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One of the first things I learned at math at secondury school is the order of operations:

  1. Things between brackets
  2. Multiplication and division
  3. Addition and substraction

(At that time exponents and roots wheren't introduced.)

Now at university, I've seen formulas like $$Z_C = 1/j\omega C$$ (for the impedance of a capacitor in an electric circuit) which is also written as $$Z_C = \frac{1}{j\omega C}$$

But since both multiplication and division has the same rank on the order of operation, I would interpreted (if I wouldn't know better) the first formula as $$Z_C = 1/j\omega C = \frac{1}{j} \omega C = \frac{\omega C}{j}$$ (which could also be written as $-j\omega C$ since $j$ is the imaginary unit, but that's beside the point).

This is just one example, another would be $$R(\lambda, T) = \frac{2\pi h c^2}{\lambda^5 (e^{hc/\lambda kT} - 1)}$$ for Plank blackbody law, where it is meant that $hc/\lambda kT = \frac{hc}{\lambda k t}$

My question is, does division has higher priority than multiplication in the order of operation? I know my question sounds really basic, but I'm never told the answer.

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    $\begingroup$ If the first is intended to be $\frac{1}{j\omega C}$ then it is very sloppy notation. Unfortunately, your best bet is to figure it out from context because it should as you say be interpreted correctly as the second, not the first. That being said, if it was intended to be the second, you would think they would have just written $\omega C/j$ instead... the fact that they didn't should be a clue that they are sloppy and might not have written correctly. $\endgroup$
    – JMoravitz
    Commented Nov 3, 2016 at 19:08
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    $\begingroup$ Just use common sense!! $\endgroup$
    – john
    Commented Nov 4, 2016 at 9:10

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One thing we always have to remember about order of operations is that it's not an arbitrary set of rules that some random people decided to demand of others. It's a description of what people actually do in practice—habits that writers of math have agreed on over the years, habits that make writing and reading math easier.

People have found that in practice, there's no lack of clarity in expressions like $1/abc$, which I would unambiguously read as $\frac1{abc}$. The reason (in my opinion) is that anybody who intended $1/a \times bc$ would never ever write $1/abc$ when there are so many other easy ways to write it ($bc/a$ being the most straightforward). The fact that we parse $1/abc$ as $\frac1{abc}$ is a consequence of this asymmetry (one form has an easy other way of writing it without using parentheses, the other doesn't) and how that affects people's writing in practice, not a calculated decision by some body of standards.

Learning "order of operations" is a way of trying to help people understand what is done in practice. It's not a law that dictates what must be done—it's an attempt to describe what is done.

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Order of operations in a FRACTION always prioritizes numerator and denominator operations FIRST.

Thus, in $\frac{1}{jωC}$ the multiplicative operations of $j*w*C$ will be prioritized. Factoring out $(\frac1j)wC$ will simply lead to an incorrect answer.

Generalizing:

For any number represented in form $\frac xy$, where $x$ and $y$ are placeholders for any number of different mathematical operations:

you must complete all operations on the numerator $x$ and the denominator $y$ FIRST.

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    $\begingroup$ I'm not sure this written description correctly parses, say, $1+2/3+4$. $\endgroup$ Commented Jun 15 at 5:58
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No, division and multiplication have equal precedence, but note that Terms are separated by operators and joined by grouping symbols, thus jωC is a single term. More precisely it's a product, which is THE RESULT of a multiplication. "Multiplication" in order of operations refers literally to multiplication signs, not Terms - if a=2 and b=3 then 1/ab=1/6 and 1/axb=3/2 - so there's no "multiplication" in that formula, only division (by a Term/product).

Here's a screenshot from Cajori (1928) saying that bc and (bc) are the same thing (and is the same usage in all modern textbooks, as you've discovered) enter image description here

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  • $\begingroup$ I think you're seeing a convention that 1/ab is sometimes interpreted as 1/(ab) and then extrapolating that there is a fundamental difference between ab and a*b. There really isn't. $\endgroup$ Commented Jun 18 at 8:16
  • $\begingroup$ @sakurashinken I think you're failing to look in any textbook at all. Did you notice how NOT following that rule gives different answers? Now look at the question - that would mean the same equation gives different answers. $\endgroup$
    – donaldp
    Commented Jun 18 at 8:55

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