Show that $\epsilon ( integer(a/\epsilon) - integer(b/\epsilon) )= a - b + O(\epsilon)$

Question

$integer(a)$ = the integer part of a real number $a$. For example, $integer (\pi) =3$. Let $\epsilon >0$ and $a<b$ real numbers.

How to show that $$\epsilon ( integer(a/\epsilon) - integer(b/\epsilon)) = a - b + O(\epsilon)$$

Answer 1

Instead of some computer language, let me use mathematical language.

Claim: $$ \epsilon\left(\left\lfloor\frac{a}{\epsilon}\right\rfloor - \left\lfloor\frac{b}{\epsilon}\right\rfloor\right) = a-b+O(\epsilon)\quad\text{as $\epsilon \to 0^+$} $$ We will use this property of the integer part: $$ x-1 < \lfloor x \rfloor \le x. $$ First an upper bound $$ \epsilon\left(\left\lfloor\frac{a}{\epsilon}\right\rfloor - \left\lfloor\frac{b}{\epsilon}\right\rfloor\right) \le \epsilon\left(\frac{a}{\epsilon} - \frac{b}{\epsilon}+1\right) = a-b+\epsilon $$ Simlarly, prove a lower bound $$ \epsilon\left(\left\lfloor\frac{a}{\epsilon}\right\rfloor - \left\lfloor\frac{b}{\epsilon}\right\rfloor\right) \ge \epsilon\left(\frac{a}{\epsilon} - 1 - \frac{b}{\epsilon}\right) = a-b-\epsilon $$ I hope you can finish the proof from these two inequalities.