Types of Relations Questions in English

(D) For $(a, b), (c, d) \in N \times N$,the relation is defined as $(a, b) R (c, d) \iff ad(b + c) = bc(a + d)$.
$1.$ Reflexive: For any $(a, b) \in N \times N$,we have $ab(b + a) = ba(a + b)$. This implies $(a, b) R (a, b)$. Thus,$R$ is reflexive.
$2.$ Symmetric: Let $(a, b) R (c, d)$. Then $ad(b + c) = bc(a + d)$. This can be rewritten as $cb(d + a) = da(c + b)$,which implies $(c, d) R (a, b)$. Thus,$R$ is symmetric.
$3.$ Transitive: Let $(a, b) R (c, d)$ and $(c, d) R (e, f)$. Then $ad(b + c) = bc(a + d)$ and $cf(d + e) = de(c + f)$.
Dividing by $abcd$ and $cdef$ respectively,we get $\frac{b+c}{bc} = \frac{a+d}{ad} \implies \frac{1}{c} + \frac{1}{b} = \frac{1}{d} + \frac{1}{a} \implies \frac{1}{a} - \frac{1}{b} = \frac{1}{c} - \frac{1}{d}$.
Similarly,$\frac{1}{c} - \frac{1}{d} = \frac{1}{e} - \frac{1}{f}$.
Therefore,$\frac{1}{a} - \frac{1}{b} = \frac{1}{e} - \frac{1}{f} \implies \frac{1}{a} + \frac{1}{f} = \frac{1}{e} + \frac{1}{b} \implies \frac{a+f}{af} = \frac{e+b}{eb} \implies eb(a+f) = af(e+b)$.
This implies $(a, b) R (e, f)$. Thus,$R$ is transitive.
Since $R$ is reflexive,symmetric,and transitive,it is an equivalence relation.

(A) For any $x \in \mathbb{R}$,$x - x + \sqrt{2} = \sqrt{2}$,which is an irrational number.
Since $xRx$ holds for all $x \in \mathbb{R}$,$R$ is reflexive.
$R$ is not symmetric because if $x = \sqrt{2}$ and $y = 1$,then $x - y + \sqrt{2} = \sqrt{2} - 1 + \sqrt{2} = 2\sqrt{2} - 1$,which is irrational,so $\sqrt{2}R1$ is true. However,$y - x + \sqrt{2} = 1 - \sqrt{2} + \sqrt{2} = 1$,which is a rational number,so $1R\sqrt{2}$ is false.
$R$ is not transitive because if $x = \sqrt{2}$,$y = 1$,and $z = 1 - \sqrt{2}$,then $xRy$ is true ($\sqrt{2} - 1 + \sqrt{2} = 2\sqrt{2} - 1$ is irrational) and $yRz$ is true ($1 - (1 - \sqrt{2}) + \sqrt{2} = 2\sqrt{2}$ is irrational),but $xRz$ is false because $x - z + \sqrt{2} = \sqrt{2} - (1 - \sqrt{2}) + \sqrt{2} = 3\sqrt{2} - 1$,which is irrational. Wait,let us re-check: $x - z + \sqrt{2} = \sqrt{2} - 1 + \sqrt{2} + \sqrt{2} = 3\sqrt{2} - 1$,which is irrational. Actually,the relation is not transitive. For example,let $x = \sqrt{2}$,$y = 0$,$z = \sqrt{2}$. Then $xRy$ is $\sqrt{2} - 0 + \sqrt{2} = 2\sqrt{2}$ (irrational),$yRz$ is $0 - \sqrt{2} + \sqrt{2} = 0$ (rational,so $yRz$ is false). Let $x = \sqrt{2}, y = 1, z = 2\sqrt{2}$. $xRy$ is $\sqrt{2}-1+\sqrt{2} = 2\sqrt{2}-1$ (irrational). $yRz$ is $1-2\sqrt{2}+\sqrt{2} = 1-\sqrt{2}$ (irrational). $xRz$ is $\sqrt{2}-2\sqrt{2}+\sqrt{2} = 0$ (rational). Thus,$R$ is not transitive.

(A) Let $A = \{1, 2, 3\}$. Consider $R = \{(1, 2)\}$ and $S = \{(2, 3)\}$. Both $R$ and $S$ are transitive relations on $A$ because they do not contain any pairs $(a, b)$ and $(b, c)$ such that $(a, c)$ is missing.
Now,$R \cup S = \{(1, 2), (2, 3)\}$.
For $R \cup S$ to be transitive,since $(1, 2) \in R \cup S$ and $(2, 3) \in R \cup S$,we must have $(1, 3) \in R \cup S$. However,$(1, 3) \notin R \cup S$.
Thus,$R \cup S$ is not necessarily transitive. Therefore,the statement '$R$ and $S$ are transitive $\implies R \cup S$ is transitive' is false.

(A) Given that $R$ is a relation from set $A$ to set $B$,denoted as $R: A \to B$.
Given that $S$ is a relation from set $B$ to set $C$,denoted as $S: B \to C$.
The composition of relations $S \circ R$ is defined as the relation from the domain of $R$ to the codomain of $S$.
Therefore,$S \circ R$ is a relation from set $A$ to set $C$,denoted as $(S \circ R): A \to C$.

(A) relation $R$ on a set $A$ is reflexive if for every element $a \in A$,the ordered pair $(a, a) \in R$.
Since the set $A$ contains $n$ elements,there must be at least $n$ ordered pairs of the form $(a, a)$ in $R$ to satisfy the reflexive property.
These $n$ pairs are $(a_1, a_1), (a_2, a_2), \dots, (a_n, a_n)$.
Therefore,the number of ordered pairs $m$ in $R$ must be greater than or equal to $n$.
Thus,$m \ge n$.

