# Finding inverse function for a mapping with a congruent class

• Sep 21st 2011, 11:40 AM
hwill205
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.

• Sep 21st 2011, 12:54 PM
HallsofIvy
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, $f^{-1}(0)= 0$, $f^{-1}(1)= 4$, $f^{-1}(2)= 1$, etc.
• Sep 21st 2011, 01:01 PM
alexmahone
Re: Finding inverse function for a mapping with a congruent class
$f(x)=2x\ (mod\ 7)$

$f^{-1}(x)=2^{-1}x\ (mod\ 7)=4x\ (mod\ 7)$
• Sep 21st 2011, 02:58 PM
hwill205
Re: Finding inverse function for a mapping with a congruent class
Quote:

Originally Posted by alexmahone
$f(x)=2x\ (mod\ 7)$

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

This is correct, but how did you go from $2^{-1}x\ (mod\ 7)$ to $4x\ (mod\ 7)$ ?
• Sep 21st 2011, 07:17 PM
alexmahone
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 $2^{-1}x\ (mod\ 7)$ to $4x\ (mod\ 7)$ ?

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

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