# Inverse of Mod Function

• alec_tronn
Thank you!In summary, the homework statement is asking for an inverse function of the mod function, but there is no such thing. The function given is the same for all values of x.f

## Homework Statement

Let S be the subset of the integers {0, 1, 2, 3, 4, 5, 6}, and let f be the function from S to S defined by f(x) = (4x+3) mod 7. Find a formula for f<superscript>-1 (the inverse of f, I don't know how to type that).

## Homework Equations

I know that the mod function is defined by floor function (a - n floor(n\a). But I'm still unable to create an inverse function with one inverse function for all values. I made a piecewise function that works for those values, but I feel like there should be a better method. Does anybody know an inverse rule for either floor or modulus?

## The Attempt at a Solution

f(x) = {2x+1, 0 <= x <= 2}
{2x-6, 3 <= x <= 6}

The 'mod' function? There is no such thing.

mod is not a function.

Just do it as for any other function:

y=4x+3, hence x= (y-3)/4, so what is division by 4 in mod 7 arithmetic?

The first thing you should do is calculate the actual values of f:
f(0)= 4(0)+ 3= 3. f(1)= 4(1)+ 3= 7= 0 mod 7. f(2)= 4(2)+ 3= 11= 4 mod 7.
f(3)= 4(3)+ 3= 15= 1 mod y. f(4)= 4(4)+ 3= 19= 5 mod 7. f(5)= 4(5)+ 3= 23= 2 mod 7. f(6)= 4(6)+ 3= 27= 6 mod 7

So you want f-1(0)= 1, f-1(1)= 3, f-1(2)= 5, f-1(3)= 0, f-1(4)= 2, f-1(5)= 4, f-1(6)= 6.

Yes, that's exactly what your function gives. Good work! That is, by the way, a single "function" and I have no problem with it. However, you might remember that -6 is 1 in mod 7. Since the first formula was given "mod 7" why not do the same with this:
f-1(x)= 2x+ 1 (mod 7) or f-1(x)= 2x- 6 (mod 7)
They are exactly the same.

The 'mod' function? There is no such thing.

mod is not a function.

My apologies... mod "operator" (or modulo operator); I was just asking if there was an inverse operator that I could use so I could switch the x and y variables and solve for x instead of switching the values around and making an entirely new function.

Yes, that's exactly what your function gives. Good work! That is, by the way, a single "function" and I have no problem with it. However, you might remember that -6 is 1 in mod 7. Since the first formula was given "mod 7" why not do the same with this:
f-1(x)= 2x+ 1 (mod 7) or f-1(x)= 2x- 6 (mod 7)
They are exactly the same.

Thanks a lot! I'm not sure that putting the (mod 7)'s in there like that truly give any more information, although I see how it looks more related to the beginning function, and I see how your solution and mine are equivalent. I'm not sure what I'll end up submitting. Thanks again.

I was wondering, though, since (a)mod(n) can be written in terms of the floor function, thus $$a - n Floor[\frac{n}{a}],$$ and therefore, can't we describe the floor function as an invertible function? I know that $$Floor[x] = -1/2 + x + \frac{ArcTan(Cot(\pi x))}{\pi}$$. 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.

Last edited: