# How to explain to a ~$12-16$ year-old student who is weak at maths, that $\ 38 \times 27 \times 14 = 27 \times 14 \times 38$?

How to explain to a ~$$12-16$$ year-old student who is weak at maths, that $$\ 38 \times 27 \times 14 = 27 \times 14 \times 38,\$$ and that $$\ 3.4 \times 10^{-6} \times 2.1 \times 10^{-5} = 3.4\times 2.1 \times 10^{-6} \times 10^{-5},\$$ and that $$\ 11^4 \times 7^7 \times 3^5 \times 5^2 = 3^5 \times 5^2 \times 7^7 \times 11^4,\$$ and that $$\ \frac{ 8 }{ 14 } \times \frac{ 3 }{ 4 } \times \frac{ 7 }{ 11 } \times \frac{ 22 }{ 9 } = \frac{ 8 \times 3 \times 7 \times 22 }{ 14 \times 4 \times 11 \times 9 } = \frac{ 7 \times 8 \times 22 \times 3 }{ 14 \times 4 \times 11 \times 9 } = \frac{ 7 }{ 14 } \times \frac{ 8 }{ 4 } \times \frac{ 22 }{ 11 } \times \frac{ 3 }{ 9 },\$$ and so on.

I have been in this situation a few times, where it was necessary - or at least extremely helpful - for them to use the fact that if the value of the product of finitely many terms remains unchanged if terms being multiplied are rearranged within the product. [Maybe there's a better way of wording this, but anyway...]

I proceeded to explain that, "you can swap any two numbers without affecting the result due to $$ab = ba,$$ and therefore we can always re-arrange the numbers in the original product $$\ 38 \times 27 \times 14\$$ in any order we want, without affecting the value of the product." However, upon reflection, I think this line of reasoning is too complicated for students who are weak at maths to understand. This is for two reasons. First, they have to understand that swapping any two numbers gives the same result, which is not necessarily obvious, and secondly, they have to understand that there is a way to keep swapping terms to get from the first product $$\ (\ 38 \times 27 \times 14 \ )\$$ to the second product $$\ (\ 27 \times 14 \times 38\ )\$$, which also is not necessarily obvious.

Although, when I was that age, I found both of those steps "obvious", even if I did not know the mathematical language to prove it formally.

And introducing rigor is not advisable for weak students either.

So is there a completely different (perhaps visual?) way to explain this phenomenon/fact that is easy for a weak student to understand, that can also be extended to the product of four, or more numbers? Or can it be explained along the lines I am trying to, but in a way that is easier to understand?

Or is this just one of those "facts" that you tell students of that age that they "just have to accept"?

• "So is there a completely different (perhaps visual?) way to explain this phenomenon/fact that is easy for a weak student to understand". It would be helpful to know what this student's visual concept of multiplication is to start with. Based on your comment to an answer below, some of these students don't know how to find the area of a rectangle. So, what should we assume they can do with multiplication? Commented Dec 12, 2022 at 16:55
• It's worth pointing out that care should be exercised with the words used. Saying that we can always rearrange the numbers in the original product could lead some to posit that $38\times 27 = 32 \times 87$. I would want to say "we can rearrange the factors in the original product" to be careful. Commented Dec 12, 2022 at 17:01
• Every time I see this question I get confused about what is really happening here. We have a young person who is already being perceived as weak at mathematics, being taught by someone who is categorizing an axiom as obvious or not obvious. Then there is the part where repeated uses of two different axioms is also categorized as obvious or not obvious. I think the students' confusion is completely reasonable and not thinking it is anything else is entirely the first step. If motivating axioms and their applications is the question, I wish this question were re-worded. Commented Dec 15, 2022 at 20:11
• However, what I do agree with you on is that my question should clarify some specific things, which I will try to do tonight. Commented Dec 15, 2022 at 22:10
• @Dominique Most students I have tutored are not awful at arithmetic in general, although all students make mistakes with arithmetic: I associate "weaker students" (at arithmetic) as the ones who make mistakes more often than "strong students". But I don't think I have encountered a student who has absolutely no conception or understanding of addition or multiplication when it comes to, for example, (real life) word problems, for example a money calculation problem. Even the weakest student I have tutored had an inkling for when to add and when to multiply. Does this address your question? Commented Feb 9, 2023 at 8:42

For up to three dimensions you could use the concept of "area of a rectangle" or "volume of a rectangular cuboid". Use only numbers (no variables) and let the student pick a convention like "the first factor is the width and the second factor is the height (and the third one the depth)". Then use lots of examples and let the student draw many rectangles (or, if material is available and the task doesn't seem silly, arrange dice or something else in rectangular patterns) and have him or her realise ("accept?") that ab = ba always, because there are the same number of unit squares in both rectangles. Extend to three dimensions and then generalise for arbitrary numbers of factors.