(D) The relation is defined as $R = \{(x, y) : x, y \in A \text{ and } |x^2 - y^2| < 16\}$.
We check the given options to see if all pairs in a set satisfy the condition $|x^2 - y^2| < 16$.
For option $A$:
$|1^2 - 1^2| = 0 < 16$ (True)
$|2^2 - 1^2| = |4 - 1| = 3 < 16$ (True)
$|3^2 - 1^2| = |9 - 1| = 8 < 16$ (True)
$|4^2 - 1^2| = |16 - 1| = 15 < 16$ (True)
$|2^2 - 3^2| = |4 - 9| = 5 < 16$ (True)
All pairs in option $A$ satisfy the condition. However,a relation $R$ is a subset of $A \times A$. The question asks for the relation $R$ itself,which contains all such pairs. Since the options provided are subsets of $R$,and option $A$ is a valid subset,we evaluate if it represents the full relation. Actually,$R$ contains many more pairs (e.g.,$(1, 2), (5, 5)$). Given the structure,option $A$ is a subset of $R$. Since none of the options represent the complete set $R$,the correct choice is $D$.

(D) By definition,the identity relation $I$ on a set $A$ is defined as $I = \{(a, a) : a \in A\}$.
Since $R$ is a transitive relation on $A$,it does not necessarily imply that $R$ must contain $I$ or be contained within $I$.
For example,if $A = \{1, 2\}$,let $R = \{(1, 1), (2, 2), (1, 2)\}$. Here,$R$ is transitive and $I = \{(1, 1), (2, 2)\}$. In this case,$I \subset R$.
However,if $R = \{(1, 1)\}$,$R$ is transitive,but $I \not\subset R$ and $R \not\subset I$.
Since no specific condition (like reflexivity or symmetry) is given for $R$ other than transitivity,none of the options $R \subset I$,$I \subset R$,or $R = I$ are universally true.
Therefore,the correct answer is $D$.

(D) $1$. Reflexive: $A$ relation $R$ on set $A$ is reflexive if $(a, a) \in R$ for all $a \in A$. Here,$(1, 1), (2, 2), (3, 3), (4, 4) \in R$. Thus,$R$ is reflexive.
$2$. Symmetric: $A$ relation $R$ is symmetric if $(a, b) \in R \implies (b, a) \in R$.
- $(1, 2) \in R$ and $(2, 1) \in R$.
- $(3, 1) \in R$ and $(1, 3) \in R$.
- All other elements $(1, 1), (2, 2), (3, 3), (4, 4)$ are symmetric to themselves. Thus,$R$ is symmetric.
$3$. Transitive: $A$ relation $R$ is transitive if $(a, b) \in R$ and $(b, c) \in R \implies (a, c) \in R$.
- Consider $(3, 1) \in R$ and $(1, 2) \in R$. For $R$ to be transitive,$(3, 2)$ must be in $R$. However,$(3, 2) \notin R$. Thus,$R$ is not transitive.
Therefore,$R$ is both reflexive and symmetric.

(B) Let $A$ be the set of natural numbers. The relation $R$ is defined as $m R n$ if $m$ is a multiple of $n$,i.e.,$m = kn$ for some $k \in \mathbb{N}$.
$1$. Reflexive: For any $m \in A$,$m = 1 \times m$,so $m$ is a multiple of $m$. Thus,$(m, m) \in R$. The relation is reflexive.
$2$. Symmetric: If $(m, n) \in R$,then $m = kn$. This does not imply $n = k'm$ for $m \neq n$. For example,$4$ is a multiple of $2$,but $2$ is not a multiple of $4$. Thus,the relation is not symmetric.
$3$. Transitive: If $(m, n) \in R$ and $(n, p) \in R$,then $m = kn$ and $n = lp$ for some $k, l \in \mathbb{N}$. Substituting $n$,we get $m = k(lp) = (kl)p$. Since $kl \in \mathbb{N}$,$m$ is a multiple of $p$. Thus,$(m, p) \in R$. The relation is transitive.
Therefore,the relation is reflexive and transitive.

(A) $1$. Reflexivity: For any $a \in N$,$a$ is divisible by $a$. Thus,$aRa$ holds for all $a \in N$. So,$R$ is reflexive.
$2$. Symmetry: Let $aRb$. This means $b$ is divisible by $a$. For example,$1R2$ holds because $2$ is divisible by $1$,but $2R1$ does not hold because $1$ is not divisible by $2$. Thus,$R$ is not symmetric.
$3$. Transitivity: Let $aRb$ and $bRc$. This means $b = ka$ and $c = mb$ for some integers $k, m \in N$. Then $c = m(ka) = (mk)a$. Since $mk$ is an integer,$c$ is divisible by $a$. Thus,$aRc$ holds. So,$R$ is transitive.
Conclusion: $R$ is reflexive and transitive,but not symmetric.

(B) By definition,a relation $R$ on a set $A$ is symmetric if and only if $(a, b) \in R \implies (b, a) \in R$ for all $a, b \in A$.
This condition is equivalent to the statement $R = R^{-1}$.
Since the problem states that $R = R^{-1}$,it directly satisfies the definition of a symmetric relation.
Therefore,$R$ is symmetric.

(C) relation $R$ on a set $A$ is symmetric if $(x, y) \in R \implies (y, x) \in R$ for all $x, y \in A$. Here,$R = \{(a, a)\}$. Since $(a, a) \in R$,its reverse $(a, a)$ is also in $R$. Thus,$R$ is symmetric.
$A$ relation $R$ on a set $A$ is anti-symmetric if $(x, y) \in R$ and $(y, x) \in R \implies x = y$ for all $x, y \in A$. Here,$(a, a) \in R$ and $(a, a) \in R$ implies $a = a$,which is true. Thus,$R$ is anti-symmetric.
Therefore,$R$ is both symmetric and anti-symmetric.

