# Prove that $\forall n\in \mathbb{N},\:\left[n+\left(-1\right)^n\right]\in \mathbb{N}$ [closed]

Prove that $\forall n\in \mathbb{N},\:\left[n+\left(-1\right)^n\right]\in \mathbb{N}$.

Hey I tried many ways to do this, like the recurrence method, and double recurrence method, but I always get stuck on $(-1)^n$ so I tried to proof that n is always bigger than $-1^n$: $n>-1^n$ for every $n>=1$ but I get stuck on the same thing. I need this just to proof that the set A= $[{n+ -1^n; n\in \mathbb{N}}]$ = $\mathbb{N}$ Thanks.

## closed as off-topic by Jack, Did, Stefan4024, Shailesh, Daniel W. FarlowOct 10 '16 at 2:49

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• Use dollar signs for math statements – G. Snapsmath Oct 9 '16 at 15:46
• There is nothing to prove. – TheGeekGreek Oct 9 '16 at 15:49
• @G.Snapsmath Yes I forgot about them. – Amine Marzouki Oct 9 '16 at 15:52
• If $\;0\notin\Bbb N\;$ , as many mathematicians consider (and I am one of them), then the claim is false, since $\;1+(-1)^1=0\notin\Bbb N\;$ . – DonAntonio Oct 9 '16 at 16:11
• @AmineMarzouki It already has been answered, no need to repeat. The following are just a few examples of books where the author doesn't consider zero a natural number: Hijab's "Introduction to Calculus and Classical Analysis", Thomas' "Calculus" (in AP-2), Adams' "Calculus", Yates', "Analytic Geometry with Calculus", Spivak's "Calculus", etc. – DonAntonio Oct 9 '16 at 17:25

We can manually check for $0$, $1$ that

$0 + 1 = 1$

$1 - 1 = 0$

Now for even $n \ge 2$, $n + 1 \in \Bbb N$ and for odd $n > 2$, $n - 1 \in \Bbb N$ by closure of $\Bbb N$ under addition.

Edit: But per your newly added paragraph, this is not the same thing as proving this set is equal to $\Bbb N$, you also need to prove that every $n \in \Bbb N$ also belongs to this set.

Update: we have shown that your set is a subset of the natural numbers. To show equality we must also show the natural numbers are a subset of your set.

So let $m \in \Bbb N$

Now if $m \ge 2$ is even, we set $m = n - 1$ which implies $m + 1 = n$. But since $m \ge 2$, by closure we know $n \in \Bbb N$ and since $m$ is even then $n$ is odd. So $m \in$ {$n + ( -1^{n}) | n \in \Bbb N$} for all odd $n \ge 2$

Similarly if $m \ge 2$ is odd, we set $m = n + 1$ which implies $m - 1 = n$. But since $m \ge 2$, by closure we know $n \in \Bbb N$ and since $m$ is odd then $n$ is even. So $m \in$ {$n + ( -1^{n}) | n \in \Bbb N$} for all even $n \in \Bbb N$.

Again we check manually for $n = 0$ and $n = 1$.

This proves that $\Bbb N$ is a subset of your set, hence they are equal.

Another way to see this is to write that for our set

$M_n =${$n + ( -1^{n}) | n \in \Bbb N$}

for all even $2j \ge 2$,

$M_{2j} =${$3, 5, 7,...$}

and for all odd $2j + 1 \ge 2$

$M_{2j + 1} =${$2, 4, 6,...$}

Then the union of these sets is {$2, 3, 4, 5, 6, 7...$} and if we add the cases of $n = 0$ and $n = 1$ then $M = \Bbb N$

Use whichever is more intuitive to you

• Why ? I thought it was the same thing. You proofed that n-1∈N and n+1∈N and if n=0 or n=1 then it's ∈N. then why we can't conclude that the sets are the same? – Amine Marzouki Oct 9 '16 at 15:58
• @Amine: If you prove that every element of $A$ is a natural number, you've only proven $A\subseteq\Bbb N$, not necessarily that $A=\Bbb N$. (This, and not the procedure described in your comment, is what Phillip has done in their answer.) – Eric Stucky Oct 9 '16 at 16:01
• @EricStucky You're right, but I though since the lim of $n + (-1)^n= \infty$ then A=N, of course with the first condition that is: every element of AA is a natural number. – Amine Marzouki Oct 9 '16 at 16:04
• No. There could be some integer $m$ which is "jumped over" in the set you have constructed. You need to prove there isn't one. – Phillip Hamilton Oct 9 '16 at 16:23
• @PhillipHamilton yep you're right. I'll think of it, do you have any suggestion ? – Amine Marzouki Oct 9 '16 at 16:25

Hint: If $a_n=\left[n+\left(-1\right)^n\right]$, then $$a_{n+2}=a_n+2$$

• would you clarify please? – Amine Marzouki Oct 9 '16 at 16:06
• @AmineMarzouki $$a_{n+2}=(n+2)+(-1)^{n+2}=(n+2)+(-1)^{n}(-1)^2$$ and use induction.... – N. S. Oct 9 '16 at 16:56

You really only need to show this holds true for $1$. You also must be assuming $0\in\mathbb{N}$. You can easily argue for natural numbers larger than $1$. since $n-1\ge 1$ for all $n > 1$.