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Edit, because I should have looked it up before I posted the question:

Is there a name for the rule $a \div (b \div c) = a \div b \times c$ ? I ran across this in Liping Ma's book, Knowing and Teaching Mathematics, and I have searched the internet for a name for this rule to no avail. It is not the distributive law, but it is rather similar. Thank you!

From Ma's book, p. 59 discussing "dividing by a number is equivalent to multiplying by its reciprocal":

"We can use the knowledge that students have learned to prove the rule that dividing by a fraction is equivalent to multiplying by its reciprocal. They have learned the commutative law. They have learned how to take off and add parentheses. They have also learned that a fraction is equivalent to to the result of a division, for example, $ \frac{1}{2} = 1 \div 2 $ . Now, using these, we can rewrite the equation this way:

$ 1\frac{3}{4} \div \frac{1}{2} \to $

$1\frac{3}{4} \div (1 \div 2) \to $

$1\frac{3}{4} \div 1 \times 2 \to $ (This is the step my question is about.)

$1\frac{3}{4} \times 2 \div 1 \to $ (and I'd like an explicit explanation of this step, too.)

$1\frac{3}{4} \times 2\to$ $1\frac{3}{4} \times (2 \div 1) \to $

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    $\begingroup$ I encourage you to write the second part as $(a \div b) \div c$ since division is not associative. $\endgroup$ – Will Jagy Sep 20 '15 at 19:36
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    $\begingroup$ An appropriate name for it is good way to make an error. $\endgroup$ – André Nicolas Sep 20 '15 at 19:39
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    $\begingroup$ I never use the division sign myself. I had to copy from the question, I did not know the Latex for division sign. On the other hand, i feel no ambiguity in $x-y-z,$ even though subtraction is not associative either. Matter of habit, I suppose. $\endgroup$ – Will Jagy Sep 20 '15 at 19:43
  • $\begingroup$ I was assuming that without parentheses, one would read from left to right, but you make a good point. I don't either except in the context of elementary (as in let's learn how to work with fractions) mathematics. Ok, I just looked it up, which I should have done in the first place. I'll add an edit to the original question, and maybe my question will be clearer. $\endgroup$ – Debra P Otto Sep 20 '15 at 20:10
  • $\begingroup$ See also matheducators.stackexchange.com/q/7837/1550. $\endgroup$ – user21820 Sep 21 '15 at 6:10
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Well the concept of inverses is fundamental so you eventually have to get it across whether or not you use the inverse notation. $\newcommand{\box}[1]{~\boxed{#1}~}$

$x \box{\times a} \box{\div a} = x$ for any $a \ne 0$ because $\box{\div a}$ exactly undoes $\box{\times a}$. $\box{\div 0}$ is not allowed simply because it is impossible to undo $\box{\times 0}$.

Also we can think of numbers as what we use to represent amounts of something, so "$a$" actually stands for "$\box{\times a}$". For example when we say "$3$ apples" we mean "apple$\box{\times 3}$". So actually what we have is $\box{\times a} \box{\div a} = \box{\times 1}$. It turns out that undoing multiplying amounts is itself undoable, so we have $\box{\div a} \box{\times a} = \box{\times 1}$ as well. Both of these are for $a \ne 0$.

Now as stated we should take "$\box{\div a}$" to mean "undo $\box{\times a}$", but it may not be obvious that this has an important implication for the question at hand as well. What is "$\box{\div (b \div c)}$"? It means precisely "undo $\box{\times (b \div c)}$" = "undo $(\box{\times b} \box{\div c})$". How to undo? Clearly "$\box{\times c} \box{\div b}$". Thus we immediately get $\box{\div (b \div c)} = \box{\times c} \box{\div b}$, which explains the original question fully.

One implicit notion used above is that we define "$\box{\times (ab)}$" to mean "$\box{\times a} \box{\times b}$". Think carefully about this. It means that the product of $a,b$ is defined as the combined action of ( multiplying by $a$ ) and ( multiplying by $b$ ). In the above we used this interpretation when we went from "$\box{\times (b \div c)}$" to "$\box{\times b} \box{\div c}$".

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  • $\begingroup$ Thank you! I am looking for mathematically correct ways to describe things at an elementary level. I was impressed with the Chinese teachers' explanations in Liping Ma's book, Knowing and Teaching Mathematics. The Chinese teachers had a good understanding of their subject, and they excelled at explaining and proving what they were teaching. I have a masters degree in mathematics, and I learned a lot from this book. Unfortunately, it isn't a math book, and even though there are examples, there is missing information for use as a curriculum guide. $\endgroup$ – Debra P Otto Sep 22 '15 at 4:52
  • $\begingroup$ @DebraPOtto: You're welcome. I know that I technically didn't answer your question since you asked for a name of the rule, though I don't see a need to name it other than as a property following from inverses. So if you think that my answer resolves your question, do mark it as the accepted answer. I empathize with your struggle to intuitively and yet correctly explain at an elementary level. It took me a few years to come to my current choice of explanation. [continued] $\endgroup$ – user21820 Sep 22 '15 at 5:41
  • $\begingroup$ @DebraPOtto: The above is just one part of it, because actually we want to give a full explanation of all the arithmetic rules. The best way in my opinion is to describe numbers as quantifiers of an action, as vaguely described at math.stackexchange.com/a/1127600/21820. The way quantifiers combine then form the intuitive reason for our arithmetic rules. For a lower level one might have to start with the positive quantities and the simpler notion of division into parts, but the more general notion is the real goal, as I briefly describe at math.stackexchange.com/a/1385576/21820. $\endgroup$ – user21820 Sep 22 '15 at 5:46
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In the same way as subtraction should be thought of as adding by the additive inverse, it is better to think of division as multiplication by the multiplicative inverse to avoid any potential confusion.