I think this is about "looking at enough examples to accept the commutative law".

• This may work for some weak students. But many weak students don't know how to work out the area of a rectangle, so considerable time would have to be spent working on this in order for your method to be plausible. Also, you haven't said how you get from the student knowing that "the commutative law works" to "you can rearrange the numbers in any order and the product is the same". Commented Dec 12, 2022 at 15:36
• If these students cannot work out the area of a rectangle, what is their mental model of multiplication? Commented Dec 12, 2022 at 18:05
• @AdamRubinson I don't think that your question is answerable until you can figure out what mental model your students are working with. If they are thinking of piles of coin, five piles of three each feels very different from three piles of five each. It is not obvious that piles can be rearranged in this way. So if that is their model, I would suggest that step 1 is to show them how to arrange those piles into rectangular arrays of coins, where commutativity is easier to spot. Commented Dec 12, 2022 at 22:14
• @Adam Rubinson: For weak students, which I interpret as those not planning to attend college except perhaps a community college (which might not be until several years after high school graduation), I wouldn't worry about it. And if you do, forget about induction -- focus on pointing out that this is actually something one might need absolute proof of, and focus on what is at stake. For example, without adding any numbers, how can we be absolutely sure that $(325+228)+(132+947)=[(325+228)+132]+947$? Probably have to describe each in words, describing explicitly what one does and in what order. Commented Dec 13, 2022 at 15:43
• Cuisenaire Rods: teaching.com.au/catalogue/mta/mta-cuisenaire-rods Commented Jan 22, 2023 at 10:47

I think having a student play with examples is a great approach, and that seems the easiest route to having them see that this is true.

If seeing multiple permutations of factors all giving the same result isn't convincing, or if they want a different way to conceptualize the situation, maybe draw a model of a product like $$2\times5\times3$$ and show that pieces can be rearranged. You'd have to choose in advance how to interpret an expression like $$a\times b\times c$$, say where there are $$c$$ objects, each containing $$a\times b$$ smaller units.

Here, each of three given rectangles has $$2\times 5$$ little squares in it. Put them together, shade in different parts (in this case columns), reform the columns into rectangles, and then interpret this new picture as a new product (using the original factors, just reordered).

I'm not totally sure how a conversation like this would go with a student, whether they would be satisfied with it. I guess it would all come down to how student and instructor decide to interpret multiplication. [E.g. If $$a$$ means "number of rows in each rectangle", $$b$$ means "number of columns in each rectangle" and $$c$$ means "number of rectangles", then you could just rotate each rectangle to start with to get other expressions like $$5\times 2 \times 3$$.]

[Edit: I can see that my color choice probably wasn't great, since it looks like red changed size.]

• 2×5×3 are arranged vertically, while 2×3×5 are arranged horizontally. They look different. In fact, they look more like 15×2 :) Commented Dec 24, 2022 at 8:46
• @RustyCore I guess you'll just have to transpose that last row of shapes into a column. Commented Dec 25, 2022 at 14:30
• For a student who cannot represent multiplication of two numbers as an area, that transposition my be the bridge they would not know how to cross. Just sayin'. And you are right with color change. Maybe there should be a "purgatory" stage where all colors change to gray. Commented Jan 17, 2023 at 23:17
• @RustyCore "For a student who cannot represent multiplication of two numbers as an area..." This example presupposes that they can, and that both teacher and student agree where the factors show up in the picture. Commented Jan 17, 2023 at 23:25

Some great answers were provided here, but I wanted to add something which I think is essential: to emphasize to the student that it is a very good thing to question this! Why am I saying that? Because indeed, for many operations in algebra the associative property does not hold! For example, when talking about division: $$8/(4/2) \neq (8/4)/2$$. So in fact I would go so far as to say, that a student who questions this may have a latent / undeveloped talent for Math! Because clearly here is an example of a single operation, namely division, where the associative property does not hold. So I would definitely compliment the student for not taking this for granted, and go on to explain that multiplication is one of the special operations where it holds that $$a\times(b \times c) = (a\times b)\times c$$ and therefore we can drop the parentheses without ambiguity.

