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On the set $A=\{1,2,3,4,5\}$, type three such relations $a,b,c$ so that $a = a^{-1}, b = b^{-1}, c = c^{-1}$. How do we write such relations on the set?

Andrés E. Caicedo
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Taylor
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  • As an aside, you could restrict yourself to only talking about permutations which are a specific type of function which are themselves specific types of relations... and talk about permutations who are self-inverses. Those will specifically be the permutations whose cyclic decomposition has only cycles of length $2$ or $1$. There are $25$ such permutations on ${1,2,3,4,5}$. Relaxing things to include any relation who is a self-inverse increases things even further... there being $2^{15}$ such relations. – JMoravitz Jun 04 '20 at 20:22
  • As an aside, a relation which is self-inverse is more commonly known as a "symmetric" relation. (Make sure that you prove that the definitions of being self-inverse and being symmetric do in fact coincide) – JMoravitz Jun 04 '20 at 20:23
  • Well, first thing is. Do you know what a relation $a$ means? Do you know ant $a^{-1}$ means? – fleablood Jun 04 '20 at 20:30
  • Thank you JMoravitz – Taylor Jun 04 '20 at 20:36

1 Answers1

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A relation on a set $A$ ($A$ to $A$) is a subset of $A\times A$.

Let $R$ be a relation from $A$ to $A$. Then the inverse of $R$ can be defined by $R^{-1}=\{(b, a)|(a, b)∈R\}$

First find $A \times A$ which is equal to $\{(1,1), (1,2), (1,3), (1,4), \cdots, (5,4), (5,5)\}$ then try to find subsets of this set to satisfy the condition.

John
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  • Such a={(1,1),(2,2)} b={(3,3)} c={(4,4)} ? – Taylor Jun 04 '20 at 20:39
  • @Taylor those are fine choices for such relations, yes. Revisit the problem when you reach the topic of functions. As an aside, there is no requirement here that the relations we list be disjoint... You could just as well have said $a={(1,1),(2,2)},b={(1,1),(2,2),(3,3)}$ and $c={(1,1),(2,2),(3,3),(4,4)}$ or any number of other things. Again, there are many correct answers here. – JMoravitz Jun 04 '20 at 20:44
  • Thank again JMoravitz.I guess I don't know how to write a relation so I can't continue the solution – Taylor Jun 04 '20 at 20:55
  • Let say $a$ is ${(1,2),(2,1) }$ consequently $a^{-1}= {(2,1),(1,2)}$ which is equal to $a$. You can find many subsets that can satisfy the requirement. – John Jun 04 '20 at 21:21
  • Thank you very much Heroshizen. – Taylor Jun 04 '20 at 21:39