Independent Events

Honestly, sometimes my first off-the-cuff measure of how well we do collectively with a topic in mathematics is how well I do with it. This is terrible reasoning, of course, but it has some uses as a kind of first measurement that needs to be independently verified. It helps me notice potential weak spots in instruction, anyway, like with independent events.

The concept of independent events is certainly a candidate for being a weak spot. Most of what I’ve seen online doesn’t really crack the surface. Students are allowed to explore the concept of independent events, but all that seems to mean is to take a situation and tell me whether the events are independent. Big whoop.

And the definition often used—that two events are independent if the occurrence of one does not affect the probability of the other—has the potential of really confusing students (and me). Consider this situation.

Are the events “selecting a letter card” and “selecting a circle card” independent events? The way a student (and I) might reason, given just the definition above, would be to think, “Well, if I pick B, that would definitely change the probability of picking a circle, because the B-card is also a circle card. And, if I pick A, that would change the probability of picking a circle card, because then I would only have three cards to choose from. So, the events are not independent.”

The above was perfectly sane reasoning; it’s just wrong because of a terrible explanation (or lack of an explanation in most cases). I thought of something maybe a little better. Here it is in sentence form, similar to how we worked on the impenetrability of trig ratios:

You have the same probability of choosing a letter card from all the cards as you have of choosing a letter card from just the circle cards.

This is what makes choosing a letter card and choosing a circle card “independent.” If I know that I’ve drawn a circle, the probability that I’ve also drawn a letter, $$\mathtt{0.5}$$, is the same as if I didn’t know I’d drawn a circle. And of course this works automatically the other way around too: if I know that I’ve drawn a letter card, the probability of drawing a circle is the same as if I didn’t know. If I reduce the sample space from all cards to circle cards, the probability of “letter” is the same.

At the heart of independent events (besides the conditional probability flavoring above) are equivalent ratios, or a proportion . . . which gives me an ultra-short way of saying it a little more mathematically:

$\mathtt{\frac{2}{4} = \frac{1}{2}}$

Or, “2 letter cards out of 4 cards in all is the same as 1 letter card out of 2 circle cards.” In the symbolism of probability, we actually write this with complex fractions (by giving each numerator and denominator above a denominator of 4), and then disguise the complex fractions with $$\mathtt{P()}$$ statements, which is all equivalent to the above proportion:

$\mathtt{\frac{\color{purple}{\frac{2}{4}}}{\color{red}{\frac{4}{4}}} = \frac{\color{red}{\frac{1}{4}}}{\color{purple}{\frac{2}{4}}} \longrightarrow \frac{\color{purple}{P(\textrm{letter})}}{\color{red}{1}} = \frac{\color{red}{P(\textrm{letter and circle})}}{\color{purple}{P(\textrm{circle})}}}$

And that gives us the definition of independence using conditional probability, from S-CP.A.3. If we remember our proportion work from way back when, then the “other” test for the independence of two events pops out of the equivalence of the products of means and extremes: $\mathtt{P(\textrm{letter}) \cdot P(\textrm{circle}) = P(\textrm{letter and circle})}$

The complex fraction part of this explanation seems to be the most important, actually. And we don’t really do a good job of letting kids in on that disguise either. But still, stapling the idea of independent events to a pair of equivalent ratios (a proportion) helps the whole idea make a lot more sense to me. And, truthfully, it makes the notion of “independence” as “not having an effect on another probability” seem almost wrong.

Update: This kind of reasoning works for the typical example of independent events. The situation involving separate spinners is fairly easy for kids to identify as being about independent events, but like a lot of other topics in mathematics education, we start off with examples that are easy and also completely misleading. Then we all opine that kids have difficulties because the material gets “harder.” Anyway, spinning a C on the first spinner and a 2 on the second spinner are independent events, but not because there are “independent” spinners.

What’s the proportion (if the events are independent) that matches the situation?

Assuming we spin the first spinner and don’t know what we get, there are $$\mathtt{3 \times 1}$$, or 3, outcomes that have 2 as the second spin, out of 12 possible outcomes. The outcomes are {(A, 2), (B, 2), (C, 2)}. But, if we know that we have spun a C first, then there is 1 outcome showing 2 on the second spinner, out of 4 possible outcomes. So, our proportion is $\mathtt{\frac{3}{12} = \frac{1}{4}}$

This is all we need to show that the two events are independent, actually. If that proportion is true, then the events are independent. But we can cue the complex fraction magic again for reinforcement: $\mathtt{\frac{\color{purple}{\frac{3}{12}}}{\color{red}{\frac{12}{12}}} = \frac{\color{red}{\frac{1}{12}}}{\color{purple}{\frac{4}{12}}} \longrightarrow \frac{\color{purple}{P(2)}}{\color{red}{1}} = \frac{\color{red}{P(\textrm{C and 2})}}{\color{purple}{P(\textrm{C})}}}$

Trig Ratios as Percents

My audience is mostly folks interested in math education in one way or another, so it’s no use starting this post off with “All you may know about trigonometry ratios is likely captured in the gibberish mnemonic SOHCAHTOA.” Your understanding of trigonometry ratios is no doubt more sophisticated than that.

