Floor Function Proof
For all the problems, the main thing you need is to write $$x = \lfloor x \rfloor + \{x\}$$ where $\lfloor x \rfloor \in \mathbb{Z}$ and $\{x\} \in [0,1)$.
For instance, for the first one
- $\lfloor m + n \rfloor = \lfloor \lfloor m \rfloor +\{m\} + n \rfloor = \lfloor \lfloor m \rfloor + n +\{m\} \rfloor = \lfloor m \rfloor + n$.
- If $$\lfloor \lfloor m \rfloor/n \rfloor = a \in \mathbb{Z}$$ then we have that $$\lfloor m \rfloor/n = a + b$$ where $b \in \left \{0,\dfrac1n, \dfrac2n, \ldots, \dfrac{(n-1)}n \right\}$. Hence, $$\lfloor m \rfloor = an + bn \implies m = an + bn + c$$ where $c \in [0,1)$. Hence, $$m/n = a + b + \dfrac{c}n$$ Note that $b + \dfrac{c}n \in \left\{ \dfrac{c}n,\dfrac{c+1}n, \ldots, \dfrac{c+(n-1)}n\right\} \subseteq [0,1)$. Hence, $$\lfloor m/n \rfloor = a = \lfloor \lfloor m \rfloor/n \rfloor$$
The first two are straightforward using the universal property of the floor function, viz. $$\rm k\le \lfloor r \rfloor \iff k\le r,\ \ \ for\ \ \ k\in \mathbb Z,\ r\in \mathbb R$$ Therefore for $\rm\:0 < n\in \mathbb Z,\ r\in \mathbb R,\ $ $$\rm\begin{eqnarray} &\rm k &\le&\:\rm\ \lfloor \lfloor r \rfloor / n\rfloor \\ \iff& \rm k &\le&\ \ \rm \lfloor r \rfloor / n \\ \iff& \rm nk &\le&\ \ \rm \lfloor r \rfloor \\ \iff& \rm nk &\le&\ \ \rm r \\ \iff& \rm k &\le&\ \ \rm r/n \\ \iff& \rm k &\le&\ \ \rm \lfloor r/n \rfloor \\ \\ \Rightarrow\ \ \rm \lfloor \lfloor r\!\!&\rm \rfloor / n\rfloor\ &=&\rm\ \ \lfloor r/n\rfloor \end{eqnarray}$$
since, having equal predecessors, these integers are equal.
If you know a little category theory you can view this universal property of floor as a right adjoint to inclusion, e.g. see Arturo's answer here, or see most any textbook on category theory. But, of course, one need not know any category theory to understand the above proof. Indeed, I've had success explaining this (and similar universal-inspired proofs) to bright high-school students.