Problem involving demonstration technique with integer $≥ 0$, such that $n2+n$ is divisible by $2$

Question

I'm learning about discrete math, more precisely about demonstration techniques. The thing is, I'm not always sure if I'm doing it the right way, I don't want to trust my intuition because maybe I'm doing it the wrong way. So I would like to know if my line of reasoning makes sense, if I am correct about the final result.

Problem:
For every integer $n ≥ 0$, then $n2+n$ is divisible by $2$.

What I tried to do:

For every integer $n ≥ 0$... That means it's any number greater than or equal to $0$, so it's infinite elements.
Then $n2+n$ is divisible by $2$...

I must replace $n$ with an integer value that matches the statement $n ≥ 0$. I can start with $0$ itself. So I multiply $0$ by $2$ and then add $+ 1$. So I get to the final result, and all I have to do is check if, in fact, this theorem is correct.

I created a table, testing 4 possible cases only, because the cases are infinite.

So, I arrived at the thesis (conclusion). And so, the theorem doesn't match, it's not true that: every integer $n ≥ 0$, then $n2+n$ is divisible by $2$.

There will always be elements that are not divisible by two, they will be the odd numbers.

I could also write using existential connective:

$∃ n ∈\Bbb Z ≥ 0 \mid n ¬ / 2$

I'm not sure how to use quantifier connectives. But the idea is that: there is an integer n greater than or equal to 0 that is not divisible by 2.

Again, I would like to know if my line of reasoning makes sense and if I am correct. Thanks.

Answer 1

I think you mean $n^2 + n$ is divisible by 2. You can use MathJax to format mathematical expressions and there is a guide here: MathJax basic tutorial and quick reference

Ok so for problems like yours you need to use Proofs by cases.

Case 1: n = 2k, $k \in \Bbb Z$ (n is even)

So $n^2 + n = (2k)^2 + 2k = 4k^2 + 2k = 2(2k^2 + k) = 2K$

$ (K = 2k^2 + k, K\in\Bbb Z)$

Case 2: n = 2k+1, $k \in \Bbb Z$ (n is odd)

So:

\begin{align} (2k+1)^2 + (2k+1) & = 4k^2 + 4k + 1 + 2k + 1\\ & = 4k^2 + 6k +2\\ & = 2(2k^2 + 3k + 1\\ & = 2K' &(K' = 2k^2 + 3k + 1), K'\in\Bbb Z)\\ \end{align}

We have proved that $n^2 + n$ is always even so it's always divisible by 2 for all $n\in\Bbb N$.