# reflexive, transitive and symmetric relations.

Problem

Let $$R:=\{(a,b) \in \mathbb{N^2}\mid a \leq b\}$$.
Is $$R$$ reflexive, symmetric, antisymmetric, transitive?

The portrayed relation is reflexive because both $$a \leq b$$ and $$b \leq a$$ works.

It is also transitive because $$a \leq b \land b \leq c \Rightarrow a \leq c$$

I'm unable to identify whether this is symmetric and/or antisymmetric.

From the looks of it, I would say that $$a \leq b \land b \leq a$$ is only true if $$a=b$$, which is the definition of antisymmetric.

Sidenote: The solution says, that this relation is only reflexive and transitive. But what about the antisymmetry I've proven?

• thanks for de-uglifying my problem – blacksmth Nov 16 '12 at 12:23

To say that $R$ is reflexive means that $aRa$ for all $a \in \mathbb N$. In this problem $R$ is reflexive because $a \leq a$ for all $a \in \mathbb N$.
To say that $R$ is symmetric means that if $aRb$ then $bRa$. In this problem $R$ is not symmetric. For example, $1 \leq 2$, but $2 \nleq 1$.
As you explained, $R$ is antisymmetric and transitive.
• Yes, $R$ is reflexive, antisymmetric, and transitive. $R$ is not symmetric. It seems like the official solution erroneously failed to mention antisymmetry. – littleO Nov 16 '12 at 12:38