Let $A = \{1,2,3,4\}$ and define the functions $f,g:A \to A$ by the rules $f(x)= 3^x\mod5$ and $g(x)=2^x\mod 5$.
The questions asks us if
$f$ is injective
$g$ is surjective
find $f(g(x))$
Clearly $f$ is not injective and $g$ is not surjective, but i cant figure out the third one
It's just a matter of understanding what these words and formulas mean! We are told that $f(x)= 3^x$ (mod 5) and x can only have 4 values so it is easy to calculate all of them:
$f(1)= 3^1= 3$ (mod 5), $f(2)= 3^2= 9= 4$ (mod 5), $f(3)= 3^3= 27= 2$ (mod 5), and $f(4)= 3^4= 81= 1$ (mod 5). Yes, f is "injective" (one to one). Two different values of x always give different values of f(x).
$g(x)= 2^x$ (mod 5) so $g(1)= 2^1= 2$ (mod 5), $g()= 2^2= 4$ (mod 5), $g(3)= 2^3= 8= 3$ (mod 5), and $g(4)= 2^4= 16= 1$ (mod 5). Yes, g is "surjective" (onto). For every number, y, in A, there is an x such that f(x)= y. g(4)= 1, g(1)= 2, g(3)= 3, and g(2)= 4.
Now, f(g(x)) is given by $f(g(1))= f(2)= 4$, $f(g(2))= f(4)= 1$, $f(g(3))= f(3)= 2$, and $f(g(4))= f(1)= 3$.