Julian's musings

Why is it that

\[\frac{3}{5}\div\frac{2}{3} = \frac{3}{5}\times\frac{3}{2},\]

or as the rule that students are frequently taught: “turn the second fraction upside-down and multiply”?

I’ve been inspired to revisit this question after listening to Ed Southall talking on Mr Barton’s Maths Podcast, where he mentioned this question.

In this post I suggest a teaching sequence which might lead to an understanding of the rule above, as well as a procedural knowledge of how to perform the rule.

Some comments on a familiar approach

I have seen textbooks and websites explain the rule for division of fractions by talking about how many times we can fit $\frac{1}{3}$ into $\frac{4}{5}$, say, but that seems to me to be quite challenging: students have to hold on to several ideas at once, and make sense of diagrammatic representations at the same time as trying to think about what division means. It also becomes very hard as the fractions become more complicated. In my experience, few students develop a solid understanding through this approach: they either get lost in the reasoning or they resort to following a rule.

This problem ties in quite neatly with some things I have recently read, in particular:

• James Tanton’s post The Unreasonableness of K-12 Mathematics, in which he gives an idealised description of the development of the concept of number.
• Liping Ma’s book “Knowing and teaching elementary mathematics”, in which US and Chinese teachers’ understanding of this rule is compared.
• John Mighton, the founder of JUMP Math, wrote The end of ignorance; he observes there that meaningful symbolic manipulation can precede both an attempt to explain an idea or technique in everyday terms, and the development of understanding; moreover, understanding can emerge from the manipulations if examples are well-chosen and students are given the opportunity to reflect.

An overview of the idea

The calculation $8-5$ means “what number $\square$ makes $\square+5=8$ true?” Similarly, when we write $12\div 3$, we mean “what number $\square$ makes $\square\times3=12$ true?” This says that division is the inverse of multiplication. (More precisely, for each non-zero number $c$, dividing by $c$ is the inverse of multiplying by $c$.) The same applies to division of fractions: $\frac{3}{5}\div\frac{2}{3}$ means “what number $\square$ makes $\square\times\frac{2}{3}=\frac{3}{5}$ true?”

Once we notice that $\frac{3}{2}\times\frac{2}{3}=1$, we can then multiply both sides of this equation by $\frac{3}{5}$ to obtain

\[\frac{3}{5}\times\frac{3}{2}\times\frac{2}{3}=\frac{3}{5}.\]

Therefore $\square$ must be $\frac{3}{5}\times\frac{3}{2}$, or

\[\frac{3}{5}\div\frac{2}{3} = \frac{3}{5}\times\frac{3}{2}.\]

This method will work for any fraction division question, and so these steps give us our familiar rule: “turn the divisor upside-down and multiply”.

A possible teaching sequence

What follows is a suggestion for how these ideas could be introduced over a sequence of lessons, which could span several months or even years. This offers students the chance to revisit the ideas again and again, thereby reinforcing them, as well as building up stronger connections and a deeper understanding. In the later steps, I assume that students can multiply fractions.

All of the questions below are available in this Word document.

Step 1: What is subtraction?

We begin by asking students what other number statements they can deduce from $3+5=8$. There are many possible answers (such as $30+50=80$), and here we highlight those obtained by rearranging the numbers. (These could be encouraged by a question such as “Using only the numbers 3, 5 and 8, what other number statements can you get from $3+5=8$?”) Three key statements are:

\[8-3=5; \qquad 8-5=3; \qquad 5+3=8\]

as well as the same statements written the other way round, such as $5=8-3$; we won’t mention these reversed statements again here.

The last of these three statements says that addition is commutative: the order of adding does not matter. The other two say that subtraction is the inverse of addition: the three problems

\[5+\square=8,\qquad \square+5=8 \qquad \text{and} \qquad 8-5=\square\]

are equivalent, as are similiar problems about $8-3=\square$. Making this connection explicit would be beneficial, especially in relation to the later parts of this sequence of steps.

Students could then be asked to write statements equivalent to statements such as $10-3=\square$ to reinforce this idea.

This idea may well have already been introduced via a bar model approach or using Cuisenaire rods or suchlike.

It is useful to recognise that it doesn’t matter whether we are working with whole numbers, directed numbers, fractions or whatever: subtraction always has this meaning, so returning to this idea periodically will benefit students’ understanding.

Step 2: And what is division?

This is the parallel of Step 1 for multiplication and division. What can be deduced from $3\times4=12$? This again leads to interesting points such as why $30\times40=120$ is an incorrect statement, whereas $30+50=80$ is correct. But for our current purposes, the key deductions are again those obtained by rearrangement:

\[12\div4=3; \qquad 12\div3=4; \qquad 4\times3=12.\]

As before, we see that multiplication is commutative and that division is the inverse of multiplication. In particular, this means that answering the question $12\div4=\square$ is the same as filling in the missing number in $4\times\square=12$; asking students to make deductions from $12\div4=\square$, as above, will reinforce this idea.

Step 3: 1 divided by a unit fraction

A key part of this approach is to learn about reciprocals of fractions. We start with the reciprocals of unit fractions.

