Marveling At The Historical

Math Oldies But Goodies

  • About This Blog

    This blog is mostly about math procedures in textbooks dated from about 1825-1900. I’m writing about them because some of the procedures are exquisite and much more powerful, and simpler, than some of the procedures in current text books. Really!

    I update this blog as frequently as possible ... every 2-3 days. And, if you are a lover of old texts and unique procedures, you might want to talk to me about them, at markdotmath@gmail.com. I’m not an antiquarian; the books I have are dusty, musty, brown-paged scribbled-in texts written by authors with insights into how math works. Unfortunately, most of their procedures have vanished. They’ve been overcome by more traditional perspectives, but you have to realize that at that time, they were teaching the traditional methods.

Archive for the ‘equations’ Category

Ted’s Question: Can I Graph a Decimal Slope?

Posted by mark schwartz on November 16, 2016

Introduction

We were working on graphing lines using the slope-intercept method.

The equation to graph was y = 4/3x + 2. Traditionally, plot the point (0, 2) first – the y-intercept and from this point, move up 4 units (positive 4 on the y-axis) while moving 3 units to the right (positive 3 on the x-axis). This finds the second point at (3, 6). This process gives an accurate line between these two points.

Ted asked “If I use my calculator to find the value for the slope, I get 1.33 … can I use 1.33 as the slope to graph the line”? Having never heard this question before, I said I wasn’t sure but let’s look at it.

The Story

As it turns out Ted is correct … 1.33 can be used but it’s important to understand how to use it.

It goes back to a basic fraction relationship. In order to preserve the relationship between the numerator and denominator, it is allowable to multiply or divide both the numerator and denominator by the same value. This is what is done when searching to either find an equivalent fraction when reducing a fraction to lowest terms or finding an equivalent fraction for adding or subtracting fractions.

Given this, it’s not that the fraction is converted to a decimal by dividing 4 by 3. Rather the mathematical operation is to divide both the numerator and denominator by 3, giving the fraction 1.33/1. When we do this conversion, we typically don’t note the denominator of 1; it simply is ignored as if it weren’t there.

So, back to plotting the equation. Again starting at (0, 2), we would move up 1.33 (move positive 1.33 on the y-axis) while moving right 1 (move positive 1 on the x-axis). This is valid and falls on the line plotted when using slope = 4/3.

Well, not exactly. Using 1.33 isn’t quite as accurate as using 4/3, simply because, in this case, it is a repeating decimal. But, even without a repeating decimal, there still is the possibility of a loss of accuracy. Of course, for classroom purposes this might be acceptable After all, we’re not designing a spacecraft that needs quite accurate calculations for design and flight.

Using this decimal idea with y = 3/5x + 2, we would have y = .6x + 2. The plot again begins at (0, 2). The issue now is the scale on the x and y axes. If these axes are laid out in .1 increments, then .6 can readily be used with the same accuracy as 3/5, but if the scale is in whole units, the .6 is an ‘eyeball’ estimate and may not be as accurate. As a reminder, in this case, when moving up .6 on the y-axis, move a corresponding 1 on the x-axis. When using a decimal, the denominator (change on the x-axis) is always 1.

However, the question was wonderful and exploring it was interesting and … well, educational.

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Heron’s Area of a Triangle

Posted by mark schwartz on October 24, 2016

Introduction

This is a brief story about a fun event that almost always happens when discussing the area of a triangle. The formula for the area is A = 1/2 bh, where ‘b’ is the base and ‘h’ is the height.

The Story

Simple enough if it’s a right triangle and the base and height can readily be seen or calculated from a2 + b2 = c2. But what happens when it’s not a right triangle? Well, one has to wiggle around a bit and do a few more calculations to determine the height, but it can be found.

But being an instructor that likes to stretch students thinking and imaginations, I draw a very scalene triangle with sides of 4, 8, and 10 and present them the task of finding the area. As I roam the room watching them work and listening to their grumblings, I ultimately have them stop and present to them Heron’s formula. I don’t bother with the derivation and for those that are interested in knowing it, I recommend Googling it.

The formula is A = 1/4 the square root of (a + b + c)( ̶ a + b + c)( a ̶ b + c)( a + b ̶ c).

This is a really nifty formula because of the pattern in it – add all the sides together, then in the next three parentheses, just negate each side, one at a time, in order. And, given that there are 4 parentheses, just divide the square root of this product by 4.

In class, I first apply the formula to a 3, 4, 5 right triangle to demonstrate how it works. Then we play with a few other right triangles to get comfortable with the formula. Then we return to that weird scalene triangle. Using Heron’s formula, we get an area of 15.1987 … and now the fun begins!

