Show that function is not turing-computable? We have two functions:

$f_1: \mathbb{N}\rightarrow \mathbb{N} \quad $
$f_2: \mathbb{N}\rightarrow \mathbb{N}$

By definition $f_1$ is turing-computable while $f_2$ is not.

Then we define a third funtion $g(n) = f_1(n) + f_2(n)$.

I want to show that $g(n)$ is not turing-computable:

So first assume that $g(n)$ is computable.

That means I can write it like this: 

$f_2(n) = g(n)-f_1(n)$

Which (is where I'm not sure) means that $f_2$ can be computed which is a contradiction to the definition, which results that $f_2$ is not turing-computable.
This proof is correct as written, well done :) Remember this pattern, because an identical proof can be used in a wide variety of similar problems such as rational/irrational numbers and polynomials/non-polynomials