Is $f(x) = x^2$ surjective on $\mathbb N$? Improving proof notation

Question

Definition of surjective:

Let $A$ and $B$ be sets and let $f: A → B$ be a function. $f$ is surjective if for each $b ∈ B$ there is some $a ∈ A$ such that $f(a) = b$

Solution attempt:

In this case, $f(x) = x^2$, $A$ and $B$ are both $\mathbb N$

$f: A → B$

$f: ℕ → ℕ$

Let $b = 3$ There is no $a ∈ ℕ$ s.t. $f(a) = a^2 = b $

Therefore $f(x) = x^2$ is not surjective on $ℕ$

Is this proof correct?, and if so, what notation can I use to make the distinction between an element of $ℕ$ as a pre-image and another as image of that element. In this case solved as "$a$" being a pre-image of $f: ℕ → ℕ$ and "$b$" the image of a on $f$.

Answer 1

Your proof is correct but depending on the level of detail expected you might want to explain why there is no $a \in \mathbb{N}$ such that $ f(a)=3$ i.e. :

assume that $\exists a \in \mathbb{N}$ such that $f(a)=3$

therefore $a^2=3$ and $a = \sqrt{3}$ or $a=-\sqrt{3}$ which is impossible because $a\in \mathbb{N}$

Answer 2

Your proof is flawless.

You have used $b$ for image and $a$ for pre-image elements of ℕ.

For natural numbers letters $m$ and $n$ are more popular than $a$ and $b$.

But that does not mean your proof is less valid.