Is there a name for the rule $a \div (b \times c) = a \div b \div c$? 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 $
 A: 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.
A: 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.
A: 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.
A: 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 "$\box{\times b} \box{\div c}$".
A: 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?
A: 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.
