The operation $(*)$ is defined as $$a*b=|a-b|, \forall a,b \in \mathbb{R},$$

and I am to prove that $(*)$ is not associative in $\mathbb{R}$, that is, to prove that it is not true in general that


I do not really know how to do it smartly. I have two options in mind: (1) I can disprove that '$(*)$ is associative in general in $\mathbb{R}$' by giving a counterexample, such as $||2-4|-6| \neq |2-|4-6||$ or (2) I can do the following:

Let us suppose on the contrary that $$||a-b|-c|=|a-|b-c||,$$


$$|a-b|-c=\pm(a-|b-c|).$$ For any of the cases, plus or minus, I can arrive at contradictions in the following conditions: (1) $a=b \neq c$ (2) $a=c\neq b$ (3) $b=c\neq a$ (4) $a=b=c$, and (5) $a \neq b \neq c.$

Now, is there a smarter or more concise way to prove that $(*)$ is not associative in $\mathbb{R}$ in general?

  • 5
    $\begingroup$ The simplest way to prove that $\forall xP(x)$ does not hold is to show that for some $a$ we have $\lnot P(a)$. Thus the counter-example you have used it's fine. $\endgroup$ – Mauro ALLEGRANZA Feb 21 '15 at 13:05
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    $\begingroup$ I think a counter example is good enough. $\endgroup$ – velut luna Feb 21 '15 at 13:05
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    $\begingroup$ As the above comments indicate, disproof is typically easier than proof. One exception ruins the whole tea-kettle. $\endgroup$ – David Wheeler Feb 22 '15 at 3:18

For $*$ to be associative over $\Bbb R$, it must be $$||a-b|-c|=|a-|b-c||\quad \forall\ a,b,c \in\Bbb R.$$ Since the above identity must hold for all $a,b,c \in\Bbb R$ then, as @Kyson commented, a counterexample like $$4=|2-6|=||2-4|-6| \neq |2-|4-6||=|2-2|=0$$ proves $*$ is not associative in general.


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