If $f: \mathbb{Z} × \mathbb{N} \to \mathbb{Z}, f(x,y)=x^2+y$. Prove if is an injection and prove if is an surjection.

Question

If $f:\mathbb{Z} × \mathbb{N} \to \mathbb{Z}$.

$f(x,y)=x^2+y$.

$f:\mathbb{Z} × \mathbb{N}$ is a relation

Prove if is an injection and prove if is an surjection.

Help, I do not know how to proof it.

Answer 1

The comments above have been rather clear. Let me give a more explicit answer.

Let us recall the definitions of injectivity and surjectivity. Let $A$ and $B$ be sets. Then, a function $f: A \to B$ is said to be injective iff:

$$\forall x,y \in A: f(x) = f(y) \implies x = y$$

and it is said to be surjective iff:

$$f(A) = B$$

where $f(A)$ is the image set of $A$ under the map $f$.

Now, we will note that for your specific case, $(x,y) \in \mathbb{Z} \times \mathbb{N}$. So, $x \in \mathbb{Z}$ and $y \in \mathbb{N}$. If we claim that the function $f: \mathbb{Z} \times \mathbb{N} \to \mathbb{Z}$ is surjective, then consider the integer $-1$. Since it is the case that:

$$\forall x \in \mathbb{Z}: x^2 \geq 0$$

$$\forall y \in \mathbb{N}: y \geq 0$$

So, their sum $x^2+y = f(x,y) \geq 0$ for any $(x,y)$. But that just means that there does not exist a pair $(x,y)$ such that $f(x,y) = -1$. Hence, this function cannot be surjective.

For injectivity, consider the integer $2$ and let $y = 1$. Then, we can see that:

$$f(-1,1)= (-1)^2+1 = 2 = 1^2+1 = f(1,1)$$

Since each image has multiple preimages under $f$, it follows that this is not an injective map.

If you do not understand the terminology that I have used above, then you are free to ask questions about it. On the other hand, if a large portion of it does look foreign, then I'd suggest that you have a look at your notes and review the relevant material before attempting such questions.