Types of Relations Questions in English

(A) The relation $R$ is defined on the set of natural numbers $N$ as $\{(a, b) : |a - b| = 3\}$.
This implies $a - b = 3$ or $b - a = 3$.
If $a - b = 3$,then $a = b + 3$. For $b = 1, 2, 3, \dots$,the pairs are $(4, 1), (5, 2), (6, 3), \dots$.
If $b - a = 3$,then $b = a + 3$. For $a = 1, 2, 3, \dots$,the pairs are $(1, 4), (2, 5), (3, 6), \dots$.
The complete relation is $\{(a, b) : |a - b| = 3\} = \{(1, 4), (4, 1), (2, 5), (5, 2), (3, 6), (6, 3), \dots\}$.

(B) The relation $R$ is defined as $R = \{(a, b) : a = 2b, a, b \in \mathbb{N}\}$.
Listing the elements of $R$,we get $R = \{(2, 1), (4, 2), (6, 3), \dots\}$.
The inverse relation $R^{-1}$ is obtained by interchanging the elements of the ordered pairs in $R$,such that $R^{-1} = \{(b, a) : (a, b) \in R\}$.
Therefore,$R^{-1} = \{(1, 2), (2, 4), (3, 6), \dots\}$.
Thus,the correct option is $B$.

(A) $1$. $A$ relation $R$ is reflexive if $(a, a) \in R$ for all $a \in A$. Here,$(1, 1), (2, 2), (3, 3) \in R$,so $R$ is reflexive.
$2$. $A$ relation $R$ is symmetric if $(a, b) \in R \implies (b, a) \in R$. Here,$(1, 2) \in R$ but $(2, 1) \notin R$,so $R$ is not symmetric.
$3$. $A$ relation $R$ is transitive if $(a, b) \in R$ and $(b, c) \in R \implies (a, c) \in R$. Checking all pairs: $(1, 2) \in R$ and $(2, 3) \in R \implies (1, 3) \in R$. Since this holds for all such pairs,$R$ is transitive.
$4$. Therefore,$R$ is reflexive and transitive but not symmetric.

(B) Let $R$ be the relation "less than" $( < )$ defined on the set of natural numbers $N$.
$1$. Reflexivity: For a relation to be reflexive,$xRx$ must hold for all $x \in N$. Since $x < x$ is false for any natural number $x$,the relation is not reflexive.
$2$. Symmetry: For a relation to be symmetric,$xRy$ must imply $yRx$. If $x < y$,it does not imply $y < x$ (e.g.,$1 < 2$ but $2 \not < 1$). Thus,the relation is not symmetric.
$3$. Transitivity: For a relation to be transitive,$xRy$ and $yRz$ must imply $xRz$. If $x < y$ and $y < z$,then it is always true that $x < z$. Therefore,the relation is transitive.
Conclusion: The relation "less than" is only transitive.

(B) Since $R$ is an equivalence relation on set $A$,it must be reflexive,symmetric,and transitive.
By the definition of a reflexive relation,for every element $a \in A$,the ordered pair $(a, a)$ must belong to $R$.
Since the set $A$ has $n$ elements,there are at least $n$ such ordered pairs of the form $(a, a)$ in $R$.
Therefore,the total number of ordered pairs in $R$ must be greater than or equal to $n$.

(A) $1$. Reflexivity: For any $x \in \mathbb{R}$,we have $x - x + \sqrt{2} = \sqrt{2}$. Since $\sqrt{2}$ is an irrational number,$xRx$ holds for all $x \in \mathbb{R}$. Thus,$R$ is reflexive.
$2$. Symmetry: For $R$ to be symmetric,$xRy$ must imply $yRx$. Let $x = \sqrt{2}$ 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,for $yRx$,we check $y - x + \sqrt{2} = 1 - \sqrt{2} + \sqrt{2} = 1$,which is a rational number. Thus,$1\not R\sqrt{2}$. Therefore,$R$ is not symmetric.
$3$. Transitivity: For $R$ to be transitive,$xRy$ and $yRz$ must imply $xRz$. Let $x = \sqrt{2}$,$y = 1$,and $z = 0$. Then $xRy = \sqrt{2} - 1 + \sqrt{2} = 2\sqrt{2} - 1$ (irrational) and $yRz = 1 - 0 + \sqrt{2} = 1 + \sqrt{2}$ (irrational). However,$xRz = \sqrt{2} - 0 + \sqrt{2} = 2\sqrt{2}$ (irrational). Let us try another set: $x = \sqrt{2}$,$y = 1$,$z = 1 + \sqrt{2}$. Here $xRy$ is irrational and $yRz = 1 - (1 + \sqrt{2}) + \sqrt{2} = 0$,which is rational. Since $yRz$ is not irrational,transitivity fails. Thus,$R$ is not transitive.

(B) relation $R$ is defined on a family of sets $X$ such that $A R B$ if $A \cap B = \emptyset$.
$1$. Reflexive: For $R$ to be reflexive,$A R A$ must hold for all $A \in X$,which means $A \cap A = \emptyset$. This is only true if $A = \emptyset$. Since it does not hold for all $A \in X$,$R$ is not reflexive.
$2$. Symmetric: If $A R B$,then $A \cap B = \emptyset$. Since intersection is commutative,$B \cap A = \emptyset$,which implies $B R A$. Thus,$R$ is symmetric.
$3$. Transitive: If $A R B$ and $B R C$,then $A \cap B = \emptyset$ and $B \cap C = \emptyset$. This does not necessarily imply $A \cap C = \emptyset$. For example,let $A = \{1\}$,$B = \{2\}$,and $C = \{1\}$. Here $A \cap B = \emptyset$ and $B \cap C = \emptyset$,but $A \cap C = \{1\} \neq \emptyset$. Thus,$R$ is not transitive.
Therefore,the relation is symmetric.

