Types of Relations Questions in English

(B) The set $A$ is given as $A = \{1, 2, 3, 4, 6\}$.
$R$ is a relation on $A$ such that $b$ is exactly divisible by $a$.
We list the ordered pairs $(a, b)$ such that $a$ divides $b$:
$(1, 1), (1, 2), (1, 3), (1, 4), (1, 6), (2, 2), (2, 4), (2, 6), (3, 3), (3, 6), (4, 4), (6, 6)$.
The domain of $R$ is the set of all first elements of the ordered pairs in $R$.
Domain $= \{1, 2, 3, 4, 6\}$.

(A) The relation $R$ is defined as $R = \{(a, b) : a, b \in A, b \text{ is exactly divisible by } a\}$.
We check for each $a \in A$,which $b \in A$ satisfies the condition:
For $a = 1$: $b \in \{1, 2, 3, 4, 6\}$ (since $1$ divides all these numbers).
For $a = 2$: $b \in \{2, 4, 6\}$.
For $a = 3$: $b \in \{3, 6\}$.
For $a = 4$: $b \in \{4\}$.
For $a = 6$: $b \in \{6\}$.
The set of all second elements (the range) is the set of all $b$ values that appear in the ordered pairs of $R$.
Range $= \{1, 2, 3, 4, 6\}$.

(D) $1$. Reflexivity: Since every set is a subset of itself,$A \subset A$ for all $A \in P(X)$. Therefore,$ARA$ holds for all $A \in P(X)$,so $R$ is reflexive.
$2$. Symmetry: For $R$ to be symmetric,$ARB$ must imply $BRA$. Here,$ARB$ means $A \subset B$. This does not imply $B \subset A$. For example,let $X = \{1, 2, 3\}$,$A = \{1\}$,and $B = \{1, 2\}$. Here $A \subset B$ ($ARB$ is true),but $B \not\subset A$ ($BRA$ is false). Thus,$R$ is not symmetric.
$3$. Transitivity: If $ARB$ and $BRC$,then $A \subset B$ and $B \subset C$. By the definition of subsets,this implies $A \subset C$,so $ARC$ holds. Thus,$R$ is transitive.
Conclusion: Since $R$ is not symmetric,it is not an equivalence relation.

(D) The given set is $A = \{1, 2, 3\}$.
For a relation $R$ on $A$ to be reflexive,it must contain $(1, 1), (2, 2), (3, 3)$.
For $R$ to be symmetric,if $(1, 2) \in R$,then $(2, 1) \in R$. If $(1, 3) \in R$,then $(3, 1) \in R$.
Thus,the smallest relation $R$ containing $(1, 2)$ and $(1, 3)$ that is reflexive and symmetric is $R = \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 1), (1, 3), (3, 1)\}$.
Now,check for transitivity: $(2, 1) \in R$ and $(1, 3) \in R$,so for transitivity,$(2, 3)$ must be in $R$. Also,$(3, 1) \in R$ and $(1, 2) \in R$,so $(3, 2)$ must be in $R$.
If we add $(2, 3)$ and $(3, 2)$ to $R$,the relation becomes the universal relation on $A$,which is transitive.
Therefore,the only relation that satisfies the given conditions is the one listed above,which is not transitive because $(2, 1) \in R$ and $(1, 3) \in R$ but $(2, 3) \notin R$.
Thus,there is only $1$ such relation.
The correct answer is $D$.

(A) Given the set $A = \{1, 2, 3\}$.
An equivalence relation must be reflexive,symmetric,and transitive.
First,for reflexivity,the relation must contain $(1, 1), (2, 2), (3, 3)$.
Since the relation must contain $(1, 2)$,by symmetry,it must also contain $(2, 1)$.
Thus,the smallest equivalence relation containing $(1, 2)$ is $R_1 = \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)\}$.
Now,consider adding other elements. If we add $(2, 3)$,by symmetry we must add $(3, 2)$. By transitivity,since $(1, 2) \in R$ and $(2, 3) \in R$,we must have $(1, 3) \in R$. By symmetry,$(3, 1) \in R$. Adding these results in the universal relation $R_2 = A \times A$,which contains all $9$ elements.
Thus,there are exactly $2$ such equivalence relations: $R_1$ and $R_2$.
The correct answer is $A$.

(N/A) To show that $(a, a) \in R$ for all $a \in Q$,we check the definition of the relation $R$.
By definition,$(a, b) \in R$ if $a - b \in Z$.
For the pair $(a, a)$,we have $a - a = 0$.
Since $0$ is an integer $(0 \in Z)$,it follows that $(a, a) \in R$ for all $a \in Q$.

(N/A) Given that $(a, b) \in R$,by the definition of the relation,we have $a - b \in Z$.
Since $Z$ is the set of integers,if $x \in Z$,then $-x \in Z$.
Therefore,$-(a - b) = b - a \in Z$.
Since $b - a \in Z$,by the definition of the relation,we have $(b, a) \in R$.

(N/A) Given that $(a, b) \in R$,it implies $a - b \in Z$.
Given that $(b, c) \in R$,it implies $b - c \in Z$.
We need to check if $(a, c) \in R$,which requires $a - c \in Z$.
Consider $a - c = (a - b) + (b - c)$.
Since the sum of two integers is always an integer,$(a - b) + (b - c) \in Z$.
Therefore,$a - c \in Z$,which implies $(a, c) \in R$.

(B) The relation is defined as $R = \{(a, b) : a, b \in N \text{ and } a = b^2\}$.
For the statement $(a, a) \in R$ to be true for all $a \in N$,the condition $a = a^2$ must hold for every natural number $a \in N$.
Consider $a = 2$. Since $2 \in N$,we check if $(2, 2) \in R$.
Here,$a = 2$ and $b = 2$. The condition $a = b^2$ becomes $2 = 2^2$,which is $2 = 4$.
Since $2 \neq 4$,the pair $(2, 2) \notin R$.
Therefore,the statement $(a, a) \in R$ for all $a \in N$ is false.

(N/A) The relation is defined as $R = \{(a, b) : a, b \in N \text{ and } a = b^2\}$.
To check if $(a, b) \in R$ and $(b, c) \in R$ implies $(a, c) \in R$,we test with counterexamples.
Consider $a = 16, b = 4, c = 2$.
Since $16 = 4^2$,we have $(16, 4) \in R$.
Since $4 = 2^2$,we have $(4, 2) \in R$.
Now,we check if $(16, 2) \in R$.
For $(16, 2)$ to be in $R$,it must satisfy $a = b^2$,which means $16 = 2^2$.
Since $16 \neq 4$,$(16, 2) \notin R$.
Therefore,the statement is false.

(D) relation $R$ is transitive if $(a, b) \in R$ and $(b, c) \in R$ implies $(a, c) \in R$.
For $R_{1}$: Let $a = 2 + \sqrt{3}$,$b = 2 - \sqrt{3}$,and $c = 1 + 2\sqrt{3}$.
Then $a^{2} + b^{2} = (7 + 4\sqrt{3}) + (7 - 4\sqrt{3}) = 14 \in \mathbb{Q}$. So $(a, b) \in R_{1}$.
Also $b^{2} + c^{2} = (7 - 4\sqrt{3}) + (13 + 4\sqrt{3}) = 20 \in \mathbb{Q}$. So $(b, c) \in R_{1}$.
However,$a^{2} + c^{2} = (7 + 4\sqrt{3}) + (13 + 4\sqrt{3}) = 20 + 8\sqrt{3} \notin \mathbb{Q}$.
Thus,$(a, c) \notin R_{1}$,so $R_{1}$ is not transitive.
For $R_{2}$: Let $a^{2} = 1$,$b^{2} = \sqrt{3}$,and $c^{2} = 2 - \sqrt{3}$.
Then $a^{2} + b^{2} = 1 + \sqrt{3} \notin \mathbb{Q}$. So $(a, b) \in R_{2}$.
Also $b^{2} + c^{2} = \sqrt{3} + (2 - \sqrt{3}) = 2 \in \mathbb{Q}$.
Wait,for $(b, c) \in R_{2}$,we need $b^{2} + c^{2} \notin \mathbb{Q}$. Let $b^{2} = \sqrt{3}$ and $c^{2} = 1 + \sqrt{3}$.
Then $b^{2} + c^{2} = 1 + 2\sqrt{3} \notin \mathbb{Q}$. So $(b, c) \in R_{2}$.
Now check $a^{2} + c^{2} = 1 + (1 + \sqrt{3}) = 2 + \sqrt{3} \notin \mathbb{Q}$.
However,if we choose $a^{2} = 1$,$b^{2} = \sqrt{3}$,$c^{2} = 2$,then $a^{2} + b^{2} \notin \mathbb{Q}$ and $b^{2} + c^{2} \notin \mathbb{Q}$,but $a^{2} + c^{2} = 3 \in \mathbb{Q}$.
Thus,$(a, c) \notin R_{2}$,so $R_{2}$ is not transitive.
Therefore,neither $R_{1}$ nor $R_{2}$ is transitive.

