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Let $|A| = 8$.

Determine the number of binary relations on $A \times A$ that satisfy the following:

A) Symmetric

B) Neither reflexive or irreflexive

C) Reflexive and symmetric

D) Irreflexive and anti-symmetric

...

I know that a relation $R$ on a set $A$ is reflexive if $(a,a) \in R$ for every element $a \in A$ and that it's symmetric if $(b,a) \in R$ whenever $(a,b) \in R$ for all $a,b \in R$.

This is what I have so far:

Reflexive: $2^{[n(n+1)]/2} = 2^{(8*9)/2} = 2^{36}$

Symmetric: $2^{(n^2 - n)} = 2^{56}$

A) = $2^{56}$

B)$2^{64}$ - $2^{57}$

C) $2^{28}$

D)...?

Any help would be greatly appreciated.

amWhy
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Terabyte
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1 Answers1

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You are correct that part C will be $\;2^{\,p}, \;p = \binom{8}{2} \implies 2^{28}$ relations that are both symmetric and reflexive.

For parts of your question which are not addressed in the related posts:

Hint: For part B, the smart thing would be to compute the total number of relations $\;2^{8^2} = 2^{64}\;$ and subtract (#reflexive + # irreflexive) relations from the total number of relations.

Recall that a relation is irreflexive if and only if for all $x \in A$, $x \not R x$, i.e. $(x, x) \notin R$.

And a relation is antisymmetric if and only if for all $x, y$ in $A$, if $(x, y) \in R$ and $(y, x) \in R$, then $x = y$.

amWhy
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  • Ok, unless I'm mistaken the number of irreflexive relations is equal to that of reflexive relations so it would be $2^{64}$ - ($2^{36}$ + $2^{36}$)?

    (sorry for all of the questions, I understand the basics but I'm currently suffering from a concussion and feeling a bit brain damaged :P).

     – Terabyte Apr 12 '13 at 04:09
  • I don't think there are that many irreflexive relations. I relation may fail to be both reflexive and irreflexive. Irreflexivity requires that NO element be related to itself. We may have a relation where some, but not all, elements are related. So it is not reflexive, because not all elements are related, but it is not irreflexive, because some elements are related. Does that make sense? – amWhy Apr 12 '13 at 04:24
  • Nice hint and explanation +1! – Amzoti Apr 12 '13 at 04:33
  • I think I get it. If it was either reflexive or irreflexive it would be $2^{n^{2}-n+1}$ but since it's neither it's $2^{n^{2}}$ - $2^{n^{2}-n+1}$? – Terabyte Apr 12 '13 at 04:39
  • Yes, I think you've got it! – amWhy Apr 12 '13 at 04:40
  • Yay. Now on to part D! I now know that irreflexive is $2^{n^{2}−n+1}$ and I believe that antisymmetric is $2^{n}∗3^{(n(n-1))/2}$ I'm a bit confused as to where to go from here though – Terabyte Apr 12 '13 at 05:00
  • Still struggling with part D. – Terabyte Apr 12 '13 at 12:34
  • Okay...let me think on it (I've only been awake for a short while ;-))...But note that in your comment above your last: we have that $2^{n^2 - n + 1}$ being the number of relations that are either reflexive or irreflexive. So the number of relations that are irreflexive is much fewer than this. The number of relations that are antisymmetric will be fewer than those that are symmetric. – amWhy Apr 12 '13 at 15:07