# Can GCD be called an operator?

Can $\gcd(a,b)$ be called a binary operator which takes operands $a$ and $b$ and returns their greatest common divisor.

And if for some operator —say $\bigotimes$ —$(a_1\bigotimes a_2 \bigotimes ... ... \bigotimes a_{n-1} \bigotimes a_n)$ $= (a_1\bigotimes a_2...\bigotimes a_m)\bigotimes (a_{m+1}\bigotimes a_{m+2}...\bigotimes a_n)$, what is this property called?

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Yes. The first expression isn't well-defined for a binary operator unless it's associative. If it is, then the property expresses a special case of associativity. – Qiaochu Yuan Dec 12 '11 at 19:21
Yes, you can call it a binary operation (if you are working in a setting where any pair of elements have a "greatest common divisor" that is somehow uniquely defined). As to your second query, that would be "associativity". – Arturo Magidin Dec 12 '11 at 19:21
@Qiaochu I might call that expression uniquely defined if there is a left-to-right reading convention. I think that might be the OP's intent, given the question. – alex.jordan Dec 12 '11 at 19:32
Indeed, it is a binary operation. The positive integers are something called a lattice under the operation of divisibility, and in lattice theory, we usually write the greatest common lower bound as $\wedge$, as in $a\wedge b$. In this lattice, $a\vee b$ is the least common upper bound, and is the least common multiple. – Thomas Andrews Dec 12 '11 at 20:36
Yes, a lattice is an abstract concept like group. It's hard to say what you mean by "help in other areas of math," but lattices come up a fair amount in lots of areas of math, including ring and group theory. – Thomas Andrews Dec 12 '11 at 21:34