证明：(lfloor{frac{n}{xy} floor}=lfloorfrac{lfloor{frac{n}{x} floor}}{y} floor)
设(n=k_1x+t_1,k_1=k_2y+t_2,显然有t_1<x,t_2<y)
(有lfloorfrac{lfloor{frac{n}{x} floor}}{y} floor=k_2)
(lfloor{frac{n}{xy} floor}=lfloorfrac{k_2xy+xt_2+t_1}{xy} floor)
(=k_2+lfloorfrac{xt_2+t_1}{xy} floor)
为了使得(frac{xt_2+t_1}{xy})的分子最大，则使(t_1=x-1,t_2=y-1)
(frac{xt_2+t_1}{xy}=frac{xy-1}{xy})
证得(lfloorfrac{xt_2+t_1}{xy} floor=0)，故(有lfloorfrac{lfloor{frac{n}{x} floor}}{y} floor=k_2=lfloor{frac{n}{xy} floor})
命题得证