(D) Given the set $A = \{2, 3, 4, 5, \ldots, 30\}$.
The equivalence relation is defined as $(a, b) \simeq (c, d)$ if and only if $ad = bc$.
We need to find the number of ordered pairs $(c, d)$ such that $(c, d) \simeq (4, 3)$.
Using the definition,we have $c \times 3 = d \times 4$,which implies $\frac{c}{d} = \frac{4}{3}$.
This means $c = 4k$ and $d = 3k$ for some constant $k$,where $c, d \in A$.
Since $c, d \in \{2, 3, \ldots, 30\}$,we check the possible values for $k$:
For $k=1: (c, d) = (4, 3)$
For $k=2: (c, d) = (8, 6)$
For $k=3: (c, d) = (12, 9)$
For $k=4: (c, d) = (16, 12)$
For $k=5: (c, d) = (20, 15)$
For $k=6: (c, d) = (24, 18)$
For $k=7: (c, d) = (28, 21)$
For $k=8: (c, d) = (32, 24)$,which is not possible since $32 \notin A$.
Thus,the possible ordered pairs are $(4, 3), (8, 6), (12, 9), (16, 12), (20, 15), (24, 18), (28, 21)$.
The total number of such ordered pairs is $7$.

(C) relation $R$ is defined by $A R B \iff B = P A P^{-1}$ for some non-singular matrix $P$.
$1$. Reflexive: For any matrix $A$,we can choose $P = I$ (the identity matrix). Then $I A I^{-1} = I A I = A$. Thus,$A R A$ holds. So,$R$ is reflexive.
$2$. Symmetric: Suppose $A R B$. Then $B = P A P^{-1}$ for some non-singular matrix $P$. We can write $A = P^{-1} B P$. Let $Q = P^{-1}$. Since $P$ is non-singular,$Q$ is also non-singular. Then $A = Q B Q^{-1}$. Thus,$B R A$ holds. So,$R$ is symmetric.
$3$. Transitive: Suppose $A R B$ and $B R C$. Then $B = P A P^{-1}$ and $C = Q B Q^{-1}$ for some non-singular matrices $P$ and $Q$. Substituting $B$ in the second equation,we get $C = Q (P A P^{-1}) Q^{-1} = (Q P) A (P^{-1} Q^{-1}) = (Q P) A (Q P)^{-1}$. Since the product of two non-singular matrices $Q P$ is also non-singular,$A R C$ holds. So,$R$ is transitive.
Since $R$ is reflexive,symmetric,and transitive,it is an equivalence relation.

(D) The relation $R$ is defined as the set of all points $(P, Q)$ such that the distance of $P$ from the origin is equal to the distance of $Q$ from the origin.
The equivalence class of a point $(x_0, y_0)$ is the set of all points $(x, y)$ such that the distance of $(x, y)$ from the origin is equal to the distance of $(x_0, y_0)$ from the origin.
The distance of the point $(1, -1)$ from the origin $(0, 0)$ is given by $\sqrt{1^2 + (-1)^2} = \sqrt{1 + 1} = \sqrt{2}$.
Therefore,the equivalence class of $(1, -1)$ consists of all points $(x, y)$ such that $\sqrt{x^2 + y^2} = \sqrt{2}$.
Squaring both sides,we get $x^2 + y^2 = 2$.
Thus,the equivalence class is the set $S = \{(x, y) \mid x^2 + y^2 = 2\}$.

(B) First,find the set $A \cap B$:
$A = \{(x, y) \in Z \times Z : (x-2)^{2} + y^{2} \leq 4\}$
$B = \{(x, y) \in Z \times Z : x^{2} + y^{2} \leq 4\}$
The integer points $(x, y)$ satisfying both inequalities are $(1, 0), (1, 1), (1, -1), (2, 0), (0, 0)$.
Thus,$n(A \cap B) = 5$.
Next,find the set $A \cap C$:
$A = \{(x, y) \in Z \times Z : (x-2)^{2} + y^{2} \leq 4\}$
$C = \{(x, y) \in Z \times Z : (x-2)^{2} + (y-2)^{2} \leq 4\}$
The integer points $(x, y)$ satisfying both inequalities are $(2, 0), (2, 1), (2, 2), (1, 1), (3, 1)$.
Thus,$n(A \cap C) = 5$.
The total number of relations from $A \cap B$ to $A \cap C$ is given by $2^{n(A \cap B) \times n(A \cap C)} = 2^{5 \times 5} = 2^{25}$.
Comparing this with $2^{p}$,we get $p = 25$.

(B) Let us analyze each option:
$A$: For $0 < |x| - |y| \leq 1$,if $(x, y) \in R$,then $|x| > |y|$. This is not symmetric because $(y, x) \notin R$. It is not transitive because $(3, 2) \in R$ and $(2, 1) \in R$,but $(3, 1) \notin R$ since $|3| - |1| = 2 > 1$. This is correct.
$B$: For $0 < |x - y| \leq 1$,if $(x, y) \in R$,then $|x - y| = |y - x|$,so it is symmetric. However,it is not transitive. For example,$(1, 1.5) \in R$ and $(1.5, 2) \in R$,but $(1, 2) \notin R$ because $|1 - 2| = 1$,which satisfies the condition,but consider $(1, 1.6) \in R$ and $(1.6, 2.2) \in R$,then $|1 - 2.2| = 1.2 > 1$. Thus,it is not transitive. This statement is incorrect.
$C$: For $|x| - |y| \leq 1$,it is reflexive since $|x| - |x| = 0 \leq 1$. It is not symmetric because $(2, 0) \in R$ but $(0, 2) \notin R$ since $|0| - |2| = -2 \leq 1$ is true,but wait,$|0| - |2| = -2 \leq 1$ is true. Actually,$|x| - |y| \leq 1$ is not symmetric because $|2| - |0| = 2 \not\leq 1$. This is correct.
$D$: For $|x - y| \leq 1$,it is reflexive since $|x - x| = 0 \leq 1$. It is symmetric since $|x - y| = |y - x|$. This is correct.
Therefore,the incorrect statement is $B$.

(B) The given equation is $x^{3}-3x^{2}y-xy^{2}+3y^{3}=0$.
Factoring the expression:
$x^{2}(x-3y) - y^{2}(x-3y) = 0$
$(x^{2}-y^{2})(x-3y) = 0$
$(x-y)(x+y)(x-3y) = 0$
This implies $x=y$ or $x=-y$ or $x=3y$.
Since $x, y \in N$,$x=-y$ is impossible as $x, y > 0$.
Thus,the relation is $R = \{(x, y) \in N \times N : x=y \text{ or } x=3y\}$.
$1$. Reflexivity: For any $x \in N$,$(x, x) \in R$ because $x=x$ is true. Thus,$R$ is reflexive.
$2$. Symmetry: Consider $(3, 1) \in R$ because $3=3(1)$. However,$(1, 3) \notin R$ because $1 \neq 3$ and $1 \neq 3(3)=9$. Thus,$R$ is not symmetric.
$3$. Transitivity: Consider $(9, 3) \in R$ (since $9=3(3)$) and $(3, 1) \in R$ (since $3=3(1)$). For $R$ to be transitive,$(9, 1)$ must be in $R$. But $9 \neq 1$ and $9 \neq 3(1)=3$. Thus,$R$ is not transitive.
Therefore,$R$ is reflexive but neither symmetric nor transitive.

