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.

Is it ̶ 3 or is it ̶ 3?

Posted by mark schwartz on August 27, 2016

Introduction

I know. The title “Is it -3 or is it -3?” looks weird but it’s not a typographical error. It’s a way to bring attention to algebraic notation. The question is: how did you read -3? Did you say “minus 3” or did you say “negative 3”? Does it make a difference?

In Day’s 1853 An Introduction to Algebra, he writes 5 pages on the topic – yes, 5 whole pages of words discussing negative quantities. He wants to make sure that students understand that the 4 basic operations in arithmetic are different from the 4 basic operations in algebra because of the introduction of negative quantities in algebra. In lengthy discussions he cites how negative quantities appear in profits of trade, ascent and decent from earth, progress of a ship relative to a latitude, and of course money. Clearly he’s conveying what I would call the algebraic trip-wire – how to handle negative quantities. This kind of lengthy discussion isn’t presented in today’s texts but rather students are presented with diagrams and number lines and visual aids to help them understand the rules. An instructor can supplement the text with their own creative explanations and demonstrations. But Day’s emphasis on this point may well be what is needed in today’s texts – a core understanding of the rationale behind the rules.

The Story

So, back to “is it  − 3 or is it – 3?”

Day’s writing prompted me to recall a question from a student. We were working with operations with signed numbers. Typically I am very careful to reference any “ ̶ “ in a problem or an answer as a negative or as a minus, depending on its use in the problem. Knowing, for example, that + ( ̶ 3) gives the same result as ̶ (+3), in the former the “ ̶ “ is understood as negative 3 but in the latter it’s understood as minus 3. As noted, it ultimately makes no difference, but a student stopped me during a discussion and pointed out that in the same problem I had referred to a term as both and it didn’t seem right to him … and in a most technical sense, he was right. I asked if he were the only one bothered by this and other students felt as he did.

I admitted to my sloppy use of the terms and we got back to discussing operations with signed numbers and then again, this student stopped me. He asked “what about – and in his words – minus a minus 5” – how come it’s plus 5?” I wrote ̶ ( ̶ 5) the board and asked him if this is what he meant and he said yes. I asked him then what operation is being indicated and he said that it indicated to subtract a negative. So, the sign inside the parenthesis isn’t a minus, rather it’s a negative sign, a sign of the number. The class was muttering about this somewhat lengthy Socratic discussion – and they participated too – which really was a very positive result of the initial question … what some might call an unintended consequence … but a good one.

And of course, there was the question of “does it make a difference what I call it if I get the right answer?” So, we played language games with various examples until there was consensus that there was a difference between “minus” as the operation of subtraction and “negative” as the sign of the number. But, for most of the class, this difference didn’t make a difference as long as they understood what the notation in the problem was asking. So, I asked them to think about this:

Don’t do this problem yet but within your group, discuss the “ ̶ “ signs in the problem 4 ̶ 6 + 2 ̶ 3 ̶ 5 + 7. Signs of the number of signs of the operation? It was fun to roam the room and listen to the within-group discussions. As expected, there were disagreements, yet those that disagreed came to understand that both were correct! It was a matter of what procedure made each person feel most comfortable.

After allowing for discussions, I asked for volunteers to go to the board and demonstrate their solution. There were two primary solutions: first, just use the order of operations and do the indicated operations from left to right, although there was some stumbling to explain how to handle “2 ̶ 3 ̶ 5”. The language used in explaining the whole problem was interesting. For example, “4 ̶ 6” equals minus 2 (not negative 2) and minus 2 and plus 2 is zero (adding two operation not two values). Then zero minus 3 (the “ ̶ “ is the sign of the operation) gave “minus 3” and the next operation was expressed as “a minus 3 and a minus 5 equals negative 8”. Think about that. Technically, the 3 and the 5 were expressed as adding two subtractions (minus wasn’t seen as an operation) yet the answer of negative 8 was correct notation. But the real thing to notice is that the answer is correct independent of technically incorrect labelling of the values.

As much as I believe in the importance of carefully using either minus or negative correctly, it clearly seems that – at least for this student and his group – knowing how to handle the negative is more important.

The second solution was given with a preface. This student rewrote the problem as 4 + ( ̶ 6) + (+2) + ( ̶ 3) + ( ̶ 5) + ( +7). She pointed out that her group saw all the signs as signs of the numbers and therefore they just added them all together. Neat.

Of course there are more ways to handle this problem but these two examples show that as long as students understand the basic rules and relationships with signed numbers, the right answer will be found. We talked about these two solutions and how to handle the signs and operations.

I then asked if all the talk we had about the difference between negative 3 and minus 3 made a difference for them. The consensus was yes and that it showed up when they were talking about the problem in their group. Apparently, it provided a clearer understanding of the difference.

There was also the comment that allowing them to challenge me (I pointed out it wasn’t challenging me but rather challenging the math content) gave them a sense that the “rules” and labels weren’t arbitrary – that there really was sense to it.

Finally, I’d like to note that hearing a student’s question as a real interest in knowing rather than a hostile kind of “whatever”, opened the door to the discussion which further opened the door for their better understanding – again an unintended positive consequence. If you have time, try it.

