Given:
\[ A = \{1, 2, 3, \ldots, 100\} \]
Step 1: Define the relation \( R \)
\[ R \Rightarrow 2x = 3y \Rightarrow y = \frac{2x}{3} \]
Therefore, \[ R = \{(3, 2), (6, 4), (9, 6), \ldots, (99, 66)\} \]
Number of elements in \( R \): \[ n(R) = 33 \]
Since \( R \subset R_1 \)
Step 2: Define \( R_1 \)
\[ R_1 = \{(3, 2), (6, 4), (9, 6), \ldots, (99, 66), (2, 3), (4, 6), (6, 9), \ldots, (66, 99)\} \]
Step 3: Minimum number of elements
\[ \text{Minimum number of elements in } R_1 = 66 \]
Final Answer:
\[ \boxed{66} \]
The relation \( R \) consists of ordered pairs \( (x, y) \) such that \( 2x = 3y \). For \( x \) and \( y \) to satisfy this relation, \( x \) and \( y \) must form pairs with specific integer values that satisfy \( 2x = 3y \).
Thus, the pairs in \( R \) are:
\[ R = \{(3, 2), (6, 4), (9, 6), (12, 8), \ldots, (99, 66)\}. \]
There are 33 such pairs in \( R \), so:
\[ n(R) = 33. \]
To make \( R_1 \) symmetric, we include both \( (x, y) \) and \( (y, x) \) for each pair in \( R \). Thus, the pairs in \( R_1 \) are:
\[ R_1 = \{(3, 2), (2, 3), (6, 4), (4, 6), (9, 6), (6, 9), \ldots, (99, 66), (66, 99)\}. \]
This doubles the number of elements:
\[ n = 2 \times 33 = 66. \]
Therefore, the minimum value of \( n \) is: \[ 66 \]
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)