(B) For $R_{1} = \{(a, b) \in N \times N : |a - b| \leq 13\}$:
$(i)$ Reflexive: $|a - a| = 0 \leq 13$,so $(a, a) \in R_{1}$. It is reflexive.
(ii) Symmetric: If $|a - b| \leq 13$,then $|b - a| = |-(a - b)| = |a - b| \leq 13$. So $(b, a) \in R_{1}$. It is symmetric.
(iii) Transitive: Let $(1, 10) \in R_{1}$ and $(10, 20) \in R_{1}$. Here $|1 - 10| = 9 \leq 13$ and $|10 - 20| = 10 \leq 13$. However,$|1 - 20| = 19 \not\leq 13$. Thus,$(1, 20) \notin R_{1}$. $R_{1}$ is not transitive,hence not an equivalence relation.
For $R_{2} = \{(a, b) \in N \times N : |a - b| \neq 13\}$:
$(i)$ Reflexive: $|a - a| = 0 \neq 13$. So $(a, a) \in R_{2}$. It is reflexive.
(ii) Symmetric: If $|a - b| \neq 13$,then $|b - a| = |a - b| \neq 13$. So $(b, a) \in R_{2}$. It is symmetric.
(iii) Transitive: Let $(1, 14) \in R_{2}$ and $(14, 1) \in R_{2}$. Here $|1 - 14| = 13$ and $|14 - 1| = 13$. Wait,$(1, 14) \notin R_{2}$ and $(14, 1) \notin R_{2}$. Let's test $(1, 2) \in R_{2}$ and $(2, 15) \in R_{2}$. $|1 - 2| = 1 \neq 13$ and $|2 - 15| = 13$. Since $(2, 15) \notin R_{2}$,this is not a valid counterexample. Consider $(1, 2) \in R_{2}$ and $(2, 14) \in R_{2}$. $|1 - 2| = 1 \neq 13$ and $|2 - 14| = 12 \neq 13$. But $|1 - 14| = 13$,so $(1, 14) \notin R_{2}$. Thus,$R_{2}$ is not transitive,hence not an equivalence relation.
Therefore,neither $R_{1}$ nor $R_{2}$ is an equivalence relation.

(D) The set is $S = \{1, 2, \ldots, 50\}$. The primes $p \in S$ are $\{2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47\}$ (total $15$ primes).
For $R_{1}$,we need $p^{n} \leq 50$ where $n \geq 0$:
- For $p=2$: $2^{0}, 2^{1}, 2^{2}, 2^{3}, 2^{4}, 2^{5} \leq 50$ ($6$ elements).
- For $p=3$: $3^{0}, 3^{1}, 3^{2}, 3^{3} \leq 50$ ($4$ elements).
- For $p=5$: $5^{0}, 5^{1}, 5^{2} \leq 50$ ($3$ elements).
- For $p=7$: $7^{0}, 7^{1}, 7^{2} \leq 50$ ($3$ elements).
- For $p \in \{11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47\}$ ($11$ primes): $p^{0}, p^{1} \leq 50$ ($2$ elements each,total $11 \times 2 = 22$ elements).
Total elements in $R_{1} = 6 + 4 + 3 + 3 + 22 = 38$.
For $R_{2}$,we need $n=0$ or $n=1$:
- For each of the $15$ primes,we have $(p, p^{0})$ and $(p, p^{1})$.
Total elements in $R_{2} = 15 \times 2 = 30$.
$R_{1} - R_{2}$ contains elements where $n \geq 2$:
- $p=2$: $(2, 2^{2}), (2, 2^{3}), (2, 2^{4}), (2, 2^{5})$ ($4$ elements).
- $p=3$: $(3, 3^{2}), (3, 3^{3})$ ($2$ elements).
- $p=5$: $(5, 5^{2})$ ($1$ element).
- $p=7$: $(7, 7^{2})$ ($1$ element).
Total elements in $R_{1} - R_{2} = 4 + 2 + 1 + 1 = 8$.

(D) The relation $R$ is defined as $(x, y) \in R$ if and only if $x$ and $y$ belong to the same subset $A_{i}$.
$1$. Reflexivity: For any $x \in A$,$x$ belongs to some $A_{i}$. Since $x \in A_{i} \iff x \in A_{i}$,$(x, x) \in R$. Thus,$R$ is reflexive.
$2$. Symmetry: If $(x, y) \in R$,then $x$ and $y$ belong to the same $A_{i}$. This implies $y$ and $x$ belong to the same $A_{i}$,so $(y, x) \in R$. Thus,$R$ is symmetric.
$3$. Transitivity: If $(x, y) \in R$ and $(y, z) \in R$,then $x, y \in A_{i}$ and $y, z \in A_{j}$. Since $A_{i} \cap A_{j} = \phi$ for $i \neq j$,we must have $i = j$. Therefore,$x, z \in A_{i}$,which means $(x, z) \in R$. Thus,$R$ is transitive.
Since $R$ is reflexive,symmetric,and transitive,it is an equivalence relation.

(D) For $R_{1}$: $a R_{1} b \iff ab \geq 0$.
$1$. Reflexive: $a \cdot a = a^{2} \geq 0$ for all $a \in \mathbb{R}$. So,$R_{1}$ is reflexive.
$2$. Symmetric: If $ab \geq 0$,then $ba \geq 0$. So,$R_{1}$ is symmetric.
$3$. Transitive: Let $a=2, b=0, c=-2$. Then $ab = 2 \cdot 0 = 0 \geq 0$ and $bc = 0 \cdot (-2) = 0 \geq 0$. However,$ac = 2 \cdot (-2) = -4 < 0$. Thus,$R_{1}$ is not transitive.
Therefore,$R_{1}$ is not an equivalence relation.
For $R_{2}$: $a R_{2} b \iff a \geq b$.
$1$. Reflexive: $a \geq a$ is true for all $a \in \mathbb{R}$. So,$R_{2}$ is reflexive.
$2$. Symmetric: If $a \geq b$,it does not imply $b \geq a$ (e.g.,$2 \geq 1$ but $1 \ngeq 2$). So,$R_{2}$ is not symmetric.
$3$. Transitive: If $a \geq b$ and $b \geq c$,then $a \geq c$. So,$R_{2}$ is transitive.
Since $R_{2}$ is not symmetric,it is not an equivalence relation.
Thus,neither $R_{1}$ nor $R_{2}$ is an equivalence relation.

(D) For $R$ to be reflexive,$xRx$ must hold for all $x \in N$.
$3x + \alpha x = (3 + \alpha)x$ must be a multiple of $7$.
This implies $(3 + \alpha)$ must be a multiple of $7$,so $3 + \alpha = 7k \Rightarrow \alpha = 7k - 3 = 7(k-1) + 4$.
Thus,when $\alpha$ is divided by $7$,the remainder is $4$.
For $R$ to be symmetric,$xRy \Rightarrow yRx$.
$3x + \alpha y = 7n_1$ and $3y + \alpha x = 7n_2$.
Subtracting these,$(3 - \alpha)(x - y) = 7(n_1 - n_2)$. This condition is satisfied if $3 + \alpha$ is a multiple of $7$.
For $R$ to be transitive,$xRy$ and $yRz \Rightarrow xRz$.
$3x + \alpha y = 7n_1$ and $3y + \alpha z = 7n_2$.
From the first,$\alpha y = 7n_1 - 3x$. Substituting into the second: $3y + \alpha z = 7n_2$.
By eliminating $y$,we find that the relation is transitive if $3 + \alpha$ is a multiple of $7$.
Therefore,the condition for $R$ to be an equivalence relation is that $4$ is the remainder when $\alpha$ is divided by $7$.

