Re: GCD(0,0)



Leroy Quet wrote:
> I notice that these math-controversy threads often get massive
> numbers of replies.
> (While more serious math posts and my games, for example,
> hardly ever get any replies.)
> So I will post this troll-bait flame-bait message to sci.math
> because I always wanted to start one of those huge threads.
> :)
>
> For n = any positive integer, it is known that
>
> GCD(n,n) = n
>
> and
>
> GCD(0,n) = n.
>
> (GCD is Greatest Common Divisor, of course.)
>
> But what is, if there is any defined value,
>
> GCD(0,0)?
>
> It certainly isn't 0 (which would fit the pattern above if
> n=0), is it?
> I would think that infinity would work as well as anything.
>
> Or is GCD(0,0) simply undefined, like 0/0?
>
>
> thanks, (half seriously, oh well, 3/4 seriously)
> Leroy Quet

Well, Leroy, it's not purely "troll bait". I had an interesting
discussion recently with someone who wrote to me about just that
question, since in my basic number theory text I'd said the gcd(0,0) is
undefined.

The short answer to your question is that "first you pick your
definition and then you get your value."

So, using the most elementary definition of gcd,

gcd(a,b) = largest integer that divides both a and b,

then gcd(0,0) is undefined (or "equal to infinity" in the sense of
being larger than every integer).

A more sophisticated , and often much more useful, definition is

gcd(a,b) = the unique nonnegative integer d so that
the ideal generated by a and b is equal to the ideal
generated by d.

With that definition it's clear that gcd(0,0)=0. Of course, this
definition assumes that one knows that the integers are a PID.

In more general rings, as long as they are PIDs, the same definition
more-or-less works, but the gcd is only defined up to multiplication by
a unit. For the ordinary integers, since the only units are +1 and -1,
we can pick a distinguished gcd by choosing the positive representative
for the ideal.

Cheers, Joe Silverman

.



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