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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0it is antisymmetric because it is never aRb and never bRa and thus never a=b? – 2012-11-22
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0Jane, is this making sense? Symmetric means for all a, b, if aRb, then bRa. Antisymmetric means that for all a, b: IF (aRb AND bRa), THEN a = b. – 2012-11-23
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1This 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. – 2012-11-23
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1@Limitless: adding different ways of saying "the same thing" is a good thing, because when first learning definitions and concepts, all it takes, sometimes, to "get it" is a simple rephrasing. So your input is most welcome! – 2012-11-23
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1Thank you, @amWhy! I love your answers, and that is why I attempt to add to them. – 2012-11-23
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0false implies false = true no? – 2012-11-23
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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