Venn diagrams have got to infiltrate the modern curricula as a way to explain probability theory, discussing sets of events. Not surprising, and we can say that they never left us really in topology and other usages of the lattice of union and intersection of sets.

Now, I have recently noticed that they are being used also for a different task, the analysis of factors of integer numbers and polynomials, and the subsequent calculation of the greatest common divisor and least common multiple.

In fact there are two different usages here: simply to represent the set of divisors of two numbers A and B and its intersection, $A \cap B$ or directly represent the aggregates (?) of prime factors of A, B and its gcd, $A \wedge B$. The first way is very in the spirit of New Math, but in order to be able to expose the LCM it needs, I believe, to use $AB=LCM \times GCD$.

The second way is more explicit, one gets the LCM, as $A \vee B$, by multiplying all the factors in the diagram and the GCD by restricting to only the "wedge" part of it.

enter image description here

But we can not see the factors as elements of a set, because of course we can have repeated factors. The problem appears when we remember that the number of subsets of a set is $2^N$, but the number of divisors of 12 = [2,2,3] needs of course to account for the repetition.

Still it looks, this second way, an useful representation and I am tempted to use it. The caveats are:

  • Could it cause to the student some contradiction with other usages of the diagrams, as e.g. in probability theory?

  • Is it actually a good tool to understand the concept? I mean, compared with the usual recipe of writing both numbers as a prime factorization and choosing the upper / lower exponent.

  • $\begingroup$ Pointers to more general studies/articles about the usage of Venn diagrams are welcome, too :-) $\endgroup$
    – arivero
    May 8, 2018 at 18:58
  • 1
    $\begingroup$ This whole discussion is missing pictures. $\endgroup$ May 9, 2018 at 13:38
  • 1
    $\begingroup$ Technically when taking this approach, these are Venn diagrams of multisets not of regular sets. This is what allows repeated factors. But in my experience students don't seem interested in the distinction. $\endgroup$
    – Aeryk
    May 9, 2018 at 21:32
  • 1
    $\begingroup$ @Aeryk The abuse of multiset vs. set is also rampant in linear algebra. $\endgroup$ May 9, 2018 at 22:12
  • 1
    $\begingroup$ It's a nice graphic but I suspect it will be a BAD idea for the audience (~5th graders). I understand the appeal of a graphical explanation. Reminds me of chem class and comparing "limiting reagent" to tires and frames on a bike. Really clicked with students--limiting reagent is a topic people struggle with. However, I think this Venn diagram thing makes more sense to people who are already sophisticated (like the teacher) than to 5th graders (to who Venn is strange). Just my caution...not based on observation. Don't do it if it becomes a derailer and you have to teach Venn diagrams! $\endgroup$
    – guest
    May 14, 2018 at 21:20

2 Answers 2


I have a worked example on MSE here in which I cite (RE: your comment on pointers to articles):

Feldman, Z. (2014). Rethinking factors. Mathematics Teaching in the Middle School, 20(4), 230-236. JSTOR (paywall).

Check out that paper (if you can gain access) and its citations; specifically, you may wish to investigate work by Zazkis and Campbell (google scholar).

I do not see any issues with students conflating the use in probability with the use for factors: it is simply a way of representing what is unique to individual numbers (with regard to their factors, or in some approaches, their prime factors) as well as what is shared by the numbers (specifically, their greatest common factor).

I think that Venn Diagrams, like all tools, are not intrinsically good, but rather only as good as they are used in practice. I happen to like them as tools for working out the LCM or GCF in cases where a non-routine problem has been posed; you can find an example in a paper I co-wrote with B Sriraman:

Sriraman, B., & Dickman, B. (2017). Mathematical pathologies as pathways into creativity. ZDM, 49(1), 137-145. Link (no paywall).

Section 4 in the aforecited/aforelinked discusses exactly how one can broach non-routine problems around LCMs, and includes (for example) the observation that one can draw a 3-Venn Diagram to represent the LCM and GCF of three whole numbers.

I close by wordily indicating how I think about LCMs in (at least) the k12 context: When I am looking at a list of whole numbers and trying to compute their LCM - and this can arise, e.g., when adding up fractions with different denominators - I think about each whole number as telling me what prime numbers I need to include. Specifically, I go through each number under consideration and look at their prime factors to see whether I can form the requisite numbers. Here is a fully worked example for how I would complete the task of finding the LCM of the first ten positive integers:

$1$: All whole numbers are divisible by one.

$2$: I need a $2$. Right now I have: $2$.

$3$: I need a $3$. Right now I have: $2 \cdot 3$.

