# Division of fractions part 3: why invert and multiply?

We ended the previous post with a bit of a cliffhanger, with two possible diagrams to represent $1\frac34 \div \frac12$:

The first of these diagrams is more familiar to students because it reflects their past work, but the second is more productive for understanding “dividing by a unit fraction is the same as multiplying by its reciprocal.”

Why is the first one more familiar? In grades 3 and 4, students study both the “how many in each (or one) group?” and “how many groups?” interpretations for division with whole numbers (see our last blog post for examples). In grade 5, they study dividing whole numbers by unit fractions and unit fractions by whole numbers. But, as we mentioned in that post, in grade 5 the “how many groups?” interpretation is easier when dividing whole numbers by unit fractions because students do not have to worry about fractions of a group. Going from $3 \div \frac12$ to $1\frac34 \div \frac12$ using this interpretation feels fairly natural:

The main intellectual work here is seeing that $\frac14$ cup is $\frac12$ of a container, but because the structure of the problem is the same and that structure can be easily seen in the diagrams, students can focus on that one new twist. The transition also helps students see that “how many groups” questions can be asked and answered when the numbers in the division are arbitrary fractions.

So the “how many groups” interpretation is useful for understanding important aspects of fraction division and has an important role in students’ learning trajectory. It enables students to see that dividing by $\frac12$ gives a result that is 2 times as great. But it doesn’t give much insight into why this should be the case when the dividend is not a whole number.

The “how much in each group” interpretation shows why. Here are diagrams using that interpretation showing $3 \div \frac12 = 2 \cdot 3$ and $1 \frac34 \div \frac12 = 2 \cdot 1 \frac34$.In fact, the structure of this context is so powerful, we can see why dividing any number by $\frac12$ would double that number: $$x \div \frac12 = 2 \cdot x = x \cdot \frac21$$

This is true for dividing by any unit fraction, for example $\frac15$:In the diagram above, we can see that $1\frac34$ is $\frac15$ of a container, so a full container is $1\frac34 \div \frac15$. Looking at the diagram, we can see why it must be that the full container is $5 \cdot 1 \frac34 = 1 \frac34 \cdot \frac51$.

With a little more work to make sense of it, we can use this interpretation to see why we multiply by the reciprocal when we divide by any fraction, for example $\frac25$:In the diagram above, we can see that $1\frac34$ is $\frac25$ of a container, so a full container is $1\frac34 \div \frac25$. We can see in the diagram that $\frac12$ of $1\frac34$ is $\frac15$ of the container, so our first step is to multiply by $\frac12$: $$1\frac34 \cdot \frac12$$

Now, just as before, to find the full container, we multiply by 5:

$\left (1\frac34 \cdot \frac12 \right) \cdot 5 = 1\frac34 \cdot \frac52$

This shows that dividing by $\frac25$ is the same as multiplying by $\frac52$!

There is nothing special about these numbers, and a similar argument can be made for dividing any number by any fraction. Now students, instead of saying “ours is not to reason why, just invert and multiply,” can say “now I know the reason why, I’ll just invert and multiply.”

Next time: Beyond diagrams.

# Fraction division part 2: Two interpretations of division

In our last post, we asked people if they could come up with a division story problem for $1\frac34 \div \frac12$. Interestingly, almost all of the responses used the “how many groups?” interpretation of division. When interpreting multiplication in terms of groups, the two factors play different roles, and so there is another interpretation of division worth exploring.

If we say that $a \times b$ means $a$ groups of $b$, then

• a division situation where $b$ and $a\times b$ are known but $a$ is unknown is called a “how many groups?” division problem
• a division situation where $a$ and $a\times b$ are known but $b$ is unknown is called a “how many (or how much) in each (or in one) group?” division problem.

[Pause here and see if you can come up with a “how much in one group?” story problem for $1 \frac34 \div \frac12$.]

How do these two interpretations of division come into play as students learn about fraction division? In grade 5, students solve problems like $6\div \frac12$ and $\frac12 \div 6$. What’s nice about problems involving a whole number divided by a unit fraction or a unit fraction divided by a whole number is that we can think of them using the same structure that we thought about division of whole numbers.

1. Kiki has 6 kg of chocolate chips. How many 2 kg packets of chocolate chips can she make?
2. Kiki has 6 kg of chocolate chips. How many $\frac12$ kg packets of chocolate chips can she make?

