# Surjective function proof [closed]

Given a function $f: \mathbf{N}_0 \to \mathbf{N}_0$, defined $$f(x) = \begin{cases} x+3 & \text{if } x \in \mathbf{N}_{\text{even}} \\ x-1 & \text{if } x \in \mathbf{N}_{\text{odd}} \end{cases}$$

I have to prove that the function is surjective.

My attempt:

Take any y ∈ N even

f(x)=y ⇒ x+3=y ⇒ x=y-3

Note x=y-3 ∈ N even and

f(x)=f(y-3)=y-3+3=y

Take any y ∈ N odd

f(x)=y ⇒ x-1=y ⇒ x=y+1

Note x=y+1 ∈ N odd and

f(x)=f(y+1)=y+1-1=y

This shows that f is surjective.

## closed as unclear what you're asking by lulu, Did, user90369, John B, kimchi loverJan 13 '18 at 21:30

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• For which $n$ is $f(n)=1$? – lulu Jan 13 '18 at 17:02
• Really? $n=2$ is even so I thought $f(2)=2+3=5$. – lulu Jan 13 '18 at 17:04
• Voting to close the question as it is not clear what you are talking about. – lulu Jan 13 '18 at 17:06

The function $f$ is not surjective: there is no $n\in\mathbb N_0$ such that $f(n)=1$.

Your proof is not correct: if $y$ is odd, then $y+1$ will not be odd. Similarly if $y$ is even, then $y-3$ won't be even.

If the function was defined from the integers to the integers $\mathbb Z\to\mathbb Z$ then it would be surjective by the following argument (which is the "corrected version" of yours). Given $y$ there are two cases:

1. if $y$ is even then $x:= y+1$ is odd and we have $f(x)=y$
2. if $y$ is odd then $x:= y-3$ is even and we have $f(x)=y$

Note however that in your exercise the function is defined $\mathbb N_0 \to\mathbb N_0$. Note that there is no $x\in\mathbb N_0$ that maps to $1$ so the function is not surjective. How to see there is no such $x$? If $x$ was odd then $f(x)=x-1$ would be even, so certainly $f(x)\neq 1$. If $x$ was even $f(x)=x+3\geq 0+3>1$ (here we use that $x\geq 0$ by definition of $x\in\mathbb N_0$)

• So the function is surjective only if the domain and the codomain are the same? – Marco Carta Jan 13 '18 at 16:59
• @MarcoCarta Not at all. – José Carlos Santos Jan 13 '18 at 17:03