Someone typically asks “how do you know this is correct?” So, this sets us up to explore how to find the area if we didn’t have Heron’s formula; so we do all the Algebra necessary to demonstrate that indeed 15.1987 is correct. In one class, this whole presentation got applause and I take that as their having had fun with it.

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Must We Filter Students Through the Math Sieve?

Posted by mark schwartz on October 19, 2016

Deborah Blum in The Best American Science Writing, 2011 (page 184) cites a California Institute of Technology science historian as saying “K-12 science classes in the United States are essentially designed as a filtration system, separating those fit for what he called ‘the priesthood of science’ from the unfit rest of us.”

I believe the same can be said for math classes. Of course, I can assume that math was included in science, but to be very specific about it, math actually seems to be a more severe filter than general science. Today, many science classes involve students in exploration and experimentation and some of the valuable lessons of accurate measurement, recording and analysis. And, some of these activities include the necessity of math. But, when doing math in a vacuum, unrelated to an activity – in essence, the math part of the activity is secondary – the filtering action seems more apparent.

For example, in today’s texts there are typically sections on “applications”. There are even entire texts dedicated to applications and these applications show the students how math is in our everyday activity – sports, statistics, banking, calculating interest, taxes, consumption, measurements of all kinds. And this is fine. But, it’s still done in the context of filtering those who have an aptitude for it from those who don’t because …

Texts still tend to present formulae and algorithms and teachers say “this is how to do it”. In essence, teachers are saying “here’s how to do it” rather than asking “how do you imagine how this can be done?”   We don’t ask students to generate their own conception of how to solve the problem, most likely because we believe they can’t or don’t. However, many math researchers of early childhood “math” capability have found that even before entering elementary school, most children are already identifying quantitative relationships, imagining algorithms that help them understand the relationships, verifying that their conception will always work, and subsequently and repeatedly, altering their algorithm if their conceptions don’t work. It’s sort of a fundamental, built-in scientific approach to what’s going on around them. So, having created their own quantitative environment, what happens not only to the environment but also – more critically – their formulating such systems when the teacher, the text, and “schooling” provides the algorithms for them? Who needs to continue exploring the pieces of the puzzle when a solution methodology is already provided? Further, if a student in elementary school proposes a solution differing from the text, is the teacher prepared to explore that proposal to its end and see if indeed it may be worthy of consideration?

When math is taught, it in essence teaches students not to think about the relationships. The tendency – and the pedagogy – is to teach students how not to think about it because we proffer the historically valid rule, procedure, formula or algorithm which allows them to get to the answer in the most efficient way (“rule” will be used from now on to summarize procedures, formula, algorithms, etc.). Why mess around with inefficient or erroneous methods? Just give them the rule and have them practice it. Well, this does two things: first of all, practice doesn’t make perfect, rather perfect practice makes perfect and second, it suppresses what seems to be a natural urge to play with the information presented and explore the quantitative relationships that might be there.

Let’s address the practice concept for a moment. A common phrase touted by math instructors is “math is not a spectator sport” or “you don’t learn math by watching others do it.” There is some validity to this, but there is also the reality that as Yogi Berra commented “you can observe a lot just by watching.” But the question is, what is it that students should be observing? Watching a math instructor use a predetermined rule to solve a pre-established problem and then ask students to mimic this activity may actually work for some students. But, in a broader sense, what is it that we want students to learn when we teach math?

This is not a simple question and doesn’t have a simple answer. Most likely, the answer is to get students to be able to do the indicated calculation or solve the problem. But is that what is intended for them to learn? Should the lesson be about applying a rule or about exploring the quantitative relationship? Rather, it’s establishing a context in which the student can imagine alternative rules and test those rules for reliability and validity. And what are we, as instructors to do, if a student discovers a less efficient but comparably valid rule? Here’s where we run into the range of expectation of the instructor as well as the training and experience of the instructor.

Going back to the premise of math learning as a filter system, it seems reasonable to assume that all students, those who can attain the priesthood and those who can’t, could manage in a system that allows and prompts for exploration, rather than being given the rules. It would still act as a filter system, but the real key is that those not destined for the priesthood would gain a better grasp of quantitative and mathematical relationships. Basically, it is math learning by doing but the “doing” is now differently defined.

Here’s something that happened in class one day. We were just beginning to work with simple equations in an introductory Algebra class. The text approached setting up the equation by making a statement which could be directly translated to an equation. This has become a typical introductory approach. For example, the student is asked “if you take a number, double it and add 1, the result will be 5. What is the number?” The expectation is that the student will write “x”, then double it by writing “2x”, then add 1 by writing “2x + 1” and then showing that 2x + 1 will have a result of 5 by writing the equation 2x + 1 = 5.

