How to explain the flipping of division by a fraction?

Question

This question is inspired by @DavidButlerUofA's discussion of "$\div \frac{2}{3}$ as $\times \frac{3}{2}$" in "Are fractions hard because they are like algebra?"

Q. How can one best convey to beginners—without algebra—the flipping of denominator fractions, say, $\frac{4}{\frac{2}{7}} = 4 \times \frac{7}{2} = 14$? In other words, what would convince a novice of the multiplication of the numerator by $7$?

The image below is my attempt, but I doubt this is convincing (the main insight being that one should partition the numerator $4$ into $7$ths—$28$ $7$ths—and then it is clear that those $28$ $7$ths contain $14$ $\frac{2}{7}$ths). Those with experience teaching fractions likely have better pedagogical routes.

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Answer 1

I'm going to focus on one aspect of the question that I think has not been fully appreciated:

How can one best convey to beginners—without algebra—the flipping of denominator fractions... what would convince a novice

Many of the ingredients of this answer are already present in some of the other answers to this question, but are rearranged here in a fashion that (I hope) is significantly novel.

What do beginners know?

Multiplication and division of fractions is, in the United States at least, typically first encountered in either third or fourth grade; that is, at age 8 or 9. Benjamin Dickman, in his answer, has discussed various representations of division (partitive/equal sharing, quotitive/measurement, missing factor). I want to take a minute to talk about representations of fractions.

Given the age of the students in question, I do not think there is much value in formal explanations, i.e. explanations that rely on symbolic manipulation, whether or not they are "algebraic" in nature. They may learn that the algorithm is the correct one, but I don't think they will be convinced by it, in the sense of believing that the algorithm has to be that way and makes sense. To convince someone that $4 \div \frac27 = \frac{4 \times 7}2$ you have to guide them to figure it out for themselves, in a way that makes the result seem completely inevitable and unsurprising. And to do that you have to try to see things from the perspective of the learner.

The fractions students most often (perhaps exclusively?) encounter are proper fractions, i.e. fractions $< 1$; quantities larger than $1$ are usually expressed as mixed numbers rather than as "improper" fractions. Students at that level typically understand fractions in the language of part/whole relationships; the most commonly used representations are:

1. a circle, divided into $n$ sectors of equal area, with $k<n$ of them shaded (corresponding to the "real-world" context of cutting up a pizza or other circular food);
2. a collection of $n$ objects (often drawn as a group of small circles contained in a rectangular frame), with $k<n$ of them shaded (corresponding to the "real-world" context of sharing a box of cookies or cupcakes);
3. a collection of $k$ units of measurement (often drawn as small scoops or cups), drawn with some kind of scale or labeling that indicates that $n$ of those units would comprise a single larger unit (corresponding to the "real-world" context of using, e.g., between three "quarter-cup scoops of flour" to measure $3/4$ of a cup of flour).

The order in which I list the representations above is not accidental; it corresponds roughly to the order in which these representations actually occur in the classroom. That is, the most commonly used representation is the "circle cut into wedges", and the least commonly used representation is the "small scoops that make up one big scoop". Since the question is about conviction, I think it is important to realize that not all representations are equally valuable for that purpose. In particular, I think the third representation above, combined with the measurement interpretation of division, is actually the most useful for the purpose of explaining how division of fractions works.

With all of that established as preamble, here is my strategy for convincing a novice that division of fractions works the way it ought to:

1. Announce the problem, but do not ask them to try to solve it yet.

Write the problem $6 \div \frac23 = ?$ in large letters on a piece of paper and tell the student that we are going to figure this out, but we're going to work out way up to it by warming up with a few simpler problems first.

2. Begin by emphasizing the measurement interpretation of division.

Since most students reflexively think of division in terms of equal-sharing, it is a good idea to start by explicitly activating the measurement interpretation, which is less common. Ask:

Suppose you want to measure six cups of flour, and all you have is a two-cup scoop. How many scoops do you need?

