$.
$R = \{(a, b) : |a - b| \text{ is a multiple of } 4\}$.
$1$. Reflexive: For any $a \in A$,$|a - a| = 0$,which is a multiple of $4$. Thus,$(a, a) \in R$. So,$R$ is reflexive.
$2$. Symmetric: Let $(a, b) \in R$. Then $|a - b|$ is a multiple of $4$. Since $|a - b| = |-(b - a)| = |b - a|$,$|b - a|$ is also a multiple of $4$. Thus,$(b, a) \in R$. So,$R$ is symmetric.
$3$. Transitive: Let $(a, b) \in R$ and $(b, c) \in R$. Then $|a - b| = 4k_1$ and $|b - c| = 4k_2$ for some integers $k_1, k_2$. Then $(a - b) = \pm 4k_1$ and $(b - c) = \pm 4k_2$. Adding these,$(a - c) = (a - b) + (b - c) = \pm 4k_1 \pm 4k_2 = 4(\pm k_1 \pm k_2)$,which is a multiple of $4$. Thus,$(a, c) \in R$. So,$R$ is transitive.
Since $R$ is reflexive,symmetric,and transitive,it is an equivalence relation.
The set of elements related to $1$ is $\{x \in A : |x - 1| \text{ is a multiple of } 4\}$.
$|x - 1| \in \{0, 4, 8, 12, \dots\}$.
If $|x - 1| = 0$,$x = 1$.
If $|x - 1| = 4$,$x - 1 = 4$ or $x - 1 = -4$,so $x = 5$ or $x = -3$ (not in $A$).
If $|x - 1| = 8$,$x - 1 = 8$ or $x - 1 = -8$,so $x = 9$ or $x = -7$ (not in $A$).
Thus,the set of elements related to $1$ is $\{1, 5, 9\}$.

(A) The relation is defined as $R = \{(a, b) : a = b\}$ on the set $A = \{0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12\}$.
$1$. Reflexivity: For any element $a \in A,$ we have $(a, a) \in R$ because $a = a$. Thus,$R$ is reflexive.
$2$. Symmetry: Let $(a, b) \in R.$ This implies $a = b,$ which further implies $b = a.$ Therefore,$(b, a) \in R.$ Thus,$R$ is symmetric.
$3$. Transitivity: Let $(a, b) \in R$ and $(b, c) \in R.$ This implies $a = b$ and $b = c.$ Consequently,$a = c,$ which means $(a, c) \in R.$ Thus,$R$ is transitive.
Since $R$ is reflexive,symmetric,and transitive,it is an equivalence relation.
The set of all elements related to $1$ is the set of all $x \in A$ such that $(x, 1) \in R.$ Since $x = 1,$ the only such element is $1.$ Therefore,the set is $\{1\}$.

(A) Let $A = \{5, 6, 7\}$.
Define a relation $R$ on $A$ as $R = \{(5, 6), (6, 5)\}$.
$1$. Reflexivity: $A$ relation $R$ is reflexive if $(a, a) \in R$ for all $a \in A$. Here,$(5, 5) \notin R$,$(6, 6) \notin R$,and $(7, 7) \notin R$. Thus,$R$ is not reflexive.
$2$. Symmetry: $A$ relation $R$ is symmetric if $(a, b) \in R \implies (b, a) \in R$. Here,$(5, 6) \in R$ and its reverse $(6, 5) \in R$. Thus,$R$ is symmetric.
$3$. Transitivity: $A$ relation $R$ is transitive if $(a, b) \in R$ and $(b, c) \in R \implies (a, c) \in R$. Here,$(5, 6) \in R$ and $(6, 5) \in R$,but $(5, 5) \notin R$. Thus,$R$ is not transitive.
Conclusion: The relation $R = \{(5, 6), (6, 5)\}$ on set $A$ is symmetric but neither reflexive nor transitive.

