# Is the relation $x+3y = 0$ antisymmetric for all real numbers?

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.

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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.

This doesn't feel significant enough to make an answer, but I'd like to put my two cents in. It is nothing more than a different statement of what you've already said: Consider what your relation is. Let your relation be denoted by $\mathcal{R}$. If we have $x\mathcal{R}y$, then $x+3y=0$. If we have $y\mathcal{R}x$, then $y+3x=0$. Since $x+3y=0$ and $y+3x=0$ are only both true when $x=y$, $\mathcal{R}$ is antisymmetric. – 000 Nov 23 '12 at 0:21