Most 8- to 9-year-old students will be able to answer "Three" immediately. To do so, they do not need to explicitly translate the problem into "Six divided by two equals what?", nor as "Two times what equals six?" In fact, making such translations may seem like an unnatural complication to a simple problem of counting. So you have to ask them to make that translation explicit, with prompts like:

• How did you know that?
• What kind of arithmetic operation are you using? (Many will say "addition", because they are simply thinking "two plus two plus two".)
• What other kinds of arithmetic can you use to describe what you just did?

This conversation does not have to be a long one, but it does need to happen, and the goal is to activate (and keep active in the student's mind) that "$6 \div 2$" can be understood as the question "How many 2-cup scoops do you need to make 6 cups?". Now you are ready to move to the next step:

3. Consider divisors that are unit fractions.

Now we want to vary the task, just slightly:

Suppose you want to measure six cups of flour, and all you have is a $1/3$ cup scoop. How many scoops do you need?

Again most students (at this age level) will answer "18" almost immediately, at least if they have some degree of automaticity with the multiplication fact $6 \times 3 = 18$. Students who do not have such automaticity may need to count by threes. Either way they will get the answer very quickly. Ask them how they know. Guide the conversation to the following, very important summary:

Because the scoops only hold $1/3$ cup, you need three scoops for each cup, so you have to multiply three times the number of cups you need.

To the extent possible, try to get the student to be the one who says this, or something like it. Don't say it for them, but do revoice their statement of it to make it more succinct and coherent, if necessary. Once students agree with this basic idea (and once said it usually seems completely obvious to them, so much so that they wonder why you were making such a big deal about it), you can move on to the final variation:

4. Consider divisors that are not unit fractions.

One last tweak to the task:

Suppose you need 6 cups of flour, and all you have is a $2/3$ cup scoop. How many scoops do you need?

Some students will immediately try to do some kind of paper-and-pencil computation, whether they know an algorithm or not. Discourage this. Ask them instead to just think about the previous problem. We already know that it will take eighteen $1/3$ cup scoops to make six cups. What if the scoops are $2/3$ cups instead? If they are still stuck, prompt: How do the new scoops compare to the old scoops? Usually students will eventually come up with an answer like this one:

Because the scoops are twice as large, we only need half as many of them, so we need just 9 scoops instead of 18.

Once they have said this, or something like it -- but not before! -- write down "$ 6 \div \frac23 = 9$".

We are not done yet, though. The most important step is the next one:

5. Summarize and generalize

It's time to look back at the expression $6 \div \frac23$ and think about what happened when we tried to solve it. Notice that there are two basic principles interacting here:

• The 3 in the denominator of the divisor acted by multiplication, because it takes three $1/3$ cup scoops to fill a single cup, so if you are using scoops of that size, you need three times as many scoops as you do cups.
• The 2 in the numerator of the divisor acted by division, because a $2/3$ cup scoop is twice as large as a $1/3$ cup scoop, so you need half as many of the larger scoops as you do of the smaller ones. In other words:

To find $6 \div \frac23$, you multiply by $3$ and divide by $2$.

Or, transcribed symbolically,

To find $6 \div \frac23$, you can compute $\frac{6 \times 3}2$

Ask some questions to get the student to restate this result in their own words:

• Where does the numerator of the $2/3$ go? (Into the denominator of the answer.)
• Where does the denominator of the $2/3$ go? (Into the numerator of the answer.)

At this point, the "rule" that "To divide by a fraction, you invert the divisor and multiply" should not seem like a "rule", but merely a summary of something that should have been obvious all along.

Final Thoughts

I've gone into a lot of detail about this, and I think it's worth pointing out that it probably takes more time to read the above description than to implement it. I've tutored dozens of kids in precisely this manner; some of them are elementary school kids learning fractions for the first time, others are high schoolers who "learned" how to divide fractions back when they were 8 or 9 but no longer remember what to do, or remember it imperfectly. I've had very good success with this method. If working with a student one-on-one, it usually takes no more than five minutes, start to finish. I think there are two reasons why it is effective:

First, it begins by announcing the problem, and then immediately putting the problem on hold and instead considering simpler problems: first division by whole numbers, and then division by unit fractions. This models an important problem-solving heuristic: When you encounter a hard problem, consider a simpler one and see if you can get any insight from it.