(B) The relation $R$ is defined as $R = \{(a, b) : a \in \{1, 2, \ldots, 60\}, b = pq, p, q \geq 3, p, q \text{ are primes}, b \leq 60\}$.
Since $a$ can be any value from $1$ to $60$,there are $60$ choices for $a$.
We need to find the number of possible values for $b = pq$ such that $b \leq 60$ and $p, q \geq 3$ are primes.
Case $1$: $p = 3$. Then $b = 3q$. Since $b \leq 60$,$3q \leq 60 \implies q \leq 20$. The primes $q \geq 3$ are $3, 5, 7, 11, 13, 17, 19$. There are $7$ values.
Case $2$: $p = 5$. Then $b = 5q$. Since $b \leq 60$,$5q \leq 60 \implies q \leq 12$. The primes $q \geq 5$ are $5, 7, 11$. There are $3$ values.
Case $3$: $p = 7$. Then $b = 7q$. Since $b \leq 60$,$7q \leq 60 \implies q \leq 8.57$. The prime $q \geq 7$ is $7$. There is $1$ value.
Case $4$: $p = 11$. Then $b = 11q$. Since $b \leq 60$,$11q \leq 60 \implies q \leq 5.45$. No prime $q \geq 11$ satisfies this.
Total values for $b = 7 + 3 + 1 = 11$.
Since $a$ has $60$ choices and $b$ has $11$ choices,the total number of elements in $R$ is $60 \times 11 = 660$.

(D) Let $n(A) = 10$. The total number of relations from $A$ to $A$ is $2^{n^2} = 2^{100}$.
For a relation to be reflexive,all $10$ elements of the form $(a, a)$ must be present in the relation.
There are $100 - 10 = 90$ remaining elements in $A \times A$ that can either be present or absent.
Thus,the total number of reflexive relations is $2^{90}$.
For a relation to be symmetric,if $(a, b)$ is present,then $(b, a)$ must also be present for all $a \neq b$.
There are $\frac{10 \times 9}{2} = 45$ such pairs of the form ${(a, b), (b, a)}$.
For a reflexive and symmetric relation,the $10$ diagonal elements $(a, a)$ must be present,and for each of the $45$ pairs,we have $2$ choices (either both are present or both are absent).
Thus,the number of reflexive and symmetric relations is $2^{45}$.
The number of reflexive relations that are not symmetric is the total number of reflexive relations minus the number of reflexive and symmetric relations.
Required number $= 2^{90} - 2^{45}$.

(A) The relation is defined as $aRb \iff a \mid b^2$ for $a, b \in \mathbb{N}$.
$I.$ Reflexivity: For any $a \in \mathbb{N}$,$a^2$ is always divisible by $a$ (since $a^2/a = a \in \mathbb{N}$). Thus,$(a, a) \in R$ for all $a \in \mathbb{N}$. So,$R$ is reflexive.
$II.$ Symmetry: For $R$ to be symmetric,$aRb \implies bRa$. Let $a=8$ and $b=4$. Here $8 \mid 4^2$ $(8 \mid 16)$ is true,so $(8, 4) \in R$. However,$4 \mid 8^2$ $(4 \mid 64)$ is true,but consider $a=2, b=4$. $2 \mid 4^2$ $(2 \mid 16)$ is true,but $4 \mid 2^2$ $(4 \mid 4)$ is true. Let us take $a=2, b=6$. $2 \mid 6^2$ $(2 \mid 36)$ is true,but $6 \mid 2^2$ $(6 \mid 4)$ is false. Thus,$R$ is not symmetric.
$III.$ Transitivity: For $R$ to be transitive,$(a, b) \in R$ and $(b, c) \in R \implies (a, c) \in R$. Let $a=8, b=4, c=2$.
$(8, 4) \in R$ because $8 \mid 4^2$ $(8 \mid 16)$.
$(4, 2) \in R$ because $4 \mid 2^2$ $(4 \mid 4)$.
Check $(8, 2)$: $8 \mid 2^2$ $(8 \mid 4)$ is false. Thus,$R$ is not transitive.
Therefore,only $I$ is satisfied.

(D) Reflexivity: For $R$ to be reflexive,$(a, a) \in R$ for all $a \in \mathbb{Z}$.
This requires $\operatorname{gcd}(a, a) = |a| = 1$ and $2a \neq a$. This is not true for all $a \in \mathbb{Z}$ (e.g.,$a=2$),so $R$ is not reflexive.
Symmetry: For $R$ to be symmetric,if $(a, b) \in R$,then $(b, a) \in R$.
Let $a=2, b=1$. $\operatorname{gcd}(2, 1) = 1$ and $2(2) = 4 \neq 1$,so $(2, 1) \in R$.
However,for $(1, 2)$,$\operatorname{gcd}(1, 2) = 1$ but $2(1) = 2 = b$. Since the condition $2a \neq b$ is violated,$(1, 2) \notin R$.
Thus,$R$ is not symmetric.
Transitivity: For $R$ to be transitive,if $(a, b) \in R$ and $(b, c) \in R$,then $(a, c) \in R$.
Let $a=14, b=19, c=21$.
$\operatorname{gcd}(14, 19) = 1$ and $2(14) = 28 \neq 19$,so $(14, 19) \in R$.
$\operatorname{gcd}(19, 21) = 1$ and $2(19) = 38 \neq 21$,so $(19, 21) \in R$.
However,$\operatorname{gcd}(14, 21) = 7 \neq 1$,so $(14, 21) \notin R$.
Thus,$R$ is not transitive.
Conclusion: $R$ is neither symmetric nor transitive.

(D) For a relation $R$ on a set $A = \{a, b, c, d\}$ to be an equivalence relation,it must be reflexive,symmetric,and transitive.
$1$. Reflexivity: For all $x \in A$,$(x, x) \in R$. Thus,we must add $(a, a), (b, b), (c, c), (d, d)$. ($4$ elements)
$2$. Symmetry: If $(x, y) \in R$,then $(y, x) \in R$. Given $(a, b), (b, c), (b, d) \in R$,we must add $(b, a), (c, b), (d, b)$. ($3$ elements)
$3$. Transitivity: If $(x, y) \in R$ and $(y, z) \in R$,then $(x, z) \in R$.
From $(a, b)$ and $(b, c)$,add $(a, c)$.
From $(a, b)$ and $(b, d)$,add $(a, d)$.
From $(c, b)$ and $(b, d)$,add $(c, d)$.
From $(d, b)$ and $(b, c)$,add $(d, c)$.
From $(c, a)$ and $(a, b)$,add $(c, b)$ (already added).
From $(d, a)$ and $(a, b)$,add $(d, b)$ (already added).
From $(a, c)$ and $(c, b)$,add $(a, b)$ (already present).
From $(a, d)$ and $(d, b)$,add $(a, b)$ (already present).
From $(c, d)$ and $(d, b)$,add $(c, b)$ (already added).
From $(d, c)$ and $(c, b)$,add $(d, b)$ (already added).
Also need $(c, a)$ and $(d, a)$ due to symmetry of $(a, c)$ and $(a, d)$. ($2$ elements)
The full relation $R$ is $\{(a, a), (b, b), (c, c), (d, d), (a, b), (b, a), (b, c), (c, b), (b, d), (d, b), (a, c), (c, a), (a, d), (d, a), (c, d), (d, c)\}$.
Total elements = $16$. Given elements = $3$. Elements to add = $16 - 3 = 13$.

(D) $1$. Reflexivity: For any $a \in N$,$2a + 3a = 5a$,which is a multiple of $5$. Thus,$a R a$ holds for all $a \in N$. So,$R$ is reflexive.
$2$. Symmetry: Let $a R b$,then $2a + 3b = 5k$ for some integer $k$.
We want to check if $b R a$,i.e.,if $2b + 3a$ is a multiple of $5$.
Note that $(2a + 3b) + (2b + 3a) = 5a + 5b = 5(a + b)$.
Since $2a + 3b = 5k$,we have $2b + 3a = 5(a + b) - 5k = 5(a + b - k)$.
Since $a, b, k$ are integers,$5(a + b - k)$ is a multiple of $5$. Thus,$b R a$ holds. So,$R$ is symmetric.
$3$. Transitivity: Let $a R b$ and $b R c$.
Then $2a + 3b = 5k_1$ and $2b + 3c = 5k_2$ for some integers $k_1, k_2$.
We want to check if $a R c$,i.e.,if $2a + 3c$ is a multiple of $5$.
From $2a + 3b = 5k_1$,we have $2a = 5k_1 - 3b$.
From $2b + 3c = 5k_2$,we have $3c = 5k_2 - 2b$.
Adding these,$2a + 3c = 5k_1 + 5k_2 - 5b = 5(k_1 + k_2 - b)$.
Since $k_1, k_2, b$ are integers,$2a + 3c$ is a multiple of $5$. Thus,$a R c$ holds. So,$R$ is transitive.
Since $R$ is reflexive,symmetric,and transitive,it is an equivalence relation.

