Show that the function $f: \mathbb {N} \to \mathbb {N}$ given by $f(n) = n + 2$ is not onto

Question

Show that the function $f: \mathbb{N} \to \mathbb{N}$ given by $f(n) = n + 2$ isn't onto (surjective)

Any advice on what to do here would be much appreciated!

This has been taken from a past college exam paper and as far as I know, we don't include 0 as a natural number

Answer 1

Hint: Can you find a preimage of $1$ by your function?

Answer 2

Just check it by definition. A function $g$ from a set $X$ to a set $Y$ is called onto (or surjective) if and only if for every $y \in Y$ there is some $x \in X$ such that $y = g(x)$. Now let $y := 1$. Then $y \in \mathbb{N}$, the codomain of $f$. But $n + 2 = y$ if and only if $n = y - 2 = -1$, which is not in the domain of $f$. So $f$ is not onto.

Answer 3

Intuitively:

$f(?)=1$

Precisely:

That $f: \mathbb N$ (domain) $\to \mathbb N$ (range) is surjective/onto means that $\forall a \in \mathbb N$ (referring to the range not the domain), $\exists n \in \mathbb N$ (referring to the domain not the range) s.t. $f(n)=a$.

So, let's come up with some $a$ s.t. no matter what $n$ you plug into $f$, you'll never get $a$! Let's try $a=1$.

Pf that $f$ is not surjective onto:

We first claim:

$$\nexists n \in \mathbb N: f(n)=1$$

Pf claim: Suppose on the contrary that

$$\exists n \in \mathbb N: f(n)=1$$

Then

$$\exists n \in \mathbb N: n+2=1 \iff n=-1 \notin \mathbb N ↯ \text{QED claim}$$

$\therefore, \exists a \in \mathbb N: \nexists n \in \mathbb N$ s.t. $f(n) = a$. By the claim, this is $a=1$. $\therefore,$ f is not surjective/onto. QED

Answer 4

We have that

$$f(n) = n + 2\ge 2$$

therefore $\not \exists m\in \mathbb{N}$ such that $f(m)=1$.