I have tried $\gcd(0,8)$ in a lot of online gcd (or hcf) calculators, but some say $\gcd(0,8)=0$, some other gives $\gcd(0,8)=8$ and some others give $\gcd(0,8)=1$. So really which one of these is correct and why there are different conventions?
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1I haven't encountered the convention of gcd(0,8) = 1. It depends on how you define the phrase "a divides b" – The Chaz 2.0 Mar 18 '11 at 02:49

try http://www.hobbyprojects.com/calculators/GCD_LCM_Calculator.html – Vafa Khalighi Mar 18 '11 at 02:53

@The Chaz: They are really the same things but with different names. see http://en.wikipedia.org/wiki/Greatest_common_divisor – Vafa Khalighi Mar 18 '11 at 04:04

Should be a, because anything is a divisor of 0. – user5280911 Nov 26 '20 at 07:48
4 Answers
Let's recall the definition of $ $ "$\rm a $ divides $\rm b$" $ $ in a ring $\rm\,Z,\, $ often written as $\rm\ a\mid b\ \ in\ Z.$
$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad \rm\ a\mid b\ \ in\ Z\ \iff\ a\,c = b\ \ $ for some $\rm\ c\in Z$
Recall also the definition of $\rm\ gcd(a,b),\,$ namely
$(1)\rm\qquad\quad \rm gcd(a,b)\mid a,b\qquad\qquad\qquad\ $ the gcd is a common divisor
$(2)\rm\qquad\quad\! \rm c\mid a,b\ \ \ \Longrightarrow\ \ c\mid gcd(a,b)\quad$ the gcd is a greatest common divisor
$\ \ \ \ $ i.e. $\rm\quad\ c\mid a,b\ \iff\ c\mid gcd(a,b)\quad\,$ expressed in $\iff$ form $ $ [put $\rm\ c = gcd(a,b)\ $ for $(1)$]
Notice $\rm\quad\, c\mid a,0\ \iff\ c\mid a\,\ $ so $\rm\ gcd(a,0)\ =\ a\ $ by the prior "iff" form of the gcd definition.
Note that $\rm\ gcd(0,8) \ne 0\,$ since $\rm\ gcd(0,8) = 0\ \Rightarrow\ 0\mid 8\ $ contra $\rm\ 0\mid x\ \iff\ x = 0.$
Note that $\rm\ gcd(0,8) \ne 1\,$ else $\rm\ 8\mid 0,8\ \Rightarrow\ 8\mid gcd(0,8) = 1\ \Rightarrow\ 1/8 \in \mathbb Z. $
Therefore it makes no sense to define $\rm\ gcd(0,8)\ $to be $\,0\,$ or $\,1\,$ since $\,0\,$ is not a common divisor of $\,0,8\,$ and $\,1\,$ is not the greatest common divisor.
The $\iff$ gcd definition is universal  it may be employed in any domain or cancellative monoid, with the convention that the gcd is defined only up to a unit factor. This $\iff$ definition is very convenient in proofs since it enables efficient simultaneous proof of both implication directions. $\ $ For example, below is a proof of this particular form for the fundamental GCD distributive law $\rm\ (ab,ac)\ =\ a\ (b,c)\ $ slightly generalized (your problem is simply $\rm\ c=0\ $ in the special case $\rm\ (a,\ \ ac)\ =\,\ a\ (1,c)\ =\ a\, $).
Theorem $\rm\quad (a,b)\ =\ (ac,bc)/c\quad$ if $\rm\ (ac,bc)\ $ exists.
Proof $\rm\quad d\mid a,b\ \iff\ dc\mid ac,bc\ \iff\ dc\mid (ac,bc)\ \iff\ d(ac,bc)/c$
See here for further discussion of this property and its relationship with Euclid's Lemma.
Recall also how this universal approach simplifies the proof of the basic GCD * LCM law:
Theorem $\rm\;\; \ (a,b) = ab/[a,b] \;\;$ if $\;\rm\ [a,b] \;$ exists.
Proof $\rm\quad d\,a,b \;\iff\; a,b\,\,ab/d \;\iff\; [a,b]\,\,ab/d \;\iff\; d\,\,ab/[a,b] \quad\;\;$
For much further discussion see my many posts on GCDs.
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Another way to look at it is by the divisibility lattice, where gcd is the greatest lower bound. So 5 is the greatest lower bound of 10 and 15 in the lattice.
The counterintuitive thing about this lattice is that the 'bottom' (the absolute lowest element) is 1 (1 divides everything), but the highest element, the one above everybody, is 0 (everybody divides 0).
So $\gcd(0, x)$ is the same as ${\rm glb}(0, x)$ and should be $x$, because $x$ is the lower bound of the two: they are not 'apart' and 0 is '$>'$ $x$ (that is the counterintuitive part).
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In fact, the top answer can be generalized slightly: if $a \mid b$, then $\gcd(a,b)=a$ (and this holds in any algebraic structure where divisibility makes sense, e.g. a commutative, cancellative monoid).
To see why, well, it's clear that $a$ is a common divisor of $a$ and $b$, and if $\alpha$ is any common divisor of $a$ and $b$, then, of course, $\alpha \mid a$. Thus, $a=\gcd(a,b)$.
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1Indeed, even more generally, it is a special case of the distributive law  see my answer. As for commutative monoids, one usually requires them to be *cancellative* in order to obtain a rich theory. – Bill Dubuque Mar 18 '11 at 18:26
It might be partly a matter of convention. However, I believe that stating that $\gcd(8,0) = 8$ is safer. In fact, $\frac{0}{8} = 0$, with no remainder. The proof of the division, indeed is that "Dividend = divider $\times$ quotient plus remainder". In our case, 0 (dividend) = 8 (divisor) x 0 (quotient). No remainder. Now, why should 8 be the GCD? Because, while the same method of proof can be used for all numbers, proving that $0$ has infinite divisors, the greatest common divisor cannot be greater than $8$, and for the reason given above, is $8$.
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