this is my first post. My homework was to check each of tables and findout they are reflexive, symmetric, antisymmetric and transitional.


I would appreciate your help. I need someone to check my results. I really don't know why I do what I do. Especially while antysymmetric and transitional.

1: reflexive:yes, symmetric:no, antisymmetric:yes and transitional:yes
2: reflexive:yes, symmetric:yes, antisymmetric:no and transitional:yes
3: reflexive:no, symmetric:yes, antisymmetric:no and transitional:no
4: reflexive:yes, symmetric:no, antisymmetric:yes and transitional:yes
5: reflexive:yes, symmetric:no, antisymmetric:yes and transitional:no

I read all infos about that relations but it is hard for me. So what's the story. If I need find out antisymmetric case I need find at least one exception? Should I do the same also with transitional?

  • $\begingroup$ if anyone can edit my post, please insert/correct image :) $\endgroup$
    – deadfish
    Oct 29, 2011 at 19:21
  • $\begingroup$ I believe you mean "transitive", not "transitional". $\endgroup$ Oct 29, 2011 at 19:59
  • $\begingroup$ yes, transitive. I haven't wrote Math problems in English :) It should be transitive but i can not change due low level and the image will disapear :O $\endgroup$
    – deadfish
    Oct 29, 2011 at 20:41

2 Answers 2


Reflexive: there are no zeros on the diagonal.
Symmetric: the table has to be symmertic.
Antisymmetric: if you reflect the table with the diagonal (I mean a mirror symetry, where the diagonal is the mirror), then 1 goes to 0 (but 0 can go to 0).
Transitive: I can't think of any smart method of checking that. You just check if the relation is transitive, so you take element#1 (and then all the rest) and look at all the ones in the row (probably in the row, but it's a matter of signs): if there is one in a column with - say - number #3 (you have to check all the 1s), you look at the row#3 and check if for every 1 in this row, there is 1 in the row#1 - it is one eye-sight, so it is not that bad.

If you want to say 'yes', you have to check everything. But if while checking you find that something is 'wrong', then you just say 'no', because one exception is absolutely enough. There is no such thing like 'yes but...' in mathematics :)

You are wrong about antisymmetric: it does not mean 'asymmetric'. It means that if there is (1,5) in the relation, then (5,1) is not. Reflexive and symmetric are OK. I'm too lazy to check the last part.


You are correct about reflexivity and symmetry. The main diagonal is all $1$’s, so the relation is reflexive, and the matrix is symmetric about the main diagonal, so the relation is symmetric.

A relation $R$ is antisymmetric if $$(a R b\text{ AND }bRa)\implies (a=b)\;.$$ For your first relation you have $1R3$ and $3R1$, and yet $3\ne 1$, so this relation is not antisymmetric. For the same reason the second and fourth relations are not antisymmetric. The third relation is not antisymmetric because $1R5$ and $5R1$, even though $1 \ne 5$. The fifth is not antisymmetric because $2R5$ and $5R2$, even though $2 \ne 5$. Thus, none of the relations is antisymmetric. The visual test is to see whether there is a pair of $1$’s that are symmetric with respect to the main diagonal; if there are $-$ and we found such a pair of every one of these five relations $-$ then the relation is not antisymmetric.

There is no simple test for transitivity. For the first relation, note that $2R3$ and $3R1$, but it’s not true that $2R1$, so the first relation is not transitive. In the third relation $1R4$ and $4R2$, but it’s not true that $1R2$, so the relation is not transitive. In the fourth relation $2R3$ and $3R1$, but it isn’t true that $2R1$, so the relation is not transitive. In the last relation $3R2$ and $2R1$, but it’s not the case that $3R1$, so the relation is not transitive.

The second relation is, as you say, transitive. One way to demonstrate this is to notice that in this relation every odd number in $\{1,2,3,4,5\}$ is related to every odd number, and every even number to every even number, but no odd number is related to any even number or vice versa. Thus, if $aRb$ and $bRc$, then either $a,b$, and $c$ are all odd, or they’re all even, and in either case it’s true that $aRc$.


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