- #1
flouran
- 64
- 0
Hi Guys,
I was wondering, since (a)mod(n) can be written in terms of the floor function, thus [tex] a - n Floor[\frac{n}{a}],[/tex] and therefore, can't we describe the floor function as an invertible function? I know that [tex]Floor[x] = -1/2 + x + \frac{ArcTan(Cot(\pi x))}{\pi} [/tex]. Thus, can't we inverse this formula? Furthermore, there is another idea that works on any function, but deals with sets. In general, given a function f: A → B and a subset S of B, then f-1(S) = {x ∈ A | f(x) ∈ S}. In the case of the floor function, if n is an integer, then floor-1({n}) = [n, n + 1). Note that these are sets, not numbers.
Please let me know as I am saying these things right off of the top of my head.
I was wondering, since (a)mod(n) can be written in terms of the floor function, thus [tex] a - n Floor[\frac{n}{a}],[/tex] and therefore, can't we describe the floor function as an invertible function? I know that [tex]Floor[x] = -1/2 + x + \frac{ArcTan(Cot(\pi x))}{\pi} [/tex]. Thus, can't we inverse this formula? Furthermore, there is another idea that works on any function, but deals with sets. In general, given a function f: A → B and a subset S of B, then f-1(S) = {x ∈ A | f(x) ∈ S}. In the case of the floor function, if n is an integer, then floor-1({n}) = [n, n + 1). Note that these are sets, not numbers.
Please let me know as I am saying these things right off of the top of my head.