Prove that, for any $a, b, c \in \mathbb{N}$, $\bigl\lfloor \frac{a}{bc} \bigr\rfloor = \Bigl\lfloor \frac{\bigl\lfloor \frac{a}{b} \bigr\rfloor}{c} \Bigr\rfloor$.

I have no idea how to prove this. Can you help me, please? Thanks!


marked as duplicate by dxiv, Namaste, MPW, Shailesh, Clayton Jun 24 '18 at 1:07

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  • $\begingroup$ It may be useful to note this would also be $$\bigg\lfloor\frac{\lfloor\frac ac\rfloor}{b}\bigg\rfloor$$ $\endgroup$ – Rhys Hughes Jun 24 '18 at 0:41
  • $\begingroup$ Please read the entirety of the following post: How to ask a good question. Then, please edit your post accordingly, Iulian. $\endgroup$ – Namaste Jun 24 '18 at 0:44

If either $b$ or $c$ divides $a$, this is trivial based off the transposability highlighted in my comment.

Suppose $\frac ab=I+F$, where $I$ is the integer part, $F$ is the fractional part. Thus, $\lfloor I+F\rfloor=I$, and so this problem summarises to proving: $$\bigg\lfloor\frac{I+F}{c}\bigg\rfloor=\bigg\lfloor\frac{I}{c}\bigg\rfloor$$

For $0\le F<1$

Can you do this?


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