Key Concepts and Formulas

• Symmetric Relation: A relation $R_1$ on a set $A$ is symmetric if for every $(x,y) \in R_1$, it must also be true that $(y,x) \in R_1$.
• Minimum Symmetric Relation: The smallest symmetric relation $R_1$ containing a given relation $R$ is formed by taking all elements of $R$ and adding the reverse of each element $(x,y) \in R$ that is not already in $R$. This can be expressed as $R_1 = R \cup \{(y,x) \mid (x,y) \in R\}$.
• Cardinality of Union: For any two sets $P$ and $Q$, the number of elements in their union is given by $|P \cup Q| = |P| + |Q| - |P \cap Q|$.

Step-by-Step Solution

Step 1: Determine the elements of the relation $R$. The set is $A = \{1, 2, 3, \ldots, 100\}$. The relation $R$ is defined by $(x, y) \in R$ if and only if $2x = 3y$, where $x, y \in A$. From $2x = 3y$, we can deduce that $x$ must be a multiple of 3 and $y$ must be a multiple of 2. Let $x = 3k$ for some integer $k$. Substituting this into the equation gives $2(3k) = 3y$, which simplifies to $6k = 3y$, so $y = 2k$. Now we need to ensure that $x$ and $y$ are within the set $A$. For $x$: $1 \le x \le 100 \implies 1 \le 3k \le 100$. Dividing by 3, we get $\frac{1}{3} \le k \le \frac{100}{3}$, so $0.33\ldots \le k \le 33.33\ldots$. Since $k$ must be an integer, $1 \le k \le 33$. For $y$: $1 \le y \le 100 \implies 1 \le 2k \le 100$. Dividing by 2, we get $\frac{1}{2} \le k \le 50$, so $0.5 \le k \le 50$. For both conditions to be met, $k$ must be an integer such that $1 \le k \le 33$. Thus, the elements of $R$ are of the form $(3k, 2k)$ for $k = 1, 2, \ldots, 33$. The number of elements in $R$ is $n(R) = 33$. The elements of $R$ are: $(3,2), (6,4), (9,6), \ldots, (99,66)$.

Step 2: Identify elements that would make $R_1$ symmetric. We are looking for a symmetric relation $R_1$ such that $R \subset R_1$ and $n(R_1)$ is minimized. The minimum symmetric relation $R_1$ is formed by $R \cup R'$, where $R' = \{(y,x) \mid (x,y) \in R\}$. The number of elements in $R_1$ is $n(R_1) = n(R \cup R') = n(R) + n(R') - n(R \cap R')$. Since $R'$ is formed by reversing all pairs in $R$, $n(R') = n(R) = 33$. We need to find $n(R \cap R')$, which represents the number of pairs $(x,y) \in R$ such that $(y,x)$ is also in $R$.

Let's consider a pair $(x,y) \in R$. This means $2x = 3y$. If $(y,x)$ is also in $R$, then $2y = 3x$. We have the system of equations:

1. $2x = 3y$
2. $2y = 3x$

From equation (1), $y = \frac{2x}{3}$. Substitute this into equation (2): $2\left(\frac{2x}{3}\right) = 3x$ $\frac{4x}{3} = 3x$ $4x = 9x$ $5x = 0$ $x = 0$. Since $x \in A = \{1, 2, \ldots, 100\}$, $x$ cannot be 0. Therefore, there is no pair $(x,y) \in R$ such that $(y,x) \in R$. This means $R \cap R' = \emptyset$, so $n(R \cap R') = 0$.

Also, we must check for pairs of the form $(x,x) \in R$. If $(x,x) \in R$, then $2x = 3x$, which implies $x=0$. Since $0 \notin A$, there are no such pairs in $R$.

Step 3: Calculate the minimum number of elements in $R_1$. Using the formula for the cardinality of the union: $n(R_1) = n(R) + n(R') - n(R \cap R')$ We found $n(R) = 33$, $n(R') = 33$, and $n(R \cap R') = 0$. So, $n(R_1) = 33 + 33 - 0 = 66$. The number of elements in $R_1$ is $n$. Therefore, the minimum value of $n$ is 66.

Common Mistakes & Tips

• Overlooking domain constraints: Ensure that all elements derived for $x$ and $y$ are within the given set $A$. This is crucial for correctly determining $n(R)$.
• Assuming symmetry for all pairs: Do not assume that if $(x,y) \in R$, then $(y,x)$ is automatically in $R$. Always verify the symmetry condition for the specific relation.
• Forgetting $R \cap R'$: While in this problem $R \cap R'$ is empty, in general, it's important to calculate it. Pairs $(x,x) \in R$ and pairs $(x,y) \in R$ where $(y,x)$ is also in $R$ contribute to $n(R \cap R')$.

Summary We first identified the elements of the relation $R$ by solving the equation $2x=3y$ under the constraint $x, y \in \{1, 2, \ldots, 100\}$. This led to $R = \{(3k, 2k) \mid k=1, \ldots, 33\}$, giving $n(R)=33$. We then analyzed the symmetry property. By solving the system $2x=3y$ and $2y=3x$, we found that no pair $(x,y) \in R$ has its reverse $(y,x)$ also in $R$, meaning $R \cap R' = \emptyset$. The minimum symmetric relation $R_1$ containing $R$ is $R \cup R'$, and its size is $n(R_1) = n(R) + n(R') - n(R \cap R') = 33 + 33 - 0 = 66$.

The final answer is $\boxed{66}$.