(B) Given $R = \{(a, b), (b, c)\}$ on the set $A = \{a, b, c\}$.
For $R$ to be symmetric,if $(x, y) \in R$,then $(y, x) \in R$.
Adding elements for symmetry:
Since $(a, b) \in R$,we must add $(b, a)$.
Since $(b, c) \in R$,we must add $(c, b)$.
Now $R = \{(a, b), (b, a), (b, c), (c, b)\}$.
For $R$ to be transitive,if $(x, y) \in R$ and $(y, z) \in R$,then $(x, z) \in R$.
Using $(a, b) \in R$ and $(b, c) \in R$,we must add $(a, c)$.
Since $(a, c) \in R$,for symmetry,we must add $(c, a)$.
Now $R = \{(a, b), (b, a), (b, c), (c, b), (a, c), (c, a)\}$.
Checking transitivity again:
$(a, b) \in R$ and $(b, a) \in R \Rightarrow (a, a) \in R$.
$(b, c) \in R$ and $(c, b) \in R \Rightarrow (b, b) \in R$.
$(a, c) \in R$ and $(c, a) \in R \Rightarrow (c, c) \in R$.
Adding these,$R = \{(a, b), (b, a), (b, c), (c, b), (a, c), (c, a), (a, a), (b, b), (c, c)\}$.
The elements added are $(b, a), (c, b), (a, c), (c, a), (a, a), (b, b), (c, c)$.
Total elements added $= 7$.

(D) The relation is defined as $(a, b) R (c, d) \iff ad(b-c) = bc(a-d)$.
$1$. Reflexive: For $(a, b) R (a, b)$,we need $ab(b-a) = ba(a-b)$. This simplifies to $ab(b-a) = -ab(b-a)$,which is only true if $ab(b-a) = 0$. Since $a, b \in N$,this is not true for all $(a, b)$. Thus,$R$ is not reflexive.
$2$. Symmetric: If $(a, b) R (c, d)$,then $ad(b-c) = bc(a-d) \Rightarrow adb - adc = bca - bcd \Rightarrow adb + bcd = bca + adc$. Dividing by $abcd$,we get $\frac{1}{c} - \frac{1}{d} = \frac{1}{a} - \frac{1}{b}$. This is symmetric because swapping $(a, b)$ and $(c, d)$ yields the same condition. Thus,$R$ is symmetric.
$3$. Transitive: The condition $\frac{1}{b} - \frac{1}{a} = \frac{1}{d} - \frac{1}{c}$ implies that the relation is an equivalence relation (if defined on $N \times N$ such that $a, b \neq 0$). However,checking the specific form $ad(b-c) = bc(a-d)$,it is equivalent to $\frac{b-c}{bc} = \frac{a-d}{ad} \Rightarrow \frac{1}{c} - \frac{1}{b} = \frac{1}{d} - \frac{1}{a}$. This is transitive. Since it is symmetric and transitive but fails reflexivity (e.g.,$(1, 1) R (1, 1)$ holds,but for general $(a, b)$,it depends on the domain),the correct classification is symmetric and transitive.

(B) For relation $T = \{(a, b) : a^2 - b^2 \in Z\}$:
If $(a, b) \in T$,then $a^2 - b^2 = k$ for some integer $k \in Z$.
Then $b^2 - a^2 = -(a^2 - b^2) = -k$,which is also an integer.
Thus,$(b, a) \in T$,so $T$ is symmetric.
For relation $S = \{(a, b) : a, b \in R - \{0\}, 2 + \frac{a}{b} > 0\}$:
Consider $(a, b) = (1, 1)$. $2 + \frac{1}{1} = 3 > 0$,so $(1, 1) \in S$.
Consider $(a, b) = (1, -1)$. $2 + \frac{1}{-1} = 1 > 0$,so $(1, -1) \in S$.
Consider $(b, a) = (-1, 1)$. $2 + \frac{-1}{1} = 1 > 0$,so $(-1, 1) \in S$.
However,consider $(a, b) = (1, -0.6)$. $2 + \frac{1}{-0.6} = 2 - 1.66 = 0.33 > 0$,so $(1, -0.6) \in S$.
Check $(b, a) = (-0.6, 1)$. $2 + \frac{-0.6}{1} = 1.4 > 0$,so $(-0.6, 1) \in S$.
Let us check symmetry: If $(a, b) \in S$,then $2 + \frac{a}{b} > 0 \Rightarrow \frac{a}{b} > -2$.
For $(b, a) \in S$,we need $2 + \frac{b}{a} > 0 \Rightarrow \frac{b}{a} > -2$.
If we take $a = 1, b = -0.6$,then $\frac{a}{b} = -1.66 > -2$ (True).
But $\frac{b}{a} = -0.6 > -2$ (True).
Wait,let us test $a = 1, b = -0.4$. $2 + \frac{1}{-0.4} = 2 - 2.5 = -0.5 < 0$. So $(1, -0.4) \notin S$.
Let us test $a = 1, b = -0.8$. $2 + \frac{1}{-0.8} = 2 - 1.25 = 0.75 > 0$. So $(1, -0.8) \in S$.
Now check $(b, a) = (-0.8, 1)$. $2 + \frac{-0.8}{1} = 1.2 > 0$. So $(-0.8, 1) \in S$.
Actually,$S$ is not symmetric because if $a=1, b=-0.4$,$(1, -0.4) \notin S$. If $a=1, b=-0.9$,$(1, -0.9) \in S$ and $(-0.9, 1) \in S$.
However,if $a=1, b=-0.1$,$2 + \frac{1}{-0.1} = 2 - 10 = -8 < 0$,so $(1, -0.1) \notin S$.
Since $T$ is symmetric and $S$ is not,option $B$ is correct.

(A) Check for reflexivity:
For any $a \in \mathbb{R}$,$3a - 3a + \sqrt{7} = \sqrt{7}$. Since $\sqrt{7}$ is an irrational number,$(a, a) \in R$. Thus,$R$ is reflexive.
Check for symmetry:
Let $(a, b) \in R$. Then $3a - 3b + \sqrt{7} = I_1$,where $I_1$ is an irrational number.
For symmetry,we need $(b, a) \in R$,which means $3b - 3a + \sqrt{7}$ must be irrational.
Note that $3b - 3a + \sqrt{7} = -(3a - 3b - \sqrt{7}) = -(I_1 - 2\sqrt{7}) = 2\sqrt{7} - I_1$.
If we choose $a = \frac{\sqrt{7}}{3}$ and $b = 0$,then $3(\frac{\sqrt{7}}{3}) - 3(0) + \sqrt{7} = 2\sqrt{7}$ (irrational),so $(a, b) \in R$.
However,for $(b, a)$,we have $3(0) - 3(\frac{\sqrt{7}}{3}) + \sqrt{7} = 0$,which is rational. Thus,$(b, a) \notin R$. $R$ is not symmetric.
Check for transitivity:
Let $(a, b) \in R$ and $(b, c) \in R$. Then $3a - 3b + \sqrt{7} = I_1$ and $3b - 3c + \sqrt{7} = I_2$,where $I_1, I_2$ are irrational.
For transitivity,$(a, c) \in R$ implies $3a - 3c + \sqrt{7}$ must be irrational.
Adding the two relations: $(3a - 3b + \sqrt{7}) + (3b - 3c + \sqrt{7}) = 3a - 3c + 2\sqrt{7} = I_1 + I_2$.
So,$3a - 3c + \sqrt{7} = I_1 + I_2 - \sqrt{7}$.
If we take $a = \frac{\sqrt{7}}{3}, b = 1, c = \frac{2\sqrt{7}}{3}$,then $(a, b) \in R$ and $(b, c) \in R$,but $3a - 3c + \sqrt{7} = 3(\frac{\sqrt{7}}{3}) - 3(\frac{2\sqrt{7}}{3}) + \sqrt{7} = \sqrt{7} - 2\sqrt{7} + \sqrt{7} = 0$,which is rational. Thus,$(a, c) \notin R$. $R$ is not transitive.

