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Question

# How many reflexive and symmetric relations are there on an n-element set?

Solution

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$\textbf{Multiplication principle: }$If one event can occur in $m$ ways AND a second event can occur in $n$ ways, then the number of ways that the two events can occur in sequence is then $m\cdot n$.

A relation $R$ on a set $A$ is $\textbf{reflexive}$ if $(a,a)\in R$ for every element $a\in A$.

A relation $R$ on a set $A$ is $\textbf{symmetric}$ if $(b,a)\in R$ whenever $(a,b) \in R$.

Moreover, a relation is reflexive and symmetric when its corresponding matrix is a matrix symmetric and contains only 1's on the main diagonal. When we know all elements on or above the main diagonal of the symmetric matrix, then the other elements of the matrix can be determined due to symmetry.

When $A$ contains $n$ elements, then the relation $R$ is represented by a $n\times n$ matrix.

Since an $n\times n$ matrix contains $n\cdot c=n^2$ elements with $n$ elements on the main diagonal, there are $n^2-n$ elements that are not on the main diagonal.

Since exactly half of the non-main diagonal elements are on either side of the main diagonal, there are $\frac{n^2-n}{2}$ elements above the main diagonal.

For each of the $\frac{n^2-n}{2}$ elements above the main diagonal, there are $2$ options: the element is a 0 or a 1.

Use the multiplication principle:

$\underbrace{2\cdot 2\cdot ...\cdot 2}_{(n^2-n)/2\text{ repetitions}}=2^{(n^2-n)/2}$

Thus there are $2^{(n^2-n)/2}$ possible ways the values above the main diagonal can be entered while the elements on the main diagonal need to be 1's and the elements below the main diagonal are determined by the values above the main diagonal.

This then implies that there are $2^{(n^2-n)/2}$ reflexive and symmetric relations.

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