Finding inverse function for a mapping with a congruent class

Problem:

F([X]7)=[2x]7

Find the inverse for this function.

So this function maps from x(mod 7) to 2x(mod 7). I was able to prove that the function is one-to-one and onto in a previous function, but I am lost after that.

Am I looking for a function that produces [1]7 when multiplied with [2x]7?

I really would appreciate any and all help.

Re: Finding inverse function for a mapping with a congruent class

No, there is no multiplication involved here. You are, as the problem says, looking for the inverse **function**.

It's not at all difficult to write out exactly what f is:

f(0)= 0, f(1)= 2, f(2)= 4, f(3)= 6, f(4)= 1, f(5)= 3, f(6)= 5.

Therefore, $\displaystyle f^{-1}(0)= 0$, $\displaystyle f^{-1}(1)= 4$, $\displaystyle f^{-1}(2)= 1$, etc.

Re: Finding inverse function for a mapping with a congruent class

$\displaystyle f(x)=2x\ (mod\ 7)$

$\displaystyle f^{-1}(x)=2^{-1}x\ (mod\ 7)=4x\ (mod\ 7)$

Re: Finding inverse function for a mapping with a congruent class

Quote:

Originally Posted by

**alexmahone** $\displaystyle f(x)=2x\ (mod\ 7)$

$\displaystyle f^{-1}(x)=2^{-1}x\ (mod\ 7)=4x\ (mod\ 7)$

This is correct, but how did you go from $\displaystyle 2^{-1}x\ (mod\ 7)$ to $\displaystyle 4x\ (mod\ 7)$ ?

Re: Finding inverse function for a mapping with a congruent class

Quote:

Originally Posted by

**hwill205** This is correct, but how did you go from $\displaystyle 2^{-1}x\ (mod\ 7)$ to $\displaystyle 4x\ (mod\ 7)$ ?

$\displaystyle 2\times 4=8\equiv 1 \pmod{7}$

So, $\displaystyle 2^{-1}=4 \pmod{7}$