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This is something thats been bugging me since my high school.

Is A/B/C = A/BC or AC/B

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  • $\begingroup$ I have restored the original question: the edits seriously distorted it by changing the one-line notation to multi-line notation. $\endgroup$ – Brian M. Scott Oct 20 '13 at 6:51
  • $\begingroup$ $A/B/C$ is just sick notation. $\endgroup$ – Michael Hoppe Oct 20 '13 at 10:53
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The expression $A/B/C$ is sufficiently ambiguous that it should not be written at all. Some people do follow the convention that it should be evaluated from left to right, making it $(A/B)/C$, but I would never count on a reader to interpret it that way with any confidence. For the same reason I would not write $A/BC$: depending on the reader’s conventions, it can be understood either as $(A/B)C$ or as $A/(BC)$, and these are not in general equal.

If you cannot write the three-storey fraction on multiple lines, you should use parentheses:

$$\frac{A}{\frac{B}C}=A\cdot\frac{C}B=\frac{AC}B\;,$$

on one line $A/(B/C)$, while

$$\frac{\frac{A}B}C=\frac{A}B\cdot\frac1C=\frac{A}{BC}\;,$$

on one line $(A/B)/C$.

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  • $\begingroup$ This is exactly my confusion ! I just do a lengthy problem and in the end I get something like (2/3)X = 5 and I am confused here should it be (5 *3 )/2 or 5/0.666 !! They are almost equal but they are not !! $\endgroup$ – user2340452 Oct 20 '13 at 6:06
  • $\begingroup$ @user2340452: If you have $\frac23x=5$, you should be multiplying both sides by $\frac32$ to get $x=\frac32\cdot5$. Now use the fact that $5=\frac51$, and you’re home free: $\frac32\cdot\frac51=\frac{3\cdot5}{2\cdot1}=\frac{15}2$. $\endgroup$ – Brian M. Scott Oct 20 '13 at 6:08
  • $\begingroup$ @user2340452: Actually, $(5*3)/2 = 5/(2/3)$. The "almost equal but not" is because $0.666$ is not $2/3$ (although it is almost equal), so making that substitution introduces error. $\endgroup$ – Hurkyl Oct 20 '13 at 6:22
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Division is actually multiplying by the opposite, meaning:

$\frac{a}{\frac{b}{c}} = a * \frac{c}{b} = \frac{ac}{b}$

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It depends if by $\frac{A}{\frac{B}{C}}$ you mean $$\frac{\quad A\quad}{\frac{B}{C}}\qquad\text{or}\qquad\frac{\frac{A}{B}}{\quad C\quad}.$$ (You meant the first one, according to the $\TeX$ you used, but it is rather hard to tell from the way that particular equation gets rendered!)

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