Paper Generator

Consider the relation defined by $0 < |x - y| \le 1$. We need to show that this relation is symmetric but not transitive. A relation is symmetric if whenever $(a, b)$ is in the relation, $(b, a)$ is also in the relation. Since $|x-y| = |y-x|$, if $0 < |x-y| \le 1$, then $0 < |y-x| \le 1$. Therefore, if $(a, b)$ is in the relation, so is $(b, a)$, so the relation is symmetric. A relation is transitive if whenever $(a, b)$ and $(b, c)$ are in the relation, $(a, c)$ is also in the relation. Let $x = 0.2, y = 0.9, z = 1.5$. Then $|x - y| = |0.2 - 0.9| = |-0.7| = 0.7$, and $|y - z| = |0.9 - 1.5| = |-0.6| = 0.6$. So $0 < |x - y| \le 1$ and $0 < |y - z| \le 1$. However, $|x - z| = |0.2 - 1.5| = |-1.3| = 1.3$, which is not less than or equal to $1$. So $(x, z)$ is not in the relation. Therefore, the relation is not transitive. So, option B is not correct.

To check if $R_1$ and $R_2$ are equivalence relations, we need to verify if they are reflexive, symmetric, and transitive. For $R_1 = \{(a, b) \in N \times N : |a - b| \le 13\}$: Reflexive: For any $a \in N$, $|a - a| = 0 \le 13$. So, $(a, a) \in R_1$. Thus, $R_1$ is reflexive. Symmetric: If $(a, b) \in R_1$, then $|a - b| \le 13$. This implies $|b - a| \le 13$, so $(b, a) \in R_1$. Thus, $R_1$ is symmetric. Transitive: Consider $a = 2, b = 11, c = 19$. Then $|2 - 11| = 9 \le 13$, so $(2, 11) \in R_1$. Also, $|11 - 19| = 8 \le 13$, so $(11, 19) \in R_1$. However, $|2 - 19| = 17 > 13$, so $(2, 19) \notin R_1$. Thus, $R_1$ is not transitive. Since $R_1$ is not transitive, $R_1$ is not an equivalence relation. For $R_2 = \{(a, b) \in N \times N : |a - b| \ne 13\}$: Reflexive: For any $a \in N$, $|a - a| = 0 \ne 13$. So, $(a, a) \in R_2$. Thus, $R_2$ is reflexive. Symmetric: If $(a, b) \in R_2$, then $|a - b| \ne 13$. This implies $|b - a| \ne 13$, so $(b, a) \in R_2$. Thus, $R_2$ is symmetric. Transitive: Consider $a = 13, b = 3, c = 26$. Then $|13 - 3| = 10 \ne 13$, so $(13, 3) \in R_2$. Also, $|3 - 26| = 23 \ne 13$, so $(3, 26) \in R_2$. However, $|13 - 26| = 13$, so $(13, 26) \notin R_2$. Thus, $R_2$ is not transitive. Since $R_2$ is not transitive, $R_2$ is not an equivalence relation. Therefore, neither $R_1$ nor $R_2$ is an equivalence relation.

$R = \{(x, y) : y \in A_i, iff x \in A_i, 1 \le i \le k\}$ (1) Reflexive: $(a, a) \Rightarrow a \in A_i$ iff $a \in A_i$ (2) Symmetric: $(a, b) \Rightarrow a \in A_i$ iff $b \in A_i$. $(b, a) \in R$ as $b \in A_i$ iff $a \in A_i$ (3) Transitive: $(a, b) \in R$ & $(b, c) \in R$. $\Rightarrow a \in A_i$ iff $b \in A_i$ & $b \in A_i$ iff $c \in A_i$. $\Rightarrow a \in A_i$ iff $c \in A_i$. $\Rightarrow (a, c) \in R$. Therefore, the relation is an equivalence relation.

