下取整/高斯 函数的性质&证明

$$Part\ 1:$$

$$已知:\lfloor x \rfloor \leq x < \lfloor x \rfloor+1,\lfloor \lfloor x \rfloor \rfloor = \lfloor x \rfloor$$

$$证明:\lfloor \frac {\lfloor \frac {x} {a} \rfloor} {b} \rfloor=\lfloor \frac {x} {a\times b} \rfloor$$

$$那么有\frac {\lfloor \frac {x} {a} \rfloor} {b} \leq \frac {x} {a\times b} < \frac {\lfloor \frac {x} {a} \rfloor} {b} + 1$$

$$\lfloor \frac {x} {a\times b} \rfloor \leq \frac {x} {a\times b} < \lfloor \frac {x} {a\times b} \rfloor + 1$$

$$得到 \lfloor \frac {x} {a\times b} \rfloor =\frac {\lfloor \frac {x} {a} \rfloor} {b} $$

$$\lfloor \lfloor \frac {x} {a\times b} \rfloor \rfloor =\lfloor \frac {\lfloor \frac {x} {a} \rfloor} {b} \rfloor = \lfloor \frac {x} {a\times b} \rfloor,命题得证$$

$$Part\ 2:$$

$$证明:\lceil \frac{a}{b} \rceil = \lfloor \frac{a+b-1}{b} \rfloor,保证\ a,b\ 为正整数$$

$$即\ \lceil \frac{a}{b} \rceil = \lfloor \frac{a-1}{b} \rfloor+1$$

$$Situation\ 1:b\mid a$$

$$显然\ \lceil \frac{a}{b} \rceil = \frac{a}{b}，\lfloor \frac{a-1}{b} \rfloor = \frac{a}{b} -1，原式成立$$

$$Situation\ 2:b\nmid a$$

$$\lceil \frac{a}{b} \rceil = \lfloor \frac{a}{b} \rfloor + 1 ,\lfloor \frac{a-1}{b} \rfloor=\lfloor \frac{a}{b} \rfloor,原式成立$$