But have you thought about trig ratios as percents? This will be enough for most of you:

sin θ = $$\mathtt{\frac{opposite}{hypotenuse} = \frac{?}{100}}$$ = percent of hypotenuse length

It makes sense when you dredge up the 6th-grade math you remember and start making connections between it and the trigonometric ratios sine, cosine, and tangent (for example). After all, opposite : hypotenuse is the sine ratio, but it’s also just a ratio. If we think of it as a percent, we could say that if the sine of a reference angle is equal to 0.75, that means that the side opposite the angle in a right triangle is 75% the length of the hypotenuse. If the cosine were 0.75, that would mean that the side adjacent to the reference angle is 75% the length of the hypotenuse, since cosine is the ratio adjacent : hypotenuse. And a tangent of 0.75 means that the opposite side is 75% the length of the adjacent side, because tangent is simply the ratio opposite : adjacent.

The percent connection (or fraction; doesn’t have to be percent) strikes me as being immediately more useful for seeing meaning in values for trigonometric ratios. They usually go by students as just values which can’t be put into a sentence—a long list of changing decimals in a lookup table. Yet, the percent connection is right there, waiting for us to combine our middle school math knowledge with new material. We could model what this process of meaning-making actually looks like, rather than just ask them to go make meaning and hope for the best.

Of course, it also helps to be able to visualize what a sine of 0.75 looks like. Try, say, $$\mathtt{49^\circ}$$ below on the unit circle and press Enter. That gives me something that looks pretty close to a sine of 0.75 (an opposite side that is $$\mathtt{\frac{3}{4}}$$ the length of the hypotenuse, right?).

θ = °

cos-sin-1
1-tan-sec
cot-1-csc

But the interactive tool, while helpful maybe, isn’t necessary, I don’t think. One can think about drawing a right triangle, say, with an adjacent side length about 80% of the hypotenuse length (a cosine of about 0.8). It will have to be longer than it is tall, relative to the reference angle, to make that work. The percent connection thus links a trigonometry ratio value to a simple and accessible visual.

An Example Problem: Testing Out the Percent Connection

The basic mathematical (as opposed to contextual) trigonometry practice problem looks like this: Determine the length of $$\mathtt{x}$$.

I can’t say the percent connection makes this a faster or more efficient process. What I would say is that knowing that the sine of 41° means the percent of the hypotenuse length represented by the opposite side length makes me feel like I know what I’m doing, other than moving numbers and symbols around. (Thinking about percents also gives us a way to estimate what my $$\mathtt{x}$$ will be, if I know that the figure is drawn to scale.)

The sine of 41° is approximately 0.65605902899. With the percent connection, I know that this means that the opposite length is about 65.61% the length of the hypotenuse. It’s hard to overstate, I think, how useful it is to be able to wrap all of this number-and-variable work into one sentence like this: 96 is about 65.61% of x. I can climb the last few steps from there, by either dividing or setting up an equation—however the work happens, I at least have some background meaning to the numbers I’m playing with.

We can continue from there, of course (as we can without the percent connection, but so rarely do because the tedium of setting up and solving for the variable has overloaded us). The tangent of 41°, approximately 0.86928673781, tells us that the opposite side is about 86.93% the length of the adjacent side.

This guy gets it, and he seems to be the only one. It shouldn’t come as any surprise that he’s an experienced mathematics teacher a computer scientist who’s never taught. But, you know, it really should surprise us. Someday.

Modulus and Hidden Symmetries

research

A really nice research paper, titled The Hidden Symmetries of the Multiplication Table was posted over in the Math Ed Community yesterday. The key ideas in the article center around (a) the standard multiplication table—with a row of numbers at the top, a column of numbers down the left, and the products of those numbers in the body of the table, and (b) modulus. In particular, what patterns emerge in the standard multiplication table when products are colored by equivalence to $$\mathtt{n \bmod k}$$ as $$\mathtt{k}$$ is varied?

The little interactive tool below shows a large multiplication table (you can figure out the dimensions), which starts by coloring those products which are equivalent to $$\mathtt{0 \bmod 12}$$, meaning those products which, when divided by 12 give a remainder of zero (in other words, multiples of 12).

mod

When you vary $$\mathtt{k}$$, you can see some other pretty cool patterns (broken up occasionally by the boring patterns produced by primes). Observing the patterns produced by varying the remainder, $$\mathtt{n}$$, is left as an exercise for the reader (and me).

Incidentally, I’ve wired up the “u” and “d” keys, for “up” and “down.” Just click in one of the boxes and press the “u” or “d” key to vary $$\mathtt{k}$$ or $$\mathtt{n}$$ without having to retype and press Return every time. And definitely go look at the paper linked above. They’ve got some other beautiful images and interesting questions.