For this missing-number problem, I would suggest asking students to work on this themselves rather than showing them how to do the first one. (I am assuming that they already know enough about fractions to work out the answers to these questions.)

\[\begin{align*} \frac{1}{2}\times \square &= 1 \\ \frac{1}{3}\times \square &= 1 \\ \frac{1}{4}\times \square &= 1 \end{align*}\]

Students should spot the pattern. Following this by asking questions such as $\frac{1}{82}\times\square = 1$ can help them to realise that they can now do some very complicated-sounding questions, even if they can’t imagine what $\frac{1}{82}$ of a cake might look like. (I was reminded of this approach by John Mighton’s book.)

Students should then connect this back to the earlier steps, by asking them to rearrange $\frac{1}{2}\times2=1$. This will allow students to (re)discover that $1\div 2=\frac{1}{2}$ (and similarly for the other statements); this can also be used to reinforce the idea that a fraction such as $\frac{1}{2}$ just means “1 divided by 2”. (The division symbol itself suggests this: $\div$ is just a fraction with dots in place of actual numbers.) Another way of rearranging the number statement gives $1\div\frac{1}{2}=2$, which could be related to the “practical” meaning of division: there are 2 halves in a whole.

Step 4: Turning a general fraction into an integer

It might be too big of a jump for some students to go straight to finding the reciprocal of a general fraction, so this step provides a structured intermediate step, once they are developing some confidence with the above idea.

Here is a second sequence of missing-number problems:

\[\begin{align*} \frac{2}{3}\times \square &= 2 \\ \frac{2}{5}\times \square &= 2 \\ \frac{3}{5}\times \square &= 3 \end{align*}\]

Once students have worked out answers to these (and perhaps adding a few more similar examples), either ask them to generalise by making up their own similar examples, or ask superficially harder questions such as $\frac{74}{133} \times \square = 74$, so that the structure becomes clear.

Asking students to rearrange these statements once again results in statements like $2\div3 = \frac{2}{3}$ (further reinforcing the division idea) and $2\div \frac{2}{3} = 3$.

Step 5: Finding reciprocals

A useful preparatory question before this step would be something like: “If you know that $96\times 48=4608$, then what is the missing number in $96\times \square = 2304$?” This recalls the idea that we can divide the product by 2 by dividing the multiplicand (or multiplier) by 2. (The use of two-digit numbers is designed to discourage students from doing a division!)

In this step, we replace the integers on the right-hand sides of the previous set of questions with 1:

\[\begin{align*} \frac{2}{3}\times \square &= 1 \\ \frac{2}{5}\times \square &= 1 \\ \frac{3}{5}\times \square &= 1 \end{align*}\]

If students cannot work out how to answer the first question, it would be helpful to remind them of their answer to $\frac{2}{3}\times \square = 2$. Tying this to the preparatory question above should help them get to the answer.

Again, students can be invited to generalise at this point, or to answer a question like the one in the previous step: $\frac{74}{133} \times \square = 1$. Also, it is helpful to then rearrange these results; we have $1\div \frac{2}{3} = \frac{3}{2}$, and we are seeing the first clear case of turning fractions upside-down.

After these, it could be interesting to also revisit unit fractions: following the same pattern that we have seen, how else could the answer to $\frac{1}{3}\times \square = 1$ be written, besides as $3$?

Step 6: Dividing fractions

Before working on the full-blown division of fractions, it would be useful to preface it by another relevant rearranging activity: how can the number statement $2\times 3\times 4=24$ be rearranged, while keeping all of the numbers involved the same? This gives rise to a number of statements, such as:

\[24\div 4 = 2\times 3; \qquad \frac{24}{2\times 3}=4; \qquad 4\times 2\times 3 = 24.\]

This may cause some difficulty and lead to some interesting class discussions.

And now we can build on the ideas developed in Step 5. How could we complete the following statements?

\[\begin{align*} \frac{2}{3}\times \frac{3}{2} \times \square &= \frac{4}{5} \\ \frac{2}{5}\times \frac{5}{2} \times \square &= \frac{1}{3} \\ \frac{3}{5}\times \frac{5}{3} \times \square &= \frac{7}{2} \\ \frac{1}{3}\times \frac{3}{1} \times \square &= \frac{3}{4} \\ 2\times \frac{1}{2} \times \square &= \frac{1}{5} \end{align*}\]

A prompting question, if needed, is “What is $\frac{2}{3}\times \frac{3}{2}$?”

And then what about these, where the two squares should be filled in following the pattern we have just seen?

\[\begin{align*} \frac{3}{4}\times \square \times \square &= \frac{2}{5} \\ \frac{2}{7}\times \square \times \square &= \frac{4}{3} \\ \frac{1}{5}\times \square \times \square &= \frac{3}{2} \\ 4\times \square \times \square &= \frac{3}{2} \end{align*}\]

Once students feel competent at these, ask how they can use these to work out:

\[\begin{align*} \frac{2}{5} &\div \frac{3}{4} \\ \frac{4}{3} &\div \frac{2}{7} \\ \frac{3}{2} &\div \frac{1}{5} \\ \frac{3}{2} &\div 4 \end{align*}\]

And with this, students have reached a point where the rule for dividing by a fraction will make some sense: we multiply the reciprocal of the divisor (so as to get 1 when it is multiplied by the divisor itself) by the dividend, which is our well-known rule.