As I moved around the room watching and helping students work through this translation, this is what I saw on one student’s paper:

P   P   P   P   X                   The answer is 2.

I asked her how she got 2 as an answer and it went something like this: I knew there were 5 pieces when I got done, so I wrote “P” five times. But since one was added, I had to take one away. So, one of the “Ps” became an “X”. Then, since the number was doubled, I had to take half of it, so half of the 4 “Ps” that are left gave me 2.”

This is perfect logic and a valid way to reason through to the answer. In essence, she saw that the process could be reversed and mapped it. It doesn’t, however, meet the intended goal of

having a student construct and then solve an equation. What is an instructor to do? Consider that in the future, this student might be asked to solve the equation 4 ─ 2(2x + 1) = 3x + 5. Can this equation be solved using this student’s strategy? Yes, but not as efficiently as the traditional equation solving strategy. What happens to this student’s sense of self, sense of algebra and equations, and sense of quantitative relationships if, as an instructor, I have to say “no, that’s not the way to do it.”?

And the issue isn’t only the student; it’s the pedagogy. It seems that the pedagogy is probably more the issue because it doesn’t allow students to try out various strategies and come to the realization that their strategy works for some equations but not all equations. They now have a choice. They can learn several strategies and tailor the strategy to the circumstance, or accept the traditional pedagogy which offers an efficient method for solving equations of all types. It may be contended that if the student builds a library of different strategies for different equations, that it may be a big library and there may be an equation not amenable to one of the strategies. I would reply that a strategy developed and employed by a student is likely to be better remembered, and modified as necessary, than one that is presented and never “owned”.

There are ways of approaching the solution of equations which allow for the type of visual representation that this student used. Further, equations can be solved using objects and images or both; no paper and pencil need be used – at least not at first. All students could be started with this student’s approach and as the equations become more sophisticated, it could be noted that an alternative strategy needs to be used for these more sophisticated types of equations. Starting with their conceptions may well result in their all coming to the conclusion that the most efficient strategy – the classic traditional strategy – is most favorable. However, consider that getting to this point would take more time, yet that time is valuable in establishing students’ capability to imagine alternative methods, compare and contrast them, and conclude which is best. Further consider that when students are taught, for example how to solve systems of equations, texts and instructors teach the substitution and the addition method, and sometimes even matrix and determinants. We bother to do this because, with some examination before plunging into the solution, it may be determined that one method is better than the other, under the circumstance. So, why not allow students to use their methods as well as they work their way through solving the problem?

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A Student’s Aha Moment

Posted by mark schwartz on October 12, 2016

Introduction

One more example of how an aggregation of student’s imaginations led the class, including the instructor, to apply Algebra to Algebra. A type of problem which frequents texts in the U.S. is the ‘mixture’ problem. Mixing solutions of different concentrations to get a third with a desired concentration; mixing different kinds of candy or nuts to get a mixture to sell at a certain price, or getting a return on investing in two accounts at different interest rates (and other applications).

The Story

Gus has on hand a 5% alcohol solution and a 20% alcohol solution. He needs 30 liters of a 10% alcohol solution. How many liters of each solution should he mix together to obtain the 30 liters?

The classical solution is to write the equation .05x + .20(30 – x) = .10 (30). Solving the equation, the outcome is 20 liters at 5% and 10 liters at 20%.

As this type of problem was discussed, one student asked “can 15+x and 15-x be used, since they add to the total of 30 and this seems easier?” I wasn’t certain if it would work nor why it seemed easier, but we explored the idea. The student presented the problem on the board as 5(15 + x) + 20(15 – x) = 10 (30). Not only did he re-craft the unknowns, but he used whole numbers, not decimals. When asked why, he simply stated that we would be getting rid of the decimals anyway. I noted that ‘getting rid of’ is not a mathematical operation, but clearly it works. The answer to this equation is -5, and some students believed this solution to be awkward for two reasons. First because x = -5, and a negative quantity doesn’t make sense and second, because finding x doesn’t finish the problem. Recall that the unknowns in the equation are 15+x and 15–x, so another step is required to come to the correct answer of 20 liters at 5% and 10 liters at 20%. Then the question came: “Isn’t there another way to do this?”

At this point, I introduced “Alligation”, a procedure described in detail in a post in this blog titled Mixing it up with Alligation. I won’t go into detail about the procedure, so look it up if you’re interested. It’s a very different approach which was popular in the 1800s but doesn’t seem to be in any current texts.