A similar example can also explain the commutativity of multiplication: $$2 \times 4 = 4 \times 2$$ but $$2 / 4 \neq 4 / 2$$. Actually I realize that this is actually more relevant to the examples given in the question, and in fact it is important to notice the implication of having these two properties together for a single operation such as multiplication. It can actually be a fun exercise for a student to prove that given:

1. $$(a \times b) \times c = a \times (b \times c)$$
2. $$a \times b = b \times a$$

then also: $$a \times b \times c = c \times b \times a$$ .

Note that given only 1. or 2., this doesn't hold! This is also a source of (justified) confusion: a lot of students implicitly think that commutativity (swapping rule) implies associativity, but those are totally independent properties.

Another thing which occasionally helps here is to ask the student the simplest possible question: "What is bothering you about this rule?". I think I've already established, there is something to be bothered about here. But the student, not always having the faculty with words to explain what is bothering him, just feels "silly" and "inadequate" to deal with the problem. In other words, he feels that it is already wrong to be bothered about something so "simple", which is absolutely bananas! The opposite is true.

• This answer is no less great than the others. Commented Feb 7, 2023 at 2:42

1. List the principle. But NOT in a symbolic (ab) manner. Just "you can change the order when multiplying." [Realize you will need to repeat it, several times. They are not programmable machines. But just keep doing so.]

-and-

1. Each time they miss it, have them do a few simpler examples. 2 x 3 x 4. 2 x 5 x 10. (in a couple orders) [After a while, they will get sick of it. And decide to remember the principle.]

So, net net, I think your instincts are correct in recognizing that weaker students won't learn axiomatically). They are different than you. But in degree...we all share traits of ability and disability in intellect. Unless you're Von Neumann, you need drill, to engrain math tricks. He of course was a Martian. But human brains are more like Harry Potter's hand. Need Umbridge's pen to furrow some lines into it/them.

P.s. I think the "like a rectangular prism volume" is low likely to succeed. Wrong pedagogic insight. It's MUCH more likely that repeated statements and practice will make things click than some magic explanation (in this case a hard one). That doesn't mean, you can't have them, can't try them. After all the "and" principle applies. And you should try lots of things. But. I would not think "key explanation" (like a lock and key) is the way to overcome these hurdles. It's more likely that familiarization is what is needed. And yes, more than you would need. More than what a high IQ student would need.

P.s.s. I don't think of rectangles with multiplication as Xander mentions. I'm not a Ph.D. mathematician. But a decent ex-STEM student. And I think of multiplication as extension of the times table (via long multiplcation). I.e. algebraically. If I really scratch my head and think of the times table itself, I probably think of repeated addition. This is NOT to say that it's the preferred way. Just that rectangles are not something everyone thinks of. I'm always pleasantly surprised by geometric drawings for Pythagoras or the like...since I don't think as geometrically. Again, not even saying this is good. Just saying the impulse to think that everyone thinks in one manner (the way one does) is a mistake. I mean...heck with exponentiation do you think first of repeated multiplication or think first of "more dimensions in Flatland"!?

Most of the answers here have been geometric. Here is an answer for three or more factors that is algebraic. It requires the student to know that multiplication is associative (more understandable to them, "we can put parentheses anywhere because of order of operations", maybe needs additional justification), the commutative law for two numbers (justified well by the rectangles in the other answers) and that two equal expressions can be substituted for one another.

Sample explanation:

Start with $$38 \times 27 \times 14$$. We are not going to calculate this, but if we were, we would need to multiply two of the numbers together first, and then multiply that result times the third number. What that means in symbols is that we might actually calculate $$(38 \times 27) \times 14$$. Because multiplication is associative, we could also do $$38 \times (27 \times 14)$$.

To summarize, $$38 \times 27 \times 14 = (38 \times 27) \times 14 = 38 \times (27 \times 14)$$. Let's say we went with $$(38 \times 27) \times 14$$, and just take a look at $$(38 \times 27)$$ on its own. Because of our commutative law, we know that $$(38 \times 27) = (27 \times 38)$$, and since we can substitute equal expressions into a third expression, our original is now $$38 \times 27 \times 14 = (38 \times 27) \times 14 = (27 \times 38) \times 14$$. We can then move the brackets again (associativity) and switch 38 and 14 to get the final answer.