For $R_1$: $aR_1b \Leftrightarrow ab \ge 0$. Reflexive: $(a,a) \in R_1$ since $a^2 \ge 0$ for all $a \in R$. Symmetric: If $(a,b) \in R_1$, then $ab \ge 0$, so $ba \ge 0$, and $(b,a) \in R_1$. Transitive: If $(a,b) \in R_1$ and $(b,c) \in R_1$, then $ab \ge 0$ and $bc \ge 0$. This does not necessarily imply $ac \ge 0$. For example, $a = -5$, $b = 0$, $c = 5$. Then $ab = 0 \ge 0$ and $bc = 0 \ge 0$, but $ac = -25 < 0$. Thus $R_1$ is not transitive, and hence not an equivalence relation. For $R_2$: $aR_2b \Leftrightarrow a \ge b$. Reflexive: $(a,a) \in R_2$ since $a \ge a$ for all $a \in R$. Symmetric: If $(a,b) \in R_2$, then $a \ge b$. This does not necessarily imply $b \ge a$. For example, $a = 2$, $b = 1$. Then $2 \ge 1$, but $1 \ngeq 2$. Thus $R_2$ is not symmetric, and hence not an equivalence relation. Therefore, neither $R_1$ nor $R_2$ is an equivalence relation.

For $R$ to be an equivalence relation, it needs to be reflexive, symmetric and transitive. Reflexive: $(a, a) \in R$ for all $a \in N$. So, $3a + \alpha a = 7k$, where $k$ is an integer. This implies $(3 + \alpha)a = 7k$. Thus, $3 + \alpha$ must be a multiple of $7$, so $3 + \alpha = 7k_1$. Therefore, $\alpha = 7k_1 - 3 = 7k_1 + 4$ for some integer $k_1$. This indicates that when $\alpha$ is divided by $7$, the remainder is $4$. Symmetric: If $(a, b) \in R$, then $(b, a) \in R$. Given $3a + (7k - 3)b = 7m$, then $3(a - b) + 7kb = 7m$. If $3(b - a) + 7ka = 7m$, then $aRb \implies bRa$. Thus, $R$ will be symmetric for $\alpha = 7k_1 - 3$. Transitive: If $(a, b) \in R$ and $(b, c) \in R$, then $(a, c) \in R$. So, $3a + (7k - 3)b = 7k_1$ and $3b + (7k_2 - 3)c = 7k_3$. Adding $3a + 7kb + (7k_2 - 3)c = 7(k_1 + k_3)$. Then $3a + (7k_2 - 3)c = 7m$. Thus, $(a, c) \in R$. Therefore, $R$ is transitive. Since $\alpha = 7k - 3 = 7k + 4$, the remainder when $\alpha$ is divided by $7$ is $4$.

Let $S = \{1, 2, 3, \dots, 60\}$. Relation $R = \{(a,b) : b=pq, p,q \geq 3, p,q \text{ are primes}\}$. Prime numbers greater than or equal to $3$ and less than or equal to $60$ are $3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59$. We need to find values of $b = p \times q \leq 60$. If $p=3$, then $q$ can be $3, 5, 7, 11, 13, 17, 19$. That gives us $7$ possibilities. If $p=5$, then $q$ can be $5, 7, 11$. That gives us $3$ possibilities. If $p=7$, then $q$ can be $7$. That gives us $1$ possibility. Total possibilities are $7+3+1 = 11$. Since $a$ can be any number in the set $S$, there are $60$ possible values for $a$ for each of the $11$ values of $b$. Therefore, total number of elements in the relation $R$ is $60 \times 11 = 660$.

Given, $(a, b)$ belongs to relation $R$ if $gcd(a, b) = 1$, $2a \neq b$. Here $gcd$ means greatest common divisor. $gcd$ of two numbers is the largest number that divides both of them. (1) For Reflexive, In $aRa$, $gcd(a, a) = a$. Therefore, this relation is not reflexive. (2) For Symmetric: Take $a = 2, b = 1 \Rightarrow gcd(2, 1) = 1$. Also $2a = 4 \neq b$. Now $gcd(b, a) = 1 \Rightarrow gcd(1, 2) = 1$ and $2b$ should not be equal to $a$. But here, $2b = 2 = a \Rightarrow R$ is not Symmetric. (3) For Transitive: Let $a = 14, b = 19, c = 21$ $gcd(a, b) = 1, 2a \neq b$ $gcd(b, c) = 1, 2b \neq c$ $gcd(a, c) = 7, 2a \neq c$ Hence not transitive. $\Rightarrow R$ is neither symmetric nor transitive.