We did discuss several other ideas and it was an enjoyable session in which the class actually reported having fun doing Algebra!

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Homework: Solve This Equation 4 Ways

Posted by mark schwartz on October 2, 2016

Introduction

I once had a class of pre-service and in-service teachers and one of our discussions led us to explore using one’s imagination to work through a math problem, rather than relying on the standard, traditional algorithm. I asked if all of them had taken Algebra and were comfortable solving equations. As it turns out they, collectively, were quite adept at it. So, the following story shows not only how adept they were but also how imaginative they were.

The Story

First let me note that they worked in groups, so the following homework assignment was started in class and I urged them to swap email addresses so they could work on it at home, which they did. Here’s the homework: you are to solve 4 ─ 2(x ─ 1) = 8 + 6(x ─ 1) 4 different ways and annotate the steps in your solution. We reviewed the traditional algorithm in class, so it didn’t count, but I did include it in the following list of solutions.

At our next class, I asked them to write their alternative solutions on the board and although there were many similar solutions, listed below are some of the unique ones. As we discussed these, the question was asked “when teaching math, should we ever have students do an exercise like this?”

Some felt we should; some felt otherwise. The critical difference was whether or not their school’s curriculum would allow for this. They liked the idea of using imagination and had fun doing the exercise but they collectively concluded that students in elementary and secondary school should stick to the traditional. I asked them that if a student, unprompted and independently, worked an equation – or solved any other mathematical expression – in an unorthodox way, what would they do?

It was a good discussion with no clear closure about what to do with the unorthodox student. Given the current press in Common Core Math, where students are to express their reasoning, it seems that accepting unorthodox solutions might be reasonable, but on the other hand a student may have an “aha” moment and can’t clearly articulated how the solution was found. It’s an interesting challenge for today’s teachers.

Here are some of the results of their work, annotated. Follow what they did because clearly imagination was in play.

Traditional:

4 ─ 2(x ─ 1) = 8 + 6(x ─ 1)

4 ─ 2x + 2 = 8 + 6x ─ 6

6 ─ 2x = 2 + 6x

4 = 8x

1/2 = x

 

1st Alternative (only a slight difference, but a difference):

4 ─ 2(x ─ 1) = 8 + 6(x ─ 1)

2 ─ (x ─ 1) = 4 + 3(x ─ 1)       divide all terms by 2

2 ─ x + 1 = 4 + 3x ─ 3             do the indicated multiplications on both sides

3 ─ x = 1 + 3x                           combine like terms on both sides

2 ─ x = 3x                               subtract 1 from both sides

2   = 4x                                    add x to both sides

1/2 = x                                       divide both sides by 4 and simplify answer

 

2nd Alternative:

4 ─ 2(x ─ 1) = 8 + 6(x ─ 1)

0 = 4+8(x─1)                           subtract 4 ─ 2(x ─ 1) from both sides

0 = 4 + 8x ─ 8                         do indicated multiplication

0 = ─4 + 8x                             combine like terms

4 = 8x                                       add 4 to both sides

1/2 = x                                       divide both sides by 8

 

3rd Alternative:

4 ─ 2(x ─ 1) = 8 + 6(x ─ 1)

4 = 8 + 8(x ─ 1)                       add 2(x ─ 1) to both sides

─4 = 8(x ─ 1)                          subtract 8 from both sides

─ 1/2 = x ─ 1                            divide both sides by 8

1/2 = x                                      add 1 to both sides

 

4th Alternative:

4 ─ 2(x ─ 1) = 8 + 6(x ─ 1)

4/(x ̶ 1) ̶ 2 = 8/(x ̶ 1) + 6    divide every term by (x ─ 1)

─ 2 = 4/(x ̶ 1) + 6                  subtract 4/(x ̶ 1) from both sides

─ 8 = 4/(x ̶ 1)                       subtract 6 from both sides)

─ 8(x ─ 1) = 4                          multiply both sides by (x ─ 1)

─ 8x + 8 = 4                            do the indicated multiplication on the left side

─ 8x = ─ 4                                 subtract 8 from both sides

x = 1/2                                     divide both sides by ─ 8

 

5th Alternative

4 ─ 2(x ─ 1) = 8 + 6(x ─ 1)

1/2 ─ (1/4)(x ─ 1) = 1 + (3/4)(x ─ 1)     divide every term by 8

1/2 = 1+ x ─ 1                                          add   (1/4)(x ─ 1) to both sides

1/2 = x                                                    combine terms on the right side

 

6th Alternative:

4 ─ 2(x ─ 1) = 8 + 6(x ─ 1)