The bottom line is that by doing this trick, we can see that when we have a bunch of numbers multiplied together, we can switch them around in any order.

• As an aside commenting on whether this answer truly meets the other, overall need of helping the student achieve proficiency in addition to understanding, I am not sure the difference between a student who can successfully recognize the applicability of the commutative law and apply it and one who cannot is that the former has a very clear justification of why this can be done for an arbitrary number of factors. Commented Dec 13, 2022 at 13:13

Many college students have a 2nd issue with a problem like this. If they see 2(3 · 4), they sometimes want to distribute, and write 2·3 · 2·4. Yikes. This one requires seeing the difference between distributive property and a · b · c. We work on that.

First, we go back to areas and volumes. My answer is therefore much like Jasper's. Yes, considerable time must be spent on these basics for further mathematical steps to have any meaning for them.

I ask: What is the area of a rectangle? Someone tells me. Most of them nod. I ask: Why? They look at me like I'm nuts. They think of it as a "formula", and they think of those as givens.

So I have them look at the ceiling tiles (because the floor tiles are 1 foot by 1 foot and don't teach area). We estimate those to be 2 feet by 4 feet. So that's 2 rows of 4 tiles each. We could count, or we could ... multiply, yes.

Once we've done one more problem like that on the board, and they are seeing that the multiplication is how many rows times how much in a row, then we go to volume. I use my hands to "create" (mime) a cardboard box on top of my desk. It's 4 feet across, 3 feet front to back, and 2 feet high. We imagine small 1 cubic foot boxes being packed inside. There are 3 rows of 4 on the bottom layer, and then there's a top layer. How many? 12 and 12 is 24 of the cubic foot boxes. Could we count vertical layers from front to back? Sure. The front 'layer' is 4 by 2, and then they go 3 back. 8 times 3 is also 24. And we could go left to right. 3 by 2 is 6, 4 times that is still 24 cubic feet. Of course.

I also reteach distributive property, using visuals. My goal is always to help them to see why.

Exercise: Following are several products of the same 3 numbers, but in different orders. Calculate these products using a calculator.

14 x 27 x 38

14 x 38 x 27

27 x 14 x 38

27 x 38 x 14

38 x 14 x 27

38 x 27 x 14

What do you notice? Pick three other numbers to test your conjecture.

Does it work with subtraction?

Introduce them some of euclid's axioms. For example:

Things that are equal to the same thing are also equal to one another .

If equals are added to equals, then the wholes are equal (Addition property of equality).

If equals are divided from equals, then the results are equal (division property of equality).

Then, show them with examples how these axioms can be used to prove the desire things ( You can give them a chance prior to this if you wish so). It would not result into any unfairness if they accept it as common sense because these are accepted axioms and not informal shortcuts.

• Why do you prefer an axiomatic approach for a student who is weak at mathematics? Commented Feb 3, 2023 at 17:19
• @DavidSteinberg Because axioms which I mentioned are not uncommon outside math world and hence would be good for newbies to accept as common sense. Commented Feb 3, 2023 at 17:36

When Xander wrote in his comment to his answer, "If these students cannot work out the area of a rectangle, what is their mental model of multiplication?", I think a more relevant comment is, "If these students cannot work out the area of a rectangle, then they shouldn't yet be trying to learn that $$ab = ba,$$ let alone $$abc = cab.$$" I should build multiplication up with geometry at the same level of multiplication that we are working on, with geometry as a helpful tool assisting in the understanding of multiplication, in a similar manner one would use money problems to assist in the teaching of addition. And I shouldn't try to cover all of "numbers" (standard form, HCF/LCM, etc) before moving onto geometry later on; with strong students this might be possible, as we can cover ground more quickly, but with weak students - which is what we are talking about here - it isn't an effective teaching strategy, I guess.

