Grant Wiggins (who is not, as far as I know, a math teacher, but someone who likes to use math as an example of things) challenged us to make a distinction between what kids understand and what they know. I think that’s a useful distinction, and a hard one, and so I was intrigued by his set of questions that he hypothesized would show whether students had procedural knowledge or conceptual understanding (question: are there such things as conceptual knowledge and procedural understanding?). Reading the questions I realized that they are certainly evocative (if they don’t cause one to shut down and say “dunno, geez, leave me alone!”) and also that if I were to use them to assess students’ conceptual understanding, I would need to be primed to listen to a lot of ideas and make sense of a lot of different ways of thinking and understanding to truly get what students get. Coming up with answers, and even with a short list of what I’d want to attend to in kids’ responses was challenging. So: crowdsource! How would you answer these questions? What would you be attending to in listening to and assessing others’ understanding based on their answers? Go!

1) “You can’t divide by zero.” Explain why not, (even though, of course, you can multiply by zero.)
First of all, what would it mean? Is anything divided by zero equal to infinity? That’s what I thought as a kid, but I’ve since encountered experiences that suggest that 3/0 = ∞ is thinking of infinity as a number and thus not really mathematically sensible. But I got to that idea by thinking of how I understood division: How many times does 0 go into 3? Or, what times 0 gives 3? But even infinity isn’t a sensible answer to those questions. The limit of the sum of a bunch of zeros as the size of the bunch approaches infinity is not 3. Infinity zeroes is not 3 or any whole number. So those are intuitive reasons why the answer isn’t zero, it’s not even defined.

Mathematically, though, the best reason I’ve heard for why 3/0 is undefined is that if it were defined, you could make any number equal any other number! For example, if 3x = 5x, does that imply 3 = 5. Nope: you have 2 possible moves you can make solving 3x = 5x. Either use the additive property of equality to write 0 = 2x, and then the division property of equality to write x = 0. 3x = 5x -> 0 = x. Or you can try to apply the division property of equality immediately: 3x = 5x -> 3x/x = 5x/x -> 3 = 5. You have two choices for not breaking math. Either accept that 3 = 5, or say that the division property of equality only holds when you’re not dividing by 0, and since 3x = 5x is only true when x = 0, 3x = 5x only implies 3 = 5 if the division property of equality allows dividing by 0.

That’s kind of related to the fact that if 3/0 = ∞ (or even if 3/0 = some special made-up number §) then that means 3 = 0 * ∞ or 3 = 0 * §… but then doesn’t 5 = 0 * ∞ also since 5/0  = ∞? And if 3 = 0 * ∞ = 5, then 3 = 5 again. Unless you want to invent this whole shadow number system so 3/0 = §3 and π/0 = §π… but then what the heck kind of arithmetic would you do with those numbers? How would you visualize them? Why would you do this? On the other hand, please see question 10.

2) “Solving problems typically requires finding equivalent statements that simplify the problem” Explain – and in so doing, define the meaning of the = sign.

Well I know an easy example of this. Say I want to find the value of x that makes this statement true: 2x + 4π + 12 = 7(x + 2π) – 5x + 10π + x

I could guess values of x and see when I got a true statement. I could plot a graph of each side and look for the point of intersection. But easiest would be if I could simplify the above statement without changing which value of x made it true.

So… if 2x + 4π + 12 = 7(x + 2π) – 5x – 10π + x
That’s equivalent to 2(x + 2π) + 12 = 7(x + 2π) – 5(x + 2π) + x, by the distributive property of equality
Which is equivalent to 2(x + 2π) + 12 = 2(x + 2π) + x, by association and the distributive property and arithmetic
Which makes it easy to quickly guess and confirm that x had better equal 12.

Of course, I made up the example so I knew just which simplifications would get me quickly to a form of the equation that was easy to solve for x. There were many points at which I could have made other decisions and written the expression in other, equivalent ways, such as by expanding any terms with parentheses, combining like terms, adding and subtracting the same thing from both sides, etc.