(D) The relation $R$ is defined on set $A = \{1, 2, 3, 4, 5\}$ as $R = \{(x, y) : |x^2 - y^2| < 16\}$.
We check all possible pairs $(x, y)$ from $A \times A$ such that $|x^2 - y^2| < 16$.
For $x=1$: $|1-1|=0 < 16, |1-4|=3 < 16, |1-9|=8 < 16, |1-16|=15 < 16, |1-25|=24 > 16$. Pairs: $(1,1), (1,2), (1,3), (1,4)$.
For $x=2$: $|4-1|=3 < 16, |4-4|=0 < 16, |4-9|=5 < 16, |4-16|=12 < 16, |4-25|=21 > 16$. Pairs: $(2,1), (2,2), (2,3), (2,4)$.
For $x=3$: $|9-1|=8 < 16, |9-4|=5 < 16, |9-9|=0 < 16, |9-16|=7 < 16, |9-25|=16 \not< 16$. Pairs: $(3,1), (3,2), (3,3), (3,4)$.
For $x=4$: $|16-1|=15 < 16, |16-4|=12 < 16, |16-9|=7 < 16, |16-16|=0 < 16, |16-25|=9 < 16$. Pairs: $(4,1), (4,2), (4,3), (4,4), (4,5)$.
For $x=5$: $|25-1|=24 > 16, |25-4|=21 > 16, |25-9|=16 \not< 16, |25-16|=9 < 16, |25-25|=0 < 16$. Pairs: $(5,4), (5,5)$.
The set $R$ contains many more elements than those listed in options $A, B,$ or $C$. Thus,the correct option is $D$.

(B) relation $R$ on a set $A$ is called reflexive if $(a, a) \in R$ for all $a \in A$.
The identity relation $I$ on set $A$ is defined as $I = \{(a, a) : a \in A\}$.
Since every element $(a, a)$ that belongs to $I$ must also belong to $R$ by the definition of a reflexive relation,it follows that $I \subseteq R$.
Therefore,the correct option is $B$.

(A) relation $R$ on set $A$ is reflexive if $(a, a) \in R$ for all $a \in A$. Since $(1, 1), (2, 2), (3, 3), (4, 4) \in R$,the relation is reflexive.
$A$ relation $R$ is symmetric if $(a, b) \in R \implies (b, a) \in R$. Here,$(1, 2) \in R$ and $(2, 1) \in R$,and $(3, 1) \in R$ and $(1, 3) \in R$. Thus,$R$ is symmetric.
$A$ relation $R$ is transitive if $(a, b) \in R$ and $(b, c) \in R \implies (a, c) \in R$. Here,$(3, 1) \in R$ and $(1, 2) \in R$,but $(3, 2) \notin R$. Therefore,$R$ is not transitive.
Since $R$ is reflexive and symmetric but not transitive,it is not an equivalence relation. Thus,the correct option is $A$ (Reflexive).

(B) Let the relation $R$ be defined on the set of integers $Z$ such that $(m, n) \in R$ if $m$ is a multiple of $n$,i.e.,$n | m$.
$1$. Reflexivity: For any integer $n \in Z$,$n$ is a multiple of $n$ (since $n = n \times 1$). Thus,$(n, n) \in R$ for all $n \in Z$. Therefore,$R$ is reflexive.
$2$. Symmetry: Consider $m = 6$ and $n = 2$. Here,$6$ is a multiple of $2$,so $(6, 2) \in R$. However,$2$ is not a multiple of $6$. Thus,$(2, 6) \notin R$. Therefore,$R$ is not symmetric.
$3$. Transitivity: Let $(m, n) \in R$ and $(n, p) \in R$. This means $n | m$ and $p | n$. By definition of divisibility,$m = kn$ and $n = lp$ for some integers $k, l$. Substituting $n$,we get $m = k(lp) = (kl)p$. Since $kl$ is an integer,$m$ is a multiple of $p$,so $(m, p) \in R$. Therefore,$R$ is transitive.
Conclusion: The relation is reflexive and transitive.

(D) $1$. Reflexivity: For any $a \in N$,$a$ divides $a$ (since $a = a \times 1$). Thus,$aRa$ holds for all $a \in N$. So,$R$ is reflexive.
$2$. Symmetry: Consider $a=1$ and $b=2$. Here,$1|2$ is true,so $1R2$ holds. However,$2|1$ is false,so $2R1$ does not hold. Thus,$R$ is not symmetric.
$3$. Transitivity: Let $aRb$ and $bRc$. This means $a|b$ and $b|c$. Since $a|b$,there exists $k_1 \in N$ such that $b = a \times k_1$. Since $b|c$,there exists $k_2 \in N$ such that $c = b \times k_2$. Substituting $b$,we get $c = (a \times k_1) \times k_2 = a \times (k_1 \times k_2)$. Since $k_1 \times k_2 \in N$,$a|c$ holds,so $aRc$ holds. Thus,$R$ is transitive.
Conclusion: The relation is reflexive and transitive but not symmetric. Therefore,the correct option is $(d)$.

