I don't think so, because it is never $aRb$ nor $bRa$ and it is never $aRa$ or $bRb$, thus it is always false, but I don't know if I understood what antisymmetric means exactly.
Is the relation $x+3y = 0$ antisymmetric for all real numbers?
1 Answers
Given the relation $\, \mathcal{R}$ defined on $\mathbb{R}$ such that $x\,\mathcal{R}\, y\,$ if and only if $x + 3y = 0$:
$\,\mathcal{R}\,$ is antisymmetric if and only if, for all $a, b\in \mathbb{R},$ whenever both $ a\,\mathcal{R}\, b\,$ and $b \mathcal{R} a$, then it must be the case that $a=b$.
Let $a, b\in \mathbb{R}$ and define $R$ such that $a \,R \,b$ if and only if $a + 3b = 0$. $a\,R\, b \;\text{ and}\;\; b \,R \,a \implies a + 3 b = 0 \,\text{ and}\;\, b + 3a = 0.$ $\iff a + 3b = b + 3a$ $\iff -2a = -2b$ $\iff a=b.$
Therefore, $\forall a, b \in \mathbb{R},\;a \, R \, b \;\text{ and}\;\; b \,R \,a \implies a=b.$
So the relation IS antisymmetric on $\mathbb{R}$ for all real numbers since for $a, b \in \mathbb{R}$ IF both $\,a \,\mathcal{R} \,b\,$ and $\,b\, \mathcal{R} \,a$, then $a = b$.
The fact that there are no pairs $a, b \in \mathbb{R}$ other than $a = b = 0$ where both $\,a\, \mathcal{R} \,b\,$ and $\,b \,\mathcal{R} \,a\,$ doesn't matter. All that matters when determining whether a relation is antisymmtric, is that if and when it happens to be the case that there are $a, b$ such that both $\,a\,\mathcal{R} \,b\,$ AND $\,b\,\mathcal{R}\, a\,$ then it must follow that $a=b$.
In short, the relation is antisymmetric.
Edit, added for clarification in response to comments:
Note that antisymmetric is not the opposite of symmetric (the term is misleading in that sense).
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0@Jane yes, false implies false evaluates to true, and true implies true evaluates to true (in the case of a = b = 0). – 2012-11-23