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?

  • $\begingroup$ thanks for de-uglifying my problem $\endgroup$
    – blacksmth
    Commented Nov 16, 2012 at 12:23

1 Answer 1


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.

  • $\begingroup$ as well as reflexive? i'm just a bit confused, because the official solution of the problem does not mention it being antisymmetric or symmetric. $\endgroup$
    – blacksmth
    Commented Nov 16, 2012 at 12:17
  • $\begingroup$ Yes, $R$ is reflexive, antisymmetric, and transitive. $R$ is not symmetric. It seems like the official solution erroneously failed to mention antisymmetry. $\endgroup$
    – littleO
    Commented Nov 16, 2012 at 12:38

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