“But perhaps, you shall not find her common, now”.

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This is the second in my series on techniques I use in the classroom. Many of the things I talk about in this series are well-known, in the sense that at least 50 people know about them. What is certain is that they are not well-known to students. This entry concerns something that I want my college students to have down pat, but on which I expect them to do rather poorly: adding and subtracting fractions.

Tricks of the Trade

(with Professor Glesser)

For simplicity, let me break up addition of fractions into three categories:

  1. Common Denominators
  2. One denominator is a multiple of the other
  3. Other (the algebraist in me wants to write, “\mathbb{Z}-linearly independent denominators”.)

Now, once you get by students wanting to do things like:

\dfrac{3}{5} + \dfrac{2}{5} = \dfrac{2+3}{5+5} = \dfrac{5}{10} (batting average addition)

then students catch on pretty quickly that the common denominator case is where they want to be.  The second category of problems, exemplified here by

\dfrac{3}{5} - \dfrac{1}{10}

is only slightly more tricky. Any student who has ever needed to make change can figure out why you should think of \dfrac{3}{5} as \dfrac{6}{10}. In any case, I am going to assume that the students in question have mastered categories 1 and 2 (both in recognizing and computing).

Let’s up the ante a bit and try a problem like:

\dfrac{3}{5} + \dfrac{2}{7}

I was taught to find common denominators, i.e., find a common multiple of 5 and 7 and multiply each by an appropriate factor to get two fractions with common denominators. Here for instance, we note that the least common multiple of 5 and 7 is 35. Multiplying as follows:

\dfrac{3}{5} \cdot \dfrac{7}{7} = \dfrac{21}{35} and \dfrac{2}{7}\cdot \dfrac{5}{5} = \dfrac{10}{21}

we get the new (easier) problem \dfrac{3}{5} + \dfrac{2}{7} = \dfrac{21}{35} + \dfrac{10}{35} and we all know that the answer is \dfrac{31}{35}. No problem, right? Well, wrong. Students routinely mess this up. First, students are usually to taught to find the least common denominator and this, for many of them, is guess-work. Second, they tend to multiply fractions incorrectly in a way they never do for category 2 problems. Third, it is just too many steps for most of them to remember and/or complete without arithmetic errors. Heck, I even messed up the above problem when I typed it up the first time.

How the Pros Do It

Let’s take a more general situation: \dfrac{a}{b} + \dfrac{c}{d}. We can always find a common denominator by multiplying the denominators¹. So, we get \dfrac{a}{b} + \dfrac{c}{d} = \dfrac{a}{b}\cdot \dfrac{d}{d} + \dfrac{c}{d}\cdot\dfrac{b}{b} = \dfrac{ad + bc}{bd}. Simply put, starting from the original problem, you cross multiply and add to get the numerator and multiply across to get the denominator.

For example, \dfrac{3}{5} + \dfrac{2}{7} becomes easy now as the numerator is just 3 \times 7 + 2 \times 5 = 31 and the denominator is 5 \times 7 = 35, so \dfrac{3}{5} + \dfrac{2}{7} = \dfrac{31}{35}

Pictorially, it looks like: Teach your students this and it will amaze you how you no longer have to interrupt the flow of a demonstration to add fractions. Show them how useful it is in the kitchen when they need to know that \dfrac{1}{2} + \dfrac{1}{3} = \dfrac{5}{6} without the benefit of pencil and paper.

What about subtraction?

No problem! A similar derivation shows that to subtract, you simply cross multiply and subtract to get the numerator and multiply to get the denominator. For example, \dfrac{4}{9} - \dfrac{2}{5} = \dfrac{4 \times 5 - 2 \times 9}{9 \times 5} = \dfrac{2}{45}

The biggest problem students face when doing subtraction is remembering which order you subtract (it didn’t matter for addition!) If you always start with the down-right arrow, there is no issue, though. It’s amazing. I’ve taught this to students who could hardly subtract fractions on paper, but after learning this trick, would do the problems in their head as fast as I can.

What about algebra?

It is really convenient with many algebra problems. For instance, \dfrac{2x - 3}{x+2} + \dfrac{3}{x^2 + x + 1} gives a lot of students fits as they can’t see how to find a common denominator. But, using this method:

{\dfrac{2x - 3}{x+2} + \dfrac{3}{x^2 + x + 1} = \dfrac{(2x-3)(x^2 + x + 1) + 3(x+2)}{(x+2)(x^2 + x + 1)}= \dfrac{2x^3 - x^2 + 2x + 3}{(x+2)(x^2 + x + 1)}}.

