I am suspecting that this is not always true. The reason is that I do not see this inequality as a standard part of the floor function properties.

Still, I could not find counter examples and I did find the following argument which suggests that it is true:

Let $\{a\} = a - \lfloor{a}\rfloor$

Let $\{b\} = b - \lfloor{b}\rfloor$

$1 > \{b\} - \{a\} > -1$

$\lfloor{a} \rfloor - \lfloor{b}\rfloor = \lfloor\lfloor{a}\rfloor - \lfloor{b}\rfloor\rfloor \ge \lfloor\lfloor{a}\rfloor - \lfloor{b}\rfloor + (\{b\} - \{a\})\rfloor = \lfloor{a - b}\rfloor$

Is it always true? Did I make a mistake in my argument?

  • $\begingroup$ This can be rewritten as $\lfloor x+y\rfloor \geq \lfloor x\rfloor +\lfloor y\rfloor$. That might make it a little more obvious. $\endgroup$ – Thomas Andrews Jun 9 '17 at 15:21
  • $\begingroup$ Thanks very much! So, it is a standard rearrangement of a standard property! :-) $\endgroup$ – Larry Freeman Jun 9 '17 at 15:22

Hint: Show $\lfloor x+y\rfloor \geq \lfloor x\rfloor +\lfloor y\rfloor$.

Then let $x=b,y=a-b$.

  • $\begingroup$ Very clear! Thanks very much. $\endgroup$ – Larry Freeman Jun 9 '17 at 15:39

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