$4$: Since its prime factorization is $2^2$, and I only have one $2$ now, I need a second one. Right now I have: $2^2 \cdot 3$.

$5$: I need a $5$. Right now I have: $2^2 \cdot 3 \cdot 5$.

$6$: Since its prime factorization is $2 \cdot 3$, all of which I already have, there is no change to my candidate for LCM.

$7$: I need a $7$. Right now I have: $2^2 \cdot 3 \cdot 5 \cdot 7$.

$8$: Since its prime factorization is $2^3$, and I only have two $2$s now, I need a third one. Right now I have: $2^3 \cdot 3 \cdot 5 \cdot 7$.

$9$: Since its prime factorization is $3^2$, and I only have one $3$ now, I need a second one. Right now I have: $2^3 \cdot 3^2 \cdot 5 \cdot 7$.

$10$: Since its prime factorization is $2 \cdot 5$, all of which I already have, there is no change to my candidate for LCM.

So: LCM$(1, 2, 3, 4, 5, 6, 7, 8, 9, 10) = 2^3 \cdot 3^2 \cdot 5 \cdot 7$.

This approach clarifies in my own thinking how the LCM is a multiple of each number (it contains all the prime numbers to their appropriate powers necessary to make each number) and it is the least such number with this property (it contains nothing superfluous: a common error in the above worked example would be to take the product of all ten numbers).

  • 1
    $\begingroup$ Can I copy your second image to the main text of my question? It can be illustrative. $\endgroup$
    – arivero
    May 9, 2018 at 13:58
  • 1
    $\begingroup$ +1. Observation: I teach a college intermediate algebra course (which counts as remedial), starting with a review of fractions, up through polynomial factoring, reducing rational expressions, and radicals (all which have factoring as the key step). Inevitably this issue of finding the LCM is totally opaque for many students, no matter how many times we review it (and I use very similar language, i.e., "Let's check that the LCM contains all these numbers as factors"). Nothing so far has gotten much traction. $\endgroup$ May 9, 2018 at 19:34
  • $\begingroup$ @arivero Yes, definitely. $\endgroup$ May 9, 2018 at 20:29
  • 2
    $\begingroup$ @DanielR.Collins Check out Zazkis' papers; no magic fix but you're in good company! I think there is some key insight related to, even needn't be divisible by four, but the reverse is necessary. Tougher topic than it's given credit for, though. $\endgroup$ May 9, 2018 at 20:34
  • $\begingroup$ I added an answer explaining the algebraic approach to the problem in your linked MSE answer. It would be helpful to link that answer to this page (I cannot do so now). $\endgroup$ May 10, 2018 at 15:01

Perhaps worth emphasis is that the use of prime factorizations (either directly or implicitly via (multiset) Venn diagrams or exponent vectors) may actually obfusctate the essence of the matter. For example, lets solve the problem linked in Benjamin's answer without using prime factorizations but instead using gcd & lcm distributive laws to pull out and cancel a common factor $12$ from $A,B$. Namely using $\gcd(A,B) = \gcd(12a,12b) = 12 \gcd(a,b),\,$ and similarly for $\rm lcm,$ yields

$$\begin{align} \gcd(A,B) &= 12,\ \ 360 = {\rm lcm}(A,B)\\[.2em] \iff\ 12\gcd(a,b) &= 12,\ \ 360 = 12\,{\rm lcm}(a,b)\\[.2em] \iff\quad\ \gcd(a, b) &= 1,\ \ \quad 30 = {\rm lcm}(a,b)\ [= ab] \end{align}\qquad$$

Thus we have reduced it to factoring $\,30 = ab\,$ into coprime factors $a,b,\,$ which is easy.

The key simplification arises from applying the $\,\rm gcd$ distributive law

$$ \gcd(ca,cb) = c \gcd(a,b)$$ and similarly for $\rm lcm$. Solutions using prime factorizations or Venn diagrams are essentially doing the same thing, but the key role played by distributivity is highly obfuscated from that viewpoint. This is pedagogically detrimental because such distributive laws are fundamental, so this innate algebraic structure should be highlighted - not obfuscated. The solution of many number theory problems works the same way as above - exploit distributivity to cancel out a common factor so reducing to the simpler case when $a,b$ are coprime (so their lcm = product). One will be at a great disadvantage if one is forced to rediscover this distributive structure every time in terms of Venn diagrams vs. simply invoking well-known fundamental gcd laws.

Thus if you use Venn diagrams or similar devices, strive to also teach the more algebraic approaches using basic gcd laws so that students will obtain a better grasp of the algebraic essence of the matter.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.