# Surjectivity of function over the naturals

I've been doing some study with function types such as injective, surjective and bijective. I came across this question $$g(x) = 2x^2 + 5$$ where $$g : \mathbb N \to \mathbb N$$. I would like some clarification. I believe the function is injective (is monotone increasing) but I am not sure whether it is surjective. Could anyone give me an explanation on whether it is or not?

## 1 Answer

You are correct in saying that this is an injective function but you should refine your reasoning to mean strictly monotone (since constant function is monotone as well but not injective).

I will not give away the answer directly but here's how you can proceed. Let $$y \in \Bbb{N}$$. See if there exists an $$x \in \Bbb{N}$$ such that $$2x^2+5=y$$. If such an $$x$$ always exists, then the function is surjective, otherwise it is not.

EDIT: Now that you know it is not surjective, here's a formal proof. For surjectivity, $$\forall y \in \Bbb{N}$$, $$\exists x \in \Bbb{N}$$ such that $$2x^2+5=y$$ $$\implies x^2=\frac{y-5}{2}$$ But $$x^2 \geq 0 \implies \frac{y-5}{2} \geq 0 \implies y-5 \geq 0 \implies y \geq 5$$. Thus for each $$y < 5$$, there is no $$x \in \Bbb{N}$$ which maps to it. Thus, not surjective.

• So, $2x^2 + 5 = 1$, there is no solution and thus is not surjective. Thanks for your explanation! Commented Apr 4, 2020 at 8:55
• You are right. No problem. I am editing the answer to include a formal proof. Commented Apr 4, 2020 at 8:58