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

(B) The relation $R$ is defined as $A R B$ if $A \subseteq B$ for all $A, B \in P(A)$.
$1$. Reflexive: Since $A \subseteq A$ for every $A \in P(A)$,the relation is reflexive.
$2$. Anti-symmetric: If $A \subseteq B$ and $B \subseteq A$,then by the definition of set equality,$A = B$. Thus,the relation is anti-symmetric.
$3$. Transitive: If $A \subseteq B$ and $B \subseteq C$,then $A \subseteq C$. Thus,the relation is transitive.
Since the relation is reflexive,anti-symmetric,and transitive,it is a partial order relation,not an equivalence relation. Therefore,the correct option is $B$.

(C) relation $R$ on a set $A$ is called antisymmetric if for all $a, b \in A$,$(a, b) \in R$ and $(b, a) \in R$ implies $a = b$.
This definition is logically equivalent to the statement: if $a \neq b$,then it is not the case that both $(a, b) \in R$ and $(b, a) \in R$.
In other words,if $a \neq b$,we cannot have both $(a, b) \in R$ and $(b, a) \in R$ simultaneously.
Thus,option $C$ correctly describes the condition for antisymmetry.

(C) Given the set $A = \{1, 2, 3, 4, 5\}$ and the relation $R = \{(x, y) | x, y \in A, x < y\}$.
$1$. Reflexivity: For $R$ to be reflexive,$(x, x) \in R$ for all $x \in A$. Since $x < x$ is false for any $x \in A$,$R$ is not reflexive.
$2$. Symmetry: For $R$ to be symmetric,if $(x, y) \in R$,then $(y, x) \in R$. If $x < y$,it does not imply $y < x$. For example,$(1, 2) \in R$ but $(2, 1) \notin R$. Thus,$R$ is not symmetric.
$3$. Transitivity: For $R$ to be transitive,if $(x, y) \in R$ and $(y, z) \in R$,then $(x, z) \in R$. If $x < y$ and $y < z$,then by the property of inequality,$x < z$. Therefore,$(x, z) \in R$. Thus,$R$ is transitive.
Hence,the correct option is $C$.

(D) Let $R$ be the relation defined by $xRy$ if $x$ is a brother of $y$.
$1$. Reflexive: $A$ person cannot be a brother of themselves. Thus,$xRx$ is false. So,it is not reflexive.
$2$. Symmetric: If $x$ is a brother of $y$,then $y$ is a brother of $x$ (assuming $y$ is male). However,if $y$ is a sister,$y$ is not a brother of $x$. In the context of the relation '$x$ is a brother of $y$',if $x$ is a brother of $y$,it does not necessarily imply $y$ is a brother of $x$ (e.g.,if $y$ is a girl).
$3$. Transitive: If $x$ is a brother of $y$ and $y$ is a brother of $z$,then $x$ is a brother of $z$. This holds true.
Given the standard interpretation of this problem in textbooks,the relation '$x$ is a brother of $y$' is typically considered neither reflexive,nor symmetric,nor transitive in a general sense. However,if we assume the set consists only of brothers,it would be symmetric and transitive. Given the options,the most appropriate answer is that it is not reflexive,symmetric,or transitive. Therefore,the correct choice is $(d)$.

(C) relation $R$ on set $A$ is reflexive if $(a, a) \in R$ for all $a \in A$. Here,$(1, 1) \notin R$,so $R$ is not reflexive.
$A$ relation $R$ is symmetric if $(a, b) \in R \implies (b, a) \in R$. Here,$(1, 2) \in R$ but $(2, 1) \notin R$,so $R$ is not symmetric.
$A$ relation $R$ is transitive if $(a, b) \in R$ and $(b, c) \in R \implies (a, c) \in R$.
Checking the elements:
For $(1, 2) \in R$,there is no element $(2, c) \in R$ except $(2, 2)$. Since $(1, 2) \in R$ and $(2, 2) \in R$,we must have $(1, 2) \in R$,which is true.
All other pairs $(2, 2), (3, 3), (4, 4)$ satisfy the condition trivially.
Thus,$R$ is transitive.

(B) The void relation $R$ on a set $A$ is defined as $R = \emptyset \subseteq A \times A$.
For a relation to be reflexive,$(a, a) \in R$ must hold for all $a \in A$. Since $R$ is empty,$(a, a) \notin R$,so it is not reflexive.
$A$ relation $R$ is symmetric if $(a, b) \in R \implies (b, a) \in R$. Since there are no elements in $R$,the condition is vacuously true.
$A$ relation $R$ is transitive if $(a, b) \in R$ and $(b, c) \in R \implies (a, c) \in R$. Since there are no elements in $R$,the condition is vacuously true.
Therefore,the void relation is symmetric and transitive.

(B) For any $a \in R$,we have $a \ge a$,so $(a, a) \in R_1$. Thus,$R_1$ is reflexive.
For symmetry,consider $(2, 1) \in R_1$ because $2 \ge 1$. However,$(1, 2) \notin R_1$ because $1 \not\ge 2$. Thus,$R_1$ is not symmetric.
For transitivity,let $(a, b) \in R_1$ and $(b, c) \in R_1$. This implies $a \ge b$ and $b \ge c$. By the transitive property of inequality,$a \ge c$,which means $(a, c) \in R_1$. Thus,$R_1$ is transitive.
Therefore,$R_1$ is reflexive and transitive but not symmetric.