4 ─ 2x = 8 + 6x                   let x = x ─ 1

─ 4 = 8x                               subtract 8 from both sides; add 2x to both sides

─ 1/2 = x                               divide both sides by 8 and simplify

─ 1/2 = x ─ 1                         let x ─ 1   = x (“reverse” first operation)

1/2 = x                                   add 1 to both sides

 

7th Alternative:

4 ─ 2(x ─ 1) = 8 + 6(x ─ 1)

4(x + 1) ─ 2(x ─ 1) (x + 1) = 8(x + 1) + 6(x ─ 1)(x + 1)     multiply every term by (x + 1)

4x + 4 ─ 2x2 + 2 = 8x + 8 + 6x2 ─ 6                                do all indicated multiplications

─ 2x2 + 4x + 6 = 6x2 +8x +2                                            combine like terms on both sides

─ x2 + 2x + 3= 3x2 +4x + 1                                              divide all terms by 2

(─x + 3)(x + 1) = (3x + 1)(x + 1)                                  factor both sides

─x + 3 = 3x + 1                                                              divide both sides by (x + 1)

3 = 4x + 1                                                                       add x to both sides

2 = 4x                                                                            subtract 1 from both sides

1/2 = x                                                                          divide both sides by 4 and simplify

 

8th Alternative:

4 ─ 2(x ─ 1) = 8 + 6(x ─ 1)

16 ─ 16(x ─ 1) + 4(x ─ 1)2 = 64 + 96(x ─ 1) + 36(x ─ 1)2                  square both sides

16 ─ 16x + 16 + 4x2 ─ 8x + 4 = 64 + 96x ─ 96 + 36x2 ─ 72x +36     do all indicated operations

4x2 ─ 24x + 36 = 36x2 + 24x + 4                                                       combine like terms

x2 ─ 6x + 9 = 9x2 + 6x + 1                                                                 divide every term by 4

0 = 8x2 + 12x ─ 8                                                             subtract x2 ─ 6x + 9 from both sides

0 = 2x2 + 3x ─ 2                                                                divide every term by 4

0 = (2x ─ 1)(x + 2)                                                           factor 2x2 + 3x ─ 2

1/2 = x                                                                               setting both factors equal to 0 gives…

(x = ─ 2 is an extraneous root

Introduced when both sides were

Squared

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Vedic Version of a Line From Two Points

Posted by mark schwartz on September 25, 2016

In Vedic Mathematics (revised edition, 1992) a very interesting algorithm is presented. It allows one to find the equation of a line in standard form by visually examining the values of the two points, doing a little mental calculation, and writing down the equation! One need not use the slope-intercept or the point-slope formula.

Given two points (a,b) and (c,d), the vedic version (pg. 343) is: x(b-d) – y(a-c) = bc – ad

A slight notation change gives the standard form (ax+by =c), thus (b-d)x – (a-c)y= bc – ad

For example, using the vedic version with (9,7) and (5,2) the equation is:

(7 – 2)x – (9 – 5)y = 7⦁5 – 9⦁2, giving 5x – 4y = 17.

I was curious about this because it looked familiar; basically, the difference in the y-values is the x-coefficient and the difference in the x-values is the y-coefficient. The constant is the ‘inner’ minus the ‘outer’, if you are familiar with FOIL. As I played with this, I realized that the vedic algorithm could be derived from combining the slope-intercept and the point-slope formulae. Starting with the point-slope formula, one gets:

(y – y1) = m(x – x1

(y – y1) = ((y2 ̶ y1)/(x2 ̶ x1)) (x – x1)

(x2 – x1) (y – y1) = (y2 – y1)(x – x1)

(x2 – x1)y – (x2 – x1)y1 = (y2 – y1)x – (y2 – y1)x1

– (y2 – y1)x + (x2 – x1)y = (x2 – x1)y1 – (y2 – y1)x1

– (y2 – y1)x + (x2 – x1)y = x2y1 – x1y1 – x1y2 + x1y1

 – (y2 – y1)x + (x2 – x1)y = x2y1 – x1y2

 (y2 – y1)x – (x2 – x1)y = x1y2 – x2y1

 -1(y1 – y2)x – (-1)(x1 – x2)y = (-1)(x2y1 – x1y2)

(y1 – y2)x – (x1 – x2)y = x2y1 – x1y2

This form (y1 – y2)x – (x1 – x2)y = x2y1 – x1y2 is the vedic form (b-d)x – (a-c)y = bc – ad.

Furthermore, this vedic form allows one to generate the equation of the line if given the slope and a point, or a point with a line perpendicular or parallel to a given line because a second point can be found from the given point and the slope.

