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Given x is a real number and c is a natural number. Prove that:

floor( floor(x) / c ) = floor( x / c )

When x > 0, I can prove it by let x = n + z where 0 < z < 1.
Then rhs = floor( n/c + z/c ) which equals to floor( floor(x) ) since z/c -> 0.
However, when x < 0, it's wrong when letting x = -( n + z ). Because it always give me a smaller negative number. Is my logic wrong here? Any hint?

Thanks,
Chan

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I suspect that you want $x=-n+z$ (no parentheses around $n+z$) as, for instance, $\text{floor}(-1.5)=-2$. –  Isaac Feb 4 '11 at 6:34
    
@Issac: Thanks, so if x = -1.67 then I would rewrite it as: -2 + 0.33? –  Chan Feb 4 '11 at 6:41
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2 Answers

up vote 1 down vote accepted

You can see it here:

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@Chandru1: Thanks but you spoiled me :(! –  Chan Feb 4 '11 at 6:42
    
@chan: Sorry chan. –  anonymous Feb 4 '11 at 7:24
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This is the special case $\rm\ m = 1\ $ in a proof I presented here. See that thread for more on the universal viewpoint that explains the simplicity of this proof. For convenience, here is the proof.

LEMMA $\rm\: \ \lfloor x/(mn)\rfloor\ =\ \lfloor{\lfloor x/m\rfloor}/n\rfloor\ \ $ for $\rm\ \ n > 0$

Proof $\rm\quad\quad\quad\quad\quad\ \ \ k\ \le \lfloor{\lfloor x/m\rfloor}/n\rfloor$

$\rm\quad\quad\quad\quad\quad\iff\quad\ \ k\ \le\ \:{\lfloor x/m\rfloor}/n$

$\rm\quad\quad\quad\quad\quad\iff\ \ nk\ \le\ \ \lfloor x/m\rfloor$

$\rm\quad\quad\quad\quad\quad\iff\ \ nk\ \le\:\ \ \ x/m$

$\rm\quad\quad\quad\quad\quad\iff\ \ \ \ k\ \le\:\ \ \ x/(mn)$

$\rm\quad\quad\quad\quad\quad\iff\ \ \ \ k\ \le\ \ \lfloor x/(mn)\rfloor $

Compare the above trivial proof to more traditional proofs, e.g. the special case $\rm\ m = 1\ $ here.

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Many thanks for a great solution. Very clever and succinct ;) ! –  Chan Feb 4 '11 at 15:34
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