Barka, Z. (2017). The Hidden Symmetries of the Multiplication Table Journal of Humanistic Mathematics, 7 (1), 189-203 DOI: 10.5642/jhummath.201701.15

From Translations to Slope

If not before, students in 8th grade learn that a translation is a rigid motion that “slides” a point or set of points a certain distance. An important idea here that could stand to be emphasized a lot more is that the translations students study are linear translations—the translations move the set of points along a line. When this is understood prior to looking at slope, it can help with a deeper understanding of slope.

We can see the start of this in action when we play with the simulation below. Type positive numbers less than ten and greater than zero (3 characters max) into the blank boxes and then click on the arrow boxes to set the directions. This will create a translation sequence starting at (0, 0). For example, 9 ↑ 3 ← will continuously translate a point up 9 and left 3 (until it goes out of view). Click on the coordinate plane to run the sequence.

When the sequence is finished, a button should appear that allows you to click to show the line along which the point was translated using a repetition of the translation sequence. Click Clear to draw a new translation sequence (or repeat the one you just did). You can watch a (near) infinite loop if you’d like to put in things like 8 ↑ 8 ↓.

What Is Slope?

The example at right shows a finished sequence of repeated $$\mathtt{(x – 4, y + 6)}$$. There’s a whole lot to unpack here, which I won’t do. But, playing around with linear translations in this way can eventually reveal that the vertical and horizontal displacements form a ratio. For example, one can say that for every vertical move up 6 $$\mathtt{(+6)}$$, there is a horizontal move left 4 $$\mathtt{(-4)}$$. This simplifies to 3 : –2, and you can extend the sequence into the 4th quadrant to show that this is the same line as –3 : 2.

Referring to lines in terms of their slope ratios is pretty close to the finish line as far as slope understanding.

Y = Mx + B

We can ask about the corresponding y-value for an x-value of 5. The answer to this becomes the solution to a proportion, which we can generalize: $\mathtt{\frac{\color{white}{-}3}{-2} = \frac{y}{5} \quad \rightarrow \quad \frac{\color{white}{-}3}{-2} = \frac{y}{x}}$

So, we can arrive at $$\mathtt{y = -\frac{3}{2}x}$$. By this point, the slope ratio is ready for a special letter, and we can move up to the slope-intercept form. There are all kinds of catches and surprises in this development: zeros, the final b translation of the entire line, etc. But it is certainly an interesting connection between geometry and algebra for middle school, the key idea being that translations always move points along a straight line.

These ideas can essentially run alongside ratio development too, regardless whether the notion of translations is developed formally (there’s not much formality to it, even in 8th grade) or informally. See the Guzinta Math: Comparing Ratios lesson app for some more ideas about connections.

Eleven Matches

Here’s an interesting problem about eleven matches, which I’ll just work out as I write this. It’s from this book:

On the table are eleven matches (or other objects). The first player picks up 1, 2, or 3 matches. The second player picks up 1, 2, or 3, and so on. The player who picks up the last match loses. (A) Can the first player always win? (B) Can he if there are 30 matches instead of eleven matches? (C) Can he in general, with $$\mathtt{n}$$ matches to be picked up 1 through $$\mathtt{p}$$ at a time ($$\mathtt{p}$$ not greater than $$\mathtt{n}$$)?

My first thought was to play a little and see what I notice. So I cooked up a tiny program to simulate a single game—here, the choices made by each player are random whole numbers between 1 and 3 inclusive. You can press Run over and over to run several random games.

It’s not entirely random, of course. Players can’t choose a number of matches greater than the current count. Also, a player can’t deliberately lose the game by choosing all the remaining matches (if the number remaining exceeds 1).

A Wishful-Thinking Simplification

I’m not sure that did much good. But I got to see different games play out—get my head around the environment I’m dealing with. My next thought was to simplify things, a wishful-thinking simplification: as a player, I would want to be left with 4 matches. That’s a sure win for me. Ah, but not just 4. Three or two would be a win as well—corresponding to $$\mathtt{m = 2}$$ matches and $$\mathtt{m = 1}$$ match (where 4 corresponds to $$\mathtt{m = 3}$$ matches).

Keep going by adding 4 to 4, 3, and 2. If I have 8, 7, or 6 matches in front of me, I can take 3, 2, or 1 to make it 5. My opponent must then take 3, 2, or 1 to leave me with 2, 3, or 4—the numbers that I have already decided are instant wins for me. Finally, one more round of adding 4: with 12, 11, or 10 matches, I can take 3, 2, or 1 to reduce the count to 9. And my opponent must reduce the number to 8, 7, or 6. With eleven matches, that means that I can always win by taking 2 to start and following the pattern above.

