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There is relation

$$R=\left\{(1,1),(1,5),(2,4),(3,3),(4,1),(4,2),(5,4)\right\}$$

What properties (see title) it have?

Hi maths people I learn for test next week. Here is my idea is it good or not?

-not reflexive because we don't have $(2,2)$ as example

-not irreflexive because we have for example $(1,1)$

-not symmetric because for example $(1,5)$ exists but no $(5,1)$

-not asymmetric because for example $(2,4)$ and $(4,2)$ exist

-not antisymmetric because for example $(2,4)$ and $(4,2)$ exist but they are not equal

But no idea is transitive very complicated.. Is trick to check it easy pls tell me?

And is my reasons good and correct?

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    $\begingroup$ Use the transitive property on $(2,4)$ and $(4,2)$. $\endgroup$ – N74 Nov 11 '17 at 22:16
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All your answers (and reasons given!) so far are correct!

Transitivity means that whenever you have $(a,b)$ and $(b,c)$, you should also have $(a,c)$. What do you think: do you have that here?

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  • $\begingroup$ Hi Bram I always see you here thank you for confirmed!!! :) $\endgroup$ – eyesima Nov 11 '17 at 20:17
  • $\begingroup$ @AndreasAlmgren Bram answered everything the OP didn't; what's missing? $\endgroup$ – Duncan Ramage Nov 11 '17 at 20:21
  • $\begingroup$ @Bram28 I was too quick... it looks just fine now :D $\endgroup$ – Andreas Nov 11 '17 at 20:21
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    $\begingroup$ @Bram28 When I use what you say I find it's no transitive because $(2,4)$ and $(4,2)$ we have $a=2$ and $b=4$ and $c=2$ then $(a,c)=(2,2)$ no exist in the set? $\endgroup$ – eyesima Nov 11 '17 at 23:10
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    $\begingroup$ @Hamudii yes, very good! You can also point to $(1,5)$ and $(5,4)$, but no $(1,4)$. $\endgroup$ – Bram28 Nov 12 '17 at 0:15

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