Second, it concludes by looking back at a single example and trying to understand the general principles that make it work. This models a second important problem-solving heuristic: When you have solved a hard problem, take a moment to look back at it and see if the perspective of hindsight reveals any general arguments to you.

Both of these heuristics are, of course, taken directly from Polya's How To Solve It. And while these heuristics are common to most problem-solving contexts, and may even be naturally-occurring for many students, it is worth recognizing that conventional school instruction (which I would caricature as "Teach the rule, then do examples, then have students do many exercises, then provide an explanation of the rule") does not provide a lot of space for this kind of slow, reflective consideration.

One final thought: You may have noticed that the OP's example was $4 \div \frac27$, and I changed that for purposes of this explanation to $6 \div \frac23$. That change was not accidental. First of all, if you want to follow a narrative like the one I sketch above, asking about scoops that hold $1/7$ of a cup is just silly. Nobody makes or uses measuring scoops like that; the artificiality of the problem stands out and is distracting. On the other hand, $1/3$ cup measuring scoops are fairly commonplace (at least in the United States). I imagine that this whole instructional sequence would have to be reconsidered from the ground up (and may be completely unworkable) if one were teaching in a context in which the metric system is consistently used. The reason for changing the $4$ in the original problem to a $6$ is that the dual roles played by $2$ in the equation $4 \div 2 = 2$ can actually be confusing. Which $2$ represents the size of groups, and which represents the number of groups? Changing the $4$ to a $6$ eliminates that symmetry, and hence the ambiguity about which number represents which quantity.

Addendum, added September 18, 2016 This afternoon I received an email from a middle school student who had found this answer online and was hoping for some further explanation. Specifically, she wanted an explanation of how to think about division when both the dividend and the divisor are fractions (in the example above, the dividend was a whole number).

First let me say that I think the existence of the email itself is proof that what seems obvious to the experienced person may not at all be obvious to the novice. My first instinct in reading the email was to respond "Well, it's exactly the same!" -- but of course if it were exactly the same then there would have been no need for that email, would there?

So let's consider a new example: Suppose we want to find $\frac{6}{10} \div \frac{2}{3}$. Here is the essential idea that you need in order to generalize the work that was already done:

$\frac{6}{10}$ is $\frac{1}{10}$ of $6$.

Let's see how that is useful here. The problem $\frac{6}{10} \div \frac{2}{3}$ can be interpreted as meaning:

If you want $\frac{6}{10}$ of a cup of flour, and you only have a $\frac{2}{3}$ cup scoop to measure it with, how many scoops do you need?

Now we have already solved the problem of measuring out $6$ cups of flour with a $\frac{2}{3}$ cup scoop, and found that we need $9$ scoops to do it. Our new problem asks us to measure out $1/10$ as much flour total, so we need $1/10$ as many scoops. So the answer is $9/10$.

Okay, now let's look back and think about what we did. We took the numerator of the first fraction ($6$) and multiplied it by the denominator of the second fraction ($3$), then divided by the numerator of the second fraction ($2$). This gave us $9$. Finally we divided by the denominator of the first fraction ($10$) to get the answer, $9/10$.

In abstract form, the result of dividing $\frac{a}{b}$ by $\frac{c}{d}$ is $\frac{a \cdot d}{b \cdot c}$. Another way of saying that is that we take the second fraction, $\frac{c}{d}$, and "flip it" to get $\frac{d}{c}$ ; then we multiply the first fraction by the upside-down second fraction. In our example, we have:

$\frac{6}{10} \div \frac{2}{3} = \frac{6}{10} \times \frac{3}{2} = \frac{6 \times 3}{10 \times 2} = \frac{18}{20}$, which can be reduced to $\frac{9}{10}$.

