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This is a followup to one of my responses to a question by my old teacher Ravi Kulkarni found at A question about groups: may I substitute a binary operation with a function?. I attempted to make a rather elaborate fine set theoretic distinction between binary operations and partial binary operations and I constructed an example of a subgroup of the real line R* = R - {0} where the binary operation was ordinary divison.

To make a long story short, it was debated whether or not algebraicists considered division a binary operation on the reals. I think I should be more explicit about my thinking on the original literature examples, which I drew from Vinberg's A Course in Algebra I thought if you restricted the domain to exclude 0, it should indeed be a binary operation since for every nonzero a,b in R , the division a/b can also be expressed by the multiplication a * (1/b); essentially the old "high school" idea that division by whole numbers is "equal" to multiplication by rational number inverses. I suddenly realized I-as sadly many of my fellow students and math practioners had taken this dogma on faith and didn't really question it.

The obvious problem here is that they can't be "the same" since multiplication on the reals is associative and therefore forms a group on R* = R - {0} and division, while clearly a binary operation on R* = R - {0}, cannot form a group since it's not associative, as many reminded me at the question.

So here's my question of the algebraists here: Where does my example break down, exactly? It seems intuitively plausible that ( R*, ) and ( R, /) -where * and / are ordinary multiplication and division-should be isomorphic to each other if the old school dogma is true. Clearly they can't be. So where exactly does the reasoning break down?

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    $\begingroup$ I have absolutely no clue of why such a thing woudd seem «intuitively plausible», really —it seems quite an outlandish idea to me (and it is difficult to point where the reasonaning breaks when you have shown no reasoning supporting that idea!) In any case, it could be useful to remove most of the unnecessary chatter in your post, reducing it to your actual question. $\endgroup$ – Mariano Suárez-Álvarez Nov 27 '13 at 4:08
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    $\begingroup$ In particular, unsusbstantiated talk of «old school dogma» rarely helps communicate sensibly anything. $\endgroup$ – Mariano Suárez-Álvarez Nov 27 '13 at 4:10
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    $\begingroup$ What were they taught? $\endgroup$ – Mariano Suárez-Álvarez Nov 27 '13 at 4:15
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    $\begingroup$ Well, that has absolutely nothing to do with the question posed in the last paragraph of yout post. $\endgroup$ – Mariano Suárez-Álvarez Nov 27 '13 at 4:20
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    $\begingroup$ It is a basic fact that for all real numbers $x$ and $y$ with y not zero it is true that $x/y=x(1/y)$. If what you want to know is why this is true, then ask that (and if that ia what you wanted to ask then pretty much all you wrote is quite inconducive to expressing it...) $\endgroup$ – Mariano Suárez-Álvarez Nov 27 '13 at 4:28
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The fact that division by $b$ is the same as multiplication by $1/b$ does not mean that division is the same as multiplication.

Similarly, the fact that adding $0$ is the same as multiplying by $1$ does not mean that addition is the same as multiplication, nor that $0=1$.

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  • $\begingroup$ Yes,but this case, I think,is different since a/b and a*(1/b)= a/b seems in principle to yield the same net division operation for any 2 nonzero real numbers a,b.There is no analogous calculation for your case. This seems special-the question is why. $\endgroup$ – Mathemagician1234 Nov 27 '13 at 4:32
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    $\begingroup$ Okay, but generalizing your example (of multiplication and division) I'm sure you can think of plenty of examples of binary operations $f$ and $g$ and a unary operation $h$, all on a set $X$, such that $f(a,b) = g(a,h(b))$ for all $a,b\in X$ but $(X,f)$ and $(X,g)$ are not isomorphic as algebraic structures. $\endgroup$ – Trevor Wilson Nov 27 '13 at 4:35
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    $\begingroup$ What? The operations are not "the same". It sounds like you still don't understand the first sentence of my answer. $\endgroup$ – Trevor Wilson Nov 27 '13 at 5:17
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    $\begingroup$ You are asking why $a/b = a \times 1/b$? One way to see it is that $a = a/1$, so $a \times 1/b = (a/1) \times (1/b) = (a \times 1)/(1 \times b) = a/b$. $\endgroup$ – Trevor Wilson Nov 27 '13 at 5:22
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    $\begingroup$ I wouldn't say it's accidental, but I suppose that's getting into philosophy. Anyway, it is definitely important to question everything in math. In your case I would have started by questioning the statement "It seems intuitively plausible that..." :-) $\endgroup$ – Trevor Wilson Nov 27 '13 at 5:34
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Well, you've already mentioned that $\mathbb{R}^{\div}=(\mathbb{R}-\{0\},/)$ fails to be associative. But even weaker, suppose $f:\mathbb{R}^{\div}\to\mathbb{R}^*$ is an isomorphism of loops (http://en.wikipedia.org/wiki/Loop_(algebra)), then we still must have (using the fact that a morphism in this setting satisfies $f(a/b)=f(a)f(b)$)

$f(1)=f(a/a)=f(a)^2\Longrightarrow f(a)\in\{\pm\sqrt{f(1)}\}$,

so the map could not be surjective. (Also, necessarily we'd have $f(1)>0$)

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  • $\begingroup$ Ok-yet we still consider the 2 operations to be "the same" in some sense and this is supported by the fact many of us apply either method to multiplication and division to real numbers and get the same answer. So what exactly gives this "sameness"? Could it be an equvalence relation? Any function induces an equivalence relation, so that would be easy to establish-but is is correct? $\endgroup$ – Mathemagician1234 Nov 27 '13 at 4:10
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    $\begingroup$ If you consider the two operations to be the same in pretty much any sense, you are simply wrong. It is difficult to guess whatbyou are after... $\endgroup$ – Mariano Suárez-Álvarez Nov 27 '13 at 4:12
  • $\begingroup$ @Mariano: So it's an accident we get the same result when applying both operations as I indicated? I doubt that. $\endgroup$ – Mathemagician1234 Nov 27 '13 at 4:24
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    $\begingroup$ @Mathemagician1234 You have not "indicated" any examples of $ab$ and $a/b$ being the same. Obviously this only happens when $a=0$ or $b=\pm1$, which probabilistically speaking is an event with measure zero, or in other words multiplication and division produce different results $100\%$ of the time. The accident you speak of is imaginary! $\endgroup$ – anon Nov 30 '13 at 9:10
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If $G$ is a group, then $(a,b)\mapsto ab^{-1}$ is a binary operation on $G$. It is not usually the same binary operation as the multiplication $(a,b)\mapsto ab$ on $G$, because of the inverse put on the second argument (the exceptions are mentioned below). In fact, it is usually not even an associative operation, let alone a group operation, let alone an isomorphic operation, let alone the same operation.

There is usually not an identity element for this new operation. If $x$ is an identity element for the new operation, then $ax^{-1}=xa^{-1}=a$ for all $a\in G$. From $ax^{-1}=a$ we see that $x=1$, the identity element of the group $G$. Then we have $a^{-1}=a$, so the existence of an identity element implies that every element of $G$ is its own inverse. Of course in that case $ab^{-1}=ab$ is the usual product on $G$.

This new operation is called division. Division equals multiplication only if every nonidentity element of the group has order $2$. In $\mathbb R\setminus\{0\}$, most elements do not have order $2$.

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