$$已知:\lfloor x \rfloor \leq x < \lfloor x \rfloor+1,\lfloor \lfloor x \rfloor \rfloor = \lfloor x \rfloor$$
$$证明:\lfloor \frac {\lfloor \frac {x} {a} \rfloor} {b} \rfloor=\lfloor \frac {x} {a\times b} \rfloor$$
$$那么有\lfloor \frac {\lfloor \frac {x} {a} \rfloor} {b} \rfloor \leq \frac {\lfloor \frac {x} {a} \rfloor} {b} < \lfloor \frac {\lfloor \frac {x} {a} \rfloor} {b} \rfloor + 1$$
$$\frac {\lfloor \frac {x} {a} \rfloor} {b} \leq \frac {x} {a\times b} < \frac {\lfloor \frac {x} {a} \rfloor} {b} + 1$$
$$\lfloor \frac {x} {a\times b} \rfloor \leq \frac {x} {a\times b} < \lfloor \frac {x} {a\times b} \rfloor + 1$$
$$得到 \lfloor \frac {x} {a\times b} \rfloor =\frac {\lfloor \frac {x} {a} \rfloor} {b} $$
$$\lfloor \lfloor \frac {x} {a\times b} \rfloor \rfloor =\lfloor \frac {\lfloor \frac {x} {a} \rfloor} {b} \rfloor = \lfloor \frac {x} {a\times b} \rfloor,命题得证$$