Answer 2

Here are a few comments, and then an attempt at two succinct answers.

In particular, I will try to answer this using a measurement interpretation, and then again with an equal sharing interpretation. I prefer the former, but include the latter for completeness.

Comments:

Some of the key terms in unpacking this are measurement, equal sharing, and missing factors, which are (up to some name-changing) the standard three interpretations of division of whole numbers, and which can be extended to discuss the division of rational numbers, as well.

I have written a bit about each of them in MESE 5648, and, as a comment on an answer from the question that inspired this one, I included a picture of $2 \div \frac{2}{3}$ using the former two interpretations:

In the measurement picture, the dividend ($2$) refers to the total number of items, the divisor ($\frac{2}{3}$) refers to the size of each single group, and the quotient ($3$) refers to the total number of groups.

In the equal sharing picture, the meanings of the divisor and quotient switch; thus, depicted is $\frac{2}{3}$ of a group, and the quotient will be the size of a single group (drawn in with dotted lines).

Attempt at a succinct "measurement" answer:

Two additional important terms are partitioning and chunking. To speak concretely, when we divide $4$ by $\frac{2}{7}$, let us use a length model (e.g., a line segment of length $4$). We wish to see how many $\frac{2}{7}$s we can measure out from this line segment.

To do so, we begin with the simpler task of how many $\frac{1}{7}$s can be measured out from the length $4$ segment.

This is most easily done by partitioning the $4$ unit intervals, each, into $7$ pieces of equal length. Note that the effect of this partitioning is that our number of intervals has been multiplied by $7$. ($\star$)

(Ensure students have ample practice dividing whole numbers by unit fractions before moving on!)

Now we move on to the case with $\frac{2}{7}$. Each of these pieces to be measured out can be found by taking our partitioned segment and its $4 \times 7 = 28$ intervals and chunking them by combining them $2$ at a time. We do this because they were of length $\frac{1}{7}$, and we want them to be of length $\frac{2}{7}$.

What is the effect of this chunking? Well, we are making each measured out piece $2$ times as long, which means the total number we can fit in will be only $\frac{1}{2}$ of what we had before, i.e., we divide the total number by $2$. ($\star$)

Finally, let us look at the two starred items from above:

The partitioning led us to multiply $4$ by $7$, and the chunking led us to divide that result by $2$. In particular, we end up with $4 \times \frac{7}{2}$ as desired.

Attempt at a succinct "equal sharing" answer:

With equal sharing, we interpret $4 \div \frac{2}{7}$ by equal sharing the dividend ($4$) among the divisor ($\frac{2}{7}$) groups, and take the quotient to be the amount in one full group.

First, let us illustrate what $\frac{2}{7}$ of a group looks like:

We want to equally share our $4$ within this $\frac{2}{7}$ of a group; as apparent in the shaded illustration above, this means distributing across the $2$ components, hence dividing $4$ by $2$ to alter our illustration as follows:

Again, the quotient is the amount in one full group; this will lead us to take our $4$ divided by $2$ (alternatively written as: $4 \times \frac{1}{2}$), and put this amount in each of the $\frac{1}{7}$ pieces for the full group:

We now see that we have multiplied the $4 \times \frac{1}{2}$ by $7$ to find the quotient.

In particular, we end up with $4 \times \frac{1}{2} \times 7 = 4 \times \frac{7}{2}$ as desired.

Post-Script:

With respect to contrasting the two interpretations, the dividend is dealt with in different ways: The measurement approach multiplies by $7$ then divides by $2$; the equal sharing approach divides by $2$ then multiplies by $7$. In each interpretation, the division by $\frac{2}{7}$ is equivalent to multiplication by $\frac{7}{2}$.

