We can use what we know about division to find the inverse of some functions. Consider the function
with a domain of 
. The rule says to multiply the
input by 3 and then add 1. Intuitively, the rule for the inverse
should just reverse this process: subtract 1, then divide by 3. In
fact, mod 7 we can divide by 3 by just multiplying by 3's
multiplicative inverse (which is 5), so this rule makes sense modulo 7
as well. We get the following rule for the inverse.
Can we do the same calculations modulo 15? As mentioned above, many mod 15 standard names, including 3, do not have multiplicative inverses. So consider the function
with domain 
. We've simply changed the modulus
from 7 to 15. Since 3 does not have a multiplicative inverse modulo
15, it's not so clear you can properly reverse the process. In fact,
this function is not invertible. For example, 
.