Given a function $F \colon \mathbb{Z} \to \mathbb{Z}$ defined by $F(n) = (-1)^n n + 3$ proof that it is surjective.
Since the definition of surjectivity says that $\forall k \in \mathbb{Z}, \ \exists n \in \mathbb{Z} \ \colon k = F(n) = (-1)^n n + 3$ I need to find an expression for $k$ in terms of $n$ and check that $n$ is an integer if $k$ is also an integer.
$k = (-1)^n n + 3 \Longleftrightarrow k - 3 = (-1)^n n$
If $n$ is even I get: $k - 3 = (-1)^n n \Longleftrightarrow k - 3 = n$ If $n$ is odd I get: $k - 3 = (-1)^n n \Longleftrightarrow -(k - 3) = n$
So, in general, $n = (-1)^{k + 1} (k - 3)$.
Since I've found and expression for $n$ in terms of $k$ and, clearly if $k$ is an integer so is $n$ this implies that $F$ is surjective.
Furthermore since this final expression gives for every $k$ an unique $n$ I can say that $F$ is bijective.
Is this argument valid? How do you prove surjectivity in general?
