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

Question

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"?

Answer 1

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".

Answer 2

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.]

Answer 3

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.

Answer 4

My advice is "and".

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-

2. 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.]

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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"!?

Answer 5

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.

Answer 6

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.

Answer 7

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 this work with addition?

Does it work with subtraction?

Answer 8

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.

Answer 9

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.

Answer 10

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.

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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).

Answer 11

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.