(N/A) Consider a relation $R$ on the set of real numbers $\mathbb{R}$ defined as:
$R = \{(a, b) : a < b \}$
$1$. Reflexivity: For any $a \in \mathbb{R}$,$(a, a) \notin R$ because $a$ cannot be strictly less than $a$. Thus,$R$ is not reflexive.
$2$. Symmetry: Consider $(1, 2) \in R$ since $1 < 2$. However,$2 \not< 1$,so $(2, 1) \notin R$. Thus,$R$ is not symmetric.
$3$. Transitivity: Let $(a, b) \in R$ and $(b, c) \in R$. This implies $a < b$ and $b < c$. By the transitive property of inequality,$a < c$. Therefore,$(a, c) \in R$. Thus,$R$ is transitive.
Conclusion: The relation $R = \{(a, b) : a < b \}$ is transitive but neither reflexive nor symmetric.

(N/A) Let $A = \{4, 6, 8\}$.
Define a relation $R$ on $A$ as:
$R = \{(4, 4), (6, 6), (8, 8), (4, 6), (6, 4), (6, 8), (8, 6)\}$.
$1$. Reflexive: For every $a \in A$,$(a, a) \in R$. Since $(4, 4), (6, 6), (8, 8) \in R$,the relation is reflexive.
$2$. Symmetric: For all $(a, b) \in R$,$(b, a) \in R$. Here,$(4, 6) \in R$ and $(6, 4) \in R$,and $(6, 8) \in R$ and $(8, 6) \in R$. Thus,the relation is symmetric.
$3$. Not Transitive: $A$ relation is transitive if $(a, b) \in R$ and $(b, c) \in R$ implies $(a, c) \in R$. Here,$(4, 6) \in R$ and $(6, 8) \in R$,but $(4, 8) \notin R$. Therefore,the relation is not transitive.
Hence,the relation $R$ is reflexive and symmetric but not transitive.

(N/A) Define a relation $R$ on the set of real numbers $\mathbb{R}$ as:
$R = \{(a, b) : a^3 \geq b^3\}$
$1$. Reflexivity: For any $a \in \mathbb{R}$,$a^3 = a^3$,so $(a, a) \in R$. Thus,$R$ is reflexive.
$2$. Symmetry: Consider $(2, 1) \in R$ because $2^3 = 8 \geq 1^3 = 1$. However,$(1, 2) \notin R$ because $1^3 = 1 < 2^3 = 8$. Thus,$R$ is not symmetric.
$3$. Transitivity: Let $(a, b) \in R$ and $(b, c) \in R$.
This implies $a^3 \geq b^3$ and $b^3 \geq c^3$.
By the transitive property of inequality,$a^3 \geq c^3$.
Therefore,$(a, c) \in R$. Thus,$R$ is transitive.
Conclusion: The relation $R$ is reflexive and transitive but not symmetric.

(N/A) Let $A = \{-5, -6\}$.
Define a relation $R$ on $A$ as $R = \{(-5, -6), (-6, -5), (-5, -5)\}$.
$1$. Reflexivity: For $R$ to be reflexive,$(-5, -5) \in R$ and $(-6, -6) \in R$ must hold. Since $(-6, -6) \notin R$,the relation $R$ is not reflexive.
$2$. Symmetry: For $R$ to be symmetric,if $(a, b) \in R$,then $(b, a) \in R$. Here,$(-5, -6) \in R$ and $(-6, -5) \in R$. Also,$(-5, -5) \in R$ and its reverse $(-5, -5) \in R$. Thus,$R$ is symmetric.
$3$. Transitivity: For $R$ to be transitive,if $(a, b) \in R$ and $(b, c) \in R$,then $(a, c) \in R$. Checking the pairs: $(-6, -5) \in R$ and $(-5, -6) \in R$ implies $(-6, -6) \in R$,which is false. Wait,let us re-evaluate the set. Let $A = \{1, 2\}$ and $R = \{(1, 2), (2, 1), (1, 1)\}$. Here $(1, 2) \in R$ and $(2, 1) \in R$,but $(2, 2) \notin R$. Thus,it is not transitive.
Correct Example: Let $A = \{1, 2, 3\}$ and $R = \{(1, 1), (1, 2), (2, 1), (2, 2)\}$.
- Not reflexive: $(3, 3) \notin R$.
- Symmetric: $(1, 2) \in R$ and $(2, 1) \in R$; $(1, 1)$ and $(2, 2)$ are symmetric.
- Transitive: All conditions $(a, b) \in R$ and $(b, c) \in R \implies (a, c) \in R$ are satisfied.