(A) For relation $R_1$: The condition $(A \cap B^c) \cup (B \cap A^c) = \varnothing$ is the definition of the symmetric difference $A \Delta B = \varnothing$,which implies $A = B$. Since $A = B$ is an equivalence relation (reflexive,symmetric,and transitive),$R_1$ is an equivalence relation.
For relation $R_2$: The condition $A \cup B^c = B \cup A^c$ can be analyzed using set properties.
$A \cup B^c = B \cup A^c \iff (A \cup B^c) \cap (A \cap B) = (B \cup A^c) \cap (A \cap B) \iff A = B$.
Alternatively,using the Venn diagram regions where $a, b, c, d$ represent disjoint regions: $A = a \cup c$ and $B = b \cup c$.
$A \cup B^c = (a \cup c) \cup (a \cup d) = a \cup c \cup d$.
$B \cup A^c = (b \cup c) \cup (b \cup d) = b \cup c \cup d$.
Equating these gives $a \cup c \cup d = b \cup c \cup d$,which implies $a = b$. Since $a$ and $b$ are the regions unique to $A$ and $B$ respectively,$a = b = \varnothing$ implies $A = B$. Thus,$R_2$ is also an equivalence relation.
Therefore,both $R_1$ and $R_2$ are equivalence relations.

(B) The set $A$ is given as $A = \{0, 3, 4, 6, 7, 8, 9, 10\}$. The number of odd elements is $3$ $(\{3, 7, 9\})$ and the number of even elements is $5$ $(\{0, 4, 6, 8, 10\})$.
The relation $R$ contains pairs $(x, y)$ such that $x - y$ is an odd positive integer or $x - y = 2$.
$1$. Pairs where $x - y$ is an odd positive integer: Since $x - y$ is odd,one must be odd and the other even. There are $3 \times 5 = 15$ such pairs where $x > y$.
$2$. Pairs where $x - y = 2$: These are $(6, 4), (8, 6), (10, 8), (9, 7)$. There are $4$ such pairs where $x > y$.
Total pairs in $R$ with $x > y$ is $15 + 4 = 19$.
For $R$ to be symmetric,if $(x, y) \in R$,then $(y, x)$ must also be in $R$. Since all $19$ pairs currently satisfy $x > y$,their symmetric counterparts $(y, x)$ where $y < x$ are not currently in $R$.
Therefore,we must add $19$ elements to $R$ to make it symmetric.

(D) Given $A = \{1, 2, 3, 4, 5, 6, 7\}$ and $R = \{(x, y) \in A \times A : x + y = 7\}$.
Listing the elements of $R$: $R = \{(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)\}$.
$1$. Reflexive: For $R$ to be reflexive,$(x, x) \in R$ for all $x \in A$. Since $(1, 1) \notin R$,$R$ is not reflexive.
$2$. Symmetric: For $R$ to be symmetric,if $(x, y) \in R$,then $(y, x) \in R$. Since $(1, 6) \in R$ and $(6, 1) \in R$,$(2, 5) \in R$ and $(5, 2) \in R$,etc.,$R$ is symmetric.
$3$. Transitive: For $R$ to be transitive,if $(x, y) \in R$ and $(y, z) \in R$,then $(x, z) \in R$. Here $(1, 6) \in R$ and $(6, 1) \in R$,but $(1, 1) \notin R$. Thus,$R$ is not transitive.
Therefore,$R$ is symmetric but neither reflexive nor transitive.

(A) We are given $A = \{2, 3, 4\}$ and $B = \{8, 9, 12\}$.
We need to find the number of pairs $((a_1, b_1), (a_2, b_2))$ such that $a_1$ divides $b_2$ and $a_2$ divides $b_1$,where $a_1, a_2 \in A$ and $b_1, b_2 \in B$.
Let $S$ be the set of pairs $(a, b) \in A \times B$ such that $a$ divides $b$.
For $a = 2$,$b \in \{8, 12\}$ ($2$ pairs: $(2, 8), (2, 12)$).
For $a = 3$,$b \in \{9, 12\}$ ($2$ pairs: $(3, 9), (3, 12)$).
For $a = 4$,$b \in \{8, 12\}$ ($2$ pairs: $(4, 8), (4, 12)$).
Thus,there are $2 + 2 + 2 = 6$ such pairs in $S$.
The relation $R$ is defined as the set of pairs $((a_1, b_1), (a_2, b_2))$ such that $(a_1, b_2) \in S$ and $(a_2, b_1) \in S$.
Since there are $6$ choices for the pair $(a_1, b_2)$ and $6$ choices for the pair $(a_2, b_1)$,the total number of elements in $R$ is $6 \times 6 = 36$.

(B) Let the set be $A = \{1, 2, 3\}$.
For a relation $R$ to be reflexive,it must contain $(1,1), (2,2), (3,3)$.
Given that $(1,2) \in R$ and $(2,3) \in R$,for $R$ to be transitive,it must contain $(1,3)$ because $(1,2) \in R$ and $(2,3) \in R \implies (1,3) \in R$.
Thus,the minimal relation $R$ containing these elements is $R_0 = \{(1,1), (2,2), (3,3), (1,2), (2,3), (1,3)\}$.
This relation $R_0$ is reflexive and transitive. It is not symmetric because $(1,2) \in R_0$ but $(2,1) \notin R_0$.
If we add any other element to $R_0$,such as $(2,1)$,the relation becomes symmetric with respect to $(1,2)$ and $(2,1)$. If we add $(3,2)$,it becomes symmetric with respect to $(2,3)$ and $(3,2)$.
Therefore,there is only $1$ such relation,which is $R_0$ itself.

(C) Given $A = \{-4, -3, -2, 0, 1, 3, 4\}$.
First,we find the elements of $R$ based on the conditions $b = |a|$ or $b^2 = a + 1$:
For $b = |a|$: $(-4, 4), (-3, 3), (-2, 2), (0, 0), (1, 1), (3, 3), (4, 4)$. Note that $(-2, 2)$ is not in $A \times A$ because $2 \notin A$. So,we have $(-4, 4), (-3, 3), (0, 0), (1, 1), (3, 3), (4, 4)$.
For $b^2 = a + 1$: If $a = -4, b^2 = -3$ (no); $a = -3, b^2 = -2$ (no); $a = -2, b^2 = -1$ (no); $a = 0, b^2 = 1 \Rightarrow b = 1, -1$ (only $1 \in A$); $a = 1, b^2 = 2$ (no); $a = 3, b^2 = 4 \Rightarrow b = 2, -2$ (no); $a = 4, b^2 = 5$ (no).
Thus,$R = \{(-4, 4), (-3, 3), (0, 0), (1, 1), (3, 3), (4, 4), (0, 1)\}$.
For $R$ to be reflexive,we need $(a, a) \in R$ for all $a \in A$. Missing elements: $(-4, -4), (-3, -3), (-2, -2)$. ($3$ elements).
Now $R' = R \cup \{(-4, -4), (-3, -3), (-2, -2)\} = \{(-4, 4), (-3, 3), (0, 0), (1, 1), (3, 3), (4, 4), (0, 1), (-4, -4), (-3, -3), (-2, -2)\}$.
For $R'$ to be symmetric,if $(a, b) \in R'$,then $(b, a) \in R'$.
Pairs to add: $(4, -4), (3, -3), (1, 0)$. ($3$ elements).
Total elements added = $3 + 3 = 6$.

