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.

Posts Tagged ‘algebra’

Concrete to Abstract

Posted by mark schwartz on April 16, 2017

Introduction

When presenting operations with signed numbers, an instructor must deal with the issue of notation as well, to allow for the plus and minus having two different meanings; this has to be addressed. I spent a long time playing with this until I found a way for students to ‘see’ the difference. Given a number line which I call a road and a car which they can drive on this number-line-road, the car can be put into drive or reverse, So the direction of moving is with reference to the car. When the car is put on the number-line-road, it can face forward (toward the positive) or back (toward the negative), so the facing is with reference to the number line. Students twiddle with this a little but eventually get it. It seems clear that facing (positive or negative) is different from moving (positive or negative).

I wrote a piece in this blog (Driving the Integer Road), a somewhat long detailed almost lesson plan which describes how all this works but I never present the traditional notation until I’ve gone through a bunch of exercises which I call facing and movement. What’s interesting about it is that students seem to appreciate and understand the differences now between plus and minus in terms of the operation or the sign of the number. Although other instructors may use something like this I haven’t seen it presented anywhere. I’d never seen it in my collection of 1800s texts either … until recently. Here’s how Durell and Robbins, in a very similar way had people walk a number line. They omitted some detailed explanation but I will discuss this.

The Story

In Durell and Robbins 1898 School Algebra Complete (pgs. 20-21) they have the students visualizing walking on a number line. The students don’t actually walk the line but only visualize it. Before this ‘exercise’, the authors point out “ … the signs + and – are employed for two purposes – first, to express positive and negative quantity; and second, to indicate the operations of addition and subtraction.” This prompts the students to pay attention to the notation in a problem. I’m now going to paraphrase what the authors did to show students the relationship between walking on a number line and the traditional +/- notation.

On a number line with A at zero, B at + and C at – , a person walking from A toward B a distance of 5 units and then walking back toward A a distance of 3 units, has in total walked a distance of 2 positive units from A, or zero. In notation, the authors write “+ 5 + (- 3)”, essentially the sum of a positive and negative quantity. They point out that this is symbolically what was done on the number line. They discuss this in terms of positive and negative distance and demonstrate that “Hence, we see that adding negative quantity is the same in effect as subtracting positive quantity; therefore in the expression 5 – 3 the minus sign used may be considered either a sign of the quantity of 3, or as a sign of operation to be performed on 3.” That’s a very powerful statement and hopefully when instructors used this text, they emphasized this point because what this really does is show that + 5 + (– 3) = +5 (+ 3) = 5 – 3. The authors don’t detail this expression; they just state it … but it can be shown by walking on their number lines

Given that the authors point out “ … the signs + and – are employed for two purpose…” when they wrote the activity as + 5 + (– 3), were consistent with their own schema by designating the – as the sign of the number and the + as the sign of the operation. That’s interesting because the ‘walker’ is facing in the negative direction with reference to the number line, while walking forward with reference to his own movement and the authors don’t mention this. Their expression of the walking activity could have been written as + 5 –  ( + 3), meaning that the walker walked in a negative direction with reference to the number line, while facing in a positive direction with reference to his own movement. This is a subtle difference but consistent with their schema.

Look at the expression + 5 – (+ 3). With the ‘walker’ starting at A and taking 5 steps toward B, this is +5. If the walker does not change the direction he’s facing, then he could – with reference to his own movement – step backwards 3 units. This is the operation and thus the – outside the parentheses. So, facing forward with reference to the number line is the sign of the number, thus +3.

“With reference to” becomes a critical phrase in parsing one’s way through this demonstration of what the authors have done. Perhaps it’s too subtle for a class discussion but from my experience, this subtlety seems to make an appearance when students talked about the fuzziness in all the operations with signed numbers.

In summary, Durell and Robbins in 1898 captured the core elements of my “Driving the Integer Road” but didn’t explore the subtleties of the notation when talking about the sign of the number and the sign of the operation. I would urge instructors to explore ways of making concrete the ‘abstract’ use of – and + for this as well as other math relationships.

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Recognize x∧2 – x – 1 = 0?

Posted by mark schwartz on February 12, 2017

Introduction

If you recognize the equation in the title then you are already familiar with the golden ratio of 1.618. There are a number of books that explore the occurrence of the golden ratio in nature, architecture, the pyramids, art, aesthetics and a variety of other suspected places. I’d like to explore something about 1.618 that I’ve not seen in any books; just something to add to the mythology of it. If you aren’t that familiar with it, just Google it and have fun.

The Story

When I was teaching a pre-service teacher’s class about math concepts, I introduced the golden ratio because it was a nice vehicle for demonstrating how a pattern can show up in a lot of places. I talked a lot about teaching students to explore for patterns because math is saturated with patterns and relationships.

I asked if anyone in the class was involved in mysticism, the occult and if they would feel comfortable talking to us about it. There was one person who spoke up. I directed the conversation to focus on the pentagram (a five pointed star inscribed in a circle) because I knew that it was a significant symbol. I asked her to talk about the significance of the pentagram and of all the stuff she talked about, the golden ratio wasn’t mentioned. I mention it because it’s a somewhat invisible yet significant characteristic of the pentagram. Let me start with the star.

