(i) It follows from $[x] \leqslant x \leqslant y<[y]+1$.
(ii) It follows from $m \leqslant x < m+1$ and the definition. This property is a commonly used technique when proving properties related to $[x]$.
(iii) It follows from $[x]+m \leqslant x+m<([x]+m)+1$ and the definition.
(iv) $x+y=[x]+[y]+\{x\}+\{y\}$ and $0 \leqslant\{x\}+\{y\}<2$. When $0 \leqslant\{x\}+\{y\}<1$, by (ii) we know $[x+y]=[x]+[y]$; when $1 \leqslant\{x\}+\{y\}<2$,
$$x+y=[x]+[y]+1+(\{x\}+\{y\}-1),$$

By (ii) we know
$$[x+y]=[x]+[y]+1$$
(v) It is obviously true when $x$ is an integer. When $x$ is not an integer, $-x=-[x]-\{x\}=$ $-[x]-1+1-\{x\}, 0<1-\{x\}<1$, by (ii) we know the conclusion holds.
(vi) By the division algorithm, there exist integers $q, r$, such that
$$\begin{aligned} {[x] } & =q m+r, \quad 0 \leqslant r<m \\ {[x] / m } & =q+r / m, \quad 0 \leqslant r / m<1 \end{aligned}$$

From this and (ii), we deduce $[[x] / m]=q$. On the other hand,
$$x / m=[x] / m+\{x\} / m=q+(\{x\}+r) / m$$

Noting that $0 \leqslant(\{x\}+r) / m<1$, from this and (ii) we deduce $[x / m]=q$. Therefore, (vi) holds.
(vii) Let the smallest integer not less than $x$ be $a$, i.e., $a-1<x \leqslant a$. Therefore, $-a \leqslant -x<-a+1$, so $-a=[-x]$, i.e., $a=-[-x]$.
(viii) and (ix) are left to the reader, using the same method as (vii).
(x) The integer closest to $x$ must be among $[x]$ and $[x]+1$. When $x+1 / 2$ is an integer, these two numbers are equidistant from $x$. It is easy to verify that $[x]+1=[x+1 / 2]$ and $[x]=$ $[x-1 / 2]=-[-x+1 / 2]$. When $x+1 / 2$ is not an integer, if $\{x\}<1 / 2$, the integer closest to $x$ is $[x]$. Since $x+1 / 2=[x]+\{x\}+1 / 2,0 \leqslant\{x\}+1 / 2<1$, by (ii) we know $[x]=[x+1 / 2]$; if $1 / 2<\{x\}<1$, the integer closest to $x$ is $[x]+1$. Since $x+1 / 2=[x]+1+\{x\}-1 / 2,0<\{x\}-1 / 2<1$, by (ii) we know $[x]+1=[x+1 / 2]$. When $x+1 / 2$ is not an integer, by (v) we know
$$\begin{aligned} {[x+1 / 2] } & =-[-x-1 / 2]-1 \\ & =-[-x+1 / 2-1]-1=-[-x+1 / 2] \end{aligned}$$

The last step uses (iii). Proof completed.
(xi) Since the integer $n \leqslant x$ is $n \leqslant[x]$, it holds.
(xii) The positive integers divisible by $a$ are $a, 2 a, 3 a, \cdots$. Suppose the number of positive integers divisible by $a$ in $1,2, \cdots, N$ is $k$, then there must be $k a \leqslant N<(k+1) a$, i.e., $k \leqslant N / a<k+1$, so it holds.

The symbol $[x]$ is very useful. Here is an example.