(A) relation $R$ is an equivalence relation if it is reflexive,symmetric,and transitive.
For $R_1: a R_1 b \Leftrightarrow |a| = |b|$.
$1$. Reflexive: $|a| = |a|$ is true for all $a \in R$,so $(a, a) \in R_1$.
$2$. Symmetric: If $|a| = |b|$,then $|b| = |a|$,so if $(a, b) \in R_1$,then $(b, a) \in R_1$.
$3$. Transitive: If $|a| = |b|$ and $|b| = |c|$,then $|a| = |c|$,so if $(a, b) \in R_1$ and $(b, c) \in R_1$,then $(a, c) \in R_1$.
Since $R_1$ satisfies all three properties,it is an equivalence relation.
$R_2$ is not symmetric ($2 \ge 1$ but $1 \not\ge 2$).
$R_3$ is not symmetric ($2 \text{ divides } 4$ but $4 \text{ does not divide } 2$).
$R_4$ is not reflexive ($a < a$ is false).

(B) Given,$m R n$ if and only if $mn \ge 0$.
Reflexivity:
For any real number $m$,we know that $m^2 \ge 0$.
Since $mm = m^2$,we have $mm \ge 0$,which implies $m R m$.
Thus,$R$ is reflexive.
Symmetry:
Let $m R n$ hold,which means $mn \ge 0$.
Since the product of real numbers is commutative,$nm = mn \ge 0$.
This implies $n R m$.
Thus,$R$ is symmetric.
Transitivity:
Let $m R n$ and $n R p$ hold,which means $mn \ge 0$ and $np \ge 0$.
If $n = 0$,then $m R 0$ and $0 R p$ are always true $(0 \ge 0)$,but $m R p$ $(mp \ge 0)$ is not necessarily true (e.g.,$m=1, n=0, p=-1$).
Wait,let us re-evaluate: $1 R 0$ $(1 \times 0 = 0 \ge 0)$ and $0 R -1$ $(0 \times -1 = 0 \ge 0)$,but $1 R -1$ $(1 \times -1 = -1 < 0)$.
Therefore,$R$ is $NOT$ transitive.
Conclusion:
Since $R$ is reflexive and symmetric but not transitive,the correct option is $B$.

(D) relation $R$ on a set $A$ is called an equivalence relation if and only if it satisfies the following three properties:
$1$. Reflexive: For every $a \in A$,$(a, a) \in R$.
$2$. 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,option $D$ is the correct answer.

(B) The correct option is $B$.
Given that $R$ and $S$ are equivalence relations on set $A$.
Since $R \subseteq A \times A$ and $S \subseteq A \times A$,it follows that $R \cap S \subseteq A \times A$,so $R \cap S$ is a relation on $A$.
$1.$ Reflexivity: Let $a \in A$. Since $R$ and $S$ are reflexive,$(a, a) \in R$ and $(a, a) \in S$. Thus,$(a, a) \in R \cap S$ for all $a \in A$. So,$R \cap S$ is reflexive.
$2.$ Symmetry: Let $(a, b) \in R \cap S$. 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: Let $(a, b) \in R \cap S$ 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.
Since $R \cap S$ is reflexive,symmetric,and transitive,it is an equivalence relation on $A$.

(B) relation $R$ is defined on the set $L$ of all straight lines in a plane as $\alpha R\beta \Leftrightarrow \alpha \perp \beta$.
$1$. Reflexive: For $R$ to be reflexive,$\alpha R\alpha$ must hold for all $\alpha \in L$. This implies $\alpha \perp \alpha$,which is false because a line cannot be perpendicular to itself. Thus,$R$ is not reflexive.
$2$. Symmetric: For $R$ to be symmetric,$\alpha R\beta$ must imply $\beta R\alpha$. If $\alpha \perp \beta$,then clearly $\beta \perp \alpha$. Thus,$R$ is symmetric.
$3$. Transitive: For $R$ to be transitive,$\alpha R\beta$ and $\beta R\gamma$ must imply $\alpha R\gamma$. If $\alpha \perp \beta$ and $\beta \perp \gamma$,then $\alpha$ and $\gamma$ are both perpendicular to $\beta$,which means $\alpha \parallel \gamma$. Therefore,$\alpha R\gamma$ is false. Thus,$R$ is not transitive.
Conclusion: $R$ is symmetric.

(D) $1$. Reflexivity: For any $(a, b) \in N \times N$,we have $a + b = b + a$,which implies $(a, b)R(a, b)$. Thus,$R$ is reflexive.
$2$. Symmetry: Let $(a, b)R(c, d)$. Then $a + d = b + c$. This can be rewritten as $c + b = d + a$,which implies $c + b = d + a$,so $(c, d)R(a, b)$. Thus,$R$ is symmetric.
$3$. Transitivity: Let $(a, b)R(c, d)$ and $(c, d)R(e, f)$. Then $a + d = b + c$ and $c + f = d + e$. Adding these two equations,we get $a + d + c + f = b + c + d + e$. Canceling $c + d$ from both sides,we obtain $a + f = b + e$,which implies $(a, b)R(e, f)$. Thus,$R$ is transitive.
Since $R$ is reflexive,symmetric,and transitive,it is an equivalence relation.

