## 2013-02-13

### Explaining Proportions

A common basic math exercise is to set up and solve a proportion (equivalence of two ratios, i.e., fractions), often in the context of some word problem. The funny thing I recently discovered (updating lecture notes for the spring term) is how there's usually a complete absence of explanation on why you're doing this, or justification for why it makes sense to do so. In fact, I flat-out couldn't find any explanation for the procedure in any of the resources that I have available to me at the moment. Here are some examples:
Writing proportions is a powerful tool for solving problems in almost every field, including business, chemistry, biology, health sciences, and engineering, as well as in daily life. Given a specified ratio (or rate) of two quantities, a proportion can be used to determine an unknown quantity. [Elayn Martin-Gay, Prealgebra & Introductory Algebra, 3rd Edition]

We can use proportions to solve applied problems by expressing a ratio in two ways, as shown below. For example, suppose that it takes 8 gal of gas to drive for 120 mi, and we want to determine how much will be required to drive for 550 mi. If we assume that the car uses gas at the same rate throughout the trip, the ratios are the same, and we can write a proportion. [Marvin Bittinger, Intermediate Algebra, 9th Edition]

Proportions are typically used when you want to solve for an unknown. Let's look back to our car example. In the last section we found we could drive 120 miles on 4 gallons of gas. We want to find out how many miles we could drive on 10 gallons of gas. This information is displayed in the table below. The value we want to determine is represented by an x in the table above. We can find this value by setting up a proportion. This is shown on the right. [Syracuse University Mathematics Tutorial, retrieved 2/13/13 -- the first of several Google searches I looked at]
In each case (and there were numerous others), that is the entirety of the explanation of why you'd want to set things up in a proportion. To my mind, each of them are extremely sketchy. And like my own lecture notes up until recently, they have a tendency to start off with a sample problem first; they say something like, "take this and set up a proportion like so", then go through the solving steps. But I've become highly sensitized to that fact that if I can't start out with a simple explanation as to why the mechanics of a certain procedure make sense (in this case, setting up the equation a certain way), then that's an indication that I don't fully understand what I need to answer questions on the subject, and need to rectify the situation.

Here's how I put it in my brief lecture notes nowadays -- Problems involving a constant rate can be set up as a proportion. For example: If 2 boxes of cereal cost $10, then how much do 6 boxes cost? One way of looking at it is this: The cost of one box is 10/2 = 5 dollars, so the cost of six boxes must be 5∙6 = 30 dollars. But another way of putting it is that, if we turn both of these into divisions, then the result is the same; i.e., 10/2 = 5 and 30/6 = 5. Therefore we could set up the original problem as a proportion, being careful to line up like units, e.g.: 10/2 = x/6 [dollars/boxes] → 60 = 2x [cross-multiply] → 30 = x [divide by 2]. And again we see that the total cost is$30.

Observations: The "constant rate" here is specifically the price point of $5 per box of cereal, which is a reasonable and common-sense assumption we're making in the solution, that the price-per-box is the same for the two transactions (barring some kind of bulk discount, say). But note that the proportion method is not the only way we could solve this problem, and in fact it has some very notable disadvantages: (a) we don't ever see the actual "constant rate" itself in the calculation ($5 per box in the example above), and (b) in the intermediary step it produces a much larger number than anything that existed in the original problem (the 60).

So for me, I frankly find the proportions method pretty unintuitive, and in my own work I rarely turn to it as a first choice in solving strategy. Particularly if I have a calculator or computer available, then I find it easier to do the divide-first-and-multiply-second method, as given initially above (and this strikes my students as far more understandable, i.e., actually seeing the constant rate price-point). Or alternatively, you could divide the box numbers first (6/2 = 3), and then intuit that the dollar amounts would have to be increased by the same factor, i.e., a product of 10∙3 = 30 in the given example (again, dealing with smaller and mentally-manipulable numbers along the way).

That said, the proportions method does indeed have some specific advantages. Ones I can think of immediately are: (a) it encapsulates the entire problem into a neat, concise, and attractively symmetric piece of equation writing; and (b) if you're working by hand, and there's going to be decimals in the final answer, then the decimal work is minimized and only appears in the very last division step (as opposed to dealing with it twice, in my divide-and-then-multiply method). This latter feature is of course devalued the more that cheap computation devices become ubiquitous, and is similar in that regard to a lot of other methods which trade off a large intermediary value so as to delay working with cranky divisions, fractions, and decimals (for example: the "calculating formula" for standard deviation, etc.). Perhaps, then, the proportions method is already something of a legacy dinosaur in that regard; I know that for my own work, I find more utility in actually seeing the constant rate I'm dealing with itself identified in the middle of the workflow.

Can you think of any other advantages to the proportions method for these types of elementary problems?

## 2013-02-02

### Graphing Mistake

You would not believe how often I see this mistake in a basic algebra class: