# Proof Verification: Show that $\Big[\frac{x+n}{m}\Big] = \bigg[\frac{[x]+n}{m}\bigg]$

Let $m,n \in \mathbb{Z}$ and let $x \in \mathbb{R}$. Let $[x]$ denote the floor function.

We will attempt to prove

$$\Big[\frac{x+n}{m}\Big] = \bigg[\frac{[x]+n}{m}\bigg]$$

Suppose without loss of generality that $m > 0$.

By definition of the floor function on $\frac{x+n}{m}$, we have

$$\Big[\frac{x+n}{m}\Big] \leq \frac{x+n}{m} \tag{1}$$

We observe that $\Big[\frac{x+n}{m}\Big]$ is the closest integer to $\frac{x+n}{m}$ satisfying $(1)$.

If $[x]+n \leq x+n$, then

$$\frac{[x]+n}{m} \leq \frac{x+n}{m} \tag{2}$$

Once more applying the definition of the floor function on $(2)$,

$$\bigg[\frac{[x]+n}{m}\bigg] \leq \frac{[x]+n}{m} \leq \frac{x+n}{m} \tag{3}$$

To prove that there are no integers in the interval $\Big(\frac{[x]+n}{m}, \frac{x+n}{m}\Big)$, specifically $\big[\frac{x+n}{m}\big]$, we suppose to the contrary that such an integer $q$ exists satisfying

$$\frac{[x]+n}{m} < q < \frac{x+n}{m} \tag{4}$$

But $(4)$ implies that $[x] < mq-n < x$ where $(mq-n) \in \mathbb{Z}$, which is a contradiction, since $[x]$ is the greatest intger less than or equal to $x$.

Because there are no integers in such an interval, $\big[\frac{x+n}{m}\big]$ must satisfy

$$\Big[\frac{x+n}{m}\Big] \leq \frac{[x]+n}{m} \leq \frac{x+n}{m} \tag{5}$$

which implies that $\Big[\frac{x+n}{m}\Big]$ is the closest integer less than or equal to $\frac{[x]+n}{m}$; that is, $\Big[\frac{x+n}{m}\Big]$ is the greatest integer less than or equal to $\frac{[x]+n}{m}$.

Therefore, $\Big[\frac{x+n}{m}\Big] = \bigg[\frac{[x]+n}{m}\bigg]$, as desired.

Is there anything wrong with the proof?

Note: A proof can also be achieved using the division algorithm.

• "Suppose wlog that m>0" That "wlog" does mean something? If not, what about m<0?. Thank you. – Piquito Feb 10 '16 at 13:03
• @Piquito Yeah, it's supposed to mean without loss of generality. In other words, if we assume $m<0$, then its proof would be similar to the case $m > 0$, just interchange the inequality signs. – Jeremiah Dunivin Feb 10 '16 at 13:05
• Very well. Thank you. – Piquito Feb 10 '16 at 13:08
• @Piquito I went ahead and changed "wlog" to "without loss of generality" since you're right in the sense that not everyone may know what it means :) – Jeremiah Dunivin Feb 10 '16 at 13:08

If you rearrange equation (4), you should get $qm-n$ in the middle, not $qm+n$.