(D) relation $R$ is defined on the set of integers $Z$ as $aRb \Leftrightarrow n | (a - b)$.
$1$. Reflexive: For any $a \in Z$,$(a - a) = 0$. Since $n | 0$ for any positive integer $n$,$aRa$ holds. Thus,$R$ is reflexive.
$2$. Symmetric: If $aRb$,then $n | (a - b)$,which means $(a - b) = nk$ for some integer $k$. Then $(b - a) = -(a - b) = -nk = n(-k)$. Since $-k$ is an integer,$n | (b - a)$,so $bRa$. Thus,$R$ is symmetric.
$3$. Transitive: 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,$(a - c) = (a - b) + (b - c) = n(k_1 + k_2)$. Since $(k_1 + k_2)$ is an integer,$n | (a - c)$,so $aRc$. Thus,$R$ is transitive.
Since $R$ is reflexive,symmetric,and transitive,it is an equivalence relation. Therefore,the correct option is $(d)$.

(C) $1$. Check for Reflexivity: $A$ relation $R$ on set $A$ is reflexive if $(a, a) \in R$ for all $a \in A$. Here,$(3, 3), (6, 6), (9, 9), (12, 12) \in R$. Thus,$R$ is reflexive.
$2$. Check for Symmetry: $A$ relation $R$ is symmetric if $(a, b) \in R \implies (b, a) \in R$. Here,$(3, 6) \in R$,but $(6, 3) \notin R$. Thus,$R$ is not symmetric.
$3$. Check for Transitivity: $A$ relation $R$ is transitive if $(a, b) \in R$ and $(b, c) \in R \implies (a, c) \in R$. Checking the pairs: $(3, 6) \in R$ and $(6, 12) \in R \implies (3, 12) \in R$ (present). $(3, 6) \in R$ and $(6, 6) \in R \implies (3, 6) \in R$ (present). All such combinations satisfy the condition. Thus,$R$ is transitive.
Conclusion: The relation is reflexive and transitive only.

(B) Let the relation $R$ be defined on the set of real numbers $\mathbb{R}$ as $xRy \iff x^2 = xy$.
$1$. Reflexivity: For any $x \in \mathbb{R}$,we check if $xRx$ holds. $xRx \iff x^2 = x \cdot x$,which is $x^2 = x^2$. This is true for all $x \in \mathbb{R}$. Thus,the relation is reflexive.
$2$. Symmetry: For $x, y \in \mathbb{R}$,if $xRy$,then $x^2 = xy$. Does this imply $yRx$,i.e.,$y^2 = yx$? If $x=1, y=0$,then $1^2 = 1 \cdot 0$ is $1=0$,which is false. If $x=2, y=4$,then $2^2 = 2 \cdot 4$ is $4=8$,which is false. Let us test $x=2, y=2$,then $4=4$ (True). But for $x=0, y=1$,$0^2 = 0 \cdot 1$ is $0=0$ (True),but $y^2 = yx$ gives $1^2 = 1 \cdot 0$,which is $1=0$ (False). Thus,it is not symmetric.
$3$. Transitivity: If $xRy$ and $yRz$,then $x^2 = xy$ and $y^2 = yz$. From $x^2 = xy$,if $x \neq 0$,then $x=y$. From $y^2 = yz$,if $y \neq 0$,then $y=z$. Thus $x=z$,so $x^2 = xz$. If $x=0$,then $0=0$,which is always true. However,consider $x=2, y=2, z=4$. $2^2 = 2 \cdot 2$ (True),$2^2 = 2 \cdot 4$ (False). The relation is not transitive.
Therefore,the relation is reflexive.

(C) Given $A = \{1, 2, 3, 4\}$ and $R = \{(1, 3), (4, 2), (2, 4), (2, 3), (3, 1)\}$.
$1$. For $R$ to be symmetric,$(a, b) \in R$ must imply $(b, a) \in R$. Here,$(2, 3) \in R$ but $(3, 2) \notin R$. Hence,$R$ is not symmetric.
$2$. For $R$ to be reflexive,$(a, a) \in R$ for all $a \in A$. Since $(1, 1) \notin R$,$R$ is not reflexive.
$3$. For $R$ to be a function,each element in $A$ must have a unique image. Here,$2$ is mapped to both $4$ and $3$ (i.e.,$(2, 4) \in R$ and $(2, 3) \in R$),so $R$ is not a function.
$4$. For $R$ to be transitive,$(a, b) \in R$ and $(b, c) \in R$ must imply $(a, c) \in R$. Here,$(1, 3) \in R$ and $(3, 1) \in R$,but $(1, 1) \notin R$. Thus,$R$ is not transitive.
Therefore,the correct statement is that $R$ is not symmetric.

(D) relation $R$ on a set $A$ with $n$ elements is reflexive if $(a, a) \in R$ for every $a \in A$.
There are $n^2$ total possible ordered pairs in the Cartesian product $A \times A$.
For a relation to be reflexive,all $n$ elements of the form $(a, a)$ must be present in the relation.
The remaining $n^2 - n$ ordered pairs can either be in the relation or not,giving $2^{n^2 - n}$ possibilities.
For a set with $n = 4$ elements,the number of reflexive relations is $2^{4^2 - 4} = 2^{16 - 4} = 2^{12}$.

