# Urgently in need of help with an algebra 1, functions problem

#### tatibarrera231

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

Consider N0->N0 as the set of naturals including 0; that is, N0=N∪{0}, and consider the function: N0→N0, defined by F(n)=n+(−1)^ n.
1. Is F surjective?
2. Is F injective?
3. Consider the following S relation in N0×N0. ∀a,b∈N0,
[aSb]⟺F(a)=F(b)
Show that S is an equivalence relation on N0, 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.

Here ill leve my thoughts for know:
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 justify 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.

Hope you guys can help me.

#### lex

##### Full Member
Write the function as
F(n)=n+1 when n even (inc. 0)
F(n)=n-1 when n odd
Now just look up and write down the exact definitions: of surjective, injective and equivalence relation
and apply them to this example. They should work out fairly easily.
(The key I think each time is to consider the case when 'm' is even and the case when 'm' is odd).

Last edited:

#### lex

##### Full Member
Any luck? Feel free to share any work.

#### pka

##### Elite Member
Consider N0->N0 as the set of naturals including 0; that is, N0=N∪{0}, and consider the function: N0→N0, defined by F(n)=n+(−1)^ n.
Is F surjective?
Is F injective?
Consider the following S relation in N0×N0. ∀a,b∈N0,
[aSb]⟺F(a)=F(b)
Show that S is an equivalence relation on N0, and determine the equivalence class of a=0.
This a pet peeve of mine: zero is a natural number, $$0\in\mathbb{N}$$.
If $$F:\mathbb{N}\to\mathbb{N}$$ by $$F(n)=n+(-1)^n$$ is injective and surjective. But you need to prove each.
If $$F(a)=F(b)$$ prove that $$a=b$$; if $$c\in\mathbb{N}$$ prove that some $$\exists t\in\mathbb{N}$$ and $$F(t)=c$$.
The relation $$\mathcal{S}$$ is easy to show is reflexive, $$F(a)=F(a)~\&~\text{ symmetric }F(a)=F(b)\;\iff\;F(b)=F(a)$$.
For transitive: if $$F(a)=F(b)~\&~F(b)=F(c)$$ does it follow that $$F(a)=F(c)~?$$

#### Jomo

##### Elite Member
1) solve f(x)=0
2) Let k>0. Solve for f(x) =k. How many x's satisfy the equation?

#### lookagain

##### Elite Member
This a pet peeve of mine: zero is a natural number, $$0\in\mathbb{N}$$.

No, my pet peeve is anyone that claims that "zero is a natural number" when
the truth is that there is no consensus. Traditionally, it is not a natural number,
as those are also commonly referred to as "counting numbers," beginning with 1.

To be clear, the problem should just come out and state "non-negative integers."

#### tatibarrera231

##### New member
Hi guys, wanted to say how greatful for all your help I am, I really am trully
greatful.
I think I finish it, but im not sure i'll leave it attached here for you guys to see, tell me what you think

#### Jomo

##### Elite Member
I'll ask you again to solve f(n) = 0!!

By the way, you are NOT asked (for 3) to show that a=a for all a in No. Rather, you are asked to show that aSa for all a in No. Same comment for the other parts.

#### lex

##### Full Member
I think I finish it, but im not sure i'll leave it attached here for you guys to see, tell me what you think
That all looks pretty good. Well done. I echo Jomo's comment relating to part 3. People are understandably uncomfortable arguing from something glaringly obvious, such as 'a=a'. We tend to lay it out a bit differently e.g. like this:

3). aSb ⇔ F(a)=F(b) ⇔ a=b
so: S is
Reflexive: ⇔ aSa ⇔ F(a)=F(a)
$$\displaystyle \therefore$$ S is reflexive
Symmetric: ⇔ aSb = bSa
aSb ⇔ F(a)=F(b) ⇔ F(b)=F(a) ⇔ bSa
$$\displaystyle \therefore$$ S is symmetric
Transitive: ⇔ aSb, bSc ⇒ aSc
aSb, bSc ⇔ F(a)=F(b), F(b)=F(c) ⇒ F(a)=F(c) ⇒ aSc
$$\displaystyle \therefore$$ S is transitive

I attach my own version which pretty much mirrors what you did.

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