From here on, you will be asked to do a lot of ‘construction’ but it doesn’t have to be elegant or perfect and hopefully my directions can be easily followed. I just believe that your participating – as I had my students do – will be entertaining and educational, not drudgery. There is some geometry and trigonometry involved so if you don’t recall stuff, I’ll provide the information needed to do the work.

You can ‘construct’ a five-pointed star, starting by sketching a circle. Don’t make it too small because some of the points and lines will need room to be labeled. When you get done, you will have constructed a pentagram.

Starting at 12 o’clock, place 5 equidistant points on the circle. Starting with the point at 12 o’clock, label that one A and then going clockwise label the others B, C, D, and E. Now, draw a straight line from A to C, then a line from B to D, and continue with C to E, D to A and E to B. You should have a somewhat respectable looking 5 point star. I want to highlight that this pentagram has a pentagon in the middle and each face of the pentagon has a triangle on it. Take the triangle at the top that has the vertex A and label the other two vertices in that triangle F and G. Still with me?

Now if your pentagram were a perfectly crafted one, the pentagon and each triangle would appear to be equilateral. The pentagon would be but not the triangles! What? – they look equilateral. How can we determine that they are not?

To do this we need to know something about the angles in triangle AFG. Allowing that AF and AG are equal to 1 (we could pick any arbitrary value and the outcome of this exercise wouldn’t change; it’s just that 1 simplifies the calculations). Geometry allows us to determine that angle A equals 36 degrees. Now drop a perpendicular from vertex A to side FG, labeling that point on FG as H. Why? Well, we just formed a right triangle AFH which allows us to use some trigonometry to determine the length of FH. The perpendicular also bisects angle A, so angle FAH is 18 degrees. The sine of this angle is FH over 1, or just FH. The sine of 18 degrees is .809 – the length of FH – so the length of FG is .618, thus demonstrating that this triangle and all the triangles in the pentagram are not equilateral.

Given this, the perimeter of triangle AFG is 1 + 1 + .618, or 2.618. If you’re not familiar with the golden ratio, let me note a few relationships. First, the square root of 2.618 is 1.618 and clearly 1.618-squared is 2.618. And, as trivial as it seems, you can see that 1 + 1.618 equals 2.618.

And now the point of this exercise! Recall that the title of this article is x2 – x – 1 = 0. If you solve this equation, you get 1.618. But substituting 2.618 for x2, 1.618 for x you now have 2.618 – 1.618 – 1 = 0, which is a true statement for the perimeter of the triangle. So, triangle AFG and all the other triangles in this pentagram contain the golden ratio. But, there’s more.

Look at the pentagon in the pentagram. I won’t ask you to do all the construction (you can do it if you like) but I’ll step through it to show that the pentagon also contains the golden ratio. Briefly, starting with point F and moving clockwise, the vertices of the pentagon are G, I, J, K. Drawing a straight line from K to G, forming triangle KFG and assigning KF and FG a value of 1, what is the value of KG? As with the star construction, geometry lets us know that angle KFG is 108 degrees, and a perpendicular dropped from F to KG (label this point M) bisects the angle, and results in a right triangle KFM with angle KFM equal to 54 degrees. Again, trigonometry lets us determine that the length of KM is .809, and thus the length of KG is 1.618.

So, what is true for each of the triangles is also true for the pentagon, although it’s not the perimeter of the pentagon, but a construct (a ‘chord’ of the circle too) within the pentagon.

And one more thing. All of this analysis came about because I was watching the National Geographic Channel and the show was all about creatures of the sea. One of the creatures shown was the star fish. If you connect the tips of the starfish arms you get a pentagon!

So, in addition to the pentagram having a powerful influence in the occult, perhaps reinforced by the presence of the golden ratio, nature again (using a little imagination) provides another possible example of the ‘hidden’ sway of 1.618.

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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 1st Day Handout to Students

Posted by mark schwartz on October 17, 2016

 

Author’s Note: The following is literally a handout given to students the first day of class. I give them time to read it and then we talk about it. The discussion set the tone for their learning and the idea of freedom was a surprising but satisfactory idea, although scary to some who expected this class to be like all previous math classes. What follows is the handout.

In the 1960s, a book titled “Freedom, not License” hit the bookstores. Briefly, it’s a story of the core philosophy of a school named Summerhill in England. The title refers to a subtle distinction between two conditions: freedom – being able to determine your own behaviors, live with the consequences, be self-determining, guided by your own internal discipline and control; and license – interpreting the circumstances in which you are allowed, permitted and “controlled” by an external authority. Actually, it’s misinterpreting the freedom as license, whereby the misinterpretation leads one to rely on external events, rather than understand the freedom to govern one’s own behavior and actions. License also is interfering with other’s freedom.

I give you freedom to succeed but it has to be your success, not driven by external rewards and punishments. I will teach well and you have to learn to learn well. Don’t rely on me to chase you down the hall demanding that you get assignments done on time. That’s your responsibility. Don’t rely on me to threaten you with loss of grade if you don’t attend class. Attendance is your responsibility. Don’t rely on me to control the classroom as is done in elementary school; hushing the noisy, punishing the “unruly”. It’s your responsibility to respect the classroom environment and not disrupt my teaching or the learning of others.

