# Reflexive/Symmetric/Antisymmetric/Transitive I am having issues identifying if the following are reflexive/symmetric/antisymmetric/transitive. Could anybody help me out? I have the book definitions but I'm confused on really the application of the definition.

• You should show some of your work. To get you started: For example the first relation is not reflexive, since if $x\neq 0$, then $x\sim x$ would imply that $xx = x^2 = 0$, which is a contradiction.
– Marc
May 27, 2014 at 19:57

You have to rewrite the question of reflexivity/symmetriy/etc... of the relations in terms of the definition of each relation. In example (a), the question "Is $\mathcal{R}$ reflexive?" can be rewritten as "is it true that, for any $x\in\mathbb{Q}$, $xx=0$?", which is obviously false. Thus, $\mathcal{R}$ is not reflexive.

In items (e) and (f), you can use the definition of the reflexivity/etc... more directly.

Recall that a relation $R$ on a set $A$ is reflexive if and only if for every $x\in A$ it holds that $x\mathrel{R}x$.

For example, is it true that for every $x\in\Bbb Q$ we have that $x\cdot x=x^2=0$? If the answer is yes, then the relation $\cal R$ is reflexive on $\Bbb Q$, otherwise the answer is no.

When you think the answer is yes, you have to prove it, by showing that indeed every element of the set satisfies the relation with itself. When it's not reflexive, you have to show a counterexample. For example, $\cal R$ is not reflexive, because $1^2\neq 0$.

The other properties: symmetry, antisymmetry and transitivity are similar in this aspect. These properties say "If certain conditions hold, then something happens". For example symmetry states that if $x\mathrel Ry$, then it is true that $y\mathrel Rx$ as well.

For each of those you have to either show that whenever the conditions required hold, then the conclusion holds as well; or you have to show that there is a counterexample.

(For example, $\cal R$ is symmetric: if $x\mathrel{\mathcal R} y$ then $xy=0$, but multiplication is commutative so $yx=0$ as well, and therefore $y\mathrel{\mathcal R}x$ holds as well.)

I will leave you to do all the others.

• So would it be true for the statement (c) that it would be none because it does not include the case where either a or b would be 0? May 27, 2014 at 20:33
• What do you mean by "none"? May 27, 2014 at 20:35
• That it would be none of the reflexive/symmetric/transitive/etc... May 27, 2014 at 20:36
• No, it would not be "none". Sit down with the definitions, and verify them, one by one. See which ones work out. It seems that you're just glossing over these things, instead of sitting carefully with the definitions. (Because you are making mistakes.) May 27, 2014 at 20:37