*University of Utah*

I am involved with a Math for America (MfA) project at the University of Utah that helps individuals with strong backgrounds in mathematics (typically a baccalaureate degree in mathematics) become secondary math teachers in high-needs schools. What has been extraordinary to me in this work is the gap between general mathematical knowledge and mathematical knowledge for teaching (MKT). This disparity has received significant attention in teacher education circles. (See, for example, Ball, Thames, and Phelps’s article, “Content Knowledge for Teaching: What Makes It Special?” in the Journal of Teacher Education 59[5], 2008.) The general theme of research in this area is that there is a difference between “doing” and “teaching” mathematics and that while teacher content knowledge is necessary for pedagogical knowledge and skill, the former does not guarantee the latter.

At the University of Utah, we are trying to develop a conceptual understanding of MKT at the secondary level and a means by which we might measure it. In particular, we are interested in identifying knowledge and skills that secondary mathematics teachers need but that are not necessarily possessed by those with degrees in mathematics. It may seem ridiculous to think that individuals with degrees in mathematics don’t know all the math they need for teaching secondary school students, but here are some concrete examples of where we see a gap:

Why is the area of a trapezoid ½h(b1+b2)?

Is there a geometric reason that the slopes of perpendicular lines are negative reciprocals of each other?

How might you explain why 5 minus -7 equals 12, or why the product of two negative numbers is positive?

Why is anything to the 0 power equal to 1?

When our MfA fellows begin the program, most can provide the algorithms or rules related to the above topics, but when pressed, they generally are not able to give student-friendly explanations that connect a tangible model to the algorithmic procedure. For example, fellows usually set up a proportion to solve the problem of finding 35% of 80 (x/80 = 35/100), but do not know of any way to make this problem simple enough for students to compute the answer in their head. (One approach is to understand that 35% is three and a half groups of 10% portions. Ten percent of 80 is 8, so three of them would be 24 and another half [4] would give 28.)

Indeed, once they examine the conceptualization, the models are not just intuitive—they actually enhance prior understandings of our fellows. The issue is that (a) these conceptualizations are vital to the work of teaching mathematics and (b) they do not seem to be developed in conjunction with typical preparation in mathematics.

It is not enough for a secondary teacher to say a negative times a negative is a positive—she must also be able to engage students in understanding why this is the case and then how this logic can be applied to other situations. In a similar vein, it is not enough that a teacher knows that a student made a mistake in simplifying an algebraic fraction; he must also be able to identify what the student was thinking in the erroneous simplification process. That way, the teacher has a better chance of helping the student connect his or her understanding of numeric fractions to algebraic fractions.

As we prepare individuals with strong backgrounds in math to become teachers, what we have learned is that advanced content knowledge in mathematics must be deliberately linked to content-specific pedagogical knowledge and skills. If that linkage is not made, advanced content knowledge stays “siloed” in the instructor, where it doesn’t do the instructor or the students much good.

Individuals wishing to teach mathematics at the secondary level need more than a strong background in advanced mathematics; they need a strong foundation in the mathematics they are going to teach. So, while it is essential that secondary math teachers understand abstract algebra, it doesn’t necessarily translate into the ability to teach basic algebra. If the truest test of understanding an idea is being able to teach it to someone else, then even some of the strongest graduating mathematics majors still have much to learn about the foundations of their chosen subject. The more they are willing to learn, the more their future students will be likely to follow suit.

*About the Author: Maggie Cummings is an instructor with the Center of Science and Mathematics Education at the University of Utah. Email: margaritacummings@gmail.com*

*Aftermath essays are intended to be editorials and do not necessarily reflect the views of the MAA.*

Very nice essay, Maggie. We are working on a related problem: communicating math to liberal arts college students. Definition-theorem-proof-example simply doesn't work for them. You can learn more about our project at the About Mathematics website. Do keep up your good work. Gerry

ReplyDeleteTo compute 35% of 80 in your head, notice that 8=2*4, so instead of multiplying by 8, double 35 first, getting 7 (ignore the factor of 10), then multiply by 4 to get 28. At the end we know that the answer should be in this order of magnitude, so this is the answer (alternatively, one can keep track of the 10's).

ReplyDeleteHi Maggie,

ReplyDeleteWould you be able to provide how teacher should specifically answer those gaps?

** Possible "spoiler" alert **

The percent problem has already been addressed by previous commenters.

The trapezoid formula seems to be clearly explained by your diagram.

What is the geometric reason why perpendicular lines must have slopes that are opposite reciprocals of each other?

I would explain why 5 minus -7 is 12 by starting with:

5 - 3 = 2.

5 - 2 = 3.

5 - 1 = 4.

5 - 0 = 5.

5 - (-1) = 6.

I think I saw the explanation for why the product of two negative numbers is positive in a book by I.M. Gelfand, but I don't recall the reasoning at this moment.

As for why anything to the 0th power is 1:

2^3 = 8,

2^2 = 4,

2^1 = 2,

so 2^0 = 1.

This is true for all postiive whole numbers. So if "a" is a whole number, then a^0 = 1. Then this can be shown for negative integers and also shown that the pattern breaks if a = 0.

Be careful about making statements like “anything to the 0 power is 1” to students. Is 0^0 = 1? No. The pattern that 0 ^ (any positive power) = 0 would lead to 0^0 = 0. So one pattern leads to an answer of 1, the other to an answer of 0! For that reason 0^0 is not defined.

ReplyDeleteI think a better reason for defining 0 ^ (any non-zero power) = 1 is so that the basic rules of exponents that work for positive exponents also work for non-negative exponents (when 0^0 is ruled out).

Concerning the statement that the slopes of perpendicular lines must be negative reciprocals (assuming the lines aren’t vertical and horizontal):

ReplyDeleteI’m not sure this is “geometric” enough, but here’s a reason using the Pythagorean Theorem:

We may assume both perpendicular lines pass through the origin (after translation if necessary). Then one line (with slope m) goes through a point of the form (1, m) and the other line (with slope n) goes through a point of the form (1, n). Set up the statement of the Pythagorean Theorem for the triangle with vertices at these two points and the origin. Simplifying leads to mn = -1 as desired.

As to the slopes of perpendicular lines being negative reciprocals, you can derive it from the dot product. If u=(u1,u2) and v=(v1,v2) are perpendicular u.v=0=u1v1+u2v2. (c.f. the law of cosines) Then solve to get v2/v1=-u1/u2. Since u2/u1 and v2/v1 are the slopes of u and v, it is proved.

ReplyDeleteFor a somewhat hand-wavey but easily 7th-grade explanation of the negative reciprocal slope thing: try constructing a square. Pick a starting place on the Cartesian plane; that's your first vertex. Go a units right and b units up; that's your second vertex. Go b units left and a units up; that's your third vertex. Go a units left and b units down; that's your fourth vertex. Go b units right and a units down; you're back at your first vertex.

ReplyDeleteLooking at any two adjacent sides, it's easy to show that their slopes are negative reciprocals.

(For 7th-graders, you'd want to use actual positive integers in the place of a and b.)

The "hand-wavey" aspect here, of course, is: how do we know it's a square? It's visually "obvious" if you draw it on graph paper, but I think that for an actual proof you'd want to use the Pythagorean Theorem, as per Wiley Williams' comment above. (Or, I suppose, the dot product, but that's usually not covered in the 7th-grade curriculum...)

- Alice