More generally: Making sense of the invert-and-multiply algorithm is very difficult, and may require a fair bit of scaffolding. Certainly it cannot be done justice in this one brief response. Instead, I hope you will consider the above as a sketch of how the topic might be broached. For students really to encapsulate this idea will likely take a fair amount of time, and probably experience with standard division (and its interpretations), whole numbers divided by unit fractions, whole numbers divided by non-unit fractions, and so forth.

Answer 3

I have no experience teaching fractions, but I think moving away from using the divide symbol makes things easier. It doesn't get used at university level (but exponents start being used, so there are still two notations).

I would do

$$3\div\frac{2}{7} = \frac{3}{\frac{2}{7}} = \frac{3}{\frac{2}{7}} \times \frac{7}{7} = \frac{3\times 7}{\frac{2}{7}\times7} = \frac{3 \times 7}{2} = 3 \times \frac{7}{2},$$

since you are assuming familiarity with multiplication of fractions. (Using $3$ instead of $4$ stops the point getting lost along the way.)

Answer 4

When manipulating fractions, students quickly get comfortable with the idea that to combine two fractions they have to manipulate to get the denominators the same. Multiplying by 2/2 or 3/3, etc doesn't change the fraction value, obviously. I use the method below to unfraction (<< is there a word for this?) the denominator -

$$\frac{3}{\frac{2}{7}}=\frac{3\times \frac{7}{2}}{\frac{2}{7}\times \frac{7}{2}}$$

"what do we multiply a number by to get 1?" Its reciprocal, of course. So we multiply numerator and denominator by the reciprocal of that denominator. Some students will be stuck on this point and not really mature to the next step, unfortunately. The ones who quickly get comfortable with this will see the step of showing the denominator multiplied by the reciprocal can be skipped in favor of just using that number (the reciprocal) to multiply the numerator.

Answer 5

I suspect the main trick will be going from division by unit fractions to division by non-unit fractions. I would begin by considering that you have 4 groups of a certain size, and you want to know how many groups you'll have if you make a new set of groups $\frac{1}{7}$ the size of the current groups. This can be simplified further by considering each group on its own, and then it becomes obvious why division by unit fractions results in multiplication. Once the unit fraction idea is developed, split it into two steps: finding how many groups there will be at $\frac{1}{7}$ the size of the current groups, and then how many groups at double the size of the new groups.

Answer 6

I was surprised to see such an old question with so many answers, all of them algebraic. Here is a proof (almost) without words that $1/\frac{a}{b} = \frac{b}{a}$, which I believe is the crux of the question. It uses the definition of multiplication as the area of a rectangle with the given side lengths.

Answer 7

If you have some apples and chop them into seven pieces, how many pieces do you have? Well, seven times as many as you had apples, of course! What if you pair those pieces two-by-two? Well, okay, you now have half as many pairs.

("X divided by Y" means "how many Ys can you get out of what you started with".)

(Did it matter that we used seven? Well, no, it could have been anything. That we used two? No, could have been any other thing. Yay, it always works! And we're doing algebra!)

Answer 8

This might be overly practical, or it might be overly algebraic, but you could maybe teach them that that's all division ever is.

First, you could show that $ 4 = \frac{4}{1} $, because any number divided by 1 is the same.

Then you could say that $\frac{2}{3} = \frac{2}{1} \times \frac{1}{3} = \frac{2}{1} \div \frac{3}{1} = 2 \div 3$

In this sense you don't teach them that they can flip a fraction to divide, so much as teach them that division is a convenience term we use for multiplying by the flip.

If you start there, then when you go to $\frac{2}{3} \div \frac{7}{4}$ it becomes more obvious that we just do the same thing to get $\frac{2}{3} \times \frac{4}{7}$, because it's what we already did for $\frac{1}{3}$ and $3$.

It's not a technique, it's a definition. That might not help, I'm not sure.

Answer 9

So far no-one has mentioned the double number line representation of division. I think this works best in an "equal sharing" context, so I'll add such an example.

Question: If I travel (with my bike) $14$ kilometers in $7/12$ of an hour, what is my average speed?