Using the same example as above, if presented the point (9,7) and the slope 5/4, the second point is (9 + 4, 7 + 5), or (13,12), as well as (9 – 4, 7 – 5), or (5,2). Consider using this vedic version.

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Unequations Buzz

Posted by mark schwartz on August 11, 2016

Introduction

Had a thought. Simple one-variable 1st degree equations, by definition, state that there is a bunch of stuff “here” that equals a bunch of stuff “there”. For example, 2(3x ̶ 1) = 5(x + 1). What is meant by “equal”? Looking at this equation, obviously the two bunches of stuff are not equal! What this statement means is that if you can find the value of the variable “x”, replace the “x’ with that value in both sides of the equation and evaluate both sides, the value on both sides of the equation sign will be equal. Thus, that’s why one solves for the value of “x”.

The fundamental rule for solving equations is “whatever you do to one side of the equation, you do to the other side.” This, in essence, maintains the equality. My thought was that rather than start with an equality and burp out the rules, start with an unequation and have students play with it to find out how to make it an equation. However, we won’t use paper and pencil; we’ll use poker chips.

The Story

In order to solve an equation of this order, students need to know a lot of stuff – identification of terms, order of operations, distributive law, the four basic operations with signed numbers and to verify their answer, substitution of a value for the unknown and of course the basic rule of “whatever you do to one side of the equation, you do to the other side.”

Solving unequations is simpler and is a kinesthetic, visual way to have students play with all those things which, in my view, expands their conception of equations. In many instances, I’ve seen students who know all the elements but somehow can’t blend them together to solve equations. Here’s how unequations work.

Each group of students (2 or 3 to a group) gets a handful of white poker chips and each chip has a positive on one side and a negative on the other. You can use other markers if you choose.

I ask them to put 1 to 5 chips in each pile but the total value in each pile can’t be the same. Two questions that always comes up are (1) can we put positives in one pile and negatives in the other and (2) can we put positives and negatives in the same pile? So, right away, they’re thinking about this exercise; they’re engaged. We have a discussion about this and although they don’t yet know what to do with these 2 piles (although some guess they’re equations), I let them determine what is allowable. So again, right away they “own” this exercise because they have determined what’s allowable. By the way, the discussion about what is allowable has many branches and typically includes a lot of “what if” banter. I just listen.

Once this is resolved, I then ask them to label the pile on the left “A” and the pile on the right “B”. This also is fun because there typically is someone who stacks the piles vertically rather than horizontally, so I simply say the pile furthest from you is A and the pile closest is B.

When everyone is ready I then ask them to do something to their pile A such that the total value in both piles is equal. This is also a fun point in the exercise for classes that allow positives in one pile and negatives in the other, but overall the buzz within each group again is one of the goals of this exercise. When this is done, I ask them to return to their original piles and then I ask them to do something to their pile B such that the total value in both piles is equal.

In both cases, I ask them if there was only 1 way to make the piles equal. Buzz, buzz again and the consensus was yes.

The next question to them was do something to both piles at the same time such that the total value in both piles is equal. This really generates buzz and questions to me, which I say I’ll answer later. The reason I won’t answer is that I want them to explore how this works. What they discover is that there is an unlimited number of ways to do this. For example, if A = 2 and B = 4, add 5 to A and 3 to B and both piles equal 7. There usually is an “aha” moment when they realize that as long as the difference between the two numbers added to A and B is 2, the total value will always be equal. Some also discover that unequal amounts can be subtracted from both piles and further that two numbers differing by 2 can result in an equal value in both piles. And there’s another “aha” moment – the total value in both piles can be negative if both were positive at first! And what’s more, zero is a valid value!

So, we played with these 3 options for a while and there was discussion all along about not only what was allowable but also the range of answers under the different conditions. Then we moved to equal piles to begin the exercise.

I ask them to adjust their piles so that there is an equal number in both piles. This then brings up the issue of their rule allowing positives in one pile and negatives in the other, if they allowed this. They realize they have to rule it out. But I then ask if they can have an equal value in both piles while having positives and negatives in the same pile. Can the total in both piles be positive or negative? Buzz, buzz and the conclusion is that it’s ok but this comes after a lot of discussion and this really gets them going about signed numbers. For example, if they are to have 3 positives in both piles to begin, they could put 4 positives and 1 negative, or 6 positives and 3 negatives or … here it goes again with an unlimited number of both as long as the total is 3.

