On a Property of the Least Common Multiple of Two Integers
Dan Kalman
The Least Common Multiple of two integers, is the least positive integer that is divisible
by both integers. This is connected by a simple formula with the greatest common
divisor of the two integers, a familiar topic from modern algebra and number theory. The
purpose of this paper is to present a proof for the connection between least common
multiple and greatest common divisor. Along the way we will see several other
properties of the least common multiple, as well as a number of examples.
Throughout the discussion, we will consider only positive integers, the set of which is
expressed as N. We also assume the notation and properties of the greatest common
divisor presented in Hungerford [1]. In particular, if a and b are positive integers, we
denote the greatest common divisor of a and b by (a,b). To begin the discussion of least
common multiple, we present the following definition.
Definition 1. If a and b are positive integers, the least common multiple of a and b,
denoted [a,b], is the least positive element of the set {z  N : a | z and b | z }.
It should be remarked that for any positive integers a and b, their product ab is always
divisible by both a and b, showing that the set of positive common multiples is not
empty. Thus, [a,b] is always defined.
As an example, [4,6] = 12. To see this, observe that the multiples of 4 and 6,
respectively, can be listed as follows:
4N = {4, 8, 12, 16, 20, 24, …}
6N = {6, 12, 18, 24, 30, 36 …}.
We see that 12 is the least entry that is common to both sets. Indeed, 4|12 and 6|12 so 12
is definitely a common multiple. On the other hand, since the only positive multiple of 6
less than 12 is 6 itself, and since 4 is not a divisor of 6, we see that no other common
multiple can be less than 12. This shows that 12 is the least common multiple.
In this example, we notice that 12 is not only less than or equal to every other common
multiple of 4 and 6, it is also a divisor of all those multiples. We state this as a theorem.
Theorem 2. Let a and b be positive integers, and let c = [a,b]. Also, let C be the
common multiples of a and b. That is, C = {z  N : a | z and b | z }. Then C = cN.
Proof. Because c is a common multiple of a and b, we know both a and b are divisors of
c. This shows they are also divisors of cn for any n in N. Put another way, every such cn
is a common multiple of a and b, and so is an element of C. Therefore, we have
established that cN  C.
For the opposite inclusion, let dC, so that d is an arbitrary common multiple of a and b.
We must show that d  cN. To that end, apply the division algorithm, expressing
d = cq + r
(1)
where q and r are non-negative integers, and 0 ≤ r < c. Now both d and c are multiples of
a, so we can write c = as and d = at for some positive integers s and t. Substitution in
equation (1) then leads to at = asq + r. Rearranging this last equation then produces
r = a (t - sq), demonstrating that a divides r. Exactly the same argument with b in place
of a shows that b divides r as well. Thus, r is a common multiple of a and b. Now since
r < d, which is the least positive common multiple, we see that r cannot be positive.
Thus, we conclude r = 0, and by equation (1), d = cq. This shows that d is an element of
cN. In summary, we have shown that d  C implies d  cN, which proves that C  cN.
This completes the proof that C = cN. ■
The preceding theorem shows that the LCM of two integers is a divisor of every other
common multiple of the integers. This is very similar to the situation with the greatest
common divisor, which is also a divisor of every other common divisor. Next we show
how factorization can be used to simplify the computation of LCM. This is provided by
the following two lemmas.
Lemma 3. If r and s are positive integers, and if (r,s) = 1, then [r,s] = rs.
Proof. Let c = [r,s] . Since rs is a common multiple of r and s, and c is the least
common multiple, we must have c ≤ rs. Now we know c is a multiple of r, so we can
write c = rt for some positive integer t. By substitution, we observe that rt ≤ rs, and
hence, t ≤ s.
Next, using the fact that c is also a multiple of s, we obtain s | c = rt, so s| rt. But recall
that s and r are relatively prime. Therefore, by Theorem 1.5 in reference [1], we
conclude that s| t. Since we already know that t ≤ s , it must follow that s = t. Finally,
using the equation c=rt, we obtain c = rs. That is what we wanted to show. ■
As an example, observe that [2,3] = 6 = 23. On the other hand, in the earlier example,
we saw [4,6]= 12  46. This is consistent with the lemma because 4 and 6 are not
relatively prime, sharing a common factor of 2. Indeed, in this example, we see that
[4,6]= 46/(4,6). As we will see below, this relationship between the least common
multiple and the greatest common divisor holds in general. But first, let us proceed to the
second lemma.
Lemma 4. If a, b, and t are positive integers, then [at,bt] = t [a,b].
Proof. Let c = [a,b] and let d = [at,bt]. Then a | c and b | c, so at | ct and bt | ct. This
shows that ct is a common multiple of at and bt, and so ct  d = [at,bt]. On the other
hand, we know at | d and bt | d, so we may write d = atx = bty for some positive integers
x and y. But then a | (d/t) and b | (d/t), so d/t is a common multiple of a and b. This
implies d/t  c, the least common multiple. Multiplying by t yields d  ct. Combined
with the earlier inequality, this shows d = ct. That is, we have derived [at,bt] = t [a,b],
as desired. ■
To illustrate this example, consider [10,35]. Factoring these two integers, we see that
[10,35] = [25,75] = 5 [2,7]. Proceeding further, since (2,7) = 1, Lemma 3 shows [2,7] =
14. In this way we easily find [10,35] = 514 = 70. This approach to calculating the
least common multiple is generally valid, and provides a proof for our main result.
Theorem 5. If a and b are positive integers, then (a,b)[a,b] = ab.
Proof. Let t = (a,b). Thus we may write a = rt and b = st for some positive integers r
and s. Moreover, as stated in problem 16 in section 1.2 of reference [1], (r,s) = 1. Now
according to Lemma 4, we can compute [a,b] = [rt,st] = t [r,s]. Next, by Lemma 3, we
see that [a,b] = trs. Now multiply both sides of this equation by t. That gives t[a,b] = trts
= ab. But t = (a,b). Therefore, we have shown that
(a,b)[a,b] = ab. This completes the proof. ■
Notice that the conclusion of Theorem 5 can be rewritten in the form
[a,b] = ab/(a,b).
We can apply this to the earlier example with a = 10 and b = 35. We obtain [10,35] =
1035/ (10,35) = 350/5 = 70.
REFERENCES
[1.] Thomas Hungerford, Abstract Algebra an Introduction, 2nd Ed., Brooks-Cole,
www.brookscole.com, 1997.