For some $x\in\Bbb{R}$ and $\{a,b\}\in\Bbb{Z}$, prove or deny: $\lfloor\frac{a+x}b\rfloor=\left\lfloor\frac{a+\lfloor x\rfloor}b\right\rfloor$

Question

For some $x \in \mathbb{R}$ and $\{ a,b \} \in \mathbb{Z}$, prove or deny that:

$$\left\lfloor \frac {a+x}b\right\rfloor=\left\lfloor \frac {a+\lfloor x\rfloor}b\right\rfloor$$

I'm trying to use $m−1<\lfloor m \rfloor \leq m$.

I've tried substituting $x-1$ and into each side but I don't see a guaranteed proof or denial

Answer 1

Simple proof. Note that $a = kb + r$, and substitute this in $\lfloor \frac{kb + r + x}{b} \rfloor = \lfloor \frac{r + x}{b} \rfloor$

Obviously, comparing $\lfloor \frac{r + x}{b} \rfloor$ and $\lfloor \frac{r + \lfloor x \rfloor}{b} \rfloor$, it is obvious that the statement is always true, as both $r + x$ and $r + \lfloor x \rfloor$ share the same integer part, so division by another integer will return the same quotient always.

A few adjustments need to be made to the question. Namely, $b > 0$.