That is to say, $a-b-c = a+(-b) + (-c)$ and $a\div b\div c = a\times b^{-1}\times c^{-1}=a\times \frac{1}{b}\times \frac{1}{c}$

As for why $a\times \frac{1}{b}\times \frac{1}{c}= \frac{a}{b\times c}$, this is an immediate consequence of how multiplication is defined for rational numbers (and fractions in general) and so likely doesn't have a name.

The definition of multiplication of two fractions is $\frac{a}{b}\times \frac{c}{d} := \frac{a\times c}{b\times d}$, so you have $(\frac{a}{1}\times \frac{1}{b})\times\frac{1}{c} = \frac{a}{b}\times\frac{1}{c}=\frac{a}{b\times c}$

You go on to say "but in fraction form..." implying you think something looks different about the case where the numbers are fractions instead, but I see no difference. The application of the rule is exactly the same in both scenarios.

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  • $\begingroup$ Thank you. I can explain it as inverses, but I was trying to find a way to say it for the level of fraction learners (not necessarily children, but people who haven't been exposed to the inverse notation). I'm still trying to make it clear and correct without being overly technical. I appreciate the feedback. $\endgroup$ – Debra P Otto Sep 20 '15 at 21:01
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I suppose that $a,b,c$ are rational (or real) numbers. In this case your starting expression is equivalent to: $$ \dfrac{a}{b\times c}=a\times \dfrac{1}{b}\times \dfrac{1}{c}=\left(a \times \dfrac{1}{b} \right)\times \dfrac{1}{c}=\dfrac{a \times \dfrac{1}{b}}{c}=\dfrac{ \dfrac{a}{b}}{c} $$ so you can see that this property does not need a special name since it is simply the application of the definition of inverse and of associativity for the product.

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Never thought about this. My work with quadratic forms gives many expressions in one line, with a selection of plus and minus signs. It never bothered me, if there is a plus sign it gets added, and all those with minus signs get subtracted. I think most people do that. So, $$ s + t - u - v + w - x - y + z = (s + t + w + z) - (u + v + x + y) $$

The analogous usage with multiplication and division signs would be $$ s \cdot t \div u \div v \cdot w \div x \div y \cdot z = \frac{s t w z}{ u v x y} $$ but I can hardly imagine anyone writing the thing on the left hand side and expecting to be understood. In a single line, we could write $s t w z/( u v x y)$ instead.

Go Figure.

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But if you really don't want to touch inverses at all, and stick to dividing by only real-world quantities (positive with physical units), then another way is that "$a \div ( b \div c )$" can represent "$a \text{ cake} \div ( b \text{ cake} \div c \text{ people})$, which is "number of people we can feed when $a$ cake is divided into pieces of the size that each of $c$ people would get for a fair division of $b$ cake". Example usage would be when we know that 2 cakes divided by 5 people gives each one a decent meal, and so if we have 3 cakes we can feed $3 \div ( 2 \div 5 )$ people. Clearly it is the same if we first took the ratio of cake, $3 \div 2$, and then multiplied by $5$, the number of people we could feed using $2$ cakes. Feeling hungry now?

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  • $\begingroup$ Thank you for an excellent example! This is exactly what I was looking for. $\endgroup$ – Debra P Otto Sep 22 '15 at 4:52
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The set of all nonzero real numbers with $\times$ is a commutative group. Dividing by a number can be defined as multiplying by its multiplicative inverse. So from those properties, we can derive that $\forall a \in \mathbb{R} - \text{{0}}\forall b \in \mathbb{R} - \text{{0}}\forall c \in \mathbb{R} - \text{{0}} a \div (b \times c) = a \div (c \times b) = a \times (c \times b)^{-1} = a \times (b^{-1} \times c^{-1}) = (a \times b^{-1}) \times c^{-1} = (a \div b) \times c^{-1} = (a \div b) \div c$. Here I'm using the operation ^-1 to denote taking the inverse of the number using the operation included in the group, not to denote real number exponentation although it turns out to get the same result as the other meaning of that notation. For high level mathematicans who are good at thinking of how to write the type of proof I just wrote, once you know that a set with a certain operation is a commutative group, it's so easy to derive that result. That's probably why mathematicians never decided the result that that holds for all commutative groups needs to be written as a theorem that they say was carefully checked by experts to be true. It's also easy to show that for real numbers, when ever one expression is defined, the other expression is also defined and equal to it even when you don't restrict $a$ to being nonzero.

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  • $\begingroup$ Thank you, Timothy! $\endgroup$ – Debra P Otto Dec 28 '18 at 22:58
  • $\begingroup$ @DebraPOtto Thanks, your comment was useful because you were the OP and now I know that the OP found this answer useful. $\endgroup$ – Timothy Dec 30 '18 at 2:48

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