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Given that $aRb$ if $2a + 3b = 5m$, $m \in I$. (1) $(a, a) \in R$ as $2a + 3a = 5a$, $a \in N$. Hence, R is reflexive. (2) If $(a, b) \in R$ then $2a + 3b = 5m$. Now, $5(a + b) = 5n$. $3a + 2b + 2a + 3b = 5n$ $\therefore 3a + 2b = 5(n - m)$. $\therefore (b, a) \in R$. $\therefore R$ is symmetric. (3) If $(a, b) \in R$ and $(b, c) \in R$ then $2a + 3b = 5m$, $2b + 3c = 5n$. $\Rightarrow 2a + 5b + 3c = 5(m + n)$. $\Rightarrow 2a + 3c = 5(m + n - b)$. $\therefore (a, c) \in R$. $\therefore R$ is transitive. Hence, R is an equivalence relation.

Let $|A|$ be the number of medals in event A, $|B|$ be the number of medals in event B, and $|C|$ be the number of medals in event C. We are given $|A| = 48$, $|B| = 25$, and $|C| = 18$. We are also given that the total number of people who received medals is $60$, so $|A \cup B \cup C| = 60$, and the number of people who received medals in all three events is $5$, so $|A \cap B \cap C| = 5$. We want to find the number of people who received medals in exactly two of the three events. Let $x$ be the number of people who received medals in exactly two events. Using the Principle of Inclusion-Exclusion, we have: $|A \cup B \cup C| = |A| + |B| + |C| - (|A \cap B| + |A \cap C| + |B \cap C|) + |A \cap B \cap C|$ $60 = 48 + 25 + 18 - (|A \cap B| + |A \cap C| + |B \cap C|) + 5$ $60 = 91 - (|A \cap B| + |A \cap C| + |B \cap C|) + 5$ $|A \cap B| + |A \cap C| + |B \cap C| = 91 + 5 - 60 = 36$ Now, let $N_{2}$ be the number of people who received medals in exactly two events. Then: $|A \cap B| + |A \cap C| + |B \cap C| = N_{2} + 3|A \cap B \cap C|$ $36 = N_{2} + 3(5)$ $N_{2} = 36 - 15 = 21$ Therefore, the number of people who received medals in exactly two of the three events is $21$.

Given the set $A = {1, 2, 3, 4, 5}$ and the relation $xRy$ if and only if $4x \le 5y$. We need to find the number of elements in R (denoted by m) and the minimum number of elements that need to be added to R to make it symmetric (denoted by n). First, let's find the relation R: $4x \le 5y \implies \frac{x}{y} \le \frac{5}{4} = 1.25$ Now, let's find the pairs $(x, y)$ that satisfy this condition: If $x = 1$, then $y$ can be $1, 2, 3, 4, 5$. If $x = 2$, then $y$ can be $2, 3, 4, 5$. If $x = 3$, then $y$ can be $3, 4, 5$. If $x = 4$, then $y$ can be $4, 5$. If $x = 5$, then $y$ can be $4, 5$. So, $R = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (2, 2), (2, 3), (2, 4), (2, 5), (3, 3), (3, 4), (3, 5), (4, 4), (4, 5), (5, 4), (5, 5)}$. The number of elements in $R$ is $m = 16$. Now, we need to find the elements to be added to R to make it symmetric. A relation is symmetric if $(x, y) \in R$ implies $(y, x) \in R$. Currently, the elements $(1, 2), (1, 3), (1, 4), (1, 5), (2, 3), (2, 4), (2, 5), (3, 4), (3, 5), (5,4) \in R$. The symmetric pairs that are missing are: $(2, 1), (3, 1), (4, 1), (5, 1), (3, 2), (4, 2), (5, 2), (4, 3), (5, 3)$. The number of elements to be added is $n = 9$. Therefore, $m + n = 16 + 9 = 25$.

To determine if the relation $R$ is reflexive, symmetric, and transitive, we analyze each property separately. Reflexive: For $R$ to be reflexive, $(f, f)$ must be in $R$. This means $f(0) = f(1)$ and $f(1) = f(0)$ must be true for all $f$. However, $f(0)$ is not necessarily equal to $f(1)$ for all functions $f$. Therefore, $R$ is not reflexive. Symmetric: If $(f, g) \in R$, then $f(0) = g(1)$ and $f(1) = g(0)$. To check for symmetry, we need to verify if $(g, f) \in R$. If $f(0) = g(1)$, then $g(1) = f(0)$. And if $f(1) = g(0)$, then $g(0) = f(1)$. This shows that if $(f, g) \in R$, then $(g, f) \in R$, so $R$ is symmetric. Transitive: If $(f, g) \in R$ and $(g, h) \in R$, then $f(0) = g(1)$, $f(1) = g(0)$, $g(0) = h(1)$, and $g(1) = h(0)$. For $R$ to be transitive, we need to check if $(f, h) \in R$, which means $f(0) = h(1)$ and $f(1) = h(0)$. We have $f(0) = g(1) = h(0)$, so $f(0) = h(0)$. Also, $f(1) = g(0) = h(1)$, so $f(1) = h(1)$. Therefore, $f(0) = g(1)$ and $g(1) = h(0)$ implies $f(0) = h(0)$ and $f(1) = g(0)$ and $g(0) = h(1)$ implies $f(1) = h(1)$. Since $f(0)$ is not necessarily equal to $h(1)$ and $f(1)$ is not necessarily equal to $h(0)$, $R$ is not transitive. Actually $f(0) = h(0)$ and $f(1) = h(1)$ so this doesn't imply that $f(0) = h(1)$ and $f(1) = h(0)$ so the relation is not transitive Therefore, the relation $R$ is symmetric but neither reflexive nor transitive.