To make $R$ symmetric, we need to add $(b, a)$ and $(c, b)$ to $R$. So, $R = \{(a, b), (b, c), (b, a), (c, b)\}$. To make $R$ transitive: Since $(a, b) \in R$ and $(b, c) \in R$, we need to add $(a, c)$ to $R$. Now, $R = \{(a, b), (b, c), (b, a), (c, b), (a, c)\}$. Since $(b, a) \in R$ and $(a, c) \in R$, we need to add $(b, c)$ to $R$. But $(b, c)$ is already in $R$. Since $(c, b) \in R$ and $(b, a) \in R$, we need to add $(c, a)$ to $R$. Now, $R = \{(a, b), (b, c), (b, a), (c, b), (a, c), (c, a)\}$. Since $(a, b) \in R$ and $(b, a) \in R$, we need to add $(a, a)$ to $R$. Since $(b, c) \in R$ and $(c, b) \in R$, we need to add $(b, b)$ to $R$. Since $(c, a) \in R$ and $(a, c) \in R$, we need to add $(c, c)$ to $R$. So, $R = \{(a, b), (b, c), (b, a), (c, b), (a, c), (c, a), (a, a), (b, b), (c, c)\}$. The elements that must be added are $(b, a), (c, b), (a, c), (c, a), (a, a), (b, b), (c, c)$. Therefore, the minimum number of elements that must be added is $7$.

Given $(a, b) R (c, d) \implies ad(b - c) = bc(a - d)$. Symmetric: $(c, d) R (a, b) \implies cb(d - a) = da(c - b)$. This is symmetric. Reflexive: $(a, b) R (a, b) \implies ab(b - a) \neq ba(a - b)$. Thus, not reflexive. Transitive: $(2, 3) R (3, 2)$ and $(3, 2) R (5, 30)$ but $((2, 3), (5, 30)) \notin R$. Thus, not transitive. Therefore, R is symmetric but neither reflexive nor transitive.

For relation $T$: if $(a, b) \in T$, then $a^2 - b^2 = I$ where $I \in Z$. Then, $(b, a)$ on relation $T$ means $b^2 - a^2 = -I$. Since $-I \in Z$, $T$ is symmetric. For relation $S = {(a, b) : a, b \in R - {0}, 2 + \frac{a}{b} > 0}$, $2 + \frac{a}{b} > 0 \implies \frac{a}{b} > -2$. If $(b, a) \in S$ then $2 + \frac{b}{a}$ is not necessarily positive. Therefore, $S$ is not symmetric.

$S = \{1, 2, 3, ..., 10\}$ $P(S) =$ power set of $S$ $AR_1B \implies (A \cap B^c) \cup (A^c \cap B) = \emptyset$ $R_1$ is reflexive, symmetric For transitive: $(A \cap B^c) \cup (A^c \cap B) = \emptyset$; $\emptyset = \emptyset \implies A = B$ $(B \cap C^c) \cup (B^c \cap C) = \emptyset \implies B = C$ $\therefore A = C \implies R_1$ is an equivalence relation. $R_2 \equiv A \cup B^c = A^c \cup B$ $R_2 \implies$ Reflexive, symmetric For transitive: $A \cup B^c = A^c \cup B \implies A = B$ $B \cup C^c = B^c \cup C \implies B = C$ $\therefore A = C \implies A \cup C^c = A^c \cup C \implies R_2$ is an equivalence relation.

Given sets are $A = {2, 3, 4}$ and $B = {8, 9, 12}$. We want to find the number of elements of the form $((a_1, b_1), (a_2, b_2))$ such that $a_1$ divides $b_2$ and $a_2$ divides $b_1$. For the first condition, $a_1$ divides $b_2$, with $a_1 \in A$ and $b_2 \in B$, we can list the pairs: $(a_1, b_2) \in {(2, 8), (2, 12), (3, 9), (3, 12), (4, 8), (4, 12)}$. This gives $6$ pairs. For the second condition, $a_2$ divides $b_1$, we can similarly list the pairs. This again has $6$ valid pairs. Now, for every pair from the first condition, we can have any pair from the second condition. This leads to $6 \times 6 = 36$ relations.

