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I couldn't find an answer on the internet but my script fails solving this. I have to know this because this is required to prove something else (it has to do with random numbers and their period).

I cannot say if they are coprime because I don't know if they are coprime because of the fact that division by $0$ doesn't work.

On the other hand I say they are not coprime because we don't get to $\text{gcd}(0,13)=1$.

But this is very confusing for me, please help!


marked as duplicate by user223391, Patrick Stevens, Bill Dubuque number-theory Nov 24 '16 at 23:09

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  • $\begingroup$ Thanks! From this I deduce that $\text{gcd}(0,13)=13$ and because it doesn't equal $1$, they are not coprime. Is that correct or not? $\endgroup$ – berndgr Nov 24 '16 at 22:49
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    $\begingroup$ $0$ and $13$ do indeed have $13$ as a common divisor. $\endgroup$ – Arthur Nov 24 '16 at 22:57

$gcd(a, b) = d \iff g |a \wedge g |b \Rightarrow g |d$

Notice that $x |0$ for any $x $ therefore

$gcd(a, 0) = a $ because $a |0 \wedge a |a $ and there cannot be a bigger integer that divides $a $. If $a \neq 1$, then 0 and $a $ are not coprime.

  • $\begingroup$ It might be worth noting that usually gcd$(a,b)$ is not defined when $a=0$ or $b=0$. $\endgroup$ – user378947 Nov 24 '16 at 23:05
  • $\begingroup$ @mathbeing even so, it would not make sense to define gcd(0,a) as 0 or 1. $\endgroup$ – RGS Nov 24 '16 at 23:08
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    $\begingroup$ It wouldn't definitely!! Actually I think that gcd$(a,0)=a$ is a good convention (maybe I would express it as gcd$(a,0)=$gcd$(a)=a$ instead but OK) for it is consistent. I just wanted to point out that one could argue as in my above comment. $\endgroup$ – user378947 Nov 24 '16 at 23:18

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