So . . . what is the pattern? It seems that if I’m Player A I want my opponent to have something like $$\mathtt{4n + 1}$$ matches. And I want $$\mathtt{4n}$$ or $$\mathtt{4n – 1}$$ or $$\mathtt{4n – 2}$$. Let’s focus on trying to control what my opponent gets.

Algebraic Thinking

Given that I’m dropped onto a random place on the number line, $$\mathtt{n}$$, how do I get to the nearest multiple of 4 (plus 1) to the left of, or at, my location? Well, I take my location, $$\mathtt{n}$$, and subtract $$\mathtt{(n \bmod 4) – 1}$$, unless $$\mathtt{(n \bmod 4) – 1}$$ is 0, in which case I just subtract 3. I’ll explain this in the future, but for now let’s change Player A’s strategy to that and keep Player B’s random to see if we can guarantee a win for Player A.

I think we nailed it in the code at the right. That answers Question A, probably a little more completely than we needed to, but it’s still answered mostly experimentally. We can worry about elegance later. There is a way that Player A can always win, when starting with eleven matches. No matter what Player B plays, as long as Player A plays $$\mathtt{(n \bmod 4) – 1}$$ (or 3, when $$\mathtt{(n \bmod 4) – 1}$$ results in $$\mathtt{-1}$$), then Player A will win.

Does the same strategy work when starting with 30 matches? My guess is that it should, since the rules haven’t changed (each player can still only pick up 1, 2, or 3 matches) and we built our strategy up from the simplest case. Let’s replicate the strategy in the code but change the count to start at 30 and see what we get.

Question B and the Rest

I think that answers Question B, mostly experimentally, just as we answered Question A. Here the code shows a bit more elegance. Instead of using an if statement, we can write Player A’s choice formula as $$\mathtt{(n+3) \bmod 4}$$, and it accomplishes the same thing as above. And, really, we can answer Question C too, at least partly and tentatively. And the answer is no. If we had to start with, say, 29 matches or 9 matches, our formula would tell us to play 0 on the first play. These are numbers that we want for our opponent, not for us, because they lead inexorably to a win for us, no matter what our opponent plays. So, starting off with $$\mathtt{4n + 1}$$ matches, when picking up 1, 2, or 3 matches, does not guarantee a win.

A good guess for a more general answer to Question C—given that the 4 in mod 4 seems to be the maximum number of matches that can be picked up, $$\mathtt{m}$$ plus 1—is that the game is not winnable if the starting number of matches is $$\mathtt{(m + 1)n + 1}$$, where $$\mathtt{m}$$ is the maximum number of matches that can be picked up and $$\mathtt{n}$$ is the set of natural numbers, $$\mathtt{\{1, 2, 3,\dots\}}$$.

I’ll leave it to the reader to poke holes in that if it’s wrong or tighten the screws if it’s essentially correct.

Evident Even to an Ass

Suppose there is a straight line segment $$\overline{\small\mathtt{AB}}$$ between you, at point $$\small\mathtt{A}$$, and some point $$\small\mathtt{B}$$ you want to reach. Is there a shorter way to point $$\small\mathtt{B}$$ that uses 2 line segment paths? If you think the answer is yes, or if you think that the answer is no but it’s not obviously no, the Epicureans have some pretty mean things to say about you—or, rather, about the Triangle Inequality Theorem:

It was the habit of the Epicureans, says Proclus, to ridicule this theorem as being evident even to an ass and requiring no proof, and their allegation that the theorem was “known” even to an ass was based on the fact that, if fodder is placed at one angular point and the ass at another, he does not, in order to get to his food, traverse the two sides of the triangle but only the one side separating them (an argument which makes Savile exclaim that its authors were “digni ipsi, qui cum Asino foenum essent”). Proclus replies truly that a mere perception of the truth of the theorem is a different thing from a scientific proof of it and a knowledge of the reason why it is true.

The Triangle Inequality Theorem says that the sum of the lengths of any two sides of a triangle is always greater than the length of the remaining side. And it is fascinating to me how seldom I have seen this theorem in textbooks phrased as a “shortest straight-line distance” statement. Most often, I see investigations about what side lengths can make up a triangle, which turns the ridiculously obvious into a complicated issue. But let’s discuss that below. For now, a proof!

It’s Okay To Be Both Obvious and Require Proof

We want to show that $$\small\mathtt{BA + AC}$$ is greater than $$\small\mathtt{BC}$$, that $$\small\mathtt{AB + BC}$$ is greater than $$\small\mathtt{AC}$$, and that $$\small\mathtt{BC + CA}$$ is greater than $$\small\mathtt{AB}$$.

Euclid starts by extending $$\overline{\small\mathtt{BA}}$$ to a point $$\small\mathtt{D}$$ such that $$\small\mathtt{DA = AC}$$ as I’ve shown in the diagram. This means that $$\small\Delta\mathtt{ADC}$$ is an isosceles triangle, and $$\small\measuredangle\mathtt{ACD}$$ and $$\small\measuredangle\mathtt{ADC}$$ are congruent. And since “the whole is greater than the part,” m$$\small\measuredangle\mathtt{BCD}$$ is greater than m$$\small\measuredangle\mathtt{ACD}$$, which also makes m$$\small\measuredangle\mathtt{BCD}$$ greater than m$$\small\measuredangle\mathtt{ADC}$$.