Freedom is a little scary if you’ve never experienced it in a classroom. But consider it a responsibility just like driving. You’re responsible for your car – for its maintenance and performance; for driving responsibly within the wide legal constraints of the speed limit, parking areas, passing, not drinking while driving, etc.

According to the Oxford English Dictionary, “education” is derived from its Latin root, “educare”.  Educare means “to rear or to bring up”.  Educare itself can be traced to the Latin root words, “e” and “ducere”.  Together, “e-ducere” means to “pull out” or “to lead forth”.  Hence we use the word “educare” to communicate the teaching method through which children and adults are encouraged to “think” and “draw out” information from within.

Notice the last three words: “information from within”. It is within you to learn well and to learn any subject well. I can help you draw it out, but the “you” is the important word in that sentence. You have to attend class, do the assignments, and act respectfully toward yourself and all others in the classroom.

Let me repeat – freedom is scary if you’ve never experienced it in the classroom. I will not check your classwork to see if you’ve done it and it is correct; answers are in the text. I will work with you if your answers are incorrect. You’re responsible for that and it will be hard for you to accept that responsibility because it will be tempting to leave class early and not do it because math makes you uncomfortable and anxious. But I can help you address the lack of math skills that lead you to feel that way.

My teaching doesn’t automatically lead to your learning. But take the freedom offered and use it; don’t let it become license that interferes with your learning.

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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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In 1877, Mr. Ray Reasons with Fractions

Posted by mark schwartz on September 8, 2016

Introduction

In Mr. Ray’s 1877 Ray’s New Intellectual Arithmetic, an elementary school text, he presents some of the problems with their solution. A sample of these are worth looking at because in every case he shows a solution method which is based on fractions and knowing how to handle a sequence of fractions. But it’s not only the sequence of fraction operations but also the logic of these fraction operations that elementary school children had to follow. This required them to think about the relationships in the problem. I’d like to further note that this method of solution for all 7 problems presented here is seen in many of the texts of that era. It really required students to understand fractions! I’m not proposing that we use this “fractional” method in lieu of solving them by either proportions – the first 4 problems – or simple equations, the last 3 problems.

The Story

All these problems are from his text. Read the solutions slowly to really enjoy the subtlety of the method.

  1. A yard of cloth costs $6, what would 2/3 of a yard cost?  (Pg. 48, # 3)

Solution: 1/3 of a yard would cost 1/3 of $6, which is $2; then, 2/3 of a yard would cost 2 times $2, which are $4.

  1. If 3 oranges are worth 15 cents, what are 2 oranges worth?  (Pg. 49, #19)

Solution: 1 orange is worth 1/3 of 15, or 5 cents; then 2 oranges are worth 2 times 5 cents, which are 10 cents.

  1. At $2/3 a yard, how much cloth can be purchased for $3/4?  (Pg. 75, # 5)

Solution: For $1/3, 1/2 a yard can be purchased, and for $1, 3/2 of a yard; then, for $1/4, 1/4 of 3/2, or 5/8 of a yard can be purchased, and for $3/4, 9/8 = 1 and 1/8.

  1. If 2/3 of a yard o cloth costs $5, what will 3/4 of a yard cost?  (Pg. 101, # 2)

Solution: The cost of 1/3 of a yard will be 1/2 of $5 = $5/2; and a yard will cost 3 times $5/2 = $15/2; then, 1/4 of a yard will cost 1/4 of $15/2 = $15/8; and 3/4 of a yard will cost 3 times $15/8 = $5 and 5/8.

Note that these 4 problems lend themselves well to being solved using proportions. What follows now are 3 more problems, which if presented in today’s texts would likely be solved with simple equations, but again Mr. Ray’s solutions are a sequence of fraction operations.

  1. If you have 8 cents and 3/4 of your money equals 2/3 of mine, how many cents have I? (Pg. 52, #17)

Solution: ¾ of 8 cents = 6 cents; then 2/3 of my money = 6 cents, 1/3 of my money is 1/2 of 6 cents = 3 cents, and all my money is 3 times 3 cents = 9 cents.

  1. Divide 15 into two parts, so that the less part may be 2/3 of the greater.  (Pg. 106, #1)

Solution: 3/3 + 2/3 = 5/3; 5/3 of the greater part = 15; then, 1/3 of the greater part is 1/5 of 15 = 3, and the greater part is 3 times 3 = 9; the less part is 15 ̶ 9 = 6.

  1. A and B mow a field in 4 days; B can mow it alone in 12 days: in what time can A mow it?  (Pg. 110, #14)

Solution: A can mow 1/4 ̶ 1/12 = 1/6 of the field in 1 day; then he can mow the whole field in 6 days.

I hope you appreciate what elementary school students had to do at that time. Since it was elementary school, they weren’t taught proportions and simple equations but they were “exercised” with fractions in a way that I believe could benefit today’s students understanding of fractions.

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