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I'm trying to prove rigorously the following:

$\lfloor x/a/b \rfloor$ = $\lfloor \lfloor x/a \rfloor /b \rfloor$ for $a,b>1$

So far I haven't gotten far. It's enough to prove this instead

$\lfloor z/c \rfloor$ = $\lfloor \lfloor z \rfloor /c \rfloor$ for $c>1$

since we can just put $z=\lfloor x/a \rfloor$ and $c=b$.

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    Just to say, it's not homework.2012-07-19
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    hint: Start with $x=qa+r$ and $q=q'b+r'$ ($0\le r and $0\le r' ) then $$x=q'ab+(ar'+r)\ \ \text{with $\ 0\le ar'+r < a(b-1)+a$}$$2012-07-19
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    @hermionestranger You might also like to note that in the statement of the problem, we require $a$ and $b$ to be integers, as the result is not true otherwise.2012-07-19
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    @OldJohn Actually, you just need $b$ to be an integer.2012-07-19
  • 0
    Also, $b>1$ is stricter than needed; $b=1$ trivially works, too.2012-07-19

3 Answers 3