Solution a194dae6-808f-4b79-b788-6a85873b8c8f

Points: 

By definition, to prove $a\mid a$, we need to find an integer $k$ such that
\[
  a=a\cdot k.
\]

Choose $k=1$. Since $1$ is an integer,
\[
  a=a\cdot 1.
\]
Therefore, $a\mid a$ for every integer $a\neq 0$.

Final answer:
a\mid a