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I have a division like this

16/8/4/2

what is the default way to do calculations when the bracket is not specified .

Method 1 :

Is it correct to go from right to left like

[16/ (8 / { 4 / 2 } )]

I tried it in calculator as 16/8/4/2 and it gave 0.25

Method 2 :

May be it went from left to right

[({16 / 8} / 4 ) / 2]

Why I am asking this is because I saw a problem in GRE book like

ab/c/cd/a

without any brackets . their solutions reads

Method 3 :

$$ (\frac{ab}{c} \times \frac{a}{cd} ) $$ $$ = \frac{a^2 \times b}{c^2 \times d} $$

I am puzzled how they assumed brackets like that without any indication in question .

what is the right way to go as per maths .

Attaching image to show the exact problem . (to show the bars even look of the same size) enter image description here

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    $\begingroup$ But here is no problem. They ask for $\frac{\frac{ab}{c}}{\frac{cd}{a}}$, and not for $ab/c/cd/a$. $\endgroup$ – Dietrich Burde Jan 29 '15 at 20:50
  • $\begingroup$ they are the same .. what if it is slanting dash or horizontal dash .. they all are of same size anyway .. so it is ab/c/cd/a $\endgroup$ – Harish Kayarohanam Jan 29 '15 at 20:54
  • $\begingroup$ No, it is not the same. They write it already as a fraction $\frac{A}{B}$ with $A=\frac{ab}{c}$ and $B=\frac{c}{cd}$, which determines the bracketing. $\endgroup$ – Dietrich Burde Jan 29 '15 at 20:56
  • $\begingroup$ how can we say that if there are no brackets ? $\endgroup$ – Harish Kayarohanam Jan 29 '15 at 20:58
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From the image I would say they mean $$ ab/c\big/cd/a=\frac{\frac{ab}{c}}{\frac{cd}{a}}=\frac{ab}{c}\frac{a}{cd}=\frac{a^2b}{c^2d}. $$ But you are right, without brackets we have several possibilities, e.g., $16/8$ divided by $4/2$ is $1$.

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  • $\begingroup$ e.g., 16/8 divided by 4/2 is 1. . I thinks this is what you have followed in your solution . it is not left to right I think .. you assumed a bracket in between .. $\endgroup$ – Harish Kayarohanam Jan 29 '15 at 20:45
  • $\begingroup$ if you are confident it is from left can you solve the variable problem with brackets step by step and show ... pls .. $\endgroup$ – Harish Kayarohanam Jan 29 '15 at 20:46
  • $\begingroup$ Yes, I assumed a bracket in the middle; sometimes the $/$ is a bit bigger there. $\endgroup$ – Dietrich Burde Jan 29 '15 at 20:46
  • $\begingroup$ In the problem every bar looked similar ... $\endgroup$ – Harish Kayarohanam Jan 29 '15 at 20:47
  • $\begingroup$ attaching an image of the problem statement . $\endgroup$ – Harish Kayarohanam Jan 29 '15 at 20:49

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