Dizzy yet? Just remember the last rung of the ladder we got to: m$$\small\measuredangle\mathtt{BCD}$$ is greater than m$$\small\measuredangle\mathtt{ADC}$$. Therefore, we can say something about the line segments that those two angles “catch.” Specifically, we can say that $$\overline{\small\mathtt{DB}}$$ is longer than $$\overline{\small\mathtt{BC}}$$, because Proposition 19. This is the same as saying that $$\small\mathtt{DA + AB > BC}$$. And since $$\small\mathtt{DA = AC}$$, we can make a substitution to get $$\small\mathtt{AC + AB > BC}$$, which is the first of the three statements above that we wanted to prove: $$\small\mathtt{BA + AC}$$ is greater than $$\small\mathtt{BC}$$. Euclid tells us that we can prove the other two statements with a similar method, and I believe him.

Internalizing the Idea That Mathematics Is Complex and Intuition-Free

In this link is an example of what we often put students through to investigate the Triangle Inequality Theorem. This is consistent with what I have seen in a lot of lesson plans.

Were the authors of this document aware that there is a very intuitive way of looking at triangle inequality? It might be enough to suppose that they weren’t and that they, like all of us, sometimes just repeat what we see or hear elsewhere.

But it’s also worth entertaining the possibility that they did know and pressed on anyway. What good reasons might they have for doing so? I think the best answer is simply Proclus’s reply above: “a mere perception of the truth of the theorem is a different thing from a scientific proof of it and a knowledge of the reason why it is true.” To which I would respond, Indeed, a different thing. Not necessarily a better thing.

If we can give students the “mere perception of the truth” of a theorem (or of any mathematical idea), we should do so, even if it doesn’t make us feel smart, leaves a lot of class time to fill, or runs counter to a set of standards. I would argue that students still have to prove those theorems and justify those ideas. But they can then do so correctly oriented to the reality of what they are doing: drawing on their own perceptions and knowledge to make their ideas plain to others.

Flattening Structures

This is a fun presentation by Po-Shen Loh, the coach of the winning 2015 (and 2016) U.S. Math Olympiad team, whom I mentioned here, and is primarily about the Catalan numbers. It’s not too advanced, but it’s by no means a basic intro to these numbers. And there’s a nice instance of what I’m calling “flattening” in the presentation.

James Tanton’s video explanation at the previous link is more straightforward if you’re interested in seeing how the formula for these numbers can be conceptualized. Regardless, I am more interested in the little problem Loh’s group goes through at the beginning of the presentation (and comes back to at the end) for the purposes of this post. The first part of it goes something like this:

Choose a point in the second quadrant of a coordinate plane, say (-2, 8). Choose another point in the fourth quadrant, say (3, -2). Making only right and down moves (east and south), one unit per move, how many possible ways are there to move from the first point to the second?

If you haven’t seen a problem like that before and would like to think about it, also think about how you would interpret the CC standard for mathematical practice of looking for and making use of structure in this problem.

Flattening. And Bifurcation

I’m also not interested in explaining the solution here, so I’ll refer you to a pretty good explanation at Better Explained. One quibble, though, about the explanation there: my sense is that an opportunity for greater clarity is missed in Approach 1 by not pointing out that
$\small\mathtt{\dbinom{n}{k} = \dbinom{n}{n – k}}$

Anyway, the move I like in the video (and at Better Explained) is the move that flattens the problem. For the example above, this involves perceiving that, however we get there, we have to make exactly 5 moves right (east) and exactly 6 moves down (south). We no longer have to deal with a two-dimensional problem, but a one-dimensional one; a list of moves: RRRRRDDDDDD (or EEEEESSSSSS). This is flatter.

There is also a bifurcation here—a splitting into two—when we perceive the two components D and R as being the only two objects we really have to deal with. Although this is essentially given in the setup (we’re told we can only make down and right moves), it requires a conscious effort on the part of novices to simplify in this way—to avoid seeing a mental mess of stairstep configurations from point $$\small\mathtt{A}$$ to point $$\small\mathtt{B}$$ as the structure of the problem. One can see this bifurcation in Tanton’s explanation of Catalan numbers too.

Tearing Down the Structure

Just as with looking for symmetry, flattening and splitting can be considered two general techniques that, in some sense, break down the structure of a problem rather than “make use of it”—they are different structures than the ones given by most problems. The authors of the Common Core no doubt meant “mathematical structure” when they wrote just “structure,” but we should be careful to mean the same thing when we teach this practice.

At any level. Consider this 6th grade proportion problem:

For every 5 baseball cards that Mickey has, Mike has 3 baseball cards. If Mike has 72 baseball cards, how many baseball cards does Mickey have?

