Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
thanks for de-uglifying my problem –  blacksmth Nov 16 '12 at 12:23

1 Answer 1

up vote 3 down vote accepted

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.

share|improve this answer
    
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. –  blacksmth Nov 16 '12 at 12:17
    
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
    
thank you very much. –  blacksmth Nov 16 '12 at 13:30

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.