Let's analyze the relation $R$. $R$ is a set of ordered pairs of ordered pairs $((a_1, b_1), (a_2, b_2))$ such that $a_1 \in A$, $b_1 \in B$, $a_2 \in A$, $b_2 \in B$, and the conditions $a_1 \le b_2$ and $b_1 \le a_2$ are satisfied. We need to find the number of such ordered pairs. However, the question is formulated wrongly, since we are looking at a relation of pairs. A proper relation is given by $R = {(a_1, b_1) : a_1 \in A, b_1 \in B}$. The number of elements in the relation is not a relation between ordered pairs, it is simply asking to find pairs $(a, b)$ such that $a \le b$. So let $S = {(a,b): a \in A, b \in B, a \le b}$ and find the number of elements in $S$. Now we enumerate: - If $a=1$, then $b$ can be 2, 4, 5, 8, 10. (5 options) - If $a=3$, then $b$ can be 4, 5, 8, 10. (4 options) - If $a=4$, then $b$ can be 4, 5, 8, 10. (4 options) - If $a=6$, then $b$ can be 8, 10. (2 options) - If $a=9$, then $b$ can be 10. (1 option) So the total number of elements in $S$ is $5+4+4+2+1 = 16$. However, we are looking for elements that obey BOTH rules. That is: Number of pairs satisfying BOTH $a_1 \le b_2$ and $b_1 \le a_2$. Let's re-read the provided solution: - For each $a_1$ in $A = {1, 3, 4, 6, 9}$, we find the possible $b_2$ values in $B = {2, 4, 5, 8, 10}$ such that $a_1 \le b_2$. - $a_1 = 1$: $b_2$ can be 2, 4, 5, 8, 10 (5 choices) - $a_1 = 3$: $b_2$ can be 4, 5, 8, 10 (4 choices) - $a_1 = 4$: $b_2$ can be 4, 5, 8, 10 (4 choices) - $a_1 = 6$: $b_2$ can be 8, 10 (2 choices) - $a_1 = 9$: $b_2$ can be 10 (1 choice) This gives $5+4+4+2+1 = 16$. - For each $b_1$ in $B = {2, 4, 5, 8, 10}$, we find the possible $a_2$ values in $A = {1, 3, 4, 6, 9}$ such that $b_1 \le a_2$. - $b_1 = 2$: $a_2$ can be 3, 4, 6, 9 (4 choices) - $b_1 = 4$: $a_2$ can be 4, 6, 9 (3 choices) - $b_1 = 5$: $a_2$ can be 6, 9 (2 choices) - $b_1 = 8$: $a_2$ can be 9 (1 choice) - $b_1 = 10$: no $a_2$ works (0 choices) This gives $4+3+2+1+0 = 10$. So, $16 \times 10 = 160$. Thus the final number of elements is 160.

Let $S = {1, 2, 3, …, 10}$. $R = {(A, B): A∩B ≠ 𝜙; A, B∈ M}$. For Reflexive, $M$ is subset of 'S'. So $𝜙 ∈ M$ for $𝜙 ∩ 𝜙 = 𝜙 ⇒$ but relation is $A ∩ B ≠ 𝜙$. So it is not reflexive. For symmetric, $A R B ⇒ A ∩ B ≠ 𝜙, ⇒ B R A ⇒ B ∩ A ≠ 𝜙$, So it is symmetric. For transitive, If $A = {(1, 2), (2, 3)}, B = {(2, 3), (3, 4)}, C = {(3, 4), (5, 6)}$. $A R B$ & $B R C$ but $A$ does not relate to $C$. So it not transitive.

Given that the number of subsets of set $A$ is 56 more than that of set $B$, we have: $2^m - 2^n = 56$ $2^n(2^{m-n} - 1) = 56$ $2^n(2^{m-n} - 1) = 2^3 \times 7$ Comparing both sides, we have: $2^n = 2^3$ and $2^{m-n} - 1 = 7$ $n = 3$ and $2^{m-n} = 8$ $n = 3$ and $2^{m-n} = 2^3$ $n = 3$ and $m-n = 3$ $n = 3$ and $m = 6$ So the coordinates of point P are $(6,3)$. The coordinates of point Q are $(-2,-3)$. The distance between $P$ and $Q$ is: $PQ = \sqrt{(6 - (-2))^2 + (3 - (-3))^2}$ $PQ = \sqrt{(6+2)^2 + (3+3)^2}$ $PQ = \sqrt{8^2 + 6^2}$ $PQ = \sqrt{64 + 36}$ $PQ = \sqrt{100}$ $PQ = 10$ Therefore, the distance between $P(6,3)$ and $Q(-2,-3)$ is 10.

Reflexive: Since $a*b - a*b = 0$ is divisible by $5$, $(a, b)R(a, b)$ holds. Symmetric: If $(a, b)R(c, d)$, then $ad - bc$ is divisible by $5$. Thus, $bc - ad$ is divisible by $5$, so $(c, d)R(a, b)$ holds. Not Transitive: Consider $(3, 1)R(10, 5)$ because $3*5 - 1*10 = 5$ is divisible by $5$. Consider $(10, 5)R(1, 1)$ because $10*1 - 5*1 = 5$ is divisible by $5$. But $(3, 1)$ is not related to $(1, 1)$ because $3*1 - 1*1 = 2$ is not divisible by $5$. Hence, $R$ is reflexive and symmetric but not transitive.

