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Usually $$\frac{a}{\frac{b}{c}} = \frac{ac}{b}$$ i.e. $b/c$ is seen as the denominator, and $a$ is the numerator.

If you have $a/b/c/d$, what do you choose to take as the denominator? $$\frac{b}{\frac{c}{d}} , \frac{c}{d},\text{ or } d\text{?}$$ and why?

And what about for $a/b/c/d/e$?

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    $\begingroup$ A common convention is that you carry out divisions left to right. So $a/b/c$ would mean $(a/b)/c$. Similarly, $a/b/c/d$ would be $((a/b)/c)/d$. But it is much better to use parentheses to avoid ambiguity. $\endgroup$ – almagest Apr 26 '16 at 10:42
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    $\begingroup$ Usually $a/b/c/d= \frac{a}{bcd}$ etc $\endgroup$ – gammatester Apr 26 '16 at 10:44
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    $\begingroup$ I wish math teachers would teach operator precedence and associativity. $\endgroup$ – Kamil Jarosz Apr 26 '16 at 11:35
  • $\begingroup$ It does not follow proper conventions of notation, and you should avoid such things. Most calculators and computers will work left to right, apply the first operation, find and answer and then divide by the next number in the line. That doesn't mean that that is necessarily correct. $\endgroup$ – Doug M Apr 26 '16 at 20:21
  • $\begingroup$ There is no correct doug, convention is simply what people do. If everyone is going to assume that a/b/c/d is equivalent to ((a/b)/c)/d) it seems reasonable that's what you should use if that's what you intend to communicate. This would be first time I've heard where if they're all multiplication or division that they don't happen from left to right, but I'm aware that different countries and communities have different conventions. In what context is that that proper convention of notation and do you have any resources that affirm that? $\endgroup$ – VoronoiPotato Apr 26 '16 at 20:53
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The trouble is $$a/(b/c) \ne (a/b)/c$$since $a/(b/c) = \frac{{ac}}{b}$ and $(a/b)/c = \frac{a}{{bc}}$. So, without brackets, the notation is ambiguous. Introducing brackets, for example, we could choose between $$a/b/c/d=a{(b/c/d)^{ - 1}}$$ and $$a/b/c/d=(a{/b/c)d^{ - 1}}$$The first leads to $$a{(b{(c/d)^{ - 1}})^{ - 1}} = a{(b{c^{ - 1}}d)^{ - 1}} = \frac{{ac}}{{bd}}$$The second leads to $$(a/b){c^{ - 1}}{d^{ - 1}} = a{b^{ - 1}}{c^{ - 1}}{d^{ - 1}} = \frac{a}{{bcd}}$$ which is less interesting, it seems to me. The first expression leads to $$a/b/c/d/e = a{(b(c/d/e))^{ - 1}} = a{\left( {\frac{{bd}}{{ce}}} \right)^{ - 1}} = \frac{{ace}}{{bd}}$$ Collecting results, the following interesting pattern emerges:$$a/b = a{b^{ - 1}} = \frac{a}{b}$$ $$a/b/c = a{b^{ - 1}}c = \frac{{ac}}{b}$$ $$a/b/c/d = a{b^{ - 1}}c{d^{ - 1}} = \frac{{ac}}{{bd}}$$ $$a/b/c/d/e = a{b^{ - 1}}c{d^{ - 1}}e = \frac{{ace}}{{bd}}$$ $$a/b/c/d/e/f = a{b^{ - 1}}c{d^{ - 1}}e{f^{ - 1}} = \frac{{ace}}{{bdf}}...$$ and so on.

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  • $\begingroup$ It's pretty simple to extend to higher levels of fractions when you realise you just alternate whether the new latter goes on the top or bottom! $\endgroup$ – Idios Apr 26 '16 at 20:34
  • $\begingroup$ @Idos - hope you can see my new edit. Thanks for the interesting question. I enjoy all things ambiguous. And on that subject I strongly recommend 'How Mathematicians Think' by Byers, W. who discusses the progress of maths in terms of ambiguity. $\endgroup$ – user328032 Apr 26 '16 at 20:41
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As @almagest points out, the convention is that operations at the same level of precedence (e.g. multiplication/division) are carried out from left to right. So $a/b/c/d$ is interpreted as $((a/b)/c)/d$, and similarly $1-2-3-4$ is interpreted as $((1-2)-3)-4$. It is best to use parenthesis in any case, or even better, to use prefix or postfix notation.

$a/(b/c)$ or $\frac{a}{\frac{b}{c}}$ is indeed $ac/b$. We also have $$\frac{a}{\frac{b}{\frac{c}{d}}} = a/(b/(c/d)) = a/(bd/c) = ac/bd$$ and so on.

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  • $\begingroup$ I think the brackets definitely help clarify the order of operations, and make them more manageable by breaking it into familiar steps. $\endgroup$ – Idios Apr 26 '16 at 20:36
  • $\begingroup$ The brackets make the meaning clear. You can choose where to put them. $\endgroup$ – user328032 Apr 26 '16 at 20:44
  • $\begingroup$ @Idios - If you appreciate the answer that I gave can you accept to close the question? That is customary on the website. Thanks again. $\endgroup$ – user328032 Apr 26 '16 at 20:47

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