This is something thats been bugging me since my high school.
Is A/B/C = A/BC or AC/B
Asked
Active
Viewed 1.6k times
0
-
I have restored the original question: the edits seriously distorted it by changing the one-line notation to multi-line notation. – Brian M. Scott Oct 20 '13 at 06:51
-
$A/B/C$ is just sick notation. – Michael Hoppe Oct 20 '13 at 10:53
3 Answers
2
Division is actually multiplying by the opposite, meaning:
$\frac{a}{\frac{b}{c}} = a * \frac{c}{b} = \frac{ac}{b}$
DanielY
- 979
2
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 $(A/B)/C$.
Brian M. Scott
- 616,228
-
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 !! – user2340452 Oct 20 '13 at 06:06
-
@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: $\frac32\cdot\frac51=\frac{3\cdot5}{2\cdot1}=\frac{15}2$. – Brian M. Scott Oct 20 '13 at 06:08
-
@user2340452: Actually, $(53)/2 = 5/(2/3)$. The "almost equal but not" is because $0.666$ is not $2/3$ (although it is almost* equal), so making that substitution introduces error. – Oct 20 '13 at 06:22
1
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!)
Adriano
- 41,576
Mariano Suárez-Álvarez
- 135,076