(B) Let $R$ be the relation defined on $P(A)$ such that $X R Y$ if and only if $X \subseteq Y$ for all $X, Y \in P(A)$.
$1$. Reflexivity: For any $X \in P(A)$,$X \subseteq X$ is always true. Thus,$R$ is reflexive.
$2$. Anti-symmetry: If $X \subseteq Y$ and $Y \subseteq X$,then by the definition of set equality,$X = Y$. Thus,$R$ is anti-symmetric.
$3$. Transitivity: If $X \subseteq Y$ and $Y \subseteq Z$,then $X \subseteq Z$. Thus,$R$ is transitive.
Since the relation is reflexive,anti-symmetric,and transitive,it is a partial order relation,not an equivalence relation. Therefore,the relation is anti-symmetric.

(A) By definition,a relation $R$ on a set $A$ is called anti-symmetric if for all $a, b \in A$,$(a, b) \in R$ and $(b, a) \in R$ implies $a = b$.
This means that if there are two distinct elements $a$ and $b$ such that $a \neq b$,then it is impossible for both $(a, b)$ and $(b, a)$ to be in $R$ simultaneously.
Therefore,the condition for anti-symmetry is $a = b$.

(C) Given the set $A = \{1, 2, 3, 4, 5\}$ and the relation $R = \{(x, y) | x, y \in A, x < y\}$.
$1$. Reflexive: $A$ relation $R$ is reflexive if $(x, x) \in R$ for all $x \in A$. Here,$x < x$ is never true,so $R$ is not reflexive.
$2$. Symmetric: $A$ relation $R$ is symmetric if $(x, y) \in R \implies (y, x) \in R$. Here,$(1, 2) \in R$ because $1 < 2$,but $(2, 1) \notin R$ because $2 \not< 1$. So,$R$ is not symmetric.
$3$. Transitive: $A$ relation $R$ is transitive if $(x, y) \in R$ and $(y, z) \in R \implies (x, z) \in R$. If $x < y$ and $y < z$,then by the property of inequality,$x < z$. Thus,$(x, z) \in R$. Therefore,$R$ is transitive.

(C) relation $R$ on a set $A$ is reflexive if $(a, a) \in R$ for all $a \in A$. Here,$A = \{1, 2, 3, 4\}$. For $R$ to be reflexive,$(1, 1), (2, 2), (3, 3), (4, 4)$ must be in $R$. Since $(1, 1) \notin R$,$R$ is not 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 $(2, 2) \in R$. This implies $(1, 2)$ must be in $R$,which is true. Checking other pairs,we find that for all $(a, b), (b, c) \in R$,$(a, c) \in R$ holds. Thus,$R$ is transitive.
Therefore,the correct option is $C$.

(B) An empty relation $\phi$ on a set $A$ is defined as $\phi = \emptyset \subseteq A \times A$.
$1$. For reflexivity,for every $a \in A$,$(a, a)$ must be in $\phi$. Since $\phi$ is empty,it contains no elements,so it is not reflexive (unless $A = \emptyset$).
$2$. For symmetry,if $(a, b) \in \phi$,then $(b, a)$ must be in $\phi$. Since there are no elements in $\phi$,the condition $(a, b) \in \phi \implies (b, a) \in \phi$ is vacuously true. Thus,it is symmetric.
$3$. For transitivity,if $(a, b) \in \phi$ and $(b, c) \in \phi$,then $(a, c)$ must be in $\phi$. Since there are no elements in $\phi$,the condition is vacuously true. Thus,it is transitive.
Therefore,the empty relation is symmetric and transitive.

(B) $1$. Reflexivity: For any $a \in R$,we know $a \ge a$ is always true. Thus,$(a, a) \in R_1$ for all $a \in R$. So,$R_1$ is reflexive.
$2$. Symmetry: If $(a, b) \in R_1$,then $a \ge b$. This does not imply $b \ge a$ (e.g.,$2 \ge 1$ is true,but $1 \ge 2$ is false). Thus,$R_1$ is not symmetric.
$3$. Transitivity: If $(a, b) \in R_1$ and $(b, c) \in R_1$,then $a \ge b$ and $b \ge c$. By the transitive property of inequality,$a \ge c$. Thus,$(a, c) \in R_1$. So,$R_1$ is transitive.
Conclusion: $R_1$ is reflexive and transitive,but not symmetric.

(D) An equivalence relation on a set $A$ must be reflexive,symmetric,and transitive.
For a relation to be reflexive,for every $x \in A$,$(x, x)$ must be in the relation.
Here,$A = \{p, q, r\}$. Thus,$(p, p), (q, q),$ and $(r, r)$ must all be present in the relation.
$1$. In $R_1$,$(q, q)$ and $(r, r)$ are missing. Thus,$R_1$ is not reflexive.
$2$. In $R_2$,$(p, p)$ is missing. Thus,$R_2$ is not reflexive.
$3$. In $R_3$,$(r, r)$ is present,but $(p, p)$ and $(q, q)$ are present,however,$(r, r)$ is present but it is not symmetric because $(p, q) \in R_3$ but $(q, p) \notin R_3$.
Since none of the given relations satisfy the conditions of reflexivity,symmetry,and transitivity,none of them are equivalence relations.
Therefore,all the given options $R_1, R_2,$ and $R_3$ are not equivalence relations. However,the question asks which is $NOT$ an equivalence relation,and since all are not,the most appropriate answer is that none of the provided sets form an equivalence relation.