Answer: We can use proportional reasoning to figure out how many kilometers in one hour this corresponds to. The situation is represented on the double number line below.

It's not easy to see the answer directly. As an intermediate step we divide by $7$ and find the distance traveled in $1/12$ of an hour.

After that we can multiply by $12$,

and get the answer $\boxed {24}$ kilometers per hour.

Comment: What we have done is of course the division $14 \div \tfrac 7 {12}$ and found it to be equivalent to "$14$ divided by $7$ and then multiplied by $12$".

Answer 10

Assuming that you've already covered

• Multiplication of fractions,
• Cancellation, and
• The idea that $a\div b$ means the solution to $bx=a$ (though not necessarily expressed symbolically; it could also be in words such as "the number that we can multiply by the divisor to get the dividend", which ought to be justifiable by reference to dividing integers into equal parts),

I think a viable way would be to present flip-and-multiply not as a rule that needs independent justification, but as a trick that happens to work for solving $bx=a$ when $a$ and $b$ are fractions:

If we want to divide $\frac38$ by $\frac25$, we're looking for a number that becomes $\frac38$ when we multiply it by $\frac25$. We can manufacture such a number by multiplying the $3$ by $5$ and the $8$ by $2$, giving $\frac{3\times 5}{8\times 2}=\frac{15}{16}$. This works because if we then multiply by $\frac25$ we get $\frac{15\times 2}{16\times 5}=\frac{30}{80}=\frac{3}{8}$.

Does this trick always work? Yes, because notice what happened to the $3$ and $8$ on the way to $\frac{30}{80}$. The $3$ got multiplied first by $5$ and then by $2$, and the $8$ got multiplied first by $2$ and then by $5$. But since the order of factors are immaterial, this means effectively both the $3$ and the $8$ got multiplied by $2\times 5$, which we're then allowed to cancel. So we do always get the original dividend back in this way.

A convenient way to remember this trick is that dividing by $\frac25$ gives the same result as multiplying by $\frac52$.

(Or in other words: It's not a definition, it's a technique).

$\frac52$ and $\frac25$ here are useful example numbers here because the eventual cancellation is then directly visible.

Answer 11

If you are looking for a concrete example, why not try rates. Namely, suppose Jimmy can drink $4$ cups of soda every $2/7$s of a second. How much soda is that per second? Well, by the time $7$ of the $2/7$s of a second have passed, he would have drunk $28$ cups. But this amount of time is $2$ seconds, so in one second he drank $14$ cups. It may help to use a clock or something for this.

Let's abstract away the specific numbers first. If you drink $x$ cups every $a/b$ time units, you can drink $b*x$ cups in $a$ time units. If the students have not yet studied algebra, demonstrate with various $x$, $a$, and $b$.

Now abstract away rates. For any situation described by multiplication, division and fractions, $\frac x{\frac ab} = x \cdot \frac ba$. See if the students can abstract this to situations such as filling a rectangle with $4$ units of water, when the base is $2/7$, and finding the height.

Finally, if you have enough time, abstract away from integers and rational numbers and discuss turning arbitrary integral domains into fields of fractions and beyond.

Answer 12

I'll mention a very simple idea which, it seems, hasn't come yet: get some intuition from addition and subtraction, to which multiplication and division are analogous. I'll run through it with the divide-by-$2/3$ example asked about.

We can intuitively see that subtracting $2-3$ is the same as adding $3-2$, because subtraction reverses addition. One needn't go into negative numbers with such an explanation as, "to subtract an amount means to add its additive inverse, which for a difference of two numbers is that difference backwards". Instead one can get by with noting, "the answer must be $3$ more than if I'd subtracted $2$, so may as well add $3$ before the subtraction, which means I'm adding $3-2$".

But however you explain it, the result is clear, exchanging the numbers we're subtracting (reverse-adding, if you'll forgive the Newspeak) has the effect of adding the post-exchange difference. Well, division reverses multiplication, so by the same logic dividing by $2/3$ should mean multiplying by $3/2$.