Notice that these are both a “how many groups?” division problem, and because there is always a whole number of unit fractions in 1, the solution will be a whole number of groups (so students do not have to worry about fractions of a group). If we write equations to represent these problems, that can also help us see the structure:

$$? \times 2 = 6$$

$$? \times \frac12 = 6$$

1. Nero had 6 cupcakes and 3 friends he wanted to share them with equally. How many cupcakes does each friend get?
2. Nero had $\frac12$ of a cupcake and 3 friends he wanted to share them with equally. How many cupcakes does each friend get?

Notice that these are both a “how many in each (or how much in one) group?” division problem, and students don’t have to worry about fractional groups because the whole number in the problem is the number of groups.

Again, with equations:

$$3 \times ? = 6$$

$$3 \times ? = \frac12$$

So in grade 5, students can build on their understanding of whole number division without having to grapple with fractional groups, so long as they understand both of these interpretations of division.

In grade 5, students also learn about fraction multiplication, so they do encounter fractions of a group, but they are not required to put these two understandings together until grade 6 when they extend their understanding of division to all fractions. This provides some scaffolding for students on their way to understanding division of fractions in general.

Let’s look at these two interpretations of division for $1\frac34 \div \frac12$.

• “How many groups?” : I need $1\frac34$ cups flour, but I only have a $\frac12$ cup measure. How many times do I have to fill the $\frac12$ measure to get $1\frac34$ cups flour? (A version of this was suggested by two different people on our last post.)
• “How much in one group?” : I have a container with $1\frac34$ cups flour. The container is $\frac12$ full. How much flour does the container have when it is full?

Here are two possible diagrams to represent these two interpretations of division:

Next time: how the different interpretations of division and diagrams can be used to understand the “invert and multiply rule” and other approaches to understanding this procedure.

# Illustrative Mathematics 6–8 Math

I can’t help writing this off-cycle blog post to celebrate the release of Illustrative Mathematics 6–8 Math  last Friday, a proud achievement of the extraordinary team of teachers, mathematicians and educators at Illustrative Mathematics (IM), one that I didn’t dream of when I started IM almost 7 years ago with a vision of building a world where all learners know, use, and enjoy mathematics.

Conceived initially as a  project at the University of Arizona to illustrate the standards with carefully vetted tasks, IM has grown into a not-for-profit company with 25 brilliant and creative employees and a registered user base some 40,000+ strong. Our partnership with Open Up Resources (OUR) to develop curriculum started a little over 2 years ago when we submitted a pilot grade 7 unit on proportional relationships to the K–12 OER Collaborative, as OUR was then known. In the fall of 2015, not understanding that it couldn’t be done, we agreed to write complete grades 6–8 curriculum ready for pilot in the 2016–17 school year.

One of the things I love about the curriculum is the careful attention to coherent sequencing of tasks, lesson plans, and units. The unit on dividing fractions is an example, appropriate to mention in the middle of this series of blog posts with Kristin Umland on the same topic. It moves carefully through the meanings of division, to the diagrams that help understand that meaning, to the formula that ultimately enables students to dispense with the diagrams. It illustrative perfectly our balanced approach to concepts and fluency. Kristin and I will be talking about that more in the next few blog posts.

# Fraction division part I: How do you know when it is division?

In her book Knowing and Teaching Elementary Mathematics, Liping Ma wrote about this question and how teachers responded to it:

Write a story problem for $1 ¾ \div ½$.

Many people find it hard to come up with a story problem that represents fraction division (including many math teachers, engineers, and mathematicians). Why is this hard to do? For many people, their schema for dividing fractions consists almost entirely of the “invert and multiply” rule. But there is much more to thinking about fraction division than that. So much in fact, that we can’t say it all in a single blog post. This is the first of several musings about fraction division.

### The trouble with English

Consider this problem:

If you have 12 liters of tea and a container holds 2 liters, how many containers can you fill?

You probably know instantly that this is a division problem and that the answer is 6, because you know your times tables, and specifically you know that $2 \times 6 = 12$. If we say that $a \times b$ means $a$ equal groups of $b$ things in group, then a division problem where $b$ and $a\times b$ are known but $a$ is unknown is called a “how many groups?” problem. Here are some other questions that ask “how many groups?”

