Wisniewool

Answered

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?

$\frac{b}{\frac{c}{d}},\frac{c}{d},\text{or}d\text{?}$

and why?

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

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 $a/b/c/d/e$?

Answer & Explanation

karburitc

Expert

2022-07-07Added 7 answers

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.

$a/(b/c)\ne (a/b)/c$

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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