Let S = {x | x is a prime number and 1 < x < 10}. Let N be the number of different relations that are both reflexive and symmetric that can be defined on S. Since there is only 4 elements in S={2,3,5,7}, SxS has $2^n$ elements, so only 16 elements. However, I find that the solution is N=64. I am confused how it can be such a high number. The only reflexive pairs that I can find are (2,2),(3,3),(5,5),(7,7), and any other symmetric pair would not be reflexive anyway. Any insight on what I'm missing here?
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2Does this answer your question? Number of relations that are both symmetric and reflexive: on a set of $n$ elements, it is $2^{\frac{n^2-n}{2}}.$ – Anne Bauval Dec 17 '22 at 23:05
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As you mention, $S \times S$ has $16$ elements; this is because $S$ has $4$ elements and $4^2 = 16$. The number of subsets of $S \times S$ is then $2^{16}$, which is what allows $N$ to be so large.
To see why $N = 64$, we first note that, as you mention, $(2,2), (3,3), (5,5), (7,7)$ must all be elements of any reflexive and symmetric relation.
Additionally, $(i,j)$ is in our relation if and only if $(j,i)$ is (this is what it means to be symmetric). As such, our relations are characterized by whether or not they include the pairs $(2,3), (2,5), (2,7), (3,5), (3,7), (5,7)$. For each of those $6$ pairs, we have $2$ choices (to include the pair or exclude the pair). This gives us $N = 2^6 = 64$ reflexive and symmetric relations.
Haydn Gwyn
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