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The statement:

"It is important to understand that the associative law is not self-evident; indeed, if $a*b=a/b$ for positive numbers $a$ and $b$ then, in general, $(a*b)*c\ne a*(b*c)$." - p. 3, A. Beardon, Algebra and Geometry.

I am unclear as to how to take the assumption $a*b=a/b$. Does he mean that we suppose, as an alternative (thought experiment), that $a*b := a/b$ and substitute $(a/b)$ for each $(a*b)$ and prove the inequality of the resulting statement? Or, should one notice that $a*b=a/b$ is true when $b=1$, but then note that the domain was assumed to be $\mathbb{R}_+$, therefore $(a*b)*c \ne a*(b*c)$ in general?

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    $\begingroup$ You take it as an example. $\endgroup$ – Lee Mosher Sep 18 '14 at 22:55
  • $\begingroup$ An addendum that could be added to most of the answers is that mathematicians very rarely write $*$ for ordinary multiplication. For expressions $X$ and $Y$, "$X$ times $X$" is commonly written $XY$, $X\cdot Y$ or $X\times y$ but seldom $X*Y$. $*$ is a programming language thing. $\endgroup$ – David Richerby Sep 19 '14 at 12:18
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    $\begingroup$ $*$ isn't multiplication here. The author is asking you to suppose you have a group where the binary operation (written $*$) is the division operation. The author could have said, "suppose $*$ is $/$, i.e., division." $\endgroup$ – Joshua Taylor Sep 19 '14 at 12:26
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You are basically right in your guesses. What is being emphasized is that not all binary operations (i.e. things that take two inputs from your set and return a third element of the set) that one can define are necessarily associative just because they are binary operations. In this example they are saying that if you define the particular example of a binary operation, to which they give the name $*$, on the set $\Bbb R^+$, as in this example, i.e. $a*b={a\over b}$, then

$$(a*b)*c={{a\over b}\over c}={a\over bc}$$

and

$$a*(b*c)={a\over {b\over c}}={ac\over b}$$

by direct computation.

At this point you might object and say "but doesn't $*$ mean multiplication?" If so, you can replace the symbol $*$ by another one if it helps you, $\oplus$ is a nice one as well, the point is it's the binary operation we've defined, no matter what it looks like on paper as a symbol.

Now, by definition two real numbers, $x,y\in\Bbb R$, the definition of equality is

$$x=y\iff x-y=0$$

and given two fractions, ${p\over q}, {r\over s}\in\Bbb R^+$ the definition of equality is

$${p\over q}={r\over s}\iff ps-qr=0.$$

In your case:

$${a\over bc}-{ac\over b}=0$$ $$\iff {a\over bc}={ac\over b}$$ $$\iff ab-abc^2=0$$ $$\iff ab=abc^2$$

for every $a,b,c\in\Bbb R^+$, however this condition clearly requires $c^2=1$ which is only true for $c=1$ over the positive real numbers, so the operation is not associative, because the relation only holds for a single positive real number, $c=1$, rather than all positive reals.

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    $\begingroup$ Very comprehensive, thank you Adam! $\endgroup$ – Joe Sep 18 '14 at 20:28
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    $\begingroup$ One of the best answers I've seen to a question. $\endgroup$ – avgvstvs Oct 7 '14 at 20:06
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Here, the $*$ represents an arbitrary binary operation, not multiplication as you think. The definition of $*$ is a map $\mathbb{R}_+ \times \mathbb{R}_+ \rightarrow \mathbb{R}_+$ that is defined by $*(a,b)=\frac{a}{b}$. Now, $a*b$ is just an alternate way of representing $*(a,b)$.

Now why isn't $*$ associative? Well, we have $(a*b)*c=(a/b)*c=(a/b)/c=\frac{a}{bc}$, and $a*(b*c)=a*(b/c)=a/(b/c)=\frac{ac}{b}$, hence $(a*b)*c \not= a*(b*c)$.

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  • $\begingroup$ Fujoyaki, not sure how it happened, but your answer, which was more timely than some others but didn't show up until much later upon a third refresh of the browswer. Very nice answer and wording, thank you! $\endgroup$ – Joe Sep 18 '14 at 20:42
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I think you're assuming that * means multiplication. But in this situation it represents a general binary operation, and in particular, division.

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  • $\begingroup$ Thank you anon, you nailed it. I could only give the answer to one (due to shortcomings in this SE system IMHO), but I surely appreciate your help (and fast!). $\endgroup$ – Joe Sep 18 '14 at 20:33
  • $\begingroup$ @Joe Ah yes, the only-one-answer "shortcoming." That is a fight you will not win. (To give the policy its due, I do think it's helped keep SE sites remain Q&A focused rather than devolving into discussion forums.) $\endgroup$ – Kyle Strand Sep 19 '14 at 0:00
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In general, without making any assumptions, $\ast$ stands just for a group operation. In your cited example, it is defined to be division. Therefore, in this case, $(a \ast b) \ast c = (a/b)/c \neq a/(b/c) = a \ast (b \ast c)$, except for trivial cases.

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    $\begingroup$ herrsimon, I am torn because you gave the answer succinctly and, as I now see, correctly, with the case worked out. I'd give you the answer too if I could. I was helped by some of the additional details Adam provided. Clearly, you also provided the answer. I wish this was just a points system and not only one getting the answer credit. All three responses were the answer, just varying levels of comprehensiveness/length. I appreciate succinct too! My intent was to give you two points, anon 1, and Adam the answer. Alas, somehow I gave you one and anon 2 and I can't yet change that. $\endgroup$ – Joe Sep 18 '14 at 20:32

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