(A) $1$. Reflexivity: For any $a \in S$,we have $1 + a \cdot a = 1 + a^2$. Since $a^2 \ge 0$,$1 + a^2 \ge 1 > 0$. Thus,$(a, a) \in R$ for all $a \in S$. So,$R$ is reflexive.
$2$. Symmetry: If $(a, b) \in R$,then $1 + ab > 0$. Since $ab = ba$,it follows that $1 + ba > 0$,which means $(b, a) \in R$. So,$R$ is symmetric.
$3$. Transitivity: Consider $a = 1$,$b = -1/2$,and $c = -1$.
For $(a, b) = (1, -1/2)$,$1 + (1)(-1/2) = 1 - 0.5 = 0.5 > 0$,so $(1, -1/2) \in R$.
For $(b, c) = (-1/2, -1)$,$1 + (-1/2)(-1) = 1 + 0.5 = 1.5 > 0$,so $(-1/2, -1) \in R$.
However,for $(a, c) = (1, -1)$,$1 + (1)(-1) = 1 - 1 = 0$,which is not $> 0$. Thus,$(1, -1) \notin R$.
Therefore,$R$ is not transitive.

(A) $1$. Reflexivity: For any $a \in \mathbb{R}$,$|a - a| = 0 \le 1$. Thus,$a \ R \ a$ holds for all $a \in \mathbb{R}$. Therefore,$R$ is reflexive.
$2$. Symmetry: If $a \ R \ b$,then $|a - b| \le 1$. Since $|a - b| = |b - a|$,it follows that $|b - a| \le 1$,which means $b \ R \ a$. Therefore,$R$ is symmetric.
$3$. Transitivity: Consider $a = 1, b = 1.5, c = 2$. Here,$|1 - 1.5| = 0.5 \le 1$ (so $1 \ R \ 1.5$) and $|1.5 - 2| = 0.5 \le 1$ (so $1.5 \ R \ 2$). However,$|1 - 2| = 1 \le 1$ is true,but consider $a = 1, b = 2, c = 3$. $|1 - 2| = 1 \le 1$ and $|2 - 3| = 1 \le 1$,but $|1 - 3| = 2 > 1$. Thus,$1 \ R \ 2$ and $2 \ R \ 3$ does not imply $1 \ R \ 3$. Therefore,$R$ is not transitive.
Conclusion: $R$ is reflexive and symmetric.

(D) $1$. Reflexivity: For any $n \in N$,$n$ is a factor of $n$ (since $n = n \times 1$). Thus,$nRn$ holds for all $n \in N$. Therefore,$R$ is reflexive.
$2$. Symmetry: Consider $n=2$ and $m=6$. Since $2|6$,$2R6$ is true. However,$6$ is not a factor of $2$ $(6 \nmid 2)$,so $6R2$ is false. Therefore,$R$ is not symmetric.
$3$. Transitivity: Let $nRm$ and $mRp$. This means $n|m$ and $m|p$. By the definition of divisibility,there exist integers $k_1, k_2$ such that $m = nk_1$ and $p = mk_2$. Substituting $m$,we get $p = (nk_1)k_2 = n(k_1k_2)$. Since $k_1k_2$ is an integer,$n|p$,which implies $nRp$. Therefore,$R$ is transitive.
Conclusion: $R$ is reflexive and transitive,but not symmetric.

(A) To determine which statement is false,we analyze each property:
$(a)$ Let $A = \{1, 2, 3\}$. Define $R = \{(1, 2)\}$ and $S = \{(2, 3)\}$. Both $R$ and $S$ are vacuously transitive. However,$R \cup S = \{(1, 2), (2, 3)\}$. Since $(1, 2) \in R \cup S$ and $(2, 3) \in R \cup S$,but $(1, 3) \notin R \cup S$,the union $R \cup S$ is not transitive. Thus,statement $(a)$ is false.
$(b)$ If $R$ and $S$ are transitive,then for any $(x, y) \in R \cap S$ and $(y, z) \in R \cap S$,we have $(x, y) \in R, (y, z) \in R \implies (x, z) \in R$ and $(x, y) \in S, (y, z) \in S \implies (x, z) \in S$. Thus $(x, z) \in R \cap S$. This is true.
$(c)$ If $R$ and $S$ are symmetric,then $(x, y) \in R \cup S \implies (x, y) \in R$ or $(x, y) \in S \implies (y, x) \in R$ or $(y, x) \in S \implies (y, x) \in R \cup S$. This is true.
$(d)$ If $R$ and $S$ are reflexive,then for all $x \in A$,$(x, x) \in R$ and $(x, x) \in S$,so $(x, x) \in R \cap S$. This is true.
Therefore,the false statement is $(a)$.

(D) The relation $R$ is defined on the set of natural numbers $N$ as $R = \{(x, y) : x, y \in N, 2x + y = 41\}$.
$1$. Reflexivity: For $R$ to be reflexive,$(x, x)$ must be in $R$ for all $x \in N$. This implies $2x + x = 41$,so $3x = 41$,which gives $x = 41/3$. Since $41/3 \notin N$,the relation is not reflexive.
$2$. Symmetry: For $R$ to be symmetric,if $(x, y) \in R$,then $(y, x)$ must be in $R$. We have $(1, 39) \in R$ because $2(1) + 39 = 41$. However,for $(39, 1)$ to be in $R$,we need $2(39) + 1 = 79$,which is not equal to $41$. Thus,$R$ is not symmetric.
$3$. Transitivity: For $R$ to be transitive,if $(x, y) \in R$ and $(y, z) \in R$,then $(x, z)$ must be in $R$. We have $(1, 39) \in R$ and $(39, -37) \notin N$. Since $y$ must be a natural number,we check pairs like $(1, 39)$ and $(20, 1)$. Here $(1, 39) \in R$ and $(20, 1) \in R$. For transitivity,$(1, 1)$ must be in $R$,but $2(1) + 1 = 3 \neq 41$. Thus,$R$ is not transitive.
Therefore,$R$ is none of these.