To be sure, the arithmetic is still hairy, but there is less writing than before and the student doesn’t have to think about this part, they can just do it. Now there are some cases where this method will result in some cancelling work. For example:

\dfrac{3x - 4}{x^2(x+3)(x-1)} + \dfrac{x-1}{x(x+3)^2(x-1)(x+5)}.

A student should learn to recognize that it is much easier to multiply the first fraction by \dfrac{(x+3)(x+5)}{(x+3)(x+5)} and the second by \dfrac{x}{x} and then adding. But, if they are just getting started, as long as they aren’t too overzealous about distributing, they will obtain using this method

\dfrac{(3x-4)x(x+3)^2(x-1)(x+5) + (x-1)^3x^2(x+3)}{x^3(x+3)^3(x-1)^2(x+5)}

and the cancellation is not too hard to find, leaving \dfrac{(3x-4)(x+3)(x+5) + x(x-1)^2}{x^2(x+3)^2(x-1)(x+5)}.

Is it really a trick?

No, not really. The formula {\dfrac{a}{b} + \dfrac{c}{d} = \dfrac{ad + bc}{bd}} actually defines fraction addition. In fact, if you’re interested in a little abstract algebra, hold on to your hats.

For the advanced reader who isn’t afraid of a little jargon

There is a lovely theorem in algebra saying that every integral domain can be embedded in a field. Huh? You ate what now?

All right, let’s break it down.

AG: An integral domain is just a commutative ring where…

Poor Reader: Whoa whoa whoa. Hold on just a second, there, professor. A what?

AG: Ah, yes, a commutative ring. It is a ring that is commutative.

PR: (eyes rolling) Yes, brilliant. What the heck is a ring? (looks nervously at his or her wedding band)

AG: It is an abstract structure composed of a non-empty set and two binary operations…um, long story short, it is a set where you can add and multiply and the distributive laws hold. The only real caveat is that sometimes multiplication isn’t commutative. That means that sometimes ab \neq ba. Matrix multiplication is like this. In fact, the set of n \times n matrices forms a ring since you can add and multiply them as usual.

PR: Riiiigggghhttt (eyes are glazed over).

AG: A commutative ring is where you don’t have the multiplication problem. Then everything works just the way you want it to.

PR: Okay, let’s say for a moment I understand what you’re talking about. What is this integral domain thing then?

AG: Good question. Remember how you solve x^2 + 3x + 2 = 0? You factor to get (x+2)(x+1) = 0 and then conclude that x = -2 or x = -1. How do you conclude that?

PR: Well, we know that if xy = 0 then x = 0 or y = 0.

AG: Precisely. Any commutative ring that has that property, we call an integral domain.

PR: That is a stupid name. I think I could do better.

AG: Yes, yes, you’re very smart; now, shut up! The theorem is that every integral domain can be embedded in a field. A field is just an integral domain where you can divide by anything other than 0 (Even in abstract algebra, except in 1 case, we are never, ever, allowed to divide by 0).

PR: What do you mean by embedded?

AG: It is a bit technical. Strictly speaking, it means that there is an injective ring homomorphism from the integral domain into a field. But just think about how every integer is also a rational number.

PR: Rational number?

AG: I’m sorry: a fraction. See if I have the integer 4, I can think of it as \dfrac{4}{1} and then it is a fraction. This is how we embed the integers into the rational numbers, which are a field since you can divide by fractions.

PR: You can divide by integers too. Aren’t they a field?

AG: No, because if you divide by an integer, you don’t normally get an integer. Fields have to be closed. By that, I mean that when you divide, the answer is still in the field.

PR: Hmmm. I don’t think I understand this.

AG: That’s okay. If you could really learn it in a blog post, you wouldn’t have to take an entire year of it in graduate school. The point of all this is to get at the following. To build a field from an integral domain, you take the elements of the integral domain and start writing down all fractions from those elements. You need to then explain how to add and multiply those fractions.² The answer is to define these as

{\dfrac{a}{b} + \dfrac{c}{d} = \dfrac{ad + bc}{bd}} and \dfrac{a}{b} \cdot \dfrac{c}{d} = \dfrac{ac}{bd},

which is precisely how you do it normally. Now, there is a lot more to proving the theorem. There are issues of showing this is well-defined, that the usual laws of arithmetic (e.g., associativity, commutativity, distributivity) hold and that you can produce an injective ring homomorphism. This shows though, that it is not just more computationally efficient to add this way, but that when we don’t show it, we are actually missing out on a core ingredient of a key theorem in abstract algebra.