• If you have 1 ½ liters of tea and a container holds ¼ liter, how many containers can you fill?
• If you have 1 ¼ liters of tea and a container holds ¾ liters, how many containers can you fill?
• If you have ¾ liter of tea and a container holds 1 ¼ liters, how many containers can you fill?

Some people think that the last one feels like a trick question because you can’t even fill one completely. Because we know the answer is less than one, we could also ask it this way:

• If you have ¾ liter of tea and a pitcher holds 1 ⅓ liters, how much of a container can you fill?

So a division problem that asks “how many groups?” is structurally the same as a division problem that asks about “how much of a group?”, but because of the way we speak about quantities greater than 1 and quantities less than 1, the language makes the structure harder to see.

What other ways might we see the parallel structure?

Diagrams:

Equations: $$? \times2 = 12, \quad ? \times \frac14 = 1\frac12, \quad ? \times \frac34 = 1\frac14, \quad ? \times 1\frac14 = \frac34.$$  The diagrams don’t have the language problem. In all cases the upper and lower braces show the relation between the size of a container and the amount you have.  Whether a whole number of containers can be filled (diagrams 1 and 2), a container plus a part of a container can be filled (diagram 3), or only a part of a container can be filled (diagram 4), the underlying story is the same.

Many people think of diagrams primarily as tools to solve problems. But sometimes diagrams can help students see structure or reveal other important aspects of the mathematics. This is an example of looking for and making use of structure (MP7).

The equations have an even clearer structure, but more abstract. They all have the structure $$\mbox{(quantity of containers)}\times\mbox{(size of a container)} = \mbox{(how much you have)}.$$

The intertwining of the abstraction of the equations and the concreteness of the diagrams is a good example of MP2 (reason abstractly and quantitatively).

Coming up next week: what else are diagrams good for?

# A world without order (of operations)

What would such a world look like? Like this:
$$(((3\times(x\times x)) – (7\times x)) + 2).$$What a world it would be! A world without ambiguity! A world where PEMDAS would just be P! A world where they would have to relocate the parenthesis keys to a more convenient location on the keyboard!

Parentheses, and order of operations, tell us how to read the meaning of an expression, how to parse it, not what to do with it. In the expression above, every matched pair of parentheses contains something of the form $$(\mbox{blob}) * (\mbox{another blob}), \qquad (\mbox{where * stands for +, -, or \times}),$$ unless the blobs are just numbers or letters, in which case we don’t surround them with parentheses. We always know exactly what things we are adding, subtracting, or multiplying. Starting with the outermost parentheses, we see it contains the sum of 2 and a blob. Looking inside that blob we see that it contains a blob minus another blob. And so on. The structure of the expression can be represented in a diagram:

So what is order of operations about, and why do we need it? Well, that’s a lot of parentheses up there, so it is useful to have some conventions about when things are understood to be a blob, without actually putting in the grouping symbols (blobbing symbols?). First, any sequence of multiplications and divisions is understood to be a blob (that’s the precedence of multiplication and division over addition and subtraction). Second, in a sequence of additions and subtractions, or of multiplications and divisions, you read from left to right. (Actually, there is disagreement about this last one in the case of multiplication and division, but never mind.) The first rule allows us to write the expression above as
$$((3\times x\times x- 7\times x) + 2).$$The second rule allows us to leave out all the remaining parentheses. And, of course, we have other conventions about representing multiplication by juxtaposition, and about exponent notation, which allow us to write
$$3x^2 – 7x + 2.$$

Calling it order of operations is problematic because it can be misconstrued as suggesting that there is a specific order in which you must perform operations. There isn’t, except insofar as you sometimes have to wait to perform an operation until you have calculated all the blobs in it. But, for example, there is no law that says you have to do the multiplications first in $101\times56-99\times56$ and, in fact, it is more efficient to factor out the $56$ and do a subtraction first. Order of operations tells us how to read this expression: it’s a difference of two products, not a product of three factors the middle one of which is a subtraction. But it doesn’t tell us how to compute it. The word “order” in “order of operations” is best understood as referring to order in the sense of hierarchy, as in the diagram above.

Outside of textbook school mathematics the order of operations is a matter of common law, not constitutional law, and it’s a bad idea to make a federal case out of it on assessments. For example, dinging a student for interpreting $x/2y$ as $x/(2y)$ rather than $(x/2)y$ would be unreasonable; many scientists would do the same thing. If there is any danger of ambiguity we should put the clarifying parentheses in.

