# Are the numbers $0$ and $13$ coprime with each other? [duplicate]

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 $\text{gcd}(0,13)=1$.

## marked as duplicate by user223391, Patrick Stevens, Bill Dubuque number-theory StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Nov 24 '16 at 23:09

• 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? – berndgr Nov 24 '16 at 22:49
• $0$ and $13$ do indeed have $13$ as a common divisor. – 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.
• It might be worth noting that usually gcd$(a,b)$ is not defined when $a=0$ or $b=0$. – user378947 Nov 24 '16 at 23:05
• 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. – user378947 Nov 24 '16 at 23:18