(C) An equivalence relation must be reflexive,symmetric,and transitive.
$1$. Reflexivity: For $R$ to be reflexive on $A = \{1, 2, 3\}$,it must contain $(1, 1), (2, 2), (3, 3)$. Since these are not in $R$,we must add $3$ pairs.
$2$. Symmetry: Given $(1, 2) \in R$,we must add $(2, 1)$. Given $(2, 3) \in R$,we must add $(3, 2)$.
$3$. Transitivity: Given $(1, 2) \in R$ and $(2, 3) \in R$,we must add $(1, 3)$ for transitivity. Since $(1, 3) \in R$,by symmetry,we must also add $(3, 1)$.
Combining these,the set of added pairs is $\{(1, 1), (2, 2), (3, 3), (2, 1), (3, 2), (1, 3), (3, 1)\}$.
The total number of ordered pairs to be added is $7$.

(D) The relation $R$ is defined on $N \times N$ 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 $ab(b + a) = ba(a + b)$,which is true. Thus,$(a, b) R (a, b)$. So,$R$ is reflexive.
$2$. Symmetric: Let $(a, b) R (c, d)$. Then $ad(b + c) = bc(a + d)$. This implies $bc(a + d) = ad(b + c)$,which can be rewritten as $cb(d + a) = da(c + b)$. Thus,$(c, d) R (a, b)$. So,$R$ 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,$\frac{1}{a} - \frac{1}{b} = \frac{1}{e} - \frac{1}{f} \implies \frac{1}{a} + \frac{1}{f} = \frac{1}{e} + \frac{1}{b} \implies \frac{a+f}{af} = \frac{b+e}{be} \implies af(b + e) = be(a + f)$.
Thus,$(a, b) R (e, f)$. So,$R$ is transitive.
Since $R$ is reflexive,symmetric,and transitive,it is an equivalence relation.

(D) An equivalence relation must be reflexive,symmetric,and transitive.
For a relation on $A = \{p, q, r\}$ to be reflexive,it must contain $(p, p), (q, q),$ and $(r, r)$.
$R_1$ does not contain $(q, q)$ and $(r, r)$,so it is not reflexive.
$R_2$ does not contain $(p, p)$,so it is not reflexive.
$R_3$ contains $(p, p), (q, q),$ and $(r, r)$,so it is reflexive. However,it is not symmetric because $(p, q) \in R_3$ but $(q, p) \notin R_3$.
Therefore,none of the given relations are equivalence relations.

(D) The relation $R$ is defined as $l_1 R l_2$ if $l_1 \parallel l_2$.
$1$. Reflexive: Since any line $l_1$ is parallel to itself $(l_1 \parallel l_1)$,the relation is reflexive.
$2$. Symmetric: If $l_1 \parallel l_2$,then $l_2 \parallel l_1$. Thus,the relation is symmetric.
$3$. Transitive: If $l_1 \parallel l_2$ and $l_2 \parallel l_3$,then $l_1 \parallel l_3$. Thus,the relation is transitive.
Since the relation is reflexive,symmetric,and transitive,it is an equivalence relation. Therefore,the correct option is $(d)$.

(A) The set $A$ has $n(A) = 3$ elements.
The Cartesian product $A \times A$ contains $n(A \times A) = n(A) \times n(A) = 3 \times 3 = 9$ elements.
$A$ relation on set $A$ is defined as any subset of $A \times A$.
The total number of subsets of a set with $m$ elements is $2^m$.
Therefore,the total number of relations on $A$ is $2^{n(A \times A)} = 2^9 = 512$.

(C) The relation $R$ is defined on the set of natural numbers $N$ as $R = \{(a, b) : a, b \in N, |a - b| = 3\}$.
This means the difference between $a$ and $b$ is $3$.
If $b = 1$,then $|a - 1| = 3 \implies a = 4$ (since $a \in N$).
If $b = 2$,then $|a - 2| = 3 \implies a = 5$.
If $b = 3$,then $|a - 3| = 3 \implies a = 6$.
If $a = 1$,then $|1 - b| = 3 \implies b = 4$.
If $a = 2$,then $|2 - b| = 3 \implies b = 5$.
Thus,the relation $R$ contains pairs such as $(1, 4), (4, 1), (2, 5), (5, 2), (3, 6), (6, 3), \dots$.
Comparing this with the given options,option $C$ is the correct representation.

