# Thread: Two consecutive numbers are relatively prime

1. ## Two consecutive numbers are relatively prime

Two consecutive numbers are relatively prime.

$(n, n+1)=1\rightarrow \alpha n+\beta (n+1)=1$

I am not sure what to do next to prove this.

2. Originally Posted by dwsmith
Two consecutive numbers are relatively prime.

$(n, n+1)=1\rightarrow \alpha n+\beta (n+1)=1$

I am not sure what to do next to prove this.
$\alpha=-1,\; \beta=1$

3. I see how that will justify the equality but is that ok to do for the proof and then just call it day and it is done?

4. Originally Posted by dwsmith
I see how that will justify the equality but is that ok to do for the proof and then just call it day and it is done?
Let $g=(n,n+1)$

$g\mid n$ and $g\mid n+1 \implies g\mid (n+1)-n = 1 \implies g=\pm1$. But since $g>0$, we get $g=1$.

5. Originally Posted by dwsmith
I see how that will justify the equality but is that ok to do for the proof and then just call it day and it is done?
The answer is yes. In general: Let $a$ and $b$ be integers. There exists two integers $x$ and $y$ such that $ax+by=1$ if and only if $(a,b)=1$. In your case, $a=n$, $x=-1$ and $b=n
+1$
, $y=1$.

However, the reply you got from "chiph588" using the greatest common divisor ( $g=(n,n+1)$) seems to me more elegant.

6. Originally Posted by melese
The answer is yes. In general: Let $a$ and $b$ be integers. There exists two integers $x$ and $y$ such that $ax+by=1$ if and only if $(a,b)=1$. In your case, $a=n$, $x=-1$ and $b=n
+1$
, $y=1$.
This is a special case of Bézout's identity.

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# is any two cosecutive natural numbers are relative primes

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