• The intention of my question was not to suggest that areas of rectangles are necessarily the "correct" primitive notion. Rather, the actual question as the important part: What mental model do the students have for multiplication? The way that you approach the various properties of multiplication depends on what they understand multiplication to mean in the first place. Geometry is only one possible approach. Commented Dec 20, 2022 at 22:58
• @XanderHenderson Please can you give me an example of what you're talking about and how it relates to the question. Commented Dec 21, 2022 at 8:39
• Let me clarify: your question seems to be "How do I convince weak students that "abc = cab = bca = acb = cba = bac? That is, how do I convince students that multiplication can be done in any order?" From an axiomatic point of view, this requires both commutativity ($ab = ba$) and associativity ($a(bc) = (ab)c$). But students who are struggling to understand the basic mechanics of multiplication aren't helped by these abstract notions. It is sufficient to say "you can do the multiplication in any order you like", and find some way to justify this, e.g. by permuting the sides of a solid. Commented Dec 21, 2022 at 15:25
• I have five piles of coins, each of which contains seven coins. I am going to make new piles as follows: take one coin from each of my original piles, and put them in a new pile. Then take a second coin from each of the original piles, and put them into a second new pile. Keep doing this. I will end up with piles of five coins (since my new piles contain one coin from each of the original plies), and there will be a total of seven piles (since each of my original piles had seven coins). This gives you commutativity. Commented Dec 21, 2022 at 15:38
• To expand this to more factors, try thinking about putting coins into bags: put $a$ coins into each of $b$ little bags, then put these $b$ little bags into a big bag. Do this $c$ times. This gives you $abc$. Now rearrange. Commented Dec 21, 2022 at 15:39

Start with simpler examples. So, start with explaining why $$\ 1 \times 2 = 2 \times 1$$, then $$\ 2 \times 3 = 3 \times 2$$, etc.

Next, proceed to $$\ 1 \times 2 \times 3 = 2 \times 3 \times 1$$, $$\ 1 \times 2 \times 4 = 2 \times 4 \times 1$$, etc.

Etc.

The examples you give are ridiculously complex and intimidating for a student who doesn't yet understand (or "believe") that multiplication is commutative. But once she gets the above simple examples and understands the concept, she should be able to easily extend it to your examples.

P.S. You repeatedly use the word weak to describe your students and even mention that "when I was that age, I found both of those steps "obvious"". Perhaps your student aren't as smart as you.

But if your goal is to be an effective teacher, perhaps it's better to abandon such beliefs and negative thoughts (satisfying as they may be for your ego):

• Concede that you are not a good teacher and seek to improve (rather than blame your students for their failure to learn).
• Believe it or not, your students probably know that you think they're stupid. This is very demoralizing and intimidating for them and makes it that much harder for them to learn (especially from you).
• You say to explain this and that, but you don't say how to explain it. The original poster was asking for ideas about how to explain things. Also, in my understanding, the complicated examples were examples of calculations where it might come in handy to apply commutativity in solving the kinds of problems students in the 12-16 age group are typically asked to do. I don't think they were intended as examples of how the poster would go about teaching commutativity (which is what the poster wanted to learn how to do). Commented Jan 16, 2023 at 5:36
• @WillOrrick: My answer is meant to complement the other answers already given here. I merely wanted to make one very important point, which is that the teacher should start with simplest possible examples, then work her way up. I did not want this important point to be obscured by other points.
– user18187
Commented Jan 16, 2023 at 6:05
• First of all, this isn't about multiplication being commutative. It's about multiplication being commutative and associative. Am I wrong about this? Either many people have made this mistake in answering my question, or I am wrong about this. For example, you cannot show that $(ab)c = a(bc)$ without the associativity axiom for multiplication, since this is the associativity axiom for multiplication which wouldn't exist if it was redundant, right? That all said, axiom stuff is not applicable to the original question as I have stated many times before. Commented Jan 16, 2023 at 10:18
• Secondly, you say, "The examples you give are ridiculously complex and intimidating for a student who doesn't yet understand (or "believe") that multiplication is commutative. But once she gets the above simple examples and understands the concept, she should be able to easily extend it to your examples". I disagree with literally all of your opinions here. They are just about the simplest examples I can think of. Can you think of simpler ones that aren't trivial? Secondly, just because they believe two or three numbers in a product can be rearranged without changing the result, does not... Commented Jan 16, 2023 at 10:21
• "Concede that you are not a good teacher". How did you determine that I'm not a good teacher? Your "answer" is riddled with unjustified assumptions. Commented Jan 16, 2023 at 14:21

You can teach/remind him to simplify $$(38×27×14)/(27×14×38)$$. If he knows the same number can be eliminated from both sides regardless of the position to simplify a fraction, he'll get 1, meaning they're equal.