If $A$ is an additive abelian group and $\alpha, \beta \in{\rm End}(A)$, show $\alpha+\beta\in{\rm End}(A)$

Say that $A$ is an additive abelian group. Say also that a group homomorphism (endomorphism) $\alpha: A \rightarrow A$ exists. Say that we call the set of all endomorphisms of $A$ as ${\rm End}(A)$.

I'm trying to show that for $\alpha, \beta \in{\rm End}(A)$, $\alpha + \beta \in{\rm End}(A)$ also.

So I define $\alpha + \beta$ as $(\alpha + \beta)(x) = \alpha(x) + \beta(x)$

I'm pretty sure that you can approach this with distributive laws: $$\alpha(\beta + \gamma) = \alpha\beta + \alpha\gamma$$

and then apply that form to the earlier definition of $\alpha + \beta$.

Any thoughts on a more formal proof?

Let $x, y\in A, \alpha, \beta\in{\rm End}(A)$.

Then $(\alpha +\beta)(x)=\alpha(x)+\beta(x)\in A$ since $\alpha(x),\beta(x)\in A$.

Also

$$\begin{align} (\color{red}{\alpha+\beta})(\color{green}{x}+\color{blue}{y}) &=\alpha(\color{green}{x}+\color{blue}{y})+\beta(\color{green}{x}+\color{blue}{y})\\ &=\alpha(\color{green}{x})+\alpha(\color{blue}{y})+\beta(\color{green}{x})+\beta(\color{blue}{y})\\ &=\alpha(\color{green}{x})+\beta(\color{green}{x})+\alpha(\color{blue}{y})+\beta(\color{blue}{y})\\ &=(\color{red}{\alpha+\beta})(\color{green}{x})+(\color{red}{\alpha +\beta})(\color{blue}{y}). \end{align}$$

Thus $\alpha+\beta\in{\rm End}(A)$.