As for what the = sign means, it means that the expressions on either side of it are equivalent aka have the same value. Because of some properties of equivalence relationships in our math, it’s “legal” to do certain kinds of moves to equal statements and you know you haven’t changed the truth of that equals sign.

I wonder what examples people might have come up with that didn’t have to do with the = sign? Or that had to do with the = sign but not solving for x?

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

Here is one reason it works. a/b ÷ c/d asks, “how many c/d’s are in a/b?” Because of what fractions mean, we can think of c/d as c 1/d’s. In other words, 3/5 is three one-fifths, 10/7 is ten copies of 1/7, etc. So… if we think of a/b ÷ c/d as asking “How many c/d’s are in a/b?” an easy way to calculate that is to ask first about how many 1/d’s are in a/b.”

Example: How many 3/5 are in 10/7? First lets find out how many fifths are in 10/7. There are 5 fifths in every unit, so there are 10/7 of 5 fifths in 10/7 of a unit. How many 11/20 are in 3/2? There are twenty 1/20s in 1, so there are 20 + 10 1/20s in 1 1/2. There are 20 * 3/2 or 30 twentieths in 3/2.

Now how do we use the fact that we know how many fifths are in 10/7 to find how many 3/5 are in 10/7? How do we use the fact we know how many 20ths are in 3/2 to know how many 11/20s are in 3/2? How do we use knowing how many 1/ds are in a/b to know how many c/ds are in a/b?

Well, if we know there are 50/7 fifths in 10/7, and we want to find how many sets of 3 of those fifths are in 10/7, we just break the 50/7 into groups of 3. Aka divide 50/7 by 3. The answer is 50/21. We got that by doing 10/7 multiplied by 5 and divided by 3, which is the same as 10/7 * 5/3.

Same with the 3/2 ÷ 11/20 example.

3/2 = One whole and a half.

There are 20 twentieths in the whole, and 10 twentieths in the half, for 30 twentieths in all.

How many sets of 11 twentieths are in 30 twentieths?

30/11 of course!

And finally: There are a/b * d 1/ds in a/b. There are c groups of 1/d in c/d. So there are (a/b * d)/c c/ds in a/b, aka a/b ÷ c/d = (a/b * d)/c = a/b * d/c.

4) Place these numbers in order of largest to smallest: .00156, 1/60, .0015, .001, .002

At least I don’t have to write an essay for this one… First of all, 1/60 = 100/6000 = (100/6)/1000 = (16 2/3)/1000 = 16.666666…/1000 = 1.66666…/100 = .1666…/10 = .01666…./1 = 0.016666…

So, in order, we have .002 > .00166666… = 1/60 > .00156 > .00150 > .0010

I think you were trying to trick me (and it did make bells go off in my head) with what to do when there are no digits after the first ones you compare. Like is .001 bigger than or smaller than .0015, since .0015 is fifteen ten-thousandths and .001 is one thousandth. And since Alum comes before Alumna when you are alphabetizing, since a blank after a letter is prior to any other letter after a letter.

But I wasn’t fooled! .0015 is five ten-thousandths bigger than .001 which is ten ten-thousandths!

5) “Multiplication is just repeated addition.” Explain why this statement is false, giving examples.

Gracious! It’s a good thing you included the “just” because it’s certainly false that multiplication is just-as-in-only repeated addition. Though repeated addition can be used to calculate and conceptualize many kinds of multiplication problems. Here are two examples where repeated addition doesn’t make sense.

-1 * -1 = 1.

Does adding -1 to itself -1 times make any sense? And if it does, would it really make sense that the result would be 1? Nah…

sqrt(2) * sqrt(2) = 2

What does it mean to add something to itself a irrational number of times? If you really get deep with it, you could define this using the distributive property as a limit: sqrt(2) * 1 + sqrt(2) * 4/10 + sqrt(2) * 1/100 + sqrt(2) * 4/1000 + … which (assuming you’re comfortable with adding something to itself part of a time) approaches 2. But then you have to be comfortable with defining sqrt(2) * sqrt(2) as a limit that approaches 2, and not just plain old 2, which seems kind of sad. Even though these days the limit and the thing itself are thought to be the same, such as 0.9999999999… = 1, I am still more comfortable with having a concept of multiplication such that sqrt(2) * sqrt(2) just is 2, no limits, based on, say, an area model.