A few final thoughts:

• thanks to Brian Bickley for suggesting the topic for this post
• there’s a nice discussion of the history of order of operations over at the Math Forum
• and bonus question: do we have to give multiplication precedence over addition? Could we do it the other way around?

# New look, new title

I have updated to a new WordPress theme, partly because I thought it was time for a makeover, and partly to see if it would cure some of the problems people have had commenting. In the process I decided to change the title of the blog. My recent writings have been about school mathematics generally, and I hope they are of use to teachers everywhere, whatever their standards. I will still write occasionally about the Common Core, and I will still answer questions over in the forums. I have changed the settings in the forums to allow anonymous posting for people who have trouble logging in. I may have to change that back again if it causes security problems. And, speaking of security, the site has been protected by SiteLock since last summer’s hacking, which means that you will occasionally encounter a captcha screen.

And, by the way, the url mathematicalmusings.org also points to this blog.

# Truth and consequences: talking about solving equations

The language we use when we talk about solving equations can be a bit of a minefield. It seems obvious to talk about an equation such as $3x + 2 = x + 5$ as saying that $3x+2$ is equal to $x + 5$, and that’s probably a good place to start. But there is a hidden assumption in there that the equation is true. In the Illustrative Mathematics middle school curriculum coming out this month we start students out with hanger diagrams to represent such equations:

The fact that the hanger is balanced embodies the hidden assumption that the equation is true. It is helpful for explaining why you have to perform the same operation on each side when solving equations; if you take two triangles from the left side you have to take two triangles from the right side as well in order to preserve the balance. This leads to a discussion of how performing the same operation on each side of an equation preserves the truth of the equal sign.

But what happens with an equation like $3x + 2 = 3x + 5$? In this case, the hanger diagram is a physical impossibility: the right hand side will always be heavier than the left hand side. I can imagine that students who have an idea of an equation as “the left hand side is equal to the right hand side” might be confused by this situation, and think this is not a proper equation. Especially when they reduce it to $2 = 5$. Students learn to say that this means there are no solutions, but it’s hard to make sense of that response rule without understanding what’s really going on with equations.

The fact is, an equation with a variable in it is neither true nor false, because it is merely a phrase in a longer sentence, such as “If $3x + 2 = x + 5$ then $x = \frac32$.” This sentence is true, but the phrases within it are not sentences and have no inherent truth or falsity. When we perform the same operation on each side of an equation, we are not only preserving the truth of the equal sign but also preserving the consequences of the equal sign. If we use if-then language when talking about equations, then we can make sense of equations with no solutions. A sentence like “If $x$ is a number satisfying $3x + 2 = 3x + 5$ then $2 = 5$” makes perfect sense. It’s the mathematical equivalent of “If the moon is green cheese, then I’m a monkey’s uncle.” It’s a way of saying the moon is not green cheese . . . or that there is no solution to the equation.

The middle schooler’s version of if-then language might not always use the words “if” and “then.” You might say “Imagine there is a number $x$ such that $3x + 2 = x + 5$. What can you say about $x$?” Just as you say “Imagine this hanger is balanced and the green triangles weigh one gram. How much do the blue squares weigh?” I think it’s a useful approach with students to remember that equations are a matter not just of truth, but of truth and consequences.

# Why is a negative times a negative a positive?

OK, I can hear the groans already. There are many contexts for answering this question and they are dubious in varying degrees because the real answer is “because I said so.” That is to say, the rule for multiplying negatives is a convention; adopted for good reasons, but a convention nonetheless. Those good reasons are mathematical: we want to make sure that when we extend multiplication and addition to negative numbers the properties of operations still apply. In particular, we want the distributive property to apply. Meditate on this:
$$3\cdot(5 + (-5)) = 3\cdot5 + 3 \cdot (-5).$$
The left side is really $3 \cdot 0$, so it had better be zero. So the right side had better be zero as well. The first term on the right side is 15, so the other term had better be $-15$. So $3 \cdot (-5) = -15$. We want the commutative law to hold, so we had better say $(-5)\cdot 3 = -15$ as well. Now meditate on
$$(-5)\cdot(3 + (-3)) = (-5)\cdot 3 + (-5)\cdot(-3).$$
The same reasoning tells us that $(-5)\cdot(-3) = 15$.