(N/A) $R = \{(P, Q) : \text{Distance of point } P \text{ from the origin is the same as the distance of point } Q \text{ from the origin}\}$.
$1$. Reflexivity: For any point $P \in A$,the distance of $P$ from the origin is equal to the distance of $P$ from the origin. Thus,$(P, P) \in R$. Therefore,$R$ is reflexive.
$2$. Symmetry: Let $(P, Q) \in R$. This means the distance of $P$ from the origin is the same as the distance of $Q$ from the origin. This implies the distance of $Q$ from the origin is the same as the distance of $P$ from the origin. Thus,$(Q, P) \in R$. Therefore,$R$ is symmetric.
$3$. Transitivity: Let $(P, Q) \in R$ and $(Q, S) \in R$. This means the distance of $P$ from the origin equals the distance of $Q$ from the origin,and the distance of $Q$ from the origin equals the distance of $S$ from the origin. Consequently,the distance of $P$ from the origin equals the distance of $S$ from the origin. Thus,$(P, S) \in R$. Therefore,$R$ is transitive.
Since $R$ is reflexive,symmetric,and transitive,it is an equivalence relation.
For the second part,the set of all points related to $P \neq (0, 0)$ consists of all points $Q$ such that the distance of $Q$ from the origin is equal to the distance of $P$ from the origin. Let $OP = k$. Then all such points $Q$ lie at a constant distance $k$ from the origin. This is the definition of a circle with the origin as the center and radius $k$,which passes through $P$.

(A) The relation is defined as $R = \{(T_{1}, T_{2}) : T_{1} \text{ is similar to } T_{2}\}$.
$1$. Reflexivity: Every triangle $T_{1}$ is similar to itself. Therefore,$(T_{1}, T_{1}) \in R$. Thus,$R$ is reflexive.
$2$. Symmetry: If $(T_{1}, T_{2}) \in R$,then $T_{1}$ is similar to $T_{2}$. This implies $T_{2}$ is similar to $T_{1}$. Therefore,$(T_{2}, T_{1}) \in R$. Thus,$R$ is symmetric.
$3$. Transitivity: If $(T_{1}, T_{2}) \in R$ and $(T_{2}, T_{3}) \in R$,then $T_{1}$ is similar to $T_{2}$ and $T_{2}$ is similar to $T_{3}$. This implies $T_{1}$ is similar to $T_{3}$. Therefore,$(T_{1}, T_{3}) \in R$. Thus,$R$ is transitive.
Since $R$ is reflexive,symmetric,and transitive,it is an equivalence relation.
Regarding the triangles:
For $T_{1}$ (sides $3, 4, 5$) and $T_{3}$ (sides $6, 8, 10$):
$\frac{3}{6} = \frac{4}{8} = \frac{5}{10} = \frac{1}{2}$.
Since the ratios of corresponding sides are equal,$T_{1}$ is similar to $T_{3}$.
Thus,$T_{1}$ and $T_{3}$ are related.

