Common multiple, and the least common multiple $\operatorname{lcm}(a,b)$, taken to be $0$ when $a = 0$ or $b = 0$

Let $a, b \in \mathbb{Z}$. An integer $m$ is a common multiple of $a$ and $b$ when $a \mid m$ and $b \mid m$ (Divisibility in $\mathbb{Z}$: $d \mid a$ when $a = dq$ for some integer $q$).

Case $a \ne 0$ and $b \ne 0$: a least positive common multiple exists. The integer $|ab|$ is a positive common multiple. It is a common multiple because $a \mid a$ gives $a \mid ab$ and hence $a \mid |ab|$, the last step because $|ab|$ is $ab$ or $-ab$ (Divisibility is reflexive and transitive on $\mathbb{Z}$, and is linear: if $d \mid a$ and $d \mid b$ then $d \mid ax + by$ for all integers $x, y$; also $d \mid a$ implies $d \mid ac$, $-d \mid a$ and $d \mid -a$, The absolute value $|a|$ of an integer), and symmetrically for $b$; and it is positive because $ab \ne 0$ (The integers have no zero divisors; multiplicative cancellation), so $|ab| \ge 0$ and $|ab| \ne 0$ (Absolute value in $\mathbb{Z}$: $|a| \ge 0$; $|a| = 0$ exactly when $a = 0$; $|-a| = |a|$; $|ab| = |a|\,|b|$; $-|a| \le a \le |a|$; and $|a| \le c$ exactly when $-c \le a \le c$). Every positive common multiple is nonnegative, hence of the form $\iota(k)$ for a unique $k \in \mathbb{N}$, where $\iota : \mathbb{N} \to \mathbb{Z}$ is the embedding of The naturals embed in the integers; so the set

$T \;:=\; \{\, k \in \mathbb{N} \;:\; \iota(k) \text{ is a positive common multiple of } a \text{ and } b \,\}$

is a nonempty subset of $\mathbb{N}$ and has a least element $k_0$ (The well-ordering principle). Since $\iota$ preserves the order, $\iota(k_0)$ is then the least positive common multiple of $a$ and $b$, and we define

$\operatorname{lcm}(a,b) \;:=\; \iota(k_0),$

the least positive common multiple. It is unique, greatest and least elements being unique by antisymmetry (The integers form a totally ordered ring, Order on the integers).

Case $a = 0$ or $b = 0$: the only common multiple is $0$. Say $a = 0$. Then $a \mid m$ reads $0 \mid m$, which holds exactly for $m = 0$ (Divisibility in $\mathbb{Z}$: $d \mid a$ when $a = dq$ for some integer $q$); and $0$ is indeed a common multiple, since every integer divides $0$. So there is no positive common multiple at all, and we set

$\operatorname{lcm}(a,b) \;:=\; 0 .$

This is not a free choice dressed as one: $0$ is the only common multiple of the pair, so any other value would name an integer that is not a common multiple.

With both cases together $\operatorname{lcm}(a,b)$ is defined for every pair, and $\operatorname{lcm}(a,b) \ge 0$ always.

• The two boundary values agree, but they have different standing. $\gcd(0,0) = 0$ is a genuine convention, adopted in Common divisor, and the greatest common divisor $\gcd(a,b)$, with the convention $\gcd(0,0) := 0$ because no greatest common divisor exists there; $\operatorname{lcm}(a,b) = 0$ at a vanishing argument is not a convention at all, since $0$ is the only common multiple available. With the two together, $\gcd(a,b)\operatorname{lcm}(a,b) = |ab|$ holds at every pair without exception (Every common multiple of $a$ and $b$ is a multiple of $\operatorname{lcm}(a,b)$, and $\gcd(a,b) \cdot \operatorname{lcm}(a,b) = |ab|$). At $a = 0$ that identity reads $\gcd(0,b) \cdot 0 = 0 = |0 \cdot b|$, which is true for every $b$.

• "Least" is least in the order of $\mathbb{Z}$, among the positive common multiples. The stronger statement, that $\operatorname{lcm}(a,b)$ divides every common multiple and not merely that it is smallest, is a theorem and is proved as the first half of Every common multiple of $a$ and $b$ is a multiple of $\operatorname{lcm}(a,b)$, and $\gcd(a,b) \cdot \operatorname{lcm}(a,b) = |ab|$. It is that divisibility form, not the minimality, that later pages use.

• Two arguments only. As with $\gcd$, no least common multiple of a longer list is defined on this page, because nothing here needs one.