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R is relation on A = {1,2,3,4}

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

Can you explain me why R is symmetric and transitive ?

If its symmetric then there is supposed to be (1,2) (2,1) and for transitive (1,2) (2,3).

Thanks.

Dimitry
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    It would help if you said a bit more about where you are having trouble. The relation satisfies the definitions, so which parts are you having trouble with? – Tobias Kildetoft Mar 20 '15 at 13:04
  • So if R would be {(1,1), (2,2), (3,3) (4,4) } it only be reflexive right? – Dimitry Mar 20 '15 at 13:08
  • Then it would also be reflexive (and still symmetric and transitive). – Tobias Kildetoft Mar 20 '15 at 13:09
  • Then it is reflexive but also symmetric and transitive. The rule $(a,b)\in R\Rightarrow (b,a)$ is not broken. The rule $(a,b)\in R\wedge(b,c)\in R\Rightarrow (a,c)\in R$ is also not broken. – Vera Mar 20 '15 at 13:11
  • If $S=R\cup{(1,2)}$ where $R$ is the relation mentioned in your comment then $S$ is reflexive, but is not symmetric. This because $(1,2)\in S$ and $(2,1)\notin S$. It still is transitive. If $T=S\cup{(2,3)$ then $T$ is reflexive but not transitive. This because $(1,2)\in T$ and $(2,3)\in T$ but $(1,3)\notin T$. – Vera Mar 20 '15 at 13:24
  • Tobi and Vera, thanks for explanation, its more clearly now. – Dimitry Mar 20 '15 at 13:27
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    @Dimitry, just read the definitions of "symmetric" and "transitive" very carefully. For example, for symetric: $\forall :a,:b:\in :X,$ if(!!!) $a:R:b:\Rightarrow :b:R:a.$. In your $R$, the rule is still working, but in empty. Just go every couple in $R$ and see if the opposite couple is also there. for example, $:\left(2,2\right)\in R$, and the "opposite" is again $\left(2,2\right) $, that we know that in $R$. – user2637293 Mar 20 '15 at 13:29
  • The confusion comes from the aRb and bRa definition, because i think a and b are supposed to be different numbers. – Dimitry Mar 20 '15 at 13:45
  • Does {(1,1), (2,2), (3,3) (4,4)} antisymmetric ? – Dimitry Mar 20 '15 at 15:05
  • My best guess is you are having a hard time understanding the definitions. Please include the definitions you are struggling with (symmetric, reflexive) in the body of your Question. – hardmath Mar 20 '15 at 15:09
  • "because i think a and b are supposed to be different numbers." This is simply not the case. The definition includes the case $a=b$. In you case you have only such pairs. – quid Mar 20 '15 at 15:35

1 Answers1

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The definition of symmetric is that for each $(a,b) \in R$ one has $(b,a) \in R$ as well.

Note that this definition includes the case $a=b$ though in this case the implication is trivial.

Now, your example is somewhat peculiar in that you only have pairs of the form $(a,a)$. But this really does not matter. For each $(a,b) \in R$ you do have $(b,a )\in R$, and that is all that is relevant.

The same reasoning applies for the "transitive." Again, some or all of $a,b,c$ can be equal.

quid
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