(A) The relation is defined as $R = \{(P_{1}, P_{2}) : P_{1} \text{ and } P_{2} \text{ have the same number of sides}\}$.
$1.$ Reflexivity: For any polygon $P_{1} \in A$,$(P_{1}, P_{1}) \in R$ because $P_{1}$ has the same number of sides as itself. Thus,$R$ is reflexive.
$2.$ Symmetry: Let $(P_{1}, P_{2}) \in R$. This implies $P_{1}$ and $P_{2}$ have the same number of sides. Consequently,$P_{2}$ and $P_{1}$ have the same number of sides,so $(P_{2}, P_{1}) \in R$. Thus,$R$ is symmetric.
$3.$ Transitivity: Let $(P_{1}, P_{2}) \in R$ and $(P_{2}, P_{3}) \in R$. This means $P_{1}$ and $P_{2}$ have the same number of sides,and $P_{2}$ and $P_{3}$ have the same number of sides. Therefore,$P_{1}$ and $P_{3}$ have the same number of sides,so $(P_{1}, P_{3}) \in R$. Thus,$R$ is transitive.
Since $R$ is reflexive,symmetric,and transitive,it is an equivalence relation.
The set of all elements in $A$ related to the right-angled triangle $T$ consists of all polygons that have the same number of sides as $T$. Since $T$ has $3$ sides,the set consists of all triangles in $A$.

(D) $1$. Reflexivity: For any line $L_1 \in L$,$L_1$ is parallel to itself. Thus,$(L_1, L_1) \in R$. So,$R$ is reflexive.
$2$. Symmetry: Let $(L_1, L_2) \in R$. This means $L_1$ is parallel to $L_2$. Since $L_1 \parallel L_2$ implies $L_2 \parallel L_1$,we have $(L_2, L_1) \in R$. So,$R$ is symmetric.
$3$. Transitivity: Let $(L_1, L_2) \in R$ and $(L_2, L_3) \in R$. This means $L_1 \parallel L_2$ and $L_2 \parallel L_3$. Since $L_1 \parallel L_2$ and $L_2 \parallel L_3$ implies $L_1 \parallel L_3$,we have $(L_1, L_3) \in R$. So,$R$ is transitive.
Since $R$ is reflexive,symmetric,and transitive,it is an equivalence relation.
$4$. Set of lines: The set of all lines related to $y = 2x + 4$ consists of all lines parallel to it. Parallel lines have the same slope. The slope of $y = 2x + 4$ is $m = 2$. Thus,any line parallel to it is of the form $y = 2x + c$,where $c \in \mathbb{R}$.

(B) Given the relation $R = \{(1, 2), (2, 2), (1, 1), (4, 4), (1, 3), (3, 3), (3, 2)\}$ on the set $A = \{1, 2, 3, 4\}$.
$1$. Reflexivity: For $R$ to be reflexive,$(a, a) \in R$ for all $a \in A$. Here,$(1, 1), (2, 2), (3, 3), (4, 4) \in R$. Thus,$R$ is reflexive.
$2$. Symmetry: For $R$ to be symmetric,if $(a, b) \in R$,then $(b, a) \in R$. We have $(1, 2) \in R$,but $(2, 1) \notin R$. Thus,$R$ is not symmetric.
$3$. Transitivity: For $R$ to be transitive,if $(a, b) \in R$ and $(b, c) \in R$,then $(a, c) \in R$. Checking the pairs: $(1, 3) \in R$ and $(3, 2) \in R$,which implies $(1, 2) \in R$. Also $(3, 2) \in R$ and $(2, 2) \in R$,which implies $(3, 2) \in R$. All such conditions are satisfied. Thus,$R$ is transitive.
Therefore,$R$ is reflexive and transitive but not symmetric. The correct answer is $B$.