Answer 13

I would start by demonstrating, with integers, that x / y = x * 1/y.

The technique for dividing by a fraction is based on that.

Answer 14

Below are a various methods of presenting fraction inversion based on various innate symmetries. Unlike some other methods, these ideas are more to the heart of the matter, so they generalize more nicely. The exposition is at the level of the teacher - who can then decide how to best scale the ideas to the appropriate level - be it primary, secondary or higher level.

1. Inversion as a special case of fraction scaling symmetry

We can invert fractions using the fact that they are preserved under equal scalings $s\neq 0$ of the numerator and denominator, i.e $\,\dfrac{a}b = \dfrac{as}{bs}.\,$ So $\,\dfrac{1}{2/3} = \dfrac{3}2\,$ follows by scaling by $3$ to (remove the denominator $3$ in the denominator). Viewed geometrically, in terms of similar triangles we have

$\qquad\qquad\qquad\qquad\quad$

by scaling the smaller triangle by $3$ to cause its magnified base $\color{#c00}{2/3}$ to have integer length $2.$

Of course the same scaling method works for your example $\,\dfrac{4}{2/7}\,$ by scaling by $s = 7$.

So once students know the fundamental scaling symmetry of fractions, the "inversion by flipping" reflection follows as a special case. If the class already knows some algebra then it is well worth emphasizing that the scaling symmetry of fractions is just a reformulation in "root language" of the scaling symmetry of the associated linear equation of which they are roots, i.e.

$$ x= \dfrac{a}b\iff b\,x = a\iff bs\, x = as\iff x = \dfrac{as}{bs},\quad {\rm f or}\ \ s\neq 0$$

This is a special (linear) case of the obvious fact that scaling a polynomial by a scalar $s\neq 0$ does not alter its roots $\,f(r)=0\iff s f(r) =0$ . For another concrete example let's consider the quadratic case. Here the "root language" analog of the above linear case $\,a/b = as/(bs)\,$ is that replacing $\,a,b,c\,$ by $\,as,bs,cs\,$ in the quadratic formula does not alter the set of roots.

2. Inverse of algebraic number by reverse-reflecting a polynomial of which it is a root

Given a linear polynomial $f(x)$ having a nonzero fraction as a root, we can scale $f$ by $x^{-1}$ to get a linear polynomial having $x^{-1}$ as a root, namely

$$ x= \dfrac{a}b\iff b\,x = a\iff ax^{-1} = b\iff x^{-1} = \dfrac{b}a$$

This generalizes to higher degrees polynomials too: $x\neq 0$ is a root of a polynomial $f(x)$ of degree $n$ $\iff x^{-1}$ is a root of $\,x^{-n} f(x)\, =: \bar f(x^{-1}),\,$ e.g. for quadratics

$$ ax^2 + bx+c = 0\iff c x^{-2} + b x^{-1} + a = 0$$

Notice that the map $\,f\mapsto \bar f$ simply reverses the coefficient sequence

$$ a,b,c\ \mapsto\ c,b,a\quad$$

Thus the flipping in fraction inversion is just a special case of the above general reversal reflection.

$\bar f(x)$ is called the reciprocal polynomial (or reflected or reversed). It arises arises frequently in algebra

3. Symmetries of equal fractions via symmetries of a square

Inversion can also be be seen as a special case of the symmetries of equal fractions, which can be viewed as symmetries of the square

$$\dfrac{A}B = \dfrac{a}b\iff \dfrac{B}A = \dfrac{b}a\iff \dfrac{B}b = \dfrac{A}a$$

which hold true because these reflections and rotations preserve the diagonal cross-products, hence preserve fraction equality (see here for further discussion). The special case $B=1$ above yields inversion of $\,A = a/b\,$ by flipping (reflection around the $x$-axis).

Answer 15

I think working with a slough of examples may be effective. I would start with patterns of whole numbers and then fractions n divided by n , then n divided by 1 and 1 divided by n. Then try a whole number n divided by another whole number m and then contrast with m divided by n. Note that the relation a / b = a * (1/b) holds throughout. As a result, something divided by m/n is the same as something multiplied by n/m.