6) A catering company rents out tables for big parties. 8 people can sit around a table. A school is giving a party for parents, siblings, students and teachers. The guest list totals 243. How many tables should the school rent?

Piece of cake. You think I’m going to be dumb about remainders or rounding or give an exact answer, but I know this trick!

243/8 = 30 tables with 3 people having no seats.

The school could rent 31 tables to be on the safe side, or they could take a risk and just rent 30 tables figuring that if 243 people say they’re coming it would be crazy to expect exactly all 243 to show up. All it would take is one family feeling under the weather and you’d have a whole unused table.

7) Most teachers assign final grades by using the mathematical mean (the “average”) to determine them. Give at least 2 reasons why the mean may not be the best measure of achievement by explaining what the mean hides.

Umm, is this a math question or an assessment question? One thing that the mean hides is whether you know some things not some different things, or whether you kind of know a lot of things. For example, what if your grade was based on 2 assessments, one on each of 2 major topics? What if you were taking a class on how to be nice and clean and you had a test on cleanliness and a test on niceness. Say that you are a really sweet, stinky person and you learned a lot about niceness and nothing about cleanliness, and so you got a 100% on your niceness test and a 0% on your cleanliness test. Your classmate is an okay person and kinda messy about personal hygiene, and got a 50% on both tests. You both have the same final grade (50%) in the class even though you have really different profiles as nice, clean people, and really different needs for remediation.

Another thing the mean hides is change over time. Let’s say you’re taking a class on learning to ride a bike, and when you start the class you can’t stay on the bike for even 1 meter. By the time you’re done, you can stay on the bike without putting a foot down for balance for 1000 meters. And say you developed that skill all of a sudden late in the class, so your distance assessments looked something like:
1, 1, 1, 1, 2, 3, 5, 5, 3, 4, 2, 3, 1, 3, 5, 2, 5, 5, 200, 500, 800, 800, 1000, 800, 1000. Your mean would be about 206 meters. But what about your friend who started out the class as a fine biker but developed vertigo towards the end of the course: 1000, 1000, 1000, 1000, 1000, 0, 0, 0, 0, 0, 0, 0. Her mean distance is about 416 meters. Does that mean she’s a better biker than you are at the end of the course?

8) Construct a mathematical equation that describes the mathematical relationship between feet and yards. HINT: all you need as parts of the equation are F, Y, =, and 3.

The length of 3 feet is as long as the length of 1 yard. That does not mean 3F = Y though because F usually refers to the number of feet and Y usually refers to the number of yards, not the length of feet and yards. Since the length of 3 feet is as long as the length of 1 yard, the number of feet in a given distance is three times the number of yards in a given distance and therefore F = 3Y. If we defined F as the length of a foot and Y as the length of a yard, it might make sense to say 3F = Y but that would not help us know how many yards long a distance we had measured as a number of feet was.

9) As you know, PEMDAS is shorthand for the order of operations for evaluating complex expressions (Parentheses, then Exponents, etc.). The order of operations is a convention. X(A + B) = XA + XB is the distributive property. It is a law. What is the difference between a convention and a law, then? Give another example of each.

Shoot… I’m surprised that X(A + B) = XA + XB is a law. I thought it was a property of our arithmetic system. Like maybe an axiom? Or a property of a ring or a field or some such. Aren’t there arithmetics without distributive properties? I think, though, that I could talk some about what is different between PEMDAS and the distributive property (although it would have been helpful to have said “the distributive property of multiplication over addition in the real numbers” or something more official like that).

