Rules for "multiplication" in D&D and Pathfinder are unusual.
This is what Pathfinder rules say about multiplication:
When you are asked to apply more than one multiplier to a roll, the multipliers are not multiplied by one another. Instead, you combine them into a single multiplier, with each extra multiple adding 1 less than its value to the first multiple.
…and this is what D&D 3.5e rules say:
When two or more multipliers apply to any abstract value (such as a modifier or a die roll), however, combine them into a single multiple, with each extra multiple adding 1 less than its value to the first multiple.
I know the D&D rule for multiplication does not apply to real-world values, only to abstract ones, and I have no idea if the problem I'm describing here ever happens in D&D. This is not true for Pathfinder, where the provided rules are true for every out-of-game, rule-induced multiplication, and the problem arises (see below).
So if you double \$A\$, then triple it, you get \$A \times \left[2+\left(3-1\right)\right] = 4A\$ instead of the expected \$6A\$ that a true multiplication would give us.
How does this work when dividing?
(e.g. when halving creation times because of two different features or feats in Pathfinder. I'm not sure halving twice ever applies to abstract values in D&D.)
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Do I just divide like in the real world, since this rule is only valid for multipliers?
$$ A \times \frac{1}{2} \times \frac{1}{2} = A \times \frac{1}{4} $$
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Do I apply the multiply rules, since division is just multiplication with fractions?
$$ A \times \frac{1}{2} \times \frac{1}{2} = A \times \left[ \frac{1}{2} + \left( \frac{1}{2} – 1 \right) \right] = A \times 0 $$
I guess not. As you can see the formula breaks for any second or later multiplier that is 1 or lower (which should not happen, with numbers > 1 the formula works equally well whatever the order of the multipliers).
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Do I find the inverse of the operation, so that when I multiply the values back it gives me the explected value??
$$ A \times \frac{1}{2} \times \frac{1}{2} = A \times \frac{1}{2 + \left(2 – 1\right)} = A \times \frac{1}{3} = B $$
$$ \text{so that } B \times 2 \times 2 = \left(A \times \frac{1}{3}\right) \times \left[2 + \left(2 – 1\right)\right] = A $$
I don't really expect the authors to have thought about this, but who knows?
Maybe you know.
…or at least I'm hoping so.
Best Answer
Multiplication of "xN" multipliers on a roll is a special case with a special rule in the game (http://www.d20pfsrd.com/basics-ability-scores/glossary#TOC-Multiplying). It doesn't even apply to all multiplication in the game, just that on rolls that use a multiplier. Division is not lined out as a special case, it works like normal math (plus rounding down of course), which should be clear from all the rules that discuss the progression from 1/2 to 1/4 (greater evasion, movement rates, etc.). They say halved and they mean halved, not "times 1/2" - mathematically equivalent in math class, but not the way the rules work.
Note that the answer to the question you link was incorrect (it has been corrected) under any interpretation of whether 'division is multiplication by a fractional numerator' because the multiplication rule strictly states it only applies to a roll that has a multiplier on it - "When you are asked to apply more than one multiplier to a roll" - not times, not distances, not things that aren't a roll. It is very clearly scoped.
Note also that many effects - like size change - work in moving up and down categories, not actual doubling or halving.