Show that:
As o_O pointed out this is not always true. So let us resrict our numbers. The case when either is zero is trivial, so fix
So as in past problems let and
So it is clear from the definition of the floor function that
So we must just examine . Using the same tecnhique we get
Now since it is possible that it follows that
. And since the naturals are closed under multiplication it follows that . Get ridding of the useless terms yields the desired result.