two-member relationship R maths

Question

I have $A=\{a,b,c,d\}$ and we have the two-member relationship :

$R=\{(a,a),(a,c),(c,c),(c,b),(b,b),(a,d),(d,d),(b,d)\}$

is it $R$ partial order?

is any way that I can solve it and give me the right answer?

I don't know how to discribe it I did with google translate ,other similar words are set-up,layout — ek.Sek, Jan 16 '21 at 18:57
I have no idea what you are asking, sorry. My only guess would be along the lines of "Is $R$ an equivalence relation?" which it clearly isn't. Perhaps try asking it in the original language? — lulu, Jan 16 '21 at 18:58
when it is like this https://ibb.co/KhSP1s1 its called R ordinance — ek.Sek, Jan 16 '21 at 19:00
@AlejandroBergasaAlonso partial order is the word I have edited — ek.Sek, Jan 16 '21 at 19:05
Well, $R$ is clearly not an equivalence relation since it is not symmetric ($(a,c)$ is in $R$ but $(c,a)$ is not) nor is it transitive ($(a,c)$ and $(c,b)$ are in $R$ but $(a,b)$ is not). — lulu, Jan 16 '21 at 19:08
yes I see it too. Does exists any softwrare or program or something else to check it ? — ek.Sek, Jan 16 '21 at 19:12

score 0 · Accepted Answer · answered Jan 16 '21 at 19:49

0

A partial order has to satisfy (1) $aRa$ for all $a$; (2) if $aRb$ and $bRc$, then $aRc$; (3) for $a\ne b$, at most one of $aRb$ and $bRa$ holds.

Your R has $aRc$ and $cRb$ but not $aRb$, so it is not a partial order (in the jargon, it is not transitive).

answered Jan 16 '21 at 19:49

almagest

is any software ,or calculator that I can do alone that? – ek.Sek Jan 16 '21 at 20:12
Not that I know of. But why do you need it? You just check (1): do we have all of $aRa,bRb,cRc,dRd$? Answer: yes. So the relation is reflexive. Do we have both $xRy$ and $yRx$ for any pair $x\ne y$. Answer: no. So the relation is anti-symmetric. But then when we check whether it is transitive, we find that it isn't. So even worst case where you find the flaw on your last check, the whole process only takes a couple of minutes. – almagest Jan 16 '21 at 20:23

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