(A) relation $R$ is an equivalence relation if it is reflexive,symmetric,and transitive.
$1$. For $R_1$: $a \, R_1 \, b \Leftrightarrow |a| = |b|$.
- Reflexive: $|a| = |a|$ is true for all $a$,so $a \, R_1 \, a$.
- Symmetric: If $|a| = |b|$,then $|b| = |a|$,so $b \, R_1 \, a$.
- Transitive: If $|a| = |b|$ and $|b| = |c|$,then $|a| = |c|$,so $a \, R_1 \, c$.
Since $R_1$ satisfies all three properties,it is an equivalence relation.
$2$. $R_2$ $(a \ge b)$ is not symmetric (e.g.,$2 \ge 1$ but $1 \not\ge 2$).
$3$. $R_3$ ($a$ is divisible by $b$) is not symmetric (e.g.,$4$ is divisible by $2$,but $2$ is not divisible by $4$).
$4$. $R_4$ $(a < b)$ is not reflexive $(a \not< a)$.
Thus,the correct option is $A$.

(C) An equivalence relation $R$ on a set $A$ satisfies three properties:
$1$. Reflexivity: $(a, a) \in R$ for all $a \in A$.
$2$. Symmetry: If $(a, b) \in R$,then $(b, a) \in R$.
$3$. Transitivity: If $(a, b) \in R$ and $(b, c) \in R$,then $(a, c) \in R$.
Now,consider $R^{-1} = \{(b, a) : (a, b) \in R\}$.
$1$. Reflexivity: Since $(a, a) \in R$,then $(a, a) \in R^{-1}$. Thus,$R^{-1}$ is reflexive.
$2$. Symmetry: If $(a, b) \in R^{-1}$,then $(b, a) \in R$. Since $R$ is symmetric,$(a, b) \in R$,which implies $(b, a) \in R^{-1}$. Thus,$R^{-1}$ is symmetric.
$3$. Transitivity: 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 $R^{-1}$ satisfies all three properties,it is an equivalence relation.

(B) relation $R$ on the set of real numbers $\mathbb{R}$ is defined by $nm \ge 0$.
$1$. Reflexive: For any $n \in \mathbb{R}$,$n \cdot n = n^2 \ge 0$. Thus,$(n, n) \in R$. So,$R$ is reflexive.
$2$. Symmetric: If $(n, m) \in R$,then $nm \ge 0$. Since $nm = mn$,it follows that $mn \ge 0$,so $(m, n) \in R$. Thus,$R$ is symmetric.
$3$. Transitive: Consider $(n, m) \in R$ and $(m, p) \in R$. This means $nm \ge 0$ and $mp \ge 0$. If $n=1, m=0, p=-1$,then $1 \cdot 0 = 0 \ge 0$ and $0 \cdot (-1) = 0 \ge 0$,but $n \cdot p = 1 \cdot (-1) = -1 < 0$. Since $(1, 0) \in R$ and $(0, -1) \in R$ but $(1, -1) \notin R$,the relation is not transitive.
Therefore,$R$ is reflexive and symmetric.

(D) relation $R$ on a set $A$ is called an equivalence relation if it satisfies the following three properties:
$1$. Reflexive: $(a, a) \in R$ for all $a \in A$.
$2$. Symmetric: If $(a, b) \in R$,then $(b, a) \in R$ for all $a, b \in A$.
$3$. Transitive: If $(a, b) \in R$ and $(b, c) \in R$,then $(a, c) \in R$ for all $a, b, c \in A$.
Since an equivalence relation must satisfy all three conditions,the correct option is $D$.

(D) Let $R$ be the relation defined on the set of integers $\mathbb{Z}$ such that $aRb$ if and only if $a \equiv b \pmod{m}$,which means $a - b$ is divisible by $m$.
$1$. Reflexivity: For any $a \in \mathbb{Z}$,$a - a = 0$,which is divisible by $m$. Thus,$aRa$ holds.
$2$. Symmetry: If $aRb$,then $a - b = km$ for some integer $k$. Then $b - a = -(km) = (-k)m$. Since $-k$ is an integer,$bRa$ holds.
$3$. Transitivity: If $aRb$ and $bRc$,then $a - b = km$ and $b - c = lm$ for some integers $k, l$. Adding these,$a - c = (k + l)m$. Since $k + l$ is an integer,$aRc$ holds.
Since the relation is reflexive,symmetric,and transitive,it is an equivalence relation.

(B) An equivalence relation on a set $A$ must satisfy three properties: reflexivity,symmetry,and transitivity.
$1$. Reflexivity: Since $R$ and $S$ are equivalence relations,$(a, a) \in R$ and $(a, a) \in S$ for all $a \in A$. Thus,$(a, a) \in R \cap S$,so $R \cap S$ is reflexive.
$2$. Symmetry: If $(a, b) \in R \cap S$,then $(a, b) \in R$ and $(a, b) \in S$. Since $R$ and $S$ are symmetric,$(b, a) \in R$ and $(b, a) \in S$. Thus,$(b, a) \in R \cap S$,so $R \cap S$ is symmetric.
$3$. Transitivity: If $(a, b) \in R \cap S$ and $(b, c) \in R \cap S$,then $(a, b) \in R, (b, c) \in R$ and $(a, b) \in S, (b, c) \in S$. Since $R$ and $S$ are transitive,$(a, c) \in R$ and $(a, c) \in S$. Thus,$(a, c) \in R \cap S$,so $R \cap S$ is transitive.
Therefore,$R \cap S$ is an equivalence relation on $A$.

