 Wisniewool

2022-07-06

Triple fractions (and more complex fractions)
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?

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

Expert

The trouble is
$a/\left(b/c\right)\ne \left(a/b\right)/c$
since $a/\left(b/c\right)=\frac{ac}{b}$ and $\left(a/b\right)/c=\frac{a}{bc}$. So, without brackets, the notation is ambiguous. Introducing brackets, for example, we could choose between
$a/b/c/d=a\left(b/c/d{\right)}^{-1}$
and
$a/b/c/d=\left(a/b/c\right){d}^{-1}$
$a\left(b\left(c/d{\right)}^{-1}{\right)}^{-1}=a\left(b{c}^{-1}d{\right)}^{-1}=\frac{ac}{bd}$
$\left(a/b\right){c}^{-1}{d}^{-1}=a{b}^{-1}{c}^{-1}{d}^{-1}=\frac{a}{bcd}$
which is less interesting, it seems to me. The first expression leads to
$a/b/c/d/e=a\left(b\left(c/d/e\right){\right)}^{-1}=a{\left(\frac{bd}{ce}\right)}^{-1}=\frac{ace}{bd}$
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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