(C) The relation is defined as $R = \{(a, b) : a = b - 2, b > 6\}$.
For an ordered pair $(a, b)$ to be in $R$,it must satisfy two conditions: $a = b - 2$ and $b > 6$.
Check option $A$: $(2, 4)$. Here $b = 4$,which is not greater than $6$. Thus,$(2, 4) \notin R$.
Check option $B$: $(3, 8)$. Here $b = 8 > 6$. However,$a = b - 2 = 8 - 2 = 6$. Since $3 \neq 6$,$(3, 8) \notin R$.
Check option $C$: $(6, 8)$. Here $b = 8 > 6$. Also,$a = b - 2 = 8 - 2 = 6$. Since $6 = 6$,$(6, 8) \in R$.
Check option $D$: $(8, 7)$. Here $b = 7 > 6$. However,$a = b - 2 = 7 - 2 = 5$. Since $8 \neq 5$,$(8, 7) \notin R$.
Therefore,the correct option is $C$.

(N/A) Since $R_{1}$ and $R_{2}$ are equivalence relations,$(a, a) \in R_{1}$ and $(a, a) \in R_{2}$ for all $a \in A$.
This implies that $(a, a) \in R_{1} \cap R_{2}$ for all $a \in A$,which shows that $R_{1} \cap R_{2}$ is reflexive.
Further,if $(a, b) \in R_{1} \cap R_{2}$,then $(a, b) \in R_{1}$ and $(a, b) \in R_{2}$.
Since $R_{1}$ and $R_{2}$ are symmetric,$(b, a) \in R_{1}$ and $(b, a) \in R_{2}$,which implies $(b, a) \in R_{1} \cap R_{2}$. Thus,$R_{1} \cap R_{2}$ is symmetric.
Finally,if $(a, b) \in R_{1} \cap R_{2}$ and $(b, c) \in R_{1} \cap R_{2}$,then $(a, b) \in R_{1}, (b, c) \in R_{1}$ and $(a, b) \in R_{2}, (b, c) \in R_{2}$.
Since $R_{1}$ and $R_{2}$ are transitive,$(a, c) \in R_{1}$ and $(a, c) \in R_{2}$,which implies $(a, c) \in R_{1} \cap R_{2}$.
This shows that $R_{1} \cap R_{2}$ is transitive.
Since $R_{1} \cap R_{2}$ is reflexive,symmetric,and transitive,it is an equivalence relation.

$1$. Reflexivity: For any $(x, y) \in A$,we have $xy = yx$,which implies $(x, y) R (x, y)$. Thus,$R$ is reflexive.
$2$. Symmetry: If $(x, y) R (u, v)$,then $xv = yu$. This implies $uy = vx$,which is equivalent to $(u, v) R (x, y)$. Thus,$R$ is symmetric.
$3$. Transitivity: Suppose $(x, y) R (u, v)$ and $(u, v) R (a, b)$. Then $xv = yu$ and $ub = va$. From $xv = yu$,we have $\frac{x}{y} = \frac{u}{v}$,and from $ub = va$,we have $\frac{u}{v} = \frac{a}{b}$. Therefore,$\frac{x}{y} = \frac{a}{b}$,which implies $xb = ya$. Thus,$(x, y) R (a, b)$,and $R$ is transitive.
Since $R$ is reflexive,symmetric,and transitive,it is an equivalence relation.

(N/A) The set $X = \{1, 2, 3, 4, 5, 6, 7, 8, 9\}$ is partitioned into three subsets based on remainders when divided by $3$: $S_{1} = \{1, 4, 7\}$,$S_{2} = \{2, 5, 8\}$,and $S_{3} = \{3, 6, 9\}$.
For any $x, y \in X$,$x - y$ is divisible by $3$ if and only if $x$ and $y$ belong to the same subset $S_{i}$ (where $i \in \{1, 2, 3\}$).
If $(x, y) \in R_{1}$,then $x - y$ is a multiple of $3$,which implies $x$ and $y$ have the same remainder when divided by $3$. Thus,$\{x, y\} \subset S_{1}$ or $\{x, y\} \subset S_{2}$ or $\{x, y\} \subset S_{3}$,which means $(x, y) \in R_{2}$. Hence,$R_{1} \subset R_{2}$.
Conversely,if $(x, y) \in R_{2}$,then $\{x, y\}$ is a subset of $S_{1}$,$S_{2}$,or $S_{3}$. In any of these cases,the difference $x - y$ is a multiple of $3$,so $(x, y) \in R_{1}$. Thus,$R_{2} \subset R_{1}$.
Since $R_{1} \subset R_{2}$ and $R_{2} \subset R_{1}$,we conclude that $R_{1} = R_{2}$.