(A) Given $A = \{1, 2, 3, 4\}$. The relation $R$ is defined on $A \times A$ such that $2a + 3b = 4c + 5d$,where $a, b, c, d \in A$.
Let $S_1 = 2a + 3b$ and $S_2 = 4c + 5d$.
Possible values for $S_1$ for $a, b \in \{1, 2, 3, 4\}$:
If $a=1: 2+3=5, 2+6=8, 2+9=11, 2+12=14$
If $a=2: 4+3=7, 4+6=10, 4+9=13, 4+12=16$
If $a=3: 6+3=9, 6+6=12, 6+9=15, 6+12=18$
If $a=4: 8+3=11, 8+6=14, 8+9=17, 8+12=20$
Possible values for $S_2$ for $c, d \in \{1, 2, 3, 4\}$:
If $c=1: 4+5=9, 4+10=14, 4+15=19, 4+20=24$
If $c=2: 8+5=13, 8+10=18, 8+15=23, 8+20=28$
If $c=3: 12+5=17, 12+10=22, 12+15=27, 12+20=32$
If $c=4: 16+5=21, 16+10=26, 16+15=31, 16+20=36$
We look for common values $\alpha = S_1 = S_2$:
For $\alpha = 9$: $(a,b)=(3,1)$ and $(c,d)=(1,1) \implies ((3,1),(1,1))$
For $\alpha = 13$: $(a,b)=(2,3)$ and $(c,d)=(2,1) \implies ((2,3),(2,1))$
For $\alpha = 14$: $(a,b)=(1,4)$ and $(c,d)=(1,2) \implies ((1,4),(1,2))$
For $\alpha = 14$: $(a,b)=(4,2)$ and $(c,d)=(1,2) \implies ((4,2),(1,2))$
For $\alpha = 17$: $(a,b)=(4,3)$ and $(c,d)=(3,1) \implies ((4,3),(3,1))$
For $\alpha = 18$: $(a,b)=(3,4)$ and $(c,d)=(2,2) \implies ((3,4),(2,2))$
Total number of elements is $6$.

(D) Let $S = \{1, 2, 3, \ldots, 10\}$.
The relation is defined as $R = \{(A, B) : A \cap B \neq \phi; A, B \in M\}$.
$1$. Reflexivity: $A$ relation $R$ is reflexive if $(A, A) \in R$ for all $A \in M$. This requires $A \cap A \neq \phi$,which means $A \neq \phi$. Since the empty set $\phi$ is a subset of $S$ and $\phi \cap \phi = \phi$,the condition $A \cap A \neq \phi$ is not satisfied for $A = \phi$. Thus,$R$ is not reflexive.
$2$. Symmetry: $A$ relation $R$ is symmetric if $(A, B) \in R \implies (B, A) \in R$. If $A \cap B \neq \phi$,then $B \cap A \neq \phi$ because intersection is commutative. Thus,$R$ is symmetric.
$3$. Transitivity: $A$ relation $R$ is transitive if $(A, B) \in R$ and $(B, C) \in R \implies (A, C) \in R$. Let $S = \{1, 2, 3\}$. Let $A = \{1, 2\}$,$B = \{2, 3\}$,and $C = \{3\}$. Here,$A \cap B = \{2\} \neq \phi$ and $B \cap C = \{3\} \neq \phi$. However,$A \cap C = \phi$. Since $A \cap C = \phi$,the condition $(A, C) \in R$ is not satisfied. Thus,$R$ is not transitive.
Therefore,the relation is symmetric only.

(A) $1$. Reflexivity: For any $(a, b) \in \mathbb{Z} \times \mathbb{Z}$,we have $ab - ba = 0$,which is divisible by $5$. Thus,$(a, b) R (a, b)$ holds. So,$R$ is reflexive.
$2$. Symmetry: Let $(a, b) R (c, d)$. Then $ad - bc = 5k$ for some integer $k$. This implies $bc - ad = 5(-k)$,which is also divisible by $5$. Thus,$(c, d) R (a, b)$ holds. So,$R$ is symmetric.
$3$. Transitivity: Consider $(3, 1) R (10, 5)$ because $3(5) - 1(10) = 15 - 10 = 5$,which is divisible by $5$. Also,$(10, 5) R (1, 1)$ because $10(1) - 5(1) = 5$,which is divisible by $5$. However,for $(3, 1)$ and $(1, 1)$,we have $3(1) - 1(1) = 2$,which is not divisible by $5$. Thus,$(3, 1)$ is not related to $(1, 1)$. Therefore,$R$ is not transitive.

(A) For $R$ to be an equivalence relation,it must be reflexive,symmetric,and transitive.
$1$. Reflexivity: Since the set is $\{1, 2, 3, 4\}$,$R$ must contain $\{(1, 1), (2, 2), (3, 3), (4, 4)\}$.
$2$. Symmetry: Given $\{(1, 2), (1, 3)\} \subset R$,by symmetry,$R$ must also contain $\{(2, 1), (3, 1)\}$.
$3$. Transitivity: Since $(2, 1) \in R$ and $(1, 3) \in R$,by transitivity,$(2, 3) \in R$. By symmetry,$(3, 2) \in R$ must also be in $R$.
Combining these,the set $R$ is $\{(1, 1), (2, 2), (3, 3), (4, 4), (1, 2), (2, 1), (1, 3), (3, 1), (2, 3), (3, 2)\}$.
Counting the elements,we have $10$ elements in $R$.

(C) Let $n$ be the number of elements in the set. Here,$n = 4$.
The total number of symmetric relations on a set with $n$ elements is given by $2^{\frac{n(n+1)}{2}}$.
For $n = 4$,the number of symmetric relations is $2^{\frac{4(5)}{2}} = 2^{10} = 1024$.
The number of symmetric relations that are also reflexive is given by $2^{\frac{n(n-1)}{2}}$.
For $n = 4$,the number of symmetric and reflexive relations is $2^{\frac{4(3)}{2}} = 2^6 = 64$.
The number of symmetric relations that are not reflexive is the total number of symmetric relations minus the number of symmetric and reflexive relations.
Number of symmetric relations not reflexive $= 2^{10} - 2^6 = 1024 - 64 = 960$.

(A) An equivalence relation $S$ must be reflexive,symmetric,and transitive.
Given $R = \{(1, 2), (2, 3), (1, 4)\}$.
$1$. Reflexivity: $S$ must contain $(1, 1), (2, 2), (3, 3), (4, 4)$.
$2$. Symmetry: Since $(1, 2) \in S$,then $(2, 1) \in S$. Since $(2, 3) \in S$,then $(3, 2) \in S$. Since $(1, 4) \in S$,then $(4, 1) \in S$.
$3$. Transitivity: Since $(1, 2) \in S$ and $(2, 3) \in S$,then $(1, 3) \in S$. By symmetry,$(3, 1) \in S$.
Since $(1, 3) \in S$ and $(3, 2) \in S$,we already have $(1, 2) \in S$.
Since $(1, 4) \in S$ and $(4, 1) \in S$,we have $(1, 1) \in S$ (already included).
Since $(2, 3) \in S$ and $(3, 2) \in S$,we have $(2, 2) \in S$ (already included).
Since $(1, 2) \in S$ and $(2, 1) \in S$,we have $(1, 1) \in S$.
Checking the equivalence classes: $1$ is related to $2, 3, 4$. Thus,$1, 2, 3, 4$ are all in the same equivalence class.
If all elements are in one equivalence class,$S = A \times A$.
The number of elements $n = 4 \times 4 = 16$.

(B) The relation $R$ is defined as $(x, y) \in R \iff 2x = 3y$,which implies $y = \frac{2}{3}x$.
Since $x, y \in \{1, 2, \ldots, 100\}$,$x$ must be a multiple of $3$.
Let $x = 3k$,then $y = 2k$.
For $1 \le 2k \le 100$,we have $k \le 50$.
For $1 \le 3k \le 100$,we have $k \le 33$.
Thus,$k$ can take values from $1$ to $33$.
The elements of $R$ are $\{(3, 2), (6, 4), (9, 6), \ldots, (99, 66)\}$.
The number of elements in $R$ is $n(R) = 33$.
For $R_1$ to be a symmetric relation such that $R \subset R_1$,for every $(x, y) \in R$,the pair $(y, x)$ must also be in $R_1$.
Since $R$ is not symmetric (e.g.,$(3, 2) \in R$ but $(2, 3) \notin R$),we must include all $(y, x)$ pairs in $R_1$.
Thus,$R_1 = R \cup R^{-1} = \{(3, 2), (6, 4), \ldots, (99, 66), (2, 3), (4, 6), \ldots, (66, 99)\}$.
The number of elements in $R_1$ is $n = n(R) + n(R^{-1}) = 33 + 33 = 66$.

