Consecutive Numbers are coprime Can anyone help me in proving that two consecutive numbers are co-primes?
My approach:
Let two consecutive numbers are $n$ and $n+1$.
Assume they are not co-primes.
Then $\gcd(n,n+1)=x$, because it can not equal to $1$, $x$ is natural and $x\gt1$
So $x$ divides $n$ as well as $n+1$.
Then $x$ also divides $n+1-n$, by general understanding.
Hence $x$ divides $1$ or $x=1$.
But we have assumed $x\gt 1$.
So by contradiction $n$ & $n+1$ are co-prime.
Is it right or is there any better way to prove that two consecutive numbers are co-prime?
 A: Your proof is correct, but there is a simpler one, that doesn't work by contradiction and doesn't need greatest common divisors explicitly.
If a prime $p$ divides $n$ then  dividing $n+1$ by $p$ leaves a remainder of $1$, so $p$ does not divide $n+1$. That means the factorizations of $n$ and $n+1$ have no prime in common, so $n$ and $n+1$ are relatively prime.
But ... this proof does use unique factorization, which is usually proved by thinking about greatest common divisors. So they are there, behind the scene. 
A: We $ $ have  $\,n\mid \color{#c00}{\overbrace{n\!+\!1}^{\large m}}-1.\ $  Your (correct) proof easily generalizes widely as follows.
Generally $\,n\mid \,k\:\color{#c00}m-1\,\Rightarrow\, k\,m-j\,n = 1,\ $ so $\,\ d\mid m,n\,\Rightarrow\, d\mid 1,\ $ so $\ \gcd(m,n) = 1$
$ $ i.e. $\ {\rm mod}\,\ n\!:\,\ m^{-1}\,$ exists $\,\Rightarrow\, \gcd(m,n) = 1\,$ (and conversely by Bezout, see here and here)
Remark $\ $ We can also use (a single step of) the Euclidean algorithm
$$ \gcd(km,n) = \gcd(jn\!+\!1,n) = \gcd(1,n) = 1\,\Rightarrow\, \gcd(m,n) = 1$$
A: Your proof looks good. Using the method of contradiction is not a bad idea here but you could have skipped that in your prove.
Given that $n$ and n+1 are two consecutive integers. Now suppose $gcd(n,n+1)=p$. Then p|n and $p|n+1$. Which implies that $p|n+1-n$ or $p|1$. There is no number which divides 1 except 1. So $p=1$ or  you can say $gcd(n,n+1)=p=1$. Which implies $n$ and $n+1$ are coprime.
Notice that I have not used contradiction anywhere. 
A: Here's a direct proof that preserves generality for all integers.
Let $n \in \mathbb{Z}$ such that $n \notin [-2, 1] $. Then suppose exists $d \in \mathbb{Z}$ such that $\gcd{(n, n+1)} = d$ and $d \geq 1$.
Then we know that $d \mid n $ and $d \mid n+1$, so we have $p, q \in \mathbb{Z}$ such that $n = d\cdot p$ and $n+1 = d\cdot q$.
We can then do the following:
$$\begin{equation}\begin{aligned}
n&=d \cdot p \\
n+1&=d \cdot p + 1 \\
(d \cdot q) &=d \cdot p +1 && \text{since } n+1=d\cdot q \\
d\cdot q -d\cdot p &= 1 \\
d\cdot(q-p)&=1
\end{aligned}\end{equation}$$
And therefore we have that $d \mid 1 $. Since we're in $\mathbb{Z}$, we know $1$ is only divisible by $\pm 1$ (i.e. $\forall x \in \mathbb{Z}$, $x \mid 1 $ is only true when $x = \pm 1$), therefore $ d = \pm 1$ and since $d \geq 1$, $d=1$.
Finally, we have that $\gcd{(n, n+1)} = d = 1 $.
