Question

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

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

A relation RR on a set AA is antisymmetric\textbf{antisymmetric} if (b,a)R(b,a)\in R and (a,b)R(a,b) \in R implies a=ba=b.

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

Moreover, a relation is not antisymmetric when its corresponding matrix is a matrix that is not symmetric, while the reflexive property requires that all main diagonal elements are 1's.

When AA contains nn elements, then the relation RR is represented by a n×nn\times n matrix.

Number of symmetric matrices that are not antisymmetric\text{\underline{\textbf{Number of symmetric matrices that are not antisymmetric}}}

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.

Since an n×nn\times n matrix contains nn=n2n\cdot n=n^2 elements with nn elements on the main diagonal, there are n2nn^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 n2n2\frac{n^2-n}{2} elements above the main diagonal.

For each of the n2n2\frac{n^2-n}{2} elements on or above the main diagonal, there are 22 options: the element is a 0 or a 1.

Use the multiplication principle:

22...2(n2n)/2 repetitions=2(n2n)/2\underbrace{2\cdot 2\cdot ...\cdot 2}_{(n^2-n)/2\text{ repetitions}}=2^{(n^2-n)/2}

Thus there are 2(n2n)/22^{(n^2-n)/2} possible symmetric matrices that are not antisymmetric but reflexive (with 1's on the main diagonal).

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