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Suppose I have the following relation $$R = \{(1,1), (2,3), (3,1)\}$$ To make it reflexive we add the following missing pairs: $$ \{(2,2), (3,3)\}$$ Now I wonder how to find the reflexive transitive closure of that relation?

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  • $\begingroup$ add ${(1,2),(2,1)}$ $\endgroup$
    – Guy
    Commented Mar 12, 2014 at 12:06
  • $\begingroup$ Do you mean that $$R = \{(1,1),(2,2),(3,3), (2,3), (3,1),(1,2),(2,1) \}$$ ? $\endgroup$
    – Bledi Boss
    Commented Mar 12, 2014 at 12:08
  • $\begingroup$ Yes. I believe that is reflexive transitive, yes? I don't remember the definition of transitive though. $\endgroup$
    – Guy
    Commented Mar 12, 2014 at 12:10
  • $\begingroup$ I'm supposing that transitive must implies that from (1,2) and (2,3), (1,3) follows ... $\endgroup$ Commented Mar 12, 2014 at 12:13

1 Answer 1

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Reflexive and transitive are two different things.
Transitive means: if $a \sim b$ and $b \sim c$ then $a \sim c$. While reflexive means that $a\sim a$ in your realtion.

You have $R = \{(1,1), (2,3), (3,1)\}$ and you want it to be closed under transitivity . Since $2\sim 3$ and $3\sim 1$ you must have $2\sim 1$ so you add $(2,1)$.


Edit:

Suppose that $R$ is a relation on a set $A$. The reflexive transitive closure of $R$ is the smallest relation $S$ on $A$ such that:

$R⊆S$;

$S$ is reflexive;

and $S$ is transitive.

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  • $\begingroup$ "Reflexive and transitive are two different things" - that's true. But there exist also something called 'reflexive transitive closure' and that's what I am looking for $\endgroup$
    – Bledi Boss
    Commented Mar 12, 2014 at 12:25
  • $\begingroup$ I edited my answer. you can see that adding $(2,1)$ does the job. $\endgroup$
    – Spock
    Commented Mar 12, 2014 at 12:59
  • $\begingroup$ I got it now :) thanks $\endgroup$
    – Bledi Boss
    Commented Mar 12, 2014 at 13:03

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