(D) $1$. Symmetric: If $R$ and $S$ are symmetric,then $(x, y) \in R \implies (y, x) \in R$ and $(x, y) \in S \implies (y, x) \in S$. If $(x, y) \in R \cup S$,then $(x, y) \in R$ or $(x, y) \in S$. Thus $(y, x) \in R$ or $(y, x) \in S$,which means $(y, x) \in R \cup S$. So $R \cup S$ is symmetric.
$2$. Transitive: If $R$ and $S$ are transitive,let $(x, y) \in R \cap S$ and $(y, z) \in R \cap S$. Then $(x, y) \in R, (y, z) \in R \implies (x, z) \in R$ (since $R$ is transitive). Similarly,$(x, y) \in S, (y, z) \in S \implies (x, z) \in S$ (since $S$ is transitive). Thus $(x, z) \in R \cap S$. So $R \cap S$ is transitive.
$3$. Reflexive: If $R$ and $S$ are reflexive on $A$,then for all $a \in A, (a, a) \in R$ and $(a, a) \in S$. Thus $(a, a) \in R \cap S$ for all $a \in A$. So $R \cap S$ is reflexive.
Therefore,all statements are true.

(B) Given $S = \{(2, 1), (3, 2), (2, 3)\}$.
First,find the inverse relation $S^{-1}$ by swapping the elements of each ordered pair in $S$:
$S^{-1} = \{(1, 2), (2, 3), (3, 2)\}$.
Now,we need to find the composition $R \circ S^{-1}$.
The composition $R \circ S^{-1}$ consists of pairs $(x, z)$ such that there exists $y$ where $(x, y) \in S^{-1}$ and $(y, z) \in R$.
$1$. For $(1, 2) \in S^{-1}$,we look for pairs in $R$ starting with $2$. We have $(2, 2) \in R$. Thus,$(1, 2) \in R \circ S^{-1}$.
$2$. For $(2, 3) \in S^{-1}$,we look for pairs in $R$ starting with $3$. We have $(3, 2) \in R$. Thus,$(2, 2) \in R \circ S^{-1}$.
$3$. For $(3, 2) \in S^{-1}$,we look for pairs in $R$ starting with $2$. We have $(2, 2) \in R$. Thus,$(3, 2) \in R \circ S^{-1}$.
Combining these,$R \circ S^{-1} = \{(1, 2), (2, 2), (3, 2)\}$.

(D) Given the relation $R = \{(x, y) | x, y \in N, 2x + y = 41\}$.
For $R$ to be reflexive,$(x, x) \in R$ for all $x \in N$. This implies $2x + x = 41$,so $3x = 41$,which gives $x = 41/3 \notin N$. Thus,$R$ is not reflexive.
For $R$ to be symmetric,if $(x, y) \in R$,then $(y, x) \in R$. Let $(1, 39) \in R$ because $2(1) + 39 = 41$. However,$(39, 1) \notin R$ because $2(39) + 1 = 79 \neq 41$. Thus,$R$ is not symmetric.
For $R$ to be transitive,if $(x, y) \in R$ and $(y, z) \in R$,then $(x, z) \in R$. Let $(1, 39) \in R$ and $(39, z) \in R$. For $(39, z) \in R$,$2(39) + z = 41$,which gives $z = 41 - 78 = -37 \notin N$. Since there is no such $z \in N$,the condition for transitivity fails. Thus,$R$ is not transitive.
Therefore,$R$ is none of these.

(B) relation $R$ on a set $L$ is defined as $\alpha R \beta$ if and only if $\alpha \perp \beta$ (line $\alpha$ is perpendicular to line $\beta$).
$1$. Reflexivity: For $R$ to be reflexive,$\alpha R \alpha$ must hold for all $\alpha \in L$. This implies $\alpha \perp \alpha$. Since a line cannot be perpendicular to itself,$R$ is not reflexive.
$2$. Symmetry: For $R$ to be symmetric,if $\alpha R \beta$,then $\beta R \alpha$ must hold. If $\alpha \perp \beta$,then clearly $\beta \perp \alpha$. Thus,$R$ is symmetric.
$3$. Transitivity: For $R$ to be transitive,if $\alpha R \beta$ and $\beta R \gamma$,then $\alpha R \gamma$ must hold. If $\alpha \perp \beta$ and $\beta \perp \gamma$,then $\alpha$ is parallel to $\gamma$ $(\alpha \parallel \gamma)$,not perpendicular. Thus,$R$ is not transitive.
Therefore,the relation $R$ is symmetric.

(D) To determine the nature of the relation $R$ defined by similarity $(a \sim b)$:
$1$. Reflexive: Every triangle $a$ is similar to itself $(a \sim a)$. Thus,$aRa$ holds. So,$R$ is reflexive.
$2$. Symmetric: If triangle $a$ is similar to triangle $b$ $(a \sim b)$,then triangle $b$ is also similar to triangle $a$ $(b \sim a)$. Thus,$aRb \implies bRa$. So,$R$ is symmetric.
$3$. Transitive: If triangle $a$ is similar to triangle $b$ $(a \sim b)$ and triangle $b$ is similar to triangle $c$ $(b \sim c)$,then triangle $a$ is similar to triangle $c$ $(a \sim c)$. Thus,$aRb$ and $bRc \implies aRc$. So,$R$ is transitive.
Since the relation $R$ is reflexive,symmetric,and transitive,it is an equivalence relation.

(B) Let $R$ be the relation defined on the set of points in a plane such that $(P, Q) \in R$ if and only if $OP = OQ$.
$1$. Reflexivity: For any point $P$,$OP = OP$ is always true. Thus,$(P, P) \in R$ for all $P$. So,$R$ is reflexive.
$2$. Symmetry: If $(P, Q) \in R$,then $OP = OQ$,which implies $OQ = OP$. Thus,$(Q, P) \in R$. So,$R$ is symmetric.
$3$. Transitivity: If $(P, Q) \in R$ and $(Q, S) \in R$,then $OP = OQ$ and $OQ = OS$. This implies $OP = OS$. Thus,$(P, S) \in R$. So,$R$ is transitive.
Since the relation is reflexive,symmetric,and transitive,it is an equivalence relation.

