Can a non distinct ordered pair in a relation R, belong to its power RR

Question

Here is the given problem :

Let R= {(1,1),(2,1),(3,2),(4,3)}

they have give that RR or RoR = {(1,1),(2,1),(3,2),(4,2)}

RR is defined as = { | a A c C b [b B R1 R2 ] } .

I understand the reason of all the ordered pairs in RR except (1,1)

how (1,1) be in RoR, Your help much appreciated.

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If $R={(1,1),(2,1),(3,2),(4,3)}$, then $R^2={(1,1),(2,1),\underline{(3,1)},(4,2)}$. Also, instead of what you wrote for the definition of $R^2$ (of which I can't make much sense), I would define $R^2={(a,b)\in R\mid \exists c[(a,c)\in R\land (c,b)\in R]}$. — Vsotvep, Aug 27 '19 at 10:35

score 1 · Accepted Answer · answered Aug 27 '19 at 09:25

1

Generally, if $aRb$ and $bRc$, then $aR^2c$. Here $1R1$ and $1R1$ gives $1R^21$.

answered Aug 27 '19 at 09:25

Wuestenfux

Thank you for the response. So here 1R1 is implicitly considered to be present twice ? – deepakguna Aug 27 '19 at 09:30
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Here $a=b=c=1$. – Wuestenfux Aug 27 '19 at 09:30
there is only one 1R1 given in the set of relation, so we can use that as two entites? – deepakguna Aug 27 '19 at 09:33
@deepakguna It's a set, multiplicity does not matter. You don't "use up" the entities by stating that $a\mathrel Rb$, it's simply a proposition: $a$ is related to $b$ by $R$ means that the pair $(a,b)$ is an element of $R$. Saying that "$(1,1)$ is an element of $R$ and $(1,1)$ is an element of $R$" is still true regardless of saying the same thing twice. – Vsotvep Aug 27 '19 at 10:29

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