It might not seem to be worthy of the 10-dollar word “bifurcation” to see in this problem that there are just two groups of cards to consider—Mike’s and Mickey’s—but that simplification can be a useful and indeed powerful step for a student to whom the above is, at first glance, a salad of “Mik–“, “baseball cards”, and numbers. Mike has 72 on one side, and Mickey has we-don’t-know on the other.

The flattening here is harder, and it really needs to be taught. While there’s nothing wrong with setting this up as the proportion $$\mathtt{\frac{5}{3} = \frac{x}{72}}$$, we can flatten this to good effect if we see the ratio comparison number in the first sentence. Rather than 5 to 3, it’s just that Mickey has $$\mathtt{\frac{5}{3}}$$, or $$\mathtt{1\frac{2}{3}}$$, times as many baseball cards as Mike. The resulting setup can even be literally flatter (and there may be no need for the variable): $$\small\mathtt{1.\overline{6} \cdot 72 = 120}$$.

Image credit: peasap.

Sequencing Props

Last time (here), we talked about III.27 from The Elements, which told us that “in equal circles angles standing on equal circumferences are equal to one another, whether they stand at the centres or at the circumferences.” Here we’ll go over the proof of the previous proposition in the sequence, III.26, which is different from III.27 in ways not easy to access based on just the description: “in equal circles equal angles stand on equal circumferences, whether they stand at the centres or at the circumferences.” So . . . let’s take a look.

We start by drawing some things and making some assumptions. First, Circles $$\small\mathtt{ABC}$$ and $$\small\mathtt{DEF}$$ are congruent, the central angles ($$\small\mathtt{\color{green}{\measuredangle{BGC}}}$$ and $$\small\mathtt{\color{green}{\measuredangle{EHF}}}$$) are congruent, and the inscribed angles ($$\small\mathtt{\color{blue}{\measuredangle{BAC}}}$$ and $$\small\mathtt{\color{blue}{\measuredangle{EDF}}}$$) are also congruent. What we want to show is that the minor arc $$\small\mathtt{BC}$$ is congruent to minor arc $$\small\mathtt{EF}$$.

Chain of Begats

Euclid proceeds by drawing $$\overline{\small\mathtt{BC}}$$ and $$\overline{\small\mathtt{EF}}$$, which I’ve already done above. He then concludes that $$\overline{\small\mathtt{BC}}$$ and $$\overline{\small\mathtt{EF}}$$ are congruent by reasoning as follows (using SAS, or I.4, which we covered here):

1. Circles $$\small\mathtt{ABC}$$ and $$\small\mathtt{DEF}$$ are congruent, so their radii are congruent.
2. This means that $$\overline{\small\mathtt{BG}}$$ is congruent to $$\overline{\small\mathtt{FH}}$$ and $$\overline{\small\mathtt{CG}}$$ is congruent to $$\overline{\small\mathtt{EH}}$$.
3. The central angles ($$\small\mathtt{\color{green}{\measuredangle{BGC}}}$$ and $$\small\mathtt{\color{green}{\measuredangle{EHF}}}$$) are congruent by assumption.
4. Thus, $$\overline{\small\mathtt{BC}} \cong \overline{\small\mathtt{EF}}$$ via SAS.

The Peril of Working Backwards

The sequence of the propositions in the Elements is important, which we’ll feel right now when I tell you that the next step in this proof relies on a previous theorem, III.24. That proof, which we’ll have to get to later, shows that when two circles have arcs (“segments,” actually) that lie on congruent line segments, those two arcs are also congruent. (There’s also some weirdness in the definition of “similar segments” used in that proof [and this one] which we may want to think about later too.)

For our purposes, this means that major arc $$\small\mathtt{BC}$$ is congruent to major arc $$\small\mathtt{EF}$$. And since the two entire circles are congruent by assumption, this means that the minor arcs must be congruent too.

So, in III.27, we showed that the angles are equal when we are provided a given that the intercepted arcs are equal. In this proof, we show that the arcs are equal when we are provided a given that the angles are also.

Sequence Is Important to Think About, Not Necessarily to Follow

Order is something I definitely think a lot about when it comes to designing curricula. And our public conversation about sequencing in mathematics education strikes me as incredibly naive. What I have to listen to more often than not is that students “need X to do Y.” I mentioned above that a proof of III.24 was used to construct a proof of III.26, but it does not necessarily follow that III.24 is required to create this proof (it might be required, though). Needing X to do Y is often defended on similarly shallow evidence. Usually, the foundation for this argument is simply the sequence that the arguer is used to. Or a shallow understanding of a topic.

While it is certainly true that some topics must appear before others, we should also think about what benefits to learning playing around with order can bring. At the moment, what seems to drive almost all of our sequencing decision-making is tradition, ignorance, and fear.