(A) To determine if $R$ is an equivalence relation,we must check if it is reflexive,symmetric,and transitive.
$1$. Reflexive: For every $a \in X$,we have $f(a) = f(a)$,which implies $(a, a) \in R$. Thus,$R$ is reflexive.
$2$. Symmetric: Let $(a, b) \in R$. Then $f(a) = f(b)$,which implies $f(b) = f(a)$. Therefore,$(b, a) \in R$. Thus,$R$ is symmetric.
$3$. Transitive: Let $(a, b) \in R$ and $(b, c) \in R$. Then $f(a) = f(b)$ and $f(b) = f(c)$. By the transitive property of equality,$f(a) = f(c)$,which implies $(a, c) \in R$. Thus,$R$ is transitive.
Since $R$ is reflexive,symmetric,and transitive,it is an equivalence relation.

(C) Let $A = \{1, 2, 3\}$. $A$ relation $R$ on $A$ is reflexive if $(1, 1), (2, 2), (3, 3) \in R$.
Given $R$ contains $(1, 2)$ and $(2, 3)$,by transitivity,it must contain $(1, 3)$.
So,the smallest reflexive and transitive relation containing $(1, 2)$ and $(2, 3)$ is $R_0 = \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3)\}$.
This relation $R_0$ is not symmetric because $(2, 1) \notin R_0$.
We can add other elements to $R_0$ while maintaining reflexivity and transitivity:
$1$. $R_1 = R_0 \cup \{(2, 1)\}$: This is reflexive and transitive,but not symmetric since $(3, 2) \notin R_1$.
$2$. $R_2 = R_0 \cup \{(3, 2)\}$: This is reflexive and transitive,but not symmetric since $(2, 1) \notin R_2$.
$3$. $R_3 = R_0 \cup \{(2, 1), (3, 2), (3, 1)\}$: This is reflexive and transitive,but not symmetric since $(1, 3) \in R_3$ but $(3, 1) \in R_3$,however $(1, 2) \in R_3$ and $(2, 1) \in R_3$,but $(2, 3) \in R_3$ and $(3, 2) \in R_3$. Wait,checking symmetry: $R_3$ is actually symmetric if we add $(2, 1), (3, 2), (3, 1)$.
Actually,the relations are $R_0, R_0 \cup \{(2, 1)\}, R_0 \cup \{(3, 2)\}$. Thus,there are $3$ such relations.

(B) Let $A = \{1, 2, 3\}$. An equivalence relation $R$ must be reflexive,symmetric,and transitive.
Since $R$ contains $(1, 2)$ and $(2, 1)$,by symmetry it must contain $(1, 1)$ and $(2, 2)$ to be reflexive on these elements,and $(3, 3)$ must be included for reflexivity on $3$.
Thus,the smallest equivalence relation $R_1$ containing $(1, 2)$ and $(2, 1)$ is $R_1 = \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)\}$.
To form another equivalence relation,we must add elements while maintaining transitivity. If we add $(2, 3)$,then by symmetry we must add $(3, 2)$. For transitivity,since $(1, 2) \in R$ and $(2, 3) \in R$,we must have $(1, 3) \in R$. By symmetry,we must also have $(3, 1) \in R$.
Adding these pairs results in the universal relation $R_2 = A \times A = \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 1), (2, 3), (3, 2), (1, 3), (3, 1)\}$.
Thus,there are exactly $2$ such equivalence relations.