• $a,b,c$ be integers with $a \ge b$ and $c > 0$

There exists integers $s,t$ such that:

$$\left\lfloor\frac{a}{c}\right\rfloor = \frac{a - s}{c}\text{ },\text{ }\left\lfloor\frac{b}{c}\right\rfloor = \frac{b-t}{c}$$

Does it follow that:

$$\left\lfloor\frac{a}{c}\right\rfloor - \left\lfloor\frac{b}{c}\right\rfloor = \left\lfloor\frac{a-b}{c}\right\rfloor \iff s \ge t$$

Here's my thinking:

If $s \ge t$, then $c > (s-t) \ge 0$ and:

$$\left\lfloor\frac{a}{c}\right\rfloor - \left\lfloor\frac{b}{c}\right\rfloor = \frac{a - b - (s-t)}{c} = \left\lfloor\frac{a-b}{c}\right\rfloor$$

If $s < t$, then $0 > (s-t) > -c$ and:

$$\frac{a - b - (s-t)}{c} = \frac{a - b - (s-t)}{c} = \left\lfloor\frac{a-b}{c}\right\rfloor + 1$$

Did I make a mistake? If I did not make a mistake, is there a simpler argument?

  • $\begingroup$ You don't explicitly say $c \gt 0$, but your proof assumes this. In my answer, I assumed it as well. However, FYI, if I didn't make a mistake, then as you can verify yourself, if $c \lt 0$, then your proposition would need to change to use $s \le t$ instead. $\endgroup$ – John Omielan Apr 4 '20 at 6:48
  • $\begingroup$ Good point. That was my assumption and it should have been called out. $\endgroup$ – Larry Freeman Apr 4 '20 at 6:52
  • $\begingroup$ One other small thing is your proposition doesn't depend on, or need, $a \ge b$ as it's only affected by whether or not $c$ is positive plus the relative sizes of the remainders $s$ and $t$. $\endgroup$ – John Omielan Apr 4 '20 at 6:55
  • $\begingroup$ Another good point! I was only considering positive values. My argument should apply just as well to negative values. Thanks for calling that out! $\endgroup$ – Larry Freeman Apr 4 '20 at 6:57

What you've done looks correct. Here's a somewhat different way to show it, although not really much simpler, but it does include a few points you don't show explicitly. Note you can use a similar proof in the case $c \lt 0$ to confirm your proposition, but you would need to change it to use $s \le t$ instead.

You have by Euclidean division that for some unique integer $m$, there is an integer $r$ such that

$$a = mc + r, \; 0 \le r \lt c \tag{1}\label{eq1A}$$

Thus, you then have

$$m = \left\lfloor\frac{a}{c}\right\rfloor = \frac{a - s}{c} \implies a = mc + s \tag{2}\label{eq2A}$$

This means that $s$ is uniquely the value of $r$ in \eqref{eq1A}. Similarly, this means for some unique integer $n$

$$\left\lfloor\frac{b}{c}\right\rfloor = \frac{b-t}{c} = n \implies b = cn + t, \; 0 \le t \lt c \tag{3}\label{eq3A}$$

As for your proposition of

$$\left\lfloor\frac{a}{c}\right\rfloor - \left\lfloor\frac{b}{c}\right\rfloor = \left\lfloor\frac{a-b}{c}\right\rfloor \iff s \ge t \tag{4}\label{eq4A}$$

note you have

$$m - n = \left\lfloor\frac{(m-n)c + s - t}{c}\right\rfloor = m - n + \left\lfloor\frac{s - t}{c}\right\rfloor \iff \left\lfloor\frac{s - t}{c}\right\rfloor = 0 \iff s \ge t \tag{5}\label{eq5A}$$


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