What is $\gcd(0,a)$, where $a$ is a positive integer?

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?

Solution 1:

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.

Solution 2:

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 counter-intuitive 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 counter-intuitive part).