# Prove that $(n, n + 1) = 1$ for all $n \gt 0$. [closed]

I was thinking of doing this by contradiction. So by supposing: $$(n, n + 1) \neq 1$$

Then trying to to show that $(n, n + 1) \gt 1$ or $(n, n + 1) \lt 1$. But I'm not sure how I can accomplish this.

I also thought about the fact that $n$ and $n + 1$ are always of different parity. So if $n$ is even, then $n + 1$ is odd. But I don't think that helps me.

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Assuming that the parentheses denote $\text{gcd}$, how could it be less than 1? – Trevor Wilson Feb 7 at 20:21

## closed as not a real question by Landscape, ncmathsadist, vonbrand, achille hui, Lord_FarinMay 10 at 11:20

It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, see the FAQ.

HINT: If $d\mid a$ and $d\mid b$, then $d \mid a-b$.
If $d$ divides $n$ and $d$ divides $n+1$, then $d$ divides $(n+1)-n=1$.
Hint $\$ Common multiples $\rm\,n\ne m\,$ of $\rm\,d\,$ are no closer than $\rm\,d\:$(why?)  Thus $\rm\,m = n\!+\!1\:\Rightarrow\: d \le \,\ldots$