(D) relation $r$ is an equivalence relation if it is reflexive,symmetric,and transitive.
$1$. Reflexive: For any $(a, b) \in N \times N$,we have $a + b = b + a$. Thus,$(a, b)r(a, b)$ holds. So,$r$ is reflexive.
$2$. Symmetric: If $(a, b)r(c, d)$,then $a + d = b + c$. This implies $c + b = d + a$,which means $(c, d)r(a, b)$. So,$r$ is symmetric.
$3$. Transitive: If $(a, b)r(c, d)$ and $(c, d)r(e, f)$,then $a + d = b + c$ and $c + f = d + e$. Adding these equations: $a + d + c + f = b + c + d + e$. Simplifying gives $a + f = b + e$,which means $(a, b)r(e, f)$. So,$r$ is transitive.
Since $r$ is reflexive,symmetric,and transitive,it is an equivalence relation.

(D) $1$. स्वतुल्यता: कोई भी रेखा $l_1$ हमेशा स्वयं के समांतर होती है,इसलिए $(l_1, l_1) in R$। अतः,$R$ स्वतुल्य है।
$2$. सममितता: यदि $l_1$,$l_2$ के समांतर है,तो $l_2$ भी $l_1$ के समांतर होगी। इसलिए,यदि $(l_1, l_2) in R$ है,तो $(l_2, l_1) in R$ होगा। अतः,$R$ सममित है।
$3$. संक्रामकता: यदि $l_1$,$l_2$ के समांतर है और $l_2$,$l_3$ के समांतर है,तो $l_1$,$l_3$ के समांतर होगी। इसलिए,यदि $(l_1, l_2) in R$ और $(l_2, l_3) in R$ है,तो $(l_1, l_3) in R$ होगा। अतः,$R$ संक्रामक है।
इस प्रकार,संबंध $R$ स्वतुल्य,सममित और संक्रामक तीनों है।

(D) $1$. Reflexivity: For every $a \in Z$,$a - a = 0$. Since $n | 0$ is true for any positive integer $n$,$aRa$ holds. Thus,$R$ is reflexive.
$2$. Symmetry: If $aRb$,then $n | (a - b)$,which means $a - b = nk$ for some integer $k$. Then $b - a = -(a - b) = n(-k)$,which implies $n | (b - a)$. Thus,$bRa$ holds,and $R$ is symmetric.
$3$. Transitivity: If $aRb$ and $bRc$,then $n | (a - b)$ and $n | (b - c)$. This implies $a - b = nk_1$ and $b - c = nk_2$ for some integers $k_1, k_2$. Adding these equations,$(a - b) + (b - c) = a - c = n(k_1 + k_2)$. Since $n | (a - c)$,$aRc$ holds,and $R$ is transitive.
Conclusion: Since $R$ is reflexive,symmetric,and transitive,it is an equivalence relation. Therefore,the correct option is $(D)$.

(D) $1$. Reflexivity: For any $x \in N$,the divisors of $x$ include at least $1$ and $x$ itself (since $x > 100$). Thus,$x$ and $x$ have at least two common divisors ($1$ and $x$). So,$(x, x) \in R$. $R$ is reflexive.
$2$. Symmetry: If $(x, y) \in R$,then $x$ and $y$ have at least two common divisors. This implies $y$ and $x$ also have at least two common divisors. So,$(y, x) \in R$. $R$ is symmetric.
$3$. Transitivity: Let $(x, y) \in R$ and $(y, z) \in R$. This means $x, y$ share at least two divisors and $y, z$ share at least two divisors. Consider $x = 102, y = 105, z = 110$. Divisors of $102$ are ${1, 2, 3, 6, 17, 34, 51, 102}$. Divisors of $105$ are ${1, 3, 5, 7, 15, 21, 35, 105}$. Common divisors of $102$ and $105$ are ${1, 3}$ (at least two). Divisors of $110$ are ${1, 2, 5, 10, 11, 22, 55, 110}$. Common divisors of $105$ and $110$ are ${1, 5}$ (at least two). However,common divisors of $102$ and $110$ are ${1, 2}$ (at least two). Let's check another case: $x = 105 (3 \times 5 \times 7), y = 110 (2 \times 5 \times 11), z = 143 (11 \times 13)$. $(105, 110)$ share ${1, 5}$. $(110, 143)$ share ${1, 11}$. But $(105, 143)$ only share ${1}$. Thus,$(105, 143) \notin R$. $R$ is not transitive.

(B) Reflexive: $A$ relation $r$ is reflexive if $(a, a) \in r$ for all $a \in R$. This means $a \cdot a = a^2$ must be an irrational number for all $a \in R$. If $a = 1$,$a^2 = 1$,which is rational. Thus,$r$ is not reflexive.
Symmetric: $A$ relation $r$ is symmetric if $(a, b) \in r \implies (b, a) \in r$. If $ab$ is an irrational number,then $ba$ is also an irrational number because multiplication is commutative in $R$. Thus,$r$ is symmetric.
Transitive: $A$ relation $r$ is transitive if $(a, b) \in r$ and $(b, c) \in r \implies (a, c) \in r$. Consider $a = 1, b = \sqrt{2}, c = 2$. Here,$ab = 1 \cdot \sqrt{2} = \sqrt{2}$ (irrational) and $bc = \sqrt{2} \cdot 2 = 2\sqrt{2}$ (irrational). However,$ac = 1 \cdot 2 = 2$,which is a rational number. Thus,$(1, \sqrt{2}) \in r$ and $(\sqrt{2}, 2) \in r$,but $(1, 2) \notin r$. Therefore,$r$ is not transitive.
Conclusion: The relation $r$ is symmetric only.