So… what’s the difference between PEMDAS and the distributive property (aka the d.p.o.m.o.a.i.t.r.n-or-something?)

Well, I think it’s that changing PEMDAS wouldn’t fundamentally change the way numbers relate to each other, how we solve problems, how we do math… just how we write it. Changing the distributive property would make it really hard to solve equation, it would change what was equivalent to what, it would make it really hard to know how to think about arithmetic with integers or fractions, etc.

Back in the day, there wasn’t really PEMDAS because we didn’t use symbols much. If someone wanted to write 3x + 7 = 10, he (usually, though sometimes she) would write something like “A quantity is tripled and seven is added to the result. The sum is ten.” I think there are good reasons that mathematicians chose to invent a symbol system where multiplication and division are done before addition and subtraction, exponentiation is done before multiplication and division, precisely because of the distributive property, and I have a hard time believing that there are clearer, more compact ways to express “A quantity is tripled and seven is added to the result. The sum is ten.” than using our PEMDAS, but I imagine that just like there are many languages and ways to express the same thought, there can be many mathematical ways to express the same fundamental relationship. And there are different ways that we express calculations. There’s a computer notation called “Reverse Polish Notation” in which you press 3,5,+ to get the computer to add three and five. If you want to do 3 – 4 + 5, you press 3 4 – 5 +. It makes it so you don’t really need parentheses. The math is exactly the same though, it’s just how you tell the computer what to do that’s different.

Now what about the distributive property? Does it have to do with PEMDAS? Or is it independent of that? Can you use the distributive property in Reverse Polish Notation or does it require a system with parentheses? Answer: Yes, the distributive property is independent of how you write it, and has to do with two things being equivalent. The distributive property of multiplication over addition tells you that X(A + B) is equivalent to XA + XB. It doesn’t matter if you write it X(A + B) = XA + XB or A,B+X* = X,A*X,B*+. (A way better explanation of this is here: http://mathforum.org/kb/message.jspa?messageID=4538619)

One reason arithmetic would be different without the distributive property is that the distributive property lets you know that -1 * -1 has to be equal to 1. Because we know -1 * 0 has to be equal to -1 * (anything that adds up to zero if you do the adding first). Whatever notation you use to tell you to do the adding first, just make it clear to people. So without using PEMDAS I could write -1 * 0 is equivalent to adding 1 and -1 and then multiplying that sum by -1, since -1 + 1 = 0. Using PEMDAS I could write -1 * 0 = -1 * (1 + -1). Now, without the distributive property, I’d be done. I wouldn’t know if -1 * 1 + -1 * -1 was equivalent to -1 (1 + -1) and so I couldn’t simplify or further calculate anything about -1 (1 + -1) other than to go back and forth between -1 (1 + -1) and -1 * 0. But with the distributive property I can keep going. If I use words and not PEMDAS I can write -1 * 0 is equivalent to adding 1 and -1 and then multiplying that sum by -1, which is equivalent, by the distributive property, to writing doing -1 * 1 and -1 * -1 and adding the results. So I know now that -1 * -1 + -1 * 1 = 1 * 0 = 0. And I know that -1 * 1 = -1 because anything times 1 is itself. So -1 * -1 + -1 = 0. The only thing I can add to -1 to get 0 is 1, so -1 * -1 must equal 1. But without the distributive property, how would I have know that!?! Without the distributive property, we can’t prove a lot of the arithmetic we take for granted (like combining like terms, using place value-based algorithms, etc.)

10) Why were imaginary numbers invented? [EXTRA CREDIT for 12th graders: Why was the calculus invented?]