Image credit: pratanti

Ceva’s Theorem

The structure of this proof, a proof of Ceva’s Theorem, comes from a neat little book called Geometry Revisited. But I’ll start with something the authors of that book ask us to recall: “that the areas of triangles with equal altitudes are proportional to the bases of the triangles.”

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The equation below the left triangle shown is of the form $$\small\mathtt{A_1 = b_1 \cdot k}$$ and the equation for the triangle to the right is $$\small\mathtt{A_2 = b_2 \cdot k}$$, where $$\small\mathtt{A}$$ stands for area (in square pixels) and $$\small\mathtt{b}$$ stands for base length (in pixels). Because the two triangles have equal heights (equal altitudes; in this case 180 pixels), we can use the constant $$\small\mathtt{k}$$ to represent $$\small\mathtt{\frac{1}{2} \cdot h}$$ in each area equation.

You can drag the bottom left vertex of the leftmost triangle or the bottom right vertex of the rightmost triangle to change the base length of either triangle. Note that the ratio of the areas is equal to the ratio of the base lengths (i.e., “the areas of triangles with equal altitudes are proportional to the bases of the triangles”). Since we have $$\small\mathtt{A_1 = b_1 \cdot k}$$ and $$\small\mathtt{A_2 = b_2 \cdot k}$$, then $$\small\mathtt{\frac{A_1}{b_1}}$$ gives us the same number as $$\small\mathtt{\frac{A_2}{b_2}}$$, namely $$\small\mathtt{k}$$.

What the Theorem Says and a Proof

Draw three line segments $$\overline{\small\mathtt{AX}}$$, $$\overline{\small\mathtt{BY}}$$, and $$\overline{\small\mathtt{CZ}}$$ inside a triangle, each from a vertex to the opposite side. Ceva’s Theorem (Ceva is pronounced CHEH-vuh) says that if the line segments meet at a point (are concurrent), then $$\small\mathtt{\frac{BX}{XC} \cdot \frac{CY}{YA} \cdot \frac{AZ}{ZB}}$$ = 1.

So, let’s see. Given what we learned above, $$\small\mathtt{\frac{BX}{XC}}$$ = $$\mathtt{\frac{A_{ABX}}{A_{AXC}}}$$. Each of these ratios is also equal to $$\mathtt{\frac{A_{PBX}}{A_{PXC}}}$$.

That last ratio is also equal to $$\mathtt{\frac{A_{ABX} – A_{PBX}}{A_{AXC} – A_{PXC}}}$$, because we can safely subtract across the numerators and then the denominators when working with equal part-to-part ratios (can you work out why?). Finally, subtracting the area of $$\small\Delta$$$$\small\mathtt{PBX}$$ from the area of $$\small\Delta$$$$\small\mathtt{ABX}$$ gives us the area of $$\small\Delta$$$$\small\mathtt{ABP}$$, or $$\small\mathtt{A_{ABP}}$$. Similarly, subtracting the area of $$\small\Delta$$$$\small\mathtt{PXC}$$ from the area of $$\small\Delta$$$$\small\mathtt{AXC}$$ gives us the area of $$\small\Delta$$$$\small\mathtt{CAP}$$, or $$\small\mathtt{A_{CAP}}$$. We have wound up with a chain of equals signs to prove Ceva’s Theorem, giving us, in the end, $$\small\mathtt{\frac{BX}{XC}}$$ = $$\small\mathtt{\frac{A_{ABP}}{A_{CAP}}}$$.

So now we do that with the other two sides. Here are all three equals-sign chains:

1. $$\small\mathtt{\frac{BX}{XC}}$$ = $$\mathtt{\frac{A_{ABX}}{A_{AXC}}}$$ = $$\mathtt{\frac{A_{PBX}}{A_{PXC}}}$$ = $$\mathtt{\frac{A_{ABX} – A_{PBX}}{A_{AXC} – A_{PXC}}}$$ = $$\mathtt{\frac{A_{ABP}}{A_{CAP}}}$$
2. $$\small\mathtt{\frac{CY}{YA}}$$ = $$\mathtt{\frac{A_{BCY}}{A_{BYA}}}$$ = $$\mathtt{\frac{A_{PCY}}{A_{PYA}}}$$ = $$\mathtt{\frac{A_{BCY} – A_{PCY}}{A_{BYA} – A_{PYA}}}$$ = $$\mathtt{\frac{A_{BCP}}{A_{ABP}}}$$
3. $$\small\mathtt{\frac{AZ}{ZB}}$$ = $$\mathtt{\frac{A_{CAZ}}{A_{CZB}}}$$ = $$\mathtt{\frac{A_{PAZ}}{A_{PZB}}}$$ = $$\mathtt{\frac{A_{CAZ} – A_{PAZ}}{A_{CZB} – A_{PZB}}}$$ = $$\mathtt{\frac{A_{CAP}}{A_{BCP}}}$$

Then just cross out stuff to see that $$\mathtt{\frac{A_{ABP}}{A_{CAP}} \cdot \frac{A_{BCP}}{A_{ABP}} \cdot \frac{A_{CAP}}{A_{BCP}}}$$ = 1, which means that $$\small\mathtt{\frac{BX}{XC} \cdot \frac{CY}{YA} \cdot \frac{AZ}{ZB}}$$ = 1.

