0

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

Blue
  • 75,673
Knocker379
  • 199

1 Answers1

2

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, $b > 0$.

egglog
  • 1,674
  • 1
    Indeed. Rounding $x$ down to the nearest integer can never cause $a+x$ to drop past an integer multiple of $b$, which is where the value of $\Bigl\lfloor \frac{\cdots}{b} \Bigr\rfloor$ would change. – Troposphere Jul 15 '21 at 18:42
  • Well put @Troposphere – egglog Jul 15 '21 at 18:43