The function $F(n)=n+(-1)^n$ on $\Bbb{N}_0$. Injective? Surjective? Equivalence classes?

Question

I have a few questions regarding functions, it's kinda urgent hope you can help me understand;

Consider $\Bbb{N}_0$ as the set of naturals including $0$; that is, $\Bbb{N}_0 = \Bbb{N}\cup\{0\}$, and consider the function: $\Bbb{N}_0 \to \Bbb{N}_0$, ​​defined by $F(n) = n + (−1) ^ n$.

1. Is $F$ surjective?

2. Is $F$ injective?

3. Consider the following $S$ relation in $\Bbb{N}_0 \times \Bbb{N}_0$. $\forall a, b \in \Bbb{N}_0$, $$[a S b] \iff F(a) = F(b)$$ Show that $S$ is an equivalence relation on $\Bbb{N}_0$, and determine the equivalence class of $a = 0$.

My biggest problem with this is that I'm not sure how to justify my answers in an analytic form.

Hope you guys can help me.

hi, I'm editing the post, thanks for the feedback so far. So someone suggested I put my non-analytic solutions so here it goes:

1.so for the first one, I think it is surjective cause it does have the same range and codomain, both being in the naturals including 0, but when you justify this analytically you have to clear x and that's where my main problem is cause I don't know how to(cause of the -1 elevated to n).

2.the same problem here I do think it's injective cause it only touches on one point of the graphic, but again analytically I'm not sure how to do it cause your spoused to do: f(a)=f(b)=> a=b, a+(−1)a=b+(−1)b, but again I'm not sure how to simplify it cause of the -1 elevated to n.

3.this one I completely don't know how to do

And correct me if I'm wrong, cause it's totally possible, and any help is greatly appreciated

#Functions #HelpMe! #Advice #Math #Urgent #Please

Answer 1

1. The easiest way to prove a function is surjective is to start with an arbitrary number $m$ in the codomain, and find a number $n$ in the domain such that $F(n)=m$. For this function, you'll need to handle two cases based on whether $m$ is positive or negative. Your proof should go "Choose any $m \in \mathbb{N}_0$. If $m$ is even, then $F(48m)=m$, while if $m$ is odd, then $F(m-5)=m$. Hence $m \in Range(F)$ for all $m \in \mathbb{N}_0$, so $F$ is surjective."

2. Again, you probably have to consider cases. One way would be to suppose $a$ is even, and show that $b$ has to be even.

3. Think about the definition of an equivalence relation. You just need to prove that $[aSb]$ satisfies those properties.

Answer 2

The function $F$ is an involution, so it is injective and surjective and the corresponding equivalence relation is equality.

Indeed, $F(F(n))=F(n+(-1)^n)=n+(-1)^n+(-1)^{n+(-1)^n}=n$ (using the fact that $n$ and $n+(-1)^n$ have opposite parities).

Clearly, then, the equivalence class of $0$ is just $\{0\}$. Each even number and the following odd number "switch places," giving the sequence $1,0,3,2,5,4,...$.