(C) relation $R$ on set $A$ is an equivalence relation if it is reflexive,symmetric,and transitive.
$1$. For reflexivity,$R$ must contain $(1, 1), (2, 2), (3, 3)$.
$2$. Given $R = \{(1, 2), (2, 3)\}$.
$3$. For symmetry,since $(1, 2) \in R$,we must add $(2, 1)$. Since $(2, 3) \in R$,we must add $(3, 2)$.
$4$. Now,$R = \{(1, 2), (2, 3), (2, 1), (3, 2), (1, 1), (2, 2), (3, 3)\}$.
$5$. For transitivity,since $(1, 2) \in R$ and $(2, 3) \in R$,we must have $(1, 3) \in R$. For symmetry,we must also add $(3, 1)$.
$6$. The final set $R$ becomes the universal relation on $A$,which is $A \times A = \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 1), (2, 3), (3, 2), (1, 3), (3, 1)\}$.
$7$. The original relation had $2$ elements. The universal relation has $3^2 = 9$ elements.
$8$. Number of elements to be added = $9 - 2 = 7$.

(A) $1$. Reflexivity: Since $(1, 1), (2, 2), (3, 3) \in R$,the relation $R$ is reflexive.
$2$. Symmetry: We observe that $(1, 2) \in R$,but $(2, 1) \notin R$. Therefore,$R$ is not symmetric.
$3$. Transitivity: For any $(a, b) \in R$ and $(b, c) \in R$,we check if $(a, c) \in R$.
- $(1, 2) \in R$ and $(2, 3) \in R \implies (1, 3) \in R$ (which is present).
- All other pairs satisfy the condition. Thus,$R$ is transitive.
Conclusion: The relation $R$ is reflexive and transitive,but not symmetric.

(A) $1$. Reflexivity: For any $a \in \mathbb{R}$,$|a - a| = 0 \le 1$. Thus,$a \, R \, a$ holds for all $a \in \mathbb{R}$. So,$R$ is reflexive.
$2$. Symmetry: If $a \, R \, b$,then $|a - b| \le 1$. Since $|a - b| = |b - a|$,it follows that $|b - a| \le 1$,which means $b \, R \, a$. So,$R$ is symmetric.
$3$. Transitivity: Consider $a = 1, b = 1.5, c = 2$. Here,$|1 - 1.5| = 0.5 \le 1$ $(1 \, R \, 1.5)$ and $|1.5 - 2| = 0.5 \le 1$ $(1.5 \, R \, 2)$. However,$|1 - 2| = 1 \le 1$ holds,but if we take $a = 1, b = 1.8, c = 2.6$,then $|1 - 1.8| = 0.8 \le 1$ and $|1.8 - 2.6| = 0.8 \le 1$,but $|1 - 2.6| = 1.6 > 1$. Thus,$R$ is not transitive.
Conclusion: $R$ is reflexive and symmetric.

(B) Let $R$ be the relation "is less than" on the set of natural numbers $N$.
For any $x, y, z \in N$,if $x < y$ and $y < z$,then $x < z$.
This implies that if $xRy$ and $yRz$,then $xRz$. Thus,the relation is transitive.
Since $x < y$ does not imply $y < x$ (e.g.,$1 < 2$ but $2 \not< 1$),the relation is not symmetric.
Since $x < x$ is false for any $x \in N$,the relation is not reflexive.
Therefore,the relation is transitive only.

(C) Given the set $A = \{2, 4, 6, 8\}$ and the relation $R = \{(2, 4), (4, 2), (4, 6), (6, 4)\}$.
$A$ relation $R$ is symmetric if for every $(a, b) \in R$,we have $(b, a) \in R$.
Checking the elements of $R$:
$1$. $(2, 4) \in R$ and $(4, 2) \in R$.
$2$. $(4, 6) \in R$ and $(6, 4) \in R$.
Since for all $(a, b) \in R$,the corresponding $(b, a)$ is also in $R$,the relation is symmetric.

(B) relation $P$ on a set $A$ is reflexive if $(x, x) \in P$ for all $x \in A$. Here,$x^2 + x^2 = 1 \Rightarrow 2x^2 = 1 \Rightarrow x = \pm \frac{1}{\sqrt{2}}$. Since this is not true for all $x \in \mathbb{R}$,$P$ is not reflexive.
$A$ relation $P$ is symmetric if $(x, y) \in P \Rightarrow (y, x) \in P$. Given $x^2 + y^2 = 1$,it follows that $y^2 + x^2 = 1$. Thus,if $(x, y) \in P$,then $(y, x) \in P$. Hence,$P$ is symmetric.
$A$ relation $P$ is transitive if $(x, y) \in P$ and $(y, z) \in P \Rightarrow (x, z) \in P$. Let $(x, y) = (1, 0)$ and $(y, z) = (0, 1)$. Both are in $P$ because $1^2 + 0^2 = 1$ and $0^2 + 1^2 = 1$. However,$(x, z) = (1, 1)$ is not in $P$ because $1^2 + 1^2 = 2 \neq 1$. Thus,$P$ is not transitive.

(D) Since $n|n$ for all $n \in N$,$R$ is reflexive.
Since $2|6$ but $6$ does not divide $2$,$R$ is not symmetric.
Let $nRm$ and $mRp$. This implies $n|m$ and $m|p$. Since divisibility is transitive,$n|p$,which implies $nRp$. Therefore,$R$ is transitive.
Thus,$R$ is reflexive and transitive,but not symmetric.

(B) Since $R$ is an equivalence relation on set $A$,it must be reflexive.
By the definition of reflexivity,$(a, a) \in R$ for all $a \in A$.
Since there are $n$ elements in set $A$,there must be at least $n$ ordered pairs of the form $(a, a)$ in $R$.
Therefore,the number of ordered pairs in $R$ is greater than or equal to $n$.