So, I ask them to consider there beginning equal value in both piles and typically they make it simple – either all positive or all negative and they do this partly – they tell me – because they don’t know what I’m going to ask them to do. At this point, the equation question arises and I have to admit that we’re headed in that direction. After playing with this for a while, the class concludes (again) that there is an unlimited number of values that can be added or subtracted to maintain the inequality.

The next step is to give each group a few blue chips. What the group is asked to do is have one person look away of shut their eyes while the others in the group do two things: (1) set up two piles with an equal number of chips in both and (2) remove a certain number of chips from one of the piles and place a blue chip in that pile. In essence, create a simple equation. When they are done setting it up, the closed-eye person is to look at what they’ve done and answer the question: what must you replace the blue chip with in order to make the piles have equal value?

Do each of these exercises until the class seems comfortable with all the ideas that got buzzed about.

At this point, if you’d like to extend this 2-pile concept to work with introducing work with equations, see Chipping Away at Equations in this blog. It links up with this posting and together it gives students a different view of equations.

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Algebraic Fishiness

Posted by mark schwartz on July 22, 2016

Introduction:

In an Algebra class, I gave the following two problems as an in-class assignment. We had just finished the material on solving equations and we had spent considerable time on what are called application problems. I decided to stretch their imaginations a bit by providing these two problems which are quite different from anything they had seen in the text. This, again, was a group activity and I roamed around the classroom watching the in-group strategies develop. If I hadn’t done this, I would have missed a group’s interesting insight.

The Story:

Here are the two problems.

  1. This problem is from Algebra by Davies, published by Barnes and Company, NY, NY, 1858.

A fish was caught whose tail weighed 9 pounds. His head weighed as much as his tail and half his body; his body weighed as much as his head and tail together: what was the weight of the fish?

  1. This problem is from An Introduction to Algebra by Colburn published by Hilliard, Gray and Co., 1839.

There is a fish whose head is 4 inches long, the tail is twice the length of the head, added to 2/5 of the length of the body, and the body is as long as the head and the tail both. What is the whole length of the fish?

Please note that these two fish problems represent a problem that was quite common in texts of that era. Conceptually, it’s the same problem, one phrased as a weight problem and the other as a length problem.

There was some grumbling when the class first read these problems. They hadn’t seen this type of self-referential relationship in a problem. They questioned if these were trick questions; some said it was the same problem so why give it twice; someone asked if it was the same fish, and I heard a few other choice comments. I suggested they take a deep breath and think about it and further, once they have solved one, let me see what you’ve done before moving on to the second problem.

On occasion, I gave Socratic help.

One group called me over and showed me the solution for the first problem, which was correct. But their setup for the problems was unique.

Their setup was:

T = 9                                               H = 4

H = T + 1/2(B)                              T = H + 2/5(B)

B = H + T                                       B = H + T

What’s unique about this is that they identified the relationship of the unknowns for both problems before setting up the equation for either of them. I asked them why they took this approach. The response was that the problems looked so much alike that they thought maybe there was one solution strategy, which as it turns out is true. This setup, in their minds, verified that point. In both cases, they started the solution with B = H + T. These are their solutions, but I tidied them up a bit (without changing any steps in the solutions) and showed them this way so it’s easier to see their parallel solutions. The complete solutions for both were:

For the weight problem:                   For the length problem:

B = H + 9                                            B = H + 4

H = 9 + 1/2(H + 9)                            T = 8 + 2/5(4 + T)

2H = 18 + H + 9                                 5T = 40 + 8 + 2T

H = 27                                                 3T = 48

T = 16

B = 27 + 9 = 36                                   B = 4 +16 = 20

Fish = 72 pounds                                Fish = 40 inches

They went a little further and decided that it was the same fish; that a 40 inch fish could weigh 72 pounds based on one of the group members being an avid fisherperson. Neat group!

In summary, this group of students used their imaginations, which resulted in their seeing not only the similarity in the problems but also the similarity in the solution strategy. It’s a testament to group activity and “aha” moments.

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Setting up Equations the Old Fashioned Way

Posted by mark schwartz on June 6, 2016

Introduction.

In Colaw and Elwood’s School Arithmetic Advanced Book (1900), they presented an interesting strategy for students to use in setting up equations, at least when first learning how to set up equations. They showed an interesting alternative.

The Story.

I noted something similar in several other texts of that time, but I don’t believe it was the universal approach, like today’s “let the unknown = x”. Colaw and Elwood don’t begin with “let x = ” but make an algebraic statement with words taken from the problem. It may not be as simple with problems that don’t translate as directly, but it is a very interesting approach.