PR: (snoring)…hmm, what, oh yes, that is a shame. Did you check your pocket for the key?

________________________________________________________________________________

¹ This will likely not give you the least common denominator, but the only cost is that you’ll have to do some cancellation at the end.

² You also need to define when two fractions are the same: \dfrac{a}{b} = \dfrac{c}{d} if and only if ad = bc.

11 thoughts on ““But perhaps, you shall not find her common, now”.”

  1. Although I’m not a mathematician, I do have an advanced degree in physics, and I would never use this method, at least not explicitly. If were unable to immediately think of the least common multiple, I would surely multiply the denominators and convert the fractions. But I would actually write that step. I would not skip to the punchline. I don’t necessarily have a problem with showing kids how this works and letting them use it. But as a rule, I don’t like shortcuts. How many of your students use this method and can actually explain why it works? And what happens if they do forget what order to subtract? Will you remind them?

    I teach geometry, so I run into this kind of problem often: the book puts a nice box around a formula like the area of a rhombus is half the product of its two diagonals, when if you know that a rhombus has perpendicular bisecting diagonals (which you do, because we learned that earlier in the year) and you know how to find the area of a triangle, the area of a rhombus is a piece of cake.

    Anyway, all I’m meaning to say is that if students are using this “shortcut” because they don’t actually understand how to add fractions with unlike denominators, they may someday be faced with a situation in which the method doesn’t help (only thing I can come up with right now is 3 fractions).

    Reply

    • Hi Stacy,
      Thanks for the feedback. To answer a couple of your questions: First, when I teach this, I teach it in a very similar way to how I blogged it. The usual reaction is students saying, “Oh. So that’s why we find common denominators.” For me, this translates into: they didn’t get the whole finding common denominators business in the first place. Now, do I believe that most of the students I show this to remember the whole reason it works? Not really. But they’re no worse off than when they didn’t understand what they doing before and now they have a faster algorithm. About subtraction, since I so consistently teach it where you do the down-right direction first, it just doesn’t come up a lot. When it does, I do remind them (unless, of course, I am teaching a class where this is precisely what they were learning, but in that case we would have done more than one or two examples in class).

      You’re definitely right about the rhombus area formula. As infrequently as someone would use it, why memorize it? It’s so much better to understand how to come up with it. Adding fractions, however, is used quite a bit more frequently and so justifies a nice formula (similar to how knowing the quadratic formula is reasonable versus always completing the square).

      The method, of course, doesn’t fail for three fractions. You simply need to add two at a time. Whether the method maintains its efficiency edge depends on the example. I can certainly come up with examples (not in categories 1 or 2) where finding the common denominator is much faster. Nonetheless, when I add/subtract fractions (even 3 of them), this is how I do it more often than not.

      Reply

  2. Pingback: Shortcuts, Mnemonics and Why They Ain’t So Bad « GL(s,R)

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  4. Hello, I found your blog via Dan, or was it Sam, or Kate? I am discovering the world of maths teaching blogs and enjoying it so much I started one as well! I wanted to say thanks for this highly mathematical post. It’s been five years that I’ve been teaching after turning down a PhD in math, and it feels good to go beyond school mathematics once in a while.

    What you said in your other post about shortcuts resonates with me and also reminds me of Lockhart’s Mathematician’s lament: why not take some parts out of the maths curriculum, since students aren’t learning any maths anyway even with all the topics in there. I look forward to reading more from you in the future.

    Reply

  5. Hello, I found your blog via Dan, or was it Sam, or Kate? I am discovering the world of maths teaching blogs and enjoying it so much I started one as well! I wanted to say thanks for this highly mathematical post. It’s been five years that I’ve been teaching after turning down a PhD in math, and it feels good to go beyond school mathematics once in a while.

    What you said in your other post about shortcuts resonates with me and also reminds me of Lockhart’s Mathematician’s Lament: why not take some parts out of the maths curriculum, since students aren’t learning any maths anyway even with all the topics in there? I look forward to reading more from you in the future.

    Reply

    • Thanks for coming by and reading. I’m trying very hard to keep things readable, but to also let my mathematician side out. While I probably err a bit too much on the latter side, it is quite nice to hear that it is still appreciated. I’m also really glad you bring up Lockhart’s lament as, subconsciously, I’m probably plagiarizing his magnificent work in just about every post.

      Reply

  6. Not only is this easy, handy, memorable, and graphable, but it also has a cool name: The Bow-Tie Method. Students love it!

    Reply

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