(C) For a relation $R$ on set $A = \{3, 6, 9, 12\}$ to be reflexive,$(a, a)$ must be in $R$ for all $a \in A$. Here,$(3, 3) \notin R$,so $R$ is not reflexive.
For $R$ to be symmetric,if $(a, b) \in R$,then $(b, a) \in R$. We have $(6, 12) \in R$ and $(12, 6) \in R$. All other elements like $(6, 6), (9, 9), (12, 12)$ are symmetric to themselves. Thus,$R$ is symmetric.
For $R$ to be transitive,if $(a, b) \in R$ and $(b, c) \in R$,then $(a, c) \in R$. Checking pairs: $(6, 12) \in R$ and $(12, 6) \in R \implies (6, 6) \in R$. $(12, 6) \in R$ and $(6, 12) \in R \implies (12, 12) \in R$. All conditions hold. Thus,$R$ is transitive.
Therefore,$R$ is symmetric and transitive but not reflexive.

(D) For a set $A$ with $n$ elements,the total number of symmetric relations is $2^{n(n+1)/2}$.
Here,$n = 3$,so the total number of symmetric relations is $2^{3(4)/2} = 2^6 = 64$.
$A$ symmetric relation must contain $(a, b)$ if it contains $(b, a)$.
The pairs $(1, 2)$ and $(2, 1)$ are already included in the relation.
This leaves the remaining pairs to be chosen: $(1, 1), (2, 2), (3, 3)$ (which are $3$ diagonal elements) and the pairs ${(1, 3), (3, 1)}$ and ${(2, 3), (3, 2)}$ (which are $2$ pairs of off-diagonal elements).
Each of the $3$ diagonal elements can either be present or absent ($2^3$ ways).
Each of the $2$ pairs of off-diagonal elements can either be present or absent ($2^2$ ways).
Total number of symmetric relations containing $(1, 2)$ and $(2, 1) = 2^3 \times 2^2 = 8 \times 4 = 32$.

(A) The relation is $R = \{ (a, b) : 1 + ab > 0 \}$.
$1$. Reflexive: For any $a \in S$,$(a, a) \in R$ if $1 + a^2 > 0$. Since $a^2 \ge 0$ for all real $a$,$1 + a^2 \ge 1 > 0$. Thus,$R$ is reflexive.
$2$. Symmetric: If $(a, b) \in R$,then $1 + ab > 0$. Since $ab = ba$,$1 + ba > 0$,which implies $(b, a) \in R$. Thus,$R$ is symmetric.
$3$. Transitive: For $R$ to be transitive,if $(a, b) \in R$ and $(b, c) \in R$,then $(a, c) \in R$. Let $a = -8, b = -2, c = 0.25$.
$1 + ab = 1 + (-8)(-2) = 17 > 0$.
$1 + bc = 1 + (-2)(0.25) = 1 - 0.5 = 0.5 > 0$.
However,$1 + ac = 1 + (-8)(0.25) = 1 - 2 = -1$,which is not $> 0$.
Thus,$R$ is not transitive.
Conclusion: The relation is reflexive and symmetric,but not transitive.

(A) The number of equivalence relations on a set with $n$ elements is equal to the number of partitions of the set,which is given by the Bell number $B_n$.
For a set $A$ with $n = 4$ elements,the number of equivalence relations is the Bell number $B_4$.
The Bell numbers are defined by the recurrence relation $B_{n+1} = \sum_{k=0}^{n} \binom{n}{k} B_k$.
Calculating the first few Bell numbers:
$B_0 = 1$
$B_1 = 1$
$B_2 = \binom{1}{0}B_0 + \binom{1}{1}B_1 = 1(1) + 1(1) = 2$
$B_3 = \binom{2}{0}B_0 + \binom{2}{1}B_1 + \binom{2}{2}B_2 = 1(1) + 2(1) + 1(2) = 5$
$B_4 = \binom{3}{0}B_0 + \binom{3}{1}B_1 + \binom{3}{2}B_2 + \binom{3}{3}B_3 = 1(1) + 3(1) + 3(2) + 1(5) = 1 + 3 + 6 + 5 = 15$.
Thus,$N = 15$.
Checking the options:
$14 \leq 15 \leq 20$ is true.
Therefore,the correct option is $A$.

(A) $1$. Reflexivity: For any matrix $A \in M$,$AA = AA$ holds true. Thus,$(A, A) \in R$. So,$R$ is reflexive.
$2$. Symmetry: If $(A, B) \in R$,then $AB = BA$. This implies $BA = AB$,which means $(B, A) \in R$. So,$R$ is symmetric.
$3$. Transitivity: Let $(A, B) \in R$ and $(B, C) \in R$. This means $AB = BA$ and $BC = CB$. Does $AC = CA$? Matrix multiplication is not generally commutative. For example,if $A$ is a diagonal matrix,$B$ is a matrix that commutes with $A$,and $C$ is a matrix that commutes with $B$,$A$ and $C$ do not necessarily commute. Thus,$R$ is not transitive.