Several example (pg. 172) are:

Problem: If 10 dollars is taken away from 5 times A’s money it would equal 20 dollars plus 3 times A’s money. The equation:

5 times A’s money ─ $10 =  $20 + 3 times A’s money

Problem: of my money is the same as subtracting  of my money from nine dollars. The equation:

The equation:   1/4 of my money = $9 ─  1/5 of my money.

Other examples are presented. They solve these equations without substituting “x’ for the words. Rather, they carry the words throughout the problem, including the dollar sign. In this case, they abbreviated “of my money’ to just “my money” but generally they don’t abbreviate the words, noting that having to write out the words every time helps students see the correctness or incorrectness of their solution. Another effect of this, I believe, is that students have to slow down to write everything out and perhaps gives them more time to reflect not only on their problem but also on their strategy for solving equations in general.

This strategy provides a step not typically available in today’s strategies, which is to set the unknown equal to “x”. Another aspect, I believe, of the Colaw/Elwood idea, is that it makes the process less abstract, which I believe is helpful. I’d also like to note that in other texts at that time, the use of the first letter of the item would be used rather than “x”. For the above problem  “5 times A’s money ─ $10 =  $20 + 3 times A’s money”, it could simply be “5a – 10 = 20 + 3a”.

You might want to give these ideas a try and see if it helps.

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Staring

Posted by mark schwartz on May 31, 2016

Introduction.

This is a short piece but one which I believe is worth reading. One thing I heard repeatedly from students in remedial/developmental courses was that doing math fast was important. They recall speeded exercises and tests and the contests to be first with the answer. This feature of the math classroom became part of the horror for students who weren’t fast. For some reason, the doing-math-fast-all-the-time myth becomes important and over-rides the essence of math. I call it a myth because although the answer to the problem posed will be right or wrong, how one gets to right or wrong is, in my mind, more important than speed … and there are as many ways to get to “right” as there are to get to “wrong”; well maybe not as many ways to get to “right” but allowing students to free themselves from the time boogie man can and does result in some very interesting outcomes.

The Story.

So, what does staring have to do with this? Regardless of the topic we were about to do, when it came time for students to do the work, I would tell them lay down their pencils and pens and then just stare at the problem for a while. Some students considered this silly, but for those who had the “hurry up” in their heads, this seemed to reframe the work for them. I prompted them to do this for every problem, not just for every exercise. I roamed around the room prompting them to do this … it wasn’t an easy thing for them to learn.

Here’s the point. Slowing people down to stare at the circumstance, be it a math problem or something else, provides time for reflection. “What’s really happening here? What are the relationships? What’s important? I wonder if I can …?” and a host of other “what” and “why” questions arise. This may cause discomfort at first, particularly if it’s never been experienced before, but it becomes a pattern of response enabling students to see that math is more than getting the answer quickly. Remarkably, many students started seeing alternative paths to the solution. They got invested in playing with possible solutions rather than searching for the correct, traditional path to solution. When one student did the following, I stopped and recorded it; it was a great example of imagination. The problem was  5 ( ̶ 2x + 9) ÷ 6 + 3 = 1/2

Before seeing his solution take a moment, if you like, and consider how to solve this the traditional way, starting by multiplying every term in the equation by the LCD 6. The order of operation rules then would direct you to do the indicated multiplication, “getting rid” of the parentheses. The rest is adding, subtracting and finally dividing to get the answer, x = 6. Now look at his step by step imaginative approach. The annotation in parentheses is his telling me what he did and why.

5 ( ̶ 2x + 9) ÷ 6 + 3 = 1/2

5 ( ̶ 2x + 9) ÷ 6  =  ̶  5/2         (subtracted 3 from both sides … wow, seriously violated order of operations)

10( ̶  2x + 9) =  ̶  30               (cross-multiplied, just as in solving proportions)

̶  2x + 9  =   ̶  3                   (divided both sides by 10 … saw it was possible and did it)

̶  2x =  ̶  12                               (subtracted 9 from both sides … standard stuff)

x = 6                                     (divided both sides by    ̶ 2 … yeh)

This student had mastered a lot of Algebra basics and used them cleverly. I asked him how he came to his solution and he said: “because you give us time to play with stuff”. One example does not prove the case for my proposition of staring, since this is only one example, but it is an interesting example of setting classroom conditions for the blossoming of imagination. Even though the traditional method may be more “efficient”, staring causes students to slow down, to reflect on and consider what’s being presented, and to think about possible alternatives to the traditional path to solution. Further, this student was very motivated to get on to the next problem, which seemed to be an unexpected outcome for many in the class. And by the way, there were those who chose to stick with the traditional. Yes, teach the traditional but allow for alternatives. Staring lets them do this and it only takes about a minute.

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