I actually know the history on this one. I’m skipping the extra credit. My understanding is that a LOOOONG time ago Italian guys wanted to come up with rules for finding intersections of cubic functions and lines. Every line intersects every cubic function, but sometimes their formula that usually worked spit out values for intersections that involved adding up expressions involving roots of negative numbers. “Yuck, our formula is broken,” many of them thought, but others thought, “Well, but what if there *were* a reasonable way to do arithmetic on these puppies?” They knew what their answers should come out to by carefully graphing and finding the intersections, and they saw ways to manipulate their answers using the usual rules of multiplication with just a light suspension of disbelief, and get to the known intersection. But… this line of reasoning was pretty well abandoned because even though it mostly didn’t break math (the distributive property, commutative and associative properties, and all the usual results of multiplication on the reals still held), having two positive numbers which, when multiplied, made a negative result, did seem to break math. And breaking math is generally frowned upon by mathematicians and math students alike.

So… why do we still use them? For some reason, someone wanted to think more about these roots, and had a reason to think of them on a Cartesian plane. They were able to show that these numbers could be thought of as vectors on a plane, and the algebraic operations on these numbers not only made sense visually on the plane but also made a hard problem (finding the results of rotating and stretching a vector) easier. So then people gradually came to accept that it was okay to break the idea of “a positive times a positive is a positive” because of thinking of multiplication in a new way, since this new way didn’t break any other ideas about multiplication, had a great visual analogy, and made life easier in an important new field of vectors.

11) What’s the difference between an “accurate” answer and “an appropriately precise” answer? (HINT: when is the answer on your calculator inappropriate?)

If my teacher gives a daily subjective participation grade out of 3 and a daily subjective homework completion grade out of 5 and I had some quizzes out of 10 and 20 points and some tests out of 100 points and my final point total is 538/635 then it is inappropriately precise to say that my grade is 89.449% and to quibble with me about whether that rounds to an A- or a B+. If we only went to the 1s place in our point calculations, then any decimal after the 1s place is meaningless, and maybe even after the 10s place. Best to round to the nearest 10 and say I got an A-.

12) “In geometry, we begin with undefined terms.” Here’s what’s odd, though: every Geometry textbook always draw points, lines, and planes in exactly the same familiar and obvious way – as if we CAN define them, at least visually. So: define “undefined term” and explain why it doesn’t mean that points and lines have to be drawn the way we draw them; nor does it mean, on the other hand, that math chaos will ensue if there are no definitions or familiar images for the basic elements.

I have no idea! I have hung out in Geometries where “planes” were not flat, they were hyperbolic or lived on spheres. I’ve met Geometries where circles and lines were equivalent, and called clines, and Geometries that included a “point at infinity” among other points. Geometry was still meaningful, that’s for sure, and the points, lines, and planes weren’t like good old Euclidean ones.

What I don’t know is why these different kinds of points, lines, and planes are “undefined”? How do we differentiate between this kind of plane and that kind of plane if they are undefined? Is the undefined part the part that makes 2 very different planes still planar in some sense?

Maybe points, lines, and planes (and other undefined terms) are like pornography, and we don’t have to define them because we know them when we see them? http://en.wikipedia.org/wiki/I_know_it_when_I_see_it

13) “In geometry we assume many axioms.” What’s the difference between valid and goofy axioms – in other words, what gives us the right to assume the axioms we do in Euclidean geometry?

I don’t know. I think that’s an open question. For a long time, didn’t we think that it would be goofy to imagine a geometry without Euclid’s 5th postulate? And then it turns out it wasn’t? Didn’t most smart people agree it was goofy to violate the “positive times a positive equals a positive” until they finally agreed that complex numbers were useful and meaningful? Isn’t negotiating whether an axiom will turn out to be goofy and valid part of what makes mathematics a living, breathing, fuzzy, human, creative domain? And what makes math so hard? You kind of are walking a line between madness and sanity when you make rigorous truly new mathematics. How do we know there will never be a mathematics in which dividing by zero is useful and meaningful? A geometry which angles can be trisected using only valid construction tools? I’m not sure we do!

Update:
David Radcliffe’s answers: https://docs.google.com/document/d/1FK8m_oVaS_UWS4grETMAfAhByN_vrQC0TR85ld5FXj8/edit
Erik Johnson’s answers: http://step1trysomething.wordpress.com/2014/04/25/answering-the-conceptual-questions/