(D) The relation $R$ is defined as $R = \{(a, b) \mid a \text{ divides } b, a \in A, b \in B\}$.
We check each element $a \in A$ to see which $b \in B$ it divides:
For $a = 2$: $2$ divides $6$ and $10$,so $(2, 6) \in R$ and $(2, 10) \in R$.
For $a = 3$: $3$ divides $3$ and $6$,so $(3, 3) \in R$ and $(3, 6) \in R$.
For $a = 4$: $4$ does not divide any element in $B$.
For $a = 5$: $5$ divides $10$,so $(5, 10) \in R$.
Thus,$R = \{(2, 6), (2, 10), (3, 3), (3, 6), (5, 10)\}$.
The inverse relation $R^{-1}$ is defined as $R^{-1} = \{(b, a) \mid (a, b) \in R\}$.
Therefore,$R^{-1} = \{(6, 2), (10, 2), (3, 3), (6, 3), (10, 5)\}$.
The number of elements in $R^{-1}$ is the same as the number of elements in $R$,which is $5$.

(B) The total number of elements in the Cartesian product $A \times A$ is $n(A \times A) = m \times m = m^2$.
For a relation $R$ on $A$ to be reflexive,it must contain all elements of the form $(a, a)$ for every $a \in A$.
Since there are $m$ such elements,these $m$ elements must be present in $R$.
The remaining elements in $A \times A$ are $m^2 - m$.
Each of these remaining $m^2 - m$ elements can either be in $R$ or not in $R$.
Therefore,the total number of ways to choose the remaining elements is $2^{m^2 - m}$.

(B) The relation $r$ on $R$ is defined as $r = \{(a,b) \in R \times R \mid a - b + \sqrt{3} \in R \setminus Q \}$.
$1$. Reflexivity: For any $a \in R$,$aRa \iff a - a + \sqrt{3} = \sqrt{3}$. Since $\sqrt{3}$ is an irrational number,$aRa$ holds for all $a \in R$. Thus,$r$ is reflexive.
$2$. Symmetry: For $r$ to be symmetric,$aRb \implies bRa$. Let $a = \sqrt{3}$ and $b = 0$. Then $a - b + \sqrt{3} = \sqrt{3} - 0 + \sqrt{3} = 2\sqrt{3}$,which is irrational. So,$(\sqrt{3}, 0) \in r$. However,$b - a + \sqrt{3} = 0 - \sqrt{3} + \sqrt{3} = 0$,which is a rational number. So,$(0, \sqrt{3}) \notin r$. Thus,$r$ is not symmetric.
$3$. Transitivity: For $r$ to be transitive,$aRb$ and $bRc \implies aRc$. Consider $a = \sqrt{3}$,$b = 0$,and $c = 2\sqrt{3}$.
$aRb \implies \sqrt{3} - 0 + \sqrt{3} = 2\sqrt{3}$ (irrational).
$bRc \implies 0 - 2\sqrt{3} + \sqrt{3} = -\sqrt{3}$ (irrational).
$aRc \implies \sqrt{3} - 2\sqrt{3} + \sqrt{3} = 0$ (rational).
Since $aRc$ is not irrational,$r$ is not transitive.
Therefore,$r$ is reflexive only.

(B) The total number of reflexive relations on a set with $n$ elements is given by the formula $2^{n^{2}-n}$.
Here,the number of elements $n = 3$.
Substituting the value of $n$ into the formula:
Number of reflexive relations $= 2^{3^{2}-3}$
$= 2^{9-3}$
$= 2^{6}$.

(A) relation $R$ on a set $A$ is reflexive if $(a, a) \in R$ for all $a \in A$.
Given set $A$ has $m$ elements,so the Cartesian product $A \times A$ has $m^2$ elements.
For a relation to be reflexive,it must contain all $m$ diagonal elements of the form $(1, 1), (2, 2), \dots, (m, m)$.
The remaining elements in $A \times A$ are $m^2 - m$.
Each of these $m^2 - m$ elements can either be present or absent in the relation.
Therefore,the total number of reflexive relations is $2^{m^2 - m}$.

(D) relation $R$ is an equivalence relation if it is reflexive,symmetric,and transitive.
$1$. Reflexivity: For any house $x \in H$,$x$ faces the same direction as itself. Thus,$(x, x) \in R$ for all $x \in H$. So,$R$ is reflexive.
$2$. Symmetry: If $(x, y) \in R$,then $x$ and $y$ face the same direction. This implies that $y$ and $x$ also face the same direction. Thus,$(y, x) \in R$. So,$R$ is symmetric.
$3$. Transitivity: If $(x, y) \in R$ and $(y, z) \in R$,then $x$ and $y$ face the same direction,and $y$ and $z$ face the same direction. This implies that $x$ and $z$ face the same direction. Thus,$(x, z) \in R$. So,$R$ is transitive.
Since $R$ is reflexive,symmetric,and transitive,it is an equivalence relation.

(B) The relation is defined as $a R b \iff \log_2(a/b) = k$,where $k \in \{0, 1, 2, \dots\}$. This implies $a/b = 2^k$,or $a = b \cdot 2^k$ for some non-negative integer $k$.
$1$. Reflexive: For any $a \in I$,$a/a = 1 = 2^0$. Since $0$ is a non-negative integer,$(a, a) \in R$. Thus,$R$ is reflexive.
$2$. Symmetric: Consider $(2, 1) \in R$ because $\log_2(2/1) = 1$,which is a non-negative integer. However,$(1, 2) \notin R$ because $\log_2(1/2) = -1$,which is not a non-negative integer. Thus,$R$ is not symmetric.
$3$. Transitive: Let $(a, b) \in R$ and $(b, c) \in R$. Then $a = b \cdot 2^{k_1}$ and $b = c \cdot 2^{k_2}$ for some non-negative integers $k_1, k_2$. Substituting $b$,we get $a = (c \cdot 2^{k_2}) \cdot 2^{k_1} = c \cdot 2^{k_1+k_2}$. Since $k_1+k_2$ is a non-negative integer,$(a, c) \in R$. Thus,$R$ is transitive.
Therefore,$R$ is reflexive and transitive but not symmetric.