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When $a, b, c$ are positive integers, is this identity below is true for all $a, b, c$?

$$\left\lfloor \frac{\left\lfloor\frac ab \right\rfloor}c \right\rfloor =\left\lfloor \frac{\left\lfloor\frac ac \right\rfloor}b \right\rfloor $$

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  • $\begingroup$ Consider adding a tag for a broader subject area to which the question belongs. Some of these tags might fit. (autocomment) $\endgroup$ – user147263 Feb 9 '16 at 20:49
  • $\begingroup$ Probably not if a = 27 and b = 8 [a/b] =4. if c = 1.9. the [[a/b]/c] = [4/1.9] = 3. And [a/c] = 15. And [[a/c]/8] = [15/8] = 2. $\endgroup$ – fleablood Feb 9 '16 at 20:58
  • $\begingroup$ @fleablood 1.9 is not a positive integer tho $\endgroup$ – Hagen von Eitzen Feb 9 '16 at 20:58
  • $\begingroup$ meh...$$$$$$$$$$ $\endgroup$ – fleablood Feb 9 '16 at 21:04
  • $\begingroup$ Sorry, it is floor function not ceiling function. $\endgroup$ – esege Feb 9 '16 at 21:05
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Using division with remainder write $a=qbc+r$ with $0\le r<bc$ and then write $r=q'b+r'$ with $0\le r'<b$, as well as $r=q''c+r''$ with $0\le r''<c$. So $a=qbc+q'b+r'=qbc+q''c+r''$. Because $0\le r<bc$ we conclude that $0\le q'<c$ and $0\le q''<b$. Then $$ \left\lfloor\frac{\left\lfloor\frac ab\right\rfloor}c\right\rfloor= \left\lfloor\frac{qc+q'}c\right\rfloor=q$$ and $$ \left\lfloor\frac{\left\lfloor\frac ac\right\rfloor}b\right\rfloor= \left\lfloor\frac{qb+q''}c\right\rfloor=q.$$ So indeed $$\left\lfloor\frac{\left\lfloor\frac ac\right\rfloor}b\right\rfloor=\left\lfloor\frac{\left\lfloor\frac ab\right\rfloor}c\right\rfloor =\left\lfloor\frac a{bc}\right\rfloor.$$

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    $\begingroup$ I confess that I'm somewhat surprised by this. I would have suspected it to be false. +1. $\endgroup$ – MPW Feb 9 '16 at 21:45
  • $\begingroup$ @MPW Me too, but the counterexamples kept failing :) $\endgroup$ – Hagen von Eitzen Feb 9 '16 at 21:47
  • $\begingroup$ Yes, that was my experience. I even tried a small spreadsheet with all a,b,c in the range 1-5. Oh well, live and learn. $\endgroup$ – MPW Feb 9 '16 at 21:48
  • $\begingroup$ In hindsight it makes sense. Ceiling works as well. ceiling = q + 1 if r > 0. I can't tell you how long I struggled without realizing q' < c and q" < b. $\endgroup$ – fleablood Feb 10 '16 at 1:26

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