Wiggins questions #3

Question 3

You are told to “invert and multiply” to solve division problems with fractions. But why does it work? Prove it.

Oh dear. If anyone on the Internet has had more to say about dividing fractions than I have, I am unaware of who that is. (And, for the record, I would like to buy that person an adult beverage!)

Unlike the division by zero stuff from question 1, this question is better tackled with informal notions than with formalities. The formalities leave one feeling cold and empty, for they don’t answer the conceptual why. The formalities will invoke the associative property of multiplication, the definition of reciprocal, inverse and the multiplicative identity, et cetera.

The conceptual why—for many of us—lies in thinking about fractions as operators, and in thinking about a particular meaning of division.

1. A meaning of division

There are two meanings for division: partitive (or sharing) and quotative (or measuring). The partitive meaning is the most common one we think of when we do whole number division. I have 12 cookies to share equally among 3 people. How many cookies does each person get? We know the number of groups (3 in this example) and we need to find the size of each group.

When dividing by a fraction, partitive division means that we know the fractional part of a group we have, and we need to find the size of a whole group.

I can mow 4 lawns with \frac{2}{3} of a tank of gas in my lawnmower is a partitive division problem because I know what \frac{2}{3} of a tank can do, and I want to find what a whole tank can do. So performing the division 4\div \frac{2}{3} will answer the question.

2. Fractions as operators

When I multiply by a fraction, I am making things larger (if the fraction is greater than 1), or smaller (if the fraction is less than 1, but still positive).

Scaling from (say) 5 to 4 requires multiplying 5 by \frac{4}{5}. Scaling from 4 to 5 requires multiplying by \frac{5}{4}. This relationship always holds—reverse the order of scaling and you need to multiply by the reciprocal.

putting it all together

Back to the lawnmower. There is some number of lawns I can mow with a full tank of gas in my lawnmower. Whatever that number is, it was scaled by \frac{2}{3} to get 4 lawns. Now we need to scale back to that number (whatever it is) in order to know the number of lawns I can mow with a full tank.

So I need to scale 4 up by \frac{3}{2}.

Now we have two solutions to the same problem. The first solution involved division. The second solution involved multiplication. They are both correct so they must have the same value. Therefore,

4\div \frac{2}{3} = 4 \cdot \frac{3}{2}

There was nothing special about the numbers chosen here, so the same argument applies to all positive values.


We have to be careful about zero. Negative numbers behave the same way as positive numbers in this case, since the associative and commutative properties of multiplication will let us isolate any values of -1 and treat everything else as a positive number.

More on partitive fraction division here.

Please note that you do not need to invert and multiply to solve fraction division problems. You can use common denominators, then divide just the resulting numerators. You can use common numerators, then use the reciprocal of the resulting denominators. Or you can just divide across as you do when you multiply fractions. The origins of the strong preference for invert-and-multiply are unclear.


7 responses to “Wiggins questions #3

  1. Christopher, my issue with this defense of the invert and multiply algorithm is not consistent with the way we think of c x d, where c is the multiplier and d is the multiplicand. So, when we multiply a number c (number to be replicated) by a number d (number of groups or replications), we represent that symbolically by c x d (c groups of d) and not the other way around. I agree that we “scale 4 up by 3/2,” but that is interpreted symbolically as 3/2 x 4. In this case, 4 x 3/2 doesn’t have a contextual interpretation that makes sense.

    Maybe, at the level that students are doing multiplication and division of fractions, they can abandon the (more) concrete interpretation of multiplication for the abstract, but I tend toward being more of a stickler in this case, because there is a bridge that we can make between the concrete interpretation and abstract symbols (common denominator algorithm or this on with the multiplier in the appropriate position). I also think that, despite the CC’s focus on standard algorithms, a defense of the invert and multiply algorithm can be taken as a free ticket to Same/change/flip-ville. Although, sadly, I don’t think the citizens there are really reading your blog.

  2. Wait!

    You see this as a defense of the standard algorithm for fraction division as a curricular topic, Adam? That was not its purpose at all. I am just answering the question here, which is a reasonable one: Is there a conceptual way to think about this algorithm? I say yes. As you rightly point out, this answer may not be appropriate for all audiences.

    But the algorithm is not at fault. There are no bad correct algorithms.

    There are, however, algorithms that are more likely to support the thinking of a typical student. Common denominator fraction division is more likely to support the thinking of the average 5th-7th grader. And it is more likely to be in the capacity of the average elementary teacher to teach conceptually. We are agreed on that.

  3. I think you may be ignoring what I think is the crux of my argument – the order of 4 and 3/2 in the operation cannot be overlooked, if conceptual is the goal, despite the level of the student. 4 / (2/3) is not equal to 3/2 x 4, at the concrete level.

    I didn’t say this is a bad algorithm, but I do think that there are certainly better ones, given the historical treatment of it (“Ours is not to reason why…,” Keep/Change/Flip, etc.). I also think that, much like the standard long division algorithm, this algorithm is unproductive. Kids memorize it without knowing why it works…and then forget it, like the division algorithm. Maybe I would have rather seen you answer the question with something along the lines of “yes, but here’s a different algorithm that grows out of a concrete representation of the operation” or something like that.

  4. Pingback: Carl's Teaching Blog | Around the Blogosphere: Fractions, Grading, And A #MTBoS30 Round-Up

  5. Maybe it’s a bit late, but I have just found your other blog (this one!).
    lawnmower, and other similar problems.
    Simple, common sense method:
    4 lawns with 2/3 of a tank, so 2 lawns with one third of a tank
    so 6 lawns with a full tank.
    if the kids are pushed into converting the problem into an immediate calculation, and not encouraged to “get the result somehow” then the path is smoothed for the rest of the math difficulties lying ahead.

    If the kids cannot see step 1 above then they are definitely not ready for fractions.
    My position is that from the outset fractions are numbers, and invented/used to enhance measurement.
    I will stop here!!!!!

  6. If it’s known that 4 lawns are cut with 2/3 of gas tank, then…

    Assuming that the lawnmower engine works, at full power, across all fuel states of the gas tank, except when empty.

    One third of gas, would theoretically, yield only 2 lawns cut. It makes sense, with half the gas (comparing 1/3 to 2/3), you would only cut half the grass. 2/3 gas must cut double the amount of grass, compared to 1/3 gas.

    Each third of gas, is individually, the same amount of litres, of course. Each third is the same size, of course. Otherwise it wouldn’t be called a third of some original amount. (obviously). Each third of gas therefore cuts same amount of grass, individually.

    When you know that one third is something, then you simply add three thirds together to get one whole. (1/3) + (1/3) + (1/3) = 3/3, otherwise known as one whole.

    2+2+2 = 6 lawns cut, with a fully operational 3/3 gas tank.

    To me this sounds like some blue collar math logic, but it works well enough in this case.

  7. Like a Number Talk, I’m seeing from the comments different ways of looking at the same problem. Could it be possible of doing that even with using the same algorithm? What would you say is the big idea?

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