Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Given $\{x \mid x > 1\}$, how do I prove that any given $x$ and $x+1$ are coprime?

share|improve this question
7  
$p \mid x,x+1 \Longrightarrow p \mid 1$. –  njguliyev Nov 2 '13 at 17:35

3 Answers 3

up vote 31 down vote accepted

If $y$ divides $x$ and $x+1$ then it divides $(x+1)-x=1$. Conclude.

share|improve this answer
1  
"If y divides x and x+1 then it divides (x+1)−x=1." Why is this true? –  Kevin Nov 2 '13 at 22:14
4  
If $y$ divides $x$ and $x+1$ then there's $a$ and $b$ such that $x=ay$ and $x+1=by$ then $(x+1)-x=(b-a)y$ so... Do you understand? –  Sami Ben Romdhane Nov 2 '13 at 22:19
1  
Yep! Thank you. –  Kevin Nov 2 '13 at 22:22
    
Nice! A one-liner kills the question! –  amWhy Mar 6 at 13:51
1  
Nice and concise. –  drhab Jun 12 at 20:18

$\gcd(x,x+1)=\gcd(x,x+1-x)=\gcd(x,1)=1$.

Hence $x$ and $x+1$ are coprime.

share|improve this answer
    
+1 for a general method –  Ramchandra Apte Nov 3 '13 at 7:14

If $x$ is a multiple of $p$, then the next multiple of $p$ is $x+p$, but that's clearly larger than $x+1$.

share|improve this answer
    
Or, $ $ mod $\,p\!:\ x\equiv 0\,\Rightarrow\, x+1\equiv 1\not\equiv 0\ $ (else $\,p\mid 1-0)\ \ \ $ –  Bill Dubuque May 31 at 16:53

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.