Students often struggle to apply abstract mathematical concepts to real-world problems. These might be word problems assigned as math homework, which often appear in a section titled Applications, at the end of a chapter in their math textbooks. These might be problems appearing in one of their other classes like Chemistry, Hospitality Management or Engineering. Or these might be problems that students are solving in their heads in their day-to-day lives, without even realizing that they are applying abstract math concepts.

Tutors may not realize that they can help students with such topics when they pop up in other disciplines, but a tutor’s expertise can be extremely valuable here. The tutor should have a firm grasp of the big picture (the abstract math concept) and the ability to recognize how and when it is applied in different contexts. The tutor has the chance to help the student realize that, “Hey, this is the same math concept, just dressed up in different ways!”

In this post, we tackle a topic that appears in each of these three realms: unit conversion.

A real-life example

An example of a typical day-to-day real-world problem involving this kind of conversion is:

If I use a pay-per-ride Metrocard, how much do I spend on transit each month? 

In order to answer this question, we obviously need to know a bit more information. Currently, (without a discount) a Metrocard swipe costs $2.75. Let’s assume that our student takes transit on average 3 times per day, including weekends. For simplicity, let’s assume that a month has 30 days. Many students would be able to answer this question correctly: total cost = $2.75 \times 3 \times 30 = $247.50 per month. What the tutor can do is demonstrate to the student that he or she is really converting units, from dollars per ride to dollars per month:

\frac{\$2.75}{\rm{ride}}=\left(\frac{\$2.75}{\rm{ride}}\right)\left(\frac{3rides}{day}\right)\left(\frac{30days}{month}\right)=\frac{\$2.75 \times 3 rides\times 30 days}{rides\cdot days \cdot month}= \frac{\$247.50}{month}

The abstract math concept(s)

Once a student has seen this procedure in action it might be easy enough for him or her to remember it, but might not be obvious exactly what’s going on here, and why it makes sense.

Arguably, there are only two abstract math principles at play here, and they’re simple:

  1. any quantity over itself (other than zero) is equal to 1,
  2. multiplication by 1 doesn’t change anything.

Often, if you ask a student what a number over itself is equal to, or what multiplying a number by 1 does, you’ll get a correct answer. But on the surface, it might not look like either of these facts have anything to do with the calculation above. Let’s examine more closely.

First, abstractly, unless x=0, if we know that x=x, then we have that \frac{x}{x} = 1 (by dividing both sides by x). If we start with x = y, where y \neq 0, then we have that \frac{x}{y} = 1. This should seem innocent enough to most students who are used to manipulating both sides of an equation. What might look a bit more weird for them is the same thing where units are included. For example, since 1 day = 1 day, we have \frac{1 day}{1 day} = 1. That is, this fraction is equal to the number 1 with no units. Additionally, since we’re assuming a month has 30 days, we have the equality 30 days = 1 month. Again, we can divide both sides by the right-hand-side, so we have \frac{30 days}{month} = 1. Again, this fraction is equal to the number 1 with no units. From this perspective, there are many different forms that the number 1 can take, and it’s up to us to decide what the most convenient form of the number 1 will be.

Next, abstractly, for any x-value, we know that x \times 1 = x. So all we’re saying in the calculation above is:

\frac{\$2.75}{\rm{ride}}=\left(\frac{\$2.75}{\rm{ride}}\right)\times 1 \times 1= \frac{\$247.50}{month}

We’ve just chosen convenient versions of the number 1 to multiply by, namely, those that make every unit we’re not interested in appear once in the numerator and once in the denominator. Often, as part of the last step, a student will have been instructed to cross out common units in the numerator and denominator. While this is common, technically okay, and easy for students to remember, I shy sometimes away from it myself since it hides the fact that the same principle is being applied again: something over itself is 1.

Tutors should emphasize these abstract math concepts to students as they work through examples together. Each time, all we’re doing is multiplying by a convenient version of the number 1, which is typically called a conversion factor. We’re not changing anything, we’re just changing how it looks.

An example from chemistry

The mass of a hydrogen atom is 1.67 \times 10^{-18} micrograms. Determine its mass in pounds.

First we’ll use one conversion factor to convert from micrograms to grams, and then another conversion factor to convert from grams to pounds. A tutor can help a student set up the string of conversion factors without even knowing the quantities himself or herself; leave blanks for the student to fill in.

1.67 \times 10^{-18} \mu g =(1.67 \times 10^{-18} \mu g)\times \left(\frac{\dots g}{\dots\mu g}\right)\times\left(\frac{\dots lb}{\dots g}\right)

Emphasize that the placement of the units in this procedure guides what the conversion factors should be.

An example from hospitality management

Express 3 ounces as a fraction of a pound.

In this case, we need to know that a 1 lb = 16 oz, so the conversion factor will be either \frac{lb}{16 oz} or \frac{16oz}{lb}. Have the student set up the expression with blanks that will need to be filled in:

3 oz = (3 oz) \times\left( \frac{\dots}{\dots}\right) = \dots lb