Given the set ${1, 2, 3, 4}$, and that $R$ is an equivalence relation. Since ${((1, 2), (1, 3))} \subset R$, $R$ must contain the pairs $(1, 2)$ and $(1, 3)$. An equivalence relation must be reflexive, symmetric, and transitive. Reflexive property: $(1, 1)$, $(2, 2)$, $(3, 3)$, $(4, 4) \in R$. Symmetric property: Since $(1, 2) \in R$, $(2, 1) \in R$. Since $(1, 3) \in R$, $(3, 1) \in R$. Transitive property: Since $(1, 2) \in R$ and $(1, 3) \in R$, it follows that $(2, 3) \in R$. By symmetric property, $(3, 2) \in R$. Therefore, the elements in $R$ are: $(1, 1)$, $(2, 2)$, $(3, 3)$, $(4, 4)$, $(1, 2)$, $(2, 1)$, $(1, 3)$, $(3, 1)$, $(2, 3)$, $(3, 2)$. Thus, the number of elements in $R$ is $10$.

To determine if the given relations $R_1$ and $R_2$ are equivalence relations, we need to check whether each of them satisfies the three defining properties of an equivalence relation: reflexivity, symmetry, and transitivity. For $R_1$: Reflexivity: For $R_1$ to be reflexive, $aR_1a$ must hold for all $a \in R$. This means $a^2 + a^2 = 1$, or $2a^2 = 1$, which implies $a^2 = \frac{1}{2}$. Since this is not true for every real number $a$, $R_1$ is not reflexive. Symmetry: If $aR_1b$, then $a^2 + b^2 = 1$. This also means $b^2 + a^2 = 1$, so $bR_1a$. Thus, $R_1$ is symmetric. Transitivity: If $aR_1b$ and $bR_1c$, then $a^2 + b^2 = 1$ and $b^2 + c^2 = 1$. Adding these, we get $a^2 + 2b^2 + c^2 = 2$. There is no guarantee that $a^2 + c^2 = 1$. Therefore, $R_1$ is not transitive. $R_1$ is not an equivalence relation. For $R_2$: Reflexivity: For any $(a, b) \in N \times N$, $a + b = b + a$. Therefore, $(a, b)R_2(a, b)$, and $R_2$ is reflexive. Symmetry: If $(a, b)R_2(c, d)$, then $a + d = b + c$. This can be reordered to $c + b = d + a$, so $(c, d)R_2(a, b)$. Thus, $R_2$ is symmetric. 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 the two equations, we get $a + d + c + f = b + c + d + e$. Simplifying, we get $a + f = b + e$, so $(a, b)R_2(e, f)$. $R_2$ is reflexive, symmetric, and transitive; therefore, it is an equivalence relation. Therefore, only $R_2$ is an equivalence relation.

Given the relation $(x_1, y_1) R (x_2, y_2)$ if $x_1 \le x_2$ or $y_1 \le y_2$. For reflexivity, we need to check if $(x_1, y_1) R (x_1, y_1)$. Since $x_1 \le x_1$ and $y_1 \le y_1$, the condition $x_1 \le x_1$ or $y_1 \le y_1$ is true. Therefore, $R$ is reflexive. For symmetry, if $(x_1, y_1) R (x_2, y_2)$, then $x_1 \le x_2$ or $y_1 \le y_2$. For $R$ to be symmetric, $(x_2, y_2) R (x_1, y_1)$ must also be true, which means $x_2 \le x_1$ or $y_2 \le y_1$. This is not necessarily true. For example, $(1, 2) R (3, 4)$ since $1 \le 3$ or $2 \le 4$, but $(3, 4) R (1, 2)$ is false because neither $3 \le 1$ nor $4 \le 2$ is true. Thus, $R$ is not symmetric. For transitivity, consider pairs $(3, 9)$, $(4, 6)$, and $(2, 7)$. $(3, 9) R (4, 6)$ because $3 \le 4$. $(4, 6) R (2, 7)$ because $6 \le 7$. However, $(3, 9) R (2, 7)$ is false because neither $3 \le 2$ nor $9 \le 7$ is true. Therefore, $R$ is not transitive. Since $R$ is reflexive but not symmetric, statement (I) is correct. Since $R$ is not transitive, statement (II) is incorrect. Thus, only statement (I) is correct.