Huzzah!

Conceptual Understanding Is Always There

In education, we seem to take some delight in shoveling a confused mix of folksy connotations into sciencey-shelled words and phrases. Some of my colleagues would call the result edujargon, though I think that word allows us to feel too smug about our own obtuseness—as though the problem is that the field of education is so darned technical. Anyway, I’ve been itching to pick on one such phrase lately: conceptual understanding

And I think I’ll start here with the on-picking. In doing so, I want to be clear that I’m not reaching for what ‘conceptual understanding’ actually means. Rather, I’d like to suggest that what emerges from the everyday way we talk about this concept is a meaning that, at best, doesn’t make good sense.

Always with the Algorithms

Take a look at the student work at the right and marvel at how his handwriting perfectly replicates a web font. Then decide what he might have done wrong.

If you showed this work to any gathering of K-8 math teachers, it wouldn’t be long before someone suggested (after emphasizing that there’s no way to tell from one example) that the student probably doesn’t have a strong ‘conceptual understanding’ of addition—or something along those lines. And a compelling reason to make this judgment, assuming that the student’s error is consistent, is that the sum is less than one of the addends. That is, the answer is impossible given the meaning of what the student is doing (combining two positive quantities), yet the student seems to be unaware that there even is a meaning to be had in adding (again, assuming the mistake is consistently made).

Conceptual understanding, then, has to do with meaning. That is certainly the takeaway I wanted you to get from the hypothetical above, but this is also a point to which others have returned when writing about this concept:

A procedure is a sequence of steps by which a frequently encountered problem may be solved. For example, many children learn a routine of “borrow and regroup” for multi-digit subtraction problems. Conceptual knowledge refers to an understanding of meaning; knowing that multiplying two negative numbers yields a positive result is not the same thing as understanding why it is true.

Thus, it is important to not only teach students how to go about solving problems, but also to teach them the frameworks in which those problems obtain meaning. Just one meaning of addition, for example, is a combination of values or quantities, and just one advantage of being familiar with that meaning is that it allows us to discern the reasonableness of sums. This is what we generally talk about when we talk about conceptual understanding.

But Let’s Go Deeper

As constructivists, though, we must all concede that human beings are meaning-making machines. They simply can’t help themselves. We see meaning in places where no given meaning is present. Pareidolia, Rorschach tests, and apophenia are some of the more easily googleable examples of our ineluctable fondness for injecting our thoughts onto a random and uncaring universe.

Of course, being a meaning-making machine does not mean that the machine works correctly all the time. Nor does this observation say anything about the quality of the meanings thus generated. What comes out is typically far from a model of logical perfection or mathematical purity.

So, why do we talk about students as having no conceptual understanding or weak conceptual understanding? What in the heck are we talking about when we say things like this? The student who solved the addition problem incorrectly above does have a conceptual understanding of addition. He has constructed a meaning for addition. It’s just that he has, with the help of his schooling, his parents, and his community, constructed the wrong meaning–an attenuated meaning.

A stellar example of what misguided conceptual understandings look like is Robert Kaplinsky’s “How Old Is the Shepherd” task. Here we have a problem whose only requirement, really, is to grasp its meaning in order to solve it:

It is also a good example to use to point out the bone-headedness of talking about conceptual understanding as something that can be weak or non-existent. Of the 75% of students who completed the task incorrectly, what did virtually all of them do? If a lack of something caused all of these students to misfire, this lack also caused all of them to misfire in the same basic direction: to take the numbers in the problem and operate on them.

No. There is conceptual understanding here—a conceptual understanding that was given to them and reinforced over and over in the years leading up to this. And there is plenty of meaning that these students are working with here. It’s just that it’s the meaning that their parents are comfortable with, not the meaning that will be of much use in their own futures.

Why This Is Important

The distinction between conceptual understanding as something you can have a certain quantity of (possibly 0) and conceptual understanding as the unavoidable result of mental activity lighting up and organizing your interior and exterior worlds is important primarily because it represents a divergence in diagnoses that ultimately leads to a divergence in treatment, as it were.

The latter view is almost certainly better for business, for example. If I can convince you that you have a hole to fill, I can more easily sell you something to fill it. It is also a view that can help steer attention away from more systematic failures of schooling—including what teachers are taught in universities—onto the isolated U.S. classroom teacher.

Conceptual understanding, as we have seen, is never not there; it is a complex and emergent property of knowledge shaped by a student’s interactions with the whole of her schooling, her parents, and her community. We should at the very least be aware that our own conceptual understanding of conceptual understanding, including the way we talk about it amongst ourselves and with our students, has consequences.