Question

The number of reflexive relations on a set \( A \) of \( n \) elements is equal to:

• A. \( 2^{n^2} \)
• B. \( n^2 \)
• C. \( 2^{n(n-1)} \)
• D. \( n^{2-n} \)

Hint:

To count reflexive relations, fix the diagonal pairs (required for reflexivity) and compute the choices for the remaining off-diagonal pairs, which is \( 2^{n(n-1)} \).

Answer 1

The Correct Option is C

Step 1: Understand the definition of a reflexive relation.
A relation \( R \) on a set \( A \) with \( n \) elements is reflexive if every element \( a \in A \) is related to itself, i.e., \( (a, a) \in R \) for all \( a \in A \). This means the diagonal pairs \( (a_1, a_1), (a_2, a_2), \ldots, (a_n, a_n) \) must be included in \( R \).
Step 2: Determine the total number of possible pairs.
The total number of ordered pairs \( (x, y) \) where \( x, y \in A \) is \( n \times n = n^2 \), as there are \( n \) choices for \( x \) and \( n \) choices for \( y \). This represents all possible elements in the relation \( R \).
Step 3: Account for the reflexive condition.
Since \( R \) must be reflexive, the \( n \) pairs \( (a_i, a_i) \) for \( i = 1 \) to \( n \) are fixed and must be included. This leaves the off-diagonal pairs \( (x, y) \) where \( x \neq y \). The number of such pairs is: \[ n^2 - n. \] These \( n^2 - n \) pairs can either be included in \( R \) or not, giving 2 choices (included or not included) for each pair.
Step 4: Calculate the number of reflexive relations.
The number of ways to choose which of the \( n^2 - n \) off-diagonal pairs are included in \( R \) is \( 2^{n^2 - n} \). Since the diagonal pairs are mandatory, the total number of reflexive relations is: \[ 2^{n^2 - n} = 2^{n(n-1)}. \]
Step 5: Verify the result.
For \( n = 1 \), set \( A = \{a\} \), only \( (a, a) \) is required, and there is 1 reflexive relation, \( 2^{1(1-1)} = 2^0 = 1 \).
For \( n = 2 \), set \( A = \{a, b\} \), diagonal pairs \( (a, a), (b, b) \) are fixed, and off-diagonal pairs \( (a, b), (b, a) \) give \( 2^{2(2-1)} = 2^2 = 4 \) relations, which matches.