Gerhard "Then They Remember Through Example" Paseman, 2015.04.11

Answer 16

As I said in my comments to the OP, I think this is just a bit of a tricky topic. A little like quantum mechanics, where you sort of have to get used to it, versus understand it immediately like kinematics. So hoping for some secret key to unleash their groking may not work (for this item, other topics there may be a nice mental key).

A couple practical ideas.

a. Have the student just use the technique, after learning it, on several simple problems. Perhaps even simple enough so that they could guess the answer without any manipulations. Seeing "that it worked" even if not sure why may be helpful motivation to accepting the technique.

b. Play it up a little as some secret sauce. "Here's a sneaky trick." Why are emotional, human creatures, not only logic processors. Feeling like you got the magic password is motivating.

Answer 17

$\times n$ is an action. It moves each point on the number line to a different point. $\div n$ is the opposite action, meaning: "$\times n$ then $\div n$" doesn't move any points at all. It's the same as doing nothing.

Let's start with an action and rewrite it in several steps:

$\div \frac{a}{b}$

the opposite of $\times \frac{a}{b}$

the opposite of "$\times a$ then $\div b$"

$\times b$ then $\div a$

$\times \frac{b}{a}$

Answer 18

I recently tried to explain the invert-and-multiply process to a handful of 5th graders and I was surprised at how it went. I'm not their teacher so I don't really know what they do in class, but I started by asking them to compute $\frac{15}{8}\div \frac{3}{2}$. I expected them to know how to invert and multiply and to use that process to get the answer. Then I planned to confuse them by asking ``Why not just divide across? Look, $15\div 3=5$ and $8\div 2=4$! Magic, huh? So how come your teachers didn't teach it that way?'' So imagine my surprise when they did divide across and they didn't know how to invert and multiply! (I should email their teachers and see if this is actually what they're supposed to know or if they had forgotten some things.) Of course, if they're comfortable with multiplying fractions (perhaps by using some visual representation like in the question) and the fact that $a/b=c$ means $a=bc$, it makes sense that they would know you can divide across. If you can get them to this point, the rest isn't so bad.

What I did next was ask them something like $\frac{15}{7}\div \frac{3}{2}$, and then mention that it sure would be nice if the left denominator were a multiple of $2$ and we could divide across, so we wrote it as $\frac{30}{14}\div \frac{3}{2}$ and did just that, getting $\frac{10}{7}$. I made no attempt to write it as $\frac{15}{7}\times \frac{2}{3}$ because that's too much information at once. This was just to show them that even when it initially looks like we can't divide across, sometimes (always?) if we're clever we can. I did several more examples like this, where only one numerator or denominator was a multiple of the other one.

Then I jumped to something like $\frac{13}{7}\div \frac{3}{2}$ and as usual I made, say, the numerators nice: $$\frac{13}{7}\div \frac{3}{2}=\frac{13\times 3}{7\times 3}\div \frac{3}{2}$$ All this time I kept telling them to resist their urge to immediately compute $13\times 3$ and $7\times 3$, and by now I think they were starting to understand that it's much easier to leave it in the above form. The next step was to get the left denominator to be a multiple of $2$, which by now was easy. $$\frac{13}{7}\div \frac{3}{2}=\frac{13\times 3}{7\times 3}\div \frac{3}{2}=\frac{13\times 3\times 2}{7\times 3\times 2}\div \frac{3}{2}$$ Dividing across gives $\frac{13\times 2}{7\times 3}=\frac{26}{21}$. Only then did I call attention to the fact that $\frac{13\times 2}{7\times 3}=\frac{13}{7}\times\frac{2}{3}$, and isn't it funny that the second fraction got flipped and the division turned into a multiplication? Then we went through all the previous examples to see if that funny thing happened in all of them too.

So it didn't go the way I anticipated, but I was happy with the result.