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.
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