podłoga, tożsamość

[m-2]

Wykazać, że

\(\displaystyle{ \left \lfloor \frac{\lfloor \frac{x}{m} \rfloor}{n} \right \rfloor = \left \lfloor \frac{x}{mn} \right \rfloor
\\ \\ x \in \mathbb{R_{+}}, \; m,n \in \mathbb{Z_{+}}}\)

proszę o wskazówki

[Crizz]

Wskazówka: podstaw \(\displaystyle{ x=mnp+q}\), gdzie \(\displaystyle{ p\in \mathbb{Z}_{+},q\in<0,mn)}\).

[m-2]

\(\displaystyle{ \left \lfloor \frac{\lfloor \frac{x}{m} \rfloor}{n} \right \rfloor = \left \lfloor \frac{\lfloor \frac{mnp+q}{m} \rfloor}{n} \right \rfloor = \left \lfloor \frac{\lfloor np +\frac{q}{m} \rfloor}{n} \right \rfloor = \left \lfloor \frac{np + \lfloor \frac{q}{m} \rfloor}{n} \right \rfloor = p + \left \lfloor \frac{\lfloor \frac{q}{m} \rfloor}{n} \right \rfloor <\\< p + \left \lfloor \frac{\lfloor \frac{mn}{m} \rfloor}{n} \right \rfloor = p + 1 \Rightarrow \left \lfloor \frac{\lfloor \frac{x}{m} \rfloor}{n} \right \rfloor = p \\ \left \lfloor \frac{x}{mn} \right \rfloor = \left \lfloor \frac{mnp+q}{mn} \right \rfloor = p + \left \lfloor \frac{q}{mn} \right \rfloor = p $, bo $q\in<0,mn)}\)

Dzięki