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Im sure most of you know about the proof of the sum of $n$ consecutive numbers $\sum_1^n = \frac{n(n+1)}{2}$.

The proof to this identity is quite common and I had no problems finding it. However I'm curious to know the proof that all the numbers obtained from $\frac{n(n+1)}{2}$ are indeed natural.

To be clear, I know that summing several natural numbers will result in a natural number and therefore if the sum equals that expression then the expression returns a natural number. But I'd like to know if there is any proof without resorting to the natural sum. I guess the $\frac{}{2}$ is kind of throwing me for a loop.

I have implemented a short python script to test that from one to a million and the identity returns natural numbers as expected, but this is obviously not mathematically rigorous.

Im not really used to doing proofs so I'm not sure where to begin. What would be then the proof for the following statement then: $\frac{n(n+1)}{2} \in \mathbb{N} \quad \forall n\in \mathbb{N}$

Thanks!

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    $\begingroup$ One of $n$, $n + 1$ is even, hence so is their product, so you can divide it by 2 and get a natural number. $\endgroup$
    – Ben Steffan
    Commented May 24 at 10:23
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    $\begingroup$ If $n$ is even, then $n/2$ is an integer. If $n$ is odd, then $n+1$ is even and $(n+1)/2$ is an integer. $\endgroup$
    – Marc Dinh
    Commented May 24 at 10:23
  • $\begingroup$ The alternative proof is to first prove, perhaps by induction that $~\displaystyle \frac{n(n+1)}{2} = \sum_{k=1}^n ~k.~$ Then to reason that given any $~a,b \in \Bbb{Z^+},~$ that $~(a+b) \in \Bbb{Z^+}.~$ Then, extend this reasoning, perhaps again by induction, to cover any $~a_1, a_2, \cdots, a_n \in \Bbb{Z^+}.$ Basic field theory of real numbers gives that for $~a,b \in \Bbb{R^+},~$ that $~(a + b) \in \Bbb{R^+}.~$ So, the missing link is that for $~a,b \in \Bbb{Z},~$ that $~(a + b) \in \Bbb{Z}.$ $\endgroup$
    – user2661923
    Commented May 24 at 11:13

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The proof to this identity is quite common and I had no problems finding it.

The identity is a proof because it allows you to prove $S(n+1) \in \mathbb{N}$ if $S(n) \in \mathbb{N}$ (where $S(n) = \frac{n(n+1)}{2}$), and since $S(0) \in \mathbb{N}$, by induction $\forall n \in \mathbb{N},S(n) \in\mathbb{N}$. Alternatively, as already suggested by the comments, either $n$ or $n+1$ is even therefore either $\dfrac{n}{2}$ or $\dfrac{n+1}{2}$ is a natural number.

I have implemented a short python script to test that from one to a million and the identity returns natural numbers as expected, but this is obviously not mathematically rigorous.

Not sure why you want to do this if you've already known the rigorous proof of $\displaystyle\sum_{i = 0}^n i = \frac{n(n+1)}{2}$

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