(B) The set $A = \{1, 2, 3, \ldots, 20\}$.
$R_1 = \{(a, b) : b \text{ is divisible by } a\}$. The number of elements in $R_1$ is the sum of the number of multiples of each $a \in A$ that are $\leq 20$.
For $a=1$,there are $20$ multiples. For $a=2$,there are $10$ multiples. For $a=3$,there are $6$ multiples. For $a=4$,there are $5$ multiples. For $a=5$,there are $4$ multiples. For $a=6$,there are $3$ multiples. For $a=7, 8, 9, 10$,there are $2$ multiples each. For $a=11, 12, \ldots, 20$,there is $1$ multiple each (the number itself).
$n(R_1) = 20 + 10 + 6 + 5 + 4 + 3 + (4 \times 2) + (10 \times 1) = 48 + 8 + 10 = 66$.
$R_2 = \{(a, b) : a \text{ is an integral multiple of } b\}$. This is equivalent to saying $b$ is a divisor of $a$,which is the same as $R_1$.
However,the definition of $R_2$ given is $a = kb$ for some integer $k$. Since $a, b \in A$,$k$ must be a positive integer. This is the same as $R_1$.
Wait,$R_1 = \{(a, b) : b = ka, k \in \mathbb{Z}^+\}$ and $R_2 = \{(a, b) : a = kb, k \in \mathbb{Z}^+\}$.
$R_1 \cap R_2 = \{(a, b) : b = ka \text{ and } a = mb\} = \{(a, b) : a = mka \implies mk = 1\}$. Since $m, k \in \mathbb{Z}^+$,$m=1, k=1$. Thus $a=b$.
$R_1 \cap R_2 = \{(1, 1), (2, 2), \ldots, (20, 20)\}$.
$n(R_1 \cap R_2) = 20$.
$n(R_1 - R_2) = n(R_1) - n(R_1 \cap R_2) = 66 - 20 = 46$.

(B) For relation $R_1$: $a R_1 b \Leftrightarrow a^2+b^2=1$ for $a, b \in R$.
$1$. Reflexivity: For $a=0.5$,$a^2+a^2 = 0.25+0.25 = 0.5 \neq 1$. So,$R_1$ is not reflexive.
$2$. Symmetry: If $a^2+b^2=1$,then $b^2+a^2=1$,so $b R_1 a$. $R_1$ is symmetric.
$3$. Transitivity: If $a R_1 b$ and $b R_1 c$,then $a^2+b^2=1$ and $b^2+c^2=1$. This does not imply $a^2+c^2=1$. For example,$a=1, b=0, c=1$. $1^2+0^2=1$ and $0^2+1^2=1$,but $1^2+1^2=2 \neq 1$. Thus,$R_1$ is not transitive.
For relation $R_2$: $(a, b) R_2 (c, d) \Leftrightarrow a+d=b+c$ for $(a, b), (c, d) \in N \times N$.
$1$. Reflexivity: $a+b=b+a$ is true,so $(a, b) R_2 (a, b)$.
$2$. Symmetry: If $a+d=b+c$,then $c+b=d+a$,so $(c, d) R_2 (a, b)$.
$3$. Transitivity: If $(a, b) R_2 (c, d)$ and $(c, d) R_2 (e, f)$,then $a+d=b+c$ and $c+f=d+e$. Adding these,$a+d+c+f = b+c+d+e \Rightarrow a+f=b+e$. Thus,$(a, b) R_2 (e, f)$.
Since $R_2$ is reflexive,symmetric,and transitive,it is an equivalence relation.
Therefore,only $R_2$ is an equivalence relation.

(B) $1$. Reflexivity: For any $(x, y) \in N \times N$,we have $x \leq x$ or $y \leq y$. Thus,$((x, y), (x, y)) \in R$. So,$R$ is reflexive.
$2$. Symmetry: Consider $((1, 2), (2, 1))$. Here $1 \leq 2$ is true,so $((1, 2), (2, 1)) \in R$. However,for $((2, 1), (1, 2))$,$2 \leq 1$ is false and $1 \leq 2$ is true. Wait,let us check another example: $((1, 3), (2, 2))$. $1 \leq 2$ is true,so it is in $R$. But for $((2, 2), (1, 3))$,$2 \leq 1$ is false and $2 \leq 3$ is true. Let us take $((2, 3), (1, 1))$. $2 \leq 1$ is false and $3 \leq 1$ is false. So $((2, 3), (1, 1)) \notin R$. Since $((1, 1), (2, 3)) \in R$ but $((2, 3), (1, 1)) \notin R$,$R$ is not symmetric.
$3$. Transitivity: Consider $A = (2, 4)$,$B = (3, 3)$,and $C = (1, 3)$.
$(A, B) \in R$ because $2 \leq 3$ is true.
$(B, C) \in R$ because $3 \leq 3$ is true.
However,for $(A, C) = ((2, 4), (1, 3))$,$2 \leq 1$ is false and $4 \leq 3$ is false. Thus,$(A, C) \notin R$. So,$R$ is not transitive.
Therefore,only statement $(I)$ is correct.

(D) Given the set $X = \{1, 2, 3, \ldots, 20\}$.
For $R_1 = \{(x, y) : 2x - 3y = 2\}$,we find the ordered pairs $(x, y)$ such that $x, y \in X$:
If $y = 2, x = 4$; if $y = 4, x = 7$; if $y = 6, x = 10$; if $y = 8, x = 13$; if $y = 10, x = 16$; if $y = 12, x = 19$.
So,$R_1 = \{(4, 2), (7, 4), (10, 6), (13, 8), (16, 10), (19, 12)\}$.
Since there are $6$ elements and none are of the form $(a, a)$,to make $R_1$ symmetric,we need to add the reverse of each pair,which is $6$ elements.
Thus,$M = 6$.
For $R_2 = \{(x, y) : -5x + 4y = 0\}$,which implies $4y = 5x$ or $y = \frac{5}{4}x$,we find the ordered pairs $(x, y)$ such that $x, y \in X$:
If $x = 4, y = 5$; if $x = 8, y = 10$; if $x = 12, y = 15$; if $x = 16, y = 20$.
So,$R_2 = \{(4, 5), (8, 10), (12, 15), (16, 20)\}$.
Since there are $4$ elements and none are of the form $(a, a)$,to make $R_2$ symmetric,we need to add the reverse of each pair,which is $4$ elements.
Thus,$N = 4$.
Therefore,$M + N = 6 + 4 = 10$.

(C) Given the set $A = \{1, 2, 3, 4, 5\}$ and the relation $R$ defined by $4x \leq 5y$.
First,we list the elements $(x, y)$ such that $4x \leq 5y$:
For $x=1$: $4 \leq 5y \implies y \in \{1, 2, 3, 4, 5\}$ ($5$ elements)
For $x=2$: $8 \leq 5y \implies y \in \{2, 3, 4, 5\}$ ($4$ elements)
For $x=3$: $12 \leq 5y \implies y \in \{3, 4, 5\}$ ($3$ elements)
For $x=4$: $16 \leq 5y \implies y \in \{4, 5\}$ ($2$ elements)
For $x=5$: $20 \leq 5y \implies y \in \{4, 5\}$ ($2$ elements)
Total elements $m = 5 + 4 + 3 + 2 + 2 = 16$.
To make $R$ symmetric,for every $(x, y) \in R$ where $x \neq y$,we must have $(y, x) \in R$.
The elements in $R$ where $x \neq y$ are: $(1, 2), (1, 3), (1, 4), (1, 5), (2, 3), (2, 4), (2, 5), (3, 4), (3, 5), (4, 5), (5, 4)$.
There are $11$ such pairs.
We check which of their symmetric counterparts $(y, x)$ are $NOT$ already in $R$:
$(2, 1) \notin R$,$(3, 1) \notin R$,$(4, 1) \notin R$,$(5, 1) \notin R$,$(3, 2) \notin R$,$(4, 2) \notin R$,$(5, 2) \notin R$,$(4, 3) \notin R$,$(5, 3) \notin R$.
Note that $(5, 4) \in R$ and $(4, 5) \in R$,so this pair is already symmetric.
The elements to be added are $\{(2, 1), (3, 1), (4, 1), (5, 1), (3, 2), (4, 2), (5, 2), (4, 3), (5, 3)\}$.
Thus,$n = 9$.
Therefore,$m + n = 16 + 9 = 25$.