four consecutive numbers including zero

Question

I would like to prove the obvious: We have four consecutive numbers,some of them are negative and some of them are positive.Then the product is equal to: zero(equivalent to the set of these four consecutive numbers contains zero) I've thought of using the Bolzano theorem on f(x)=x(x+1)(x+2)(x+3) but it didn't work.

Answer 1

Outline: 1)$n < 0;$ 2) $n+3 > 0; $ 3) one of the remaining terms ($n+1$ or $n+2$) must be 0; therefore 4) product must be zero.

1) 2) and 3) will be proven below in excruciating detail.

=======

Four consecutive numbers: $n, n+1, n+2, n+3$. Or generally $n + i; 0\le i \le 3$.

Some are negative: $n + i < 0$ for some $i$. So $n < -i$. As $i \ge 0$, $n < -i < 0$.

Some are positive: $n + i > 0$ for some $i$. $n > -i$. As $i \le 3$ $n > -3$.

So $-3 < n < 0$ and $0 < n+3 < 3$.

Let $k = \min(|n|, n+3)$. Then $k \in \mathbb N$ and $0< k < 3$. So $n +k \in \{n,n+1, n+2, n+3\}=\{n+3-3, n+3-2, n+3-1, n+3-0\}$. So $n+3 - k\in \{n,n+1,n+2,n+3\}$.

If $k = |n|$ then $n+k = n-n = 0$ and $\prod(n+i) = 0$.

If $k = n+3$ then $n+3 - k = (n+3)-(n+3) = 0$ and $\prod(n+i) = 0$.

===

Answer 2

If you have four consecutive integers and they include both negative and positive integers,then $0$ must come in between and anything $\times0=0.$

Just because you wanted a more 'mathematical' proof:-

If $n<0$ and $n+k>0$ are any two of the numbers we have, then we can say $n,n+1,...n+k,...n,n+1,...n+k,...$ all belong the numbers we have. Let $n+k=a$. Can you see that $n+(k−a)$ is also one of the numbers?

However, I feel we are using the same logic.