Injection/Surjection of a quadratic function

Question

Consider the function $f: \mathbb{N} \to \mathbb{N}$ (where $\mathbb{N}$ is the set of all natural numbers, zero included) defined as follows $$f(x) = (x+3)^{2} - 9.$$ Is the function injective and/or surjective?

How can I prove it?

I know that a function is injective if for all $x,y\in\mathbb{N}$ s.t. $f(x)=f(y)$ then $x=y$. So, if I put $(x+3)^2-9=(y+3)^2-9$, how can I obtain $x=y$? Furthermore, how can I find the inverse of $f(x)$?

Answer 1

Note that the function $f\colon \mathbb{N} \to \mathbb{N}$ is not surjective. Indeed, there does not exist $x\in\mathbb{N}$ such that $$ f(x)= (x+3)^{2} - 9=2. $$ If there was such an $x$, then $\sqrt{11}$ would be an integer a contradiction. It is injective. Indeed $$ (x+3)^{2} - 9=(y+3)^{2} - 9\implies |x+3|=|y+3| \implies x=y $$ since $x,y\geq 0$.

Answer 2

You can easily verify that it is injective but not surjective.

Indeed

$f: \mathbb{R^+} \to \mathbb{R^+}$ is injective and strictly increasing

$f(1)=7$ and $f(2)=16$ thus $\nexists x$ such that $f(x)=8$

Answer 3

$$ f(x) = f(y) \iff \\ (x+3)^2 - 9 = (y+3)^2 -9 \iff \\ (x+3)^2 = (y+3)^2 \iff \\ x+3 = y+3 \quad \vee \quad x+3 = -(y+3) $$ As $x$ and $y$ are non-negative, what holds for $x+3$ and $y+3$? And what can be inferred?

Answer 4

I like using $n,m$ for naturals. Here $f: \mathbb{N} \to \mathbb{N}$ such that $n \to (n+3)^2-9$

1. Injective $\implies$ no two naturals have the same image $\implies$ $f(n_1) \neq f(n_2)$ for any DISTINCT $n_1$ and $n_2$, meaning $n_1 \neq n_2$

Can two different inputs produce the same output? Are there two distinct members of $\mathbb{N}$, $\ $ $n_1$ and $n_2$ $\ $ such that $(n_1+3)^{2} - 9=(n_2+3)^2-9 \ $? It takes one counter example to show if it's not.

2. Surjective $\implies$ every natural is contained in the range of this function$\implies$ $f(n)$ takes on all values of $\mathbb{N}$

Does the range of this function contain every natural number with only natural numbers as input? Is there an $m \in \mathbb{N}$ such that $(m+3)^2-9=2 \ $for instance? It takes one counter example to show if it's not.

Answer 5

1) Injective:

$y = (x+3)^2 -9 = x(x+6)$ , $x \in \mathbb{N}$.

This function is strictly increasing , hence injective.

For $x_1 < x_2$ : $y_1 = x_1(x_1+6) \lt x_2(x_2+6) =y_2.$

2) Not surjective:

$y(0) = 0$; $y(1) = 7$.

$1,2,3,4,5,6 $ are not image points of f.