Posted in Historical Math, basic math operations, algebra, remedial/developmental math, math instruction | Tagged: , , , , , | Leave a Comment »

The Importance of a Clearly Stated Algorithm

Posted by mark schwartz on August 22, 2016

Introduction

I posted a piece earlier in this blog titled Sheldon’s Compound Proportions. It describes what Sheldon labels the “cause and effect” method for solving compound proportions, which as far as I can tell, aren’t in todays’ texts. His work was in 1886. You might want to take a look at his idea because this posting talks about other compound proportion procedures at that time and I did it to emphasize the importance of a clearly stated procedure for doing an operation.

The Story

I strolled through my collection of old texts and in quite a few of them found the same prescription for solving compound proportions not using cause and effect. I picked 5 which cover about a 20 year span from 1864 to 1883. They all have the same procedure and what I suspect is that it was the established and accepted solution method at that time. As in todays’ texts, it was just a simple matter of “borrowing” a basic algorithm from someone else’s work. There are other texts of that era which reference Sheldon’s cause and effect method and a few of them introduce it along with the procedure I’ll cite below.

The point is that his method is a much clearer statement of how to handle the information in a compound proportion problem. Further, what I’m suggesting is that we should carefully examine some of our current traditional algorithms to see if the reason students have trouble with them is because of the way they are worded and presented. For example, finding the lowest common denominator (LCD) in order to add/subtract fractions doesn’t require the extended way it’s been typically taught. In fact, I have seen some texts introducing a method which doesn’t require finding an LCD at all. Certain mixture problems can more readily be solved with an 1864 method Mixing it up with Alligation, posted earlier in this blog.

By the way, the 5 texts in which I found this procedure are all arithmetic texts, which indicates to me that this somewhat sophisticated idea of compound proportion was taught in elementary school. I’ll give you example problems from an old text to indicate that, in my view, it was a very handy procedure for the real world experience at that time. Today we call these “application” problems.

Here’s the rule as stated in Greenleaf’s 1881 The Complete Arithmetic, page 235 (the other 4 books are cited below and present the same rule).

Rule for Compound Proportions

“Make that number which is like the answer the third term. Form a ratio of each pair of the remaining numbers of the same kind according to the rule for simple proportion, as if the answer depended on them alone. Divide the product of the means by the product of the given extreme, and the quotient is the fourth term, or answer.”

Embedded in this is reference to “…the rule for simple proportion …” which Greenleaf provides on page 233 and it is:

Rule for Simple Proportions

“Make that number which is of the same kind as the answer the third term. If from the nature of the question the answer is to be larger than the third term, make the larger of the remaining numbers the second and the smaller the first term; but if the answer is to be smaller than the third term, make the second term smaller than the first. Divide the product of the means by the given extreme, and the quotient is the fourth term, or answer.”

Students had to be able to apply this latter rule for simple proportion before being presented compound proportion. There is no conflict between the two rules; in fact, there is some overlap. For simple proportions, the rule directs the student to understand “the nature of the question …” and use that to determine which values go in which of the 4 places in the proportion. The students had to be able to assess and estimate if the answer was going to be larger or smaller and place the correct terms in the first and second places. Wow! There is a lot of estimating and juggling of values and basically it seems that all of this effort is aimed at what we would say today as determining whether it’s a direct or inverse proportion. With problems with simple values, this is a somewhat manageable issue.

For example, a problem from the text is “If a man travel 319 miles in 11 days, how far will he travel in 47 days?” Using the rule for simple proportion, the setup would be:

11/47 = 319/x    (the rule doesn’t use “x”, but I did for demonstration purposes)

The solution is (47×319) ÷ 11 = 1363

However, in today’s approach to simple proportion, the setup (in most cases) simply follows from the order of the information in the problem, giving:

319/11 = x/47

This gives the same answer but notice that the rule states “Divide the product of the means by the given extreme …” and that doesn’t apply here. So, the 1881 rule is quite constraining when it comes to writing the proportion, when indeed there are several ways to set up the proportion for the problem.

Again, there is nothing wrong about the simple or compound proportion rules as provided by Greenleaf. The issue is that the rules are somewhat convoluted and constraining. If a student doesn’t learn this algorithm and follow it precisely, the likelihood is that the correct answer won’t be found. There are a lot of words referring to the terms and judgements that a student must make about which terms go where in the proportion. Further, look at what happens with a compound proportion problem, again from Greenleaf (#67, page 236):

“If 12 men in 15 days can build a wall 30 feet long, 6 feet high, and 3 feet thick, working 12 hours a day, in what time will 30 men build a wall 300 feet long, 8 feet high, and 6 feet thick, working 8 hours a day?”

Now, where does a student begin sorting through all this information if they use the rule above for simple proportion? What’s the “nature of the question”? For example, the rule states “…make the larger of the remaining numbers …” and how is a student to know which number is to be selected? I can visualize the instructor explaining in excruciating terms how all this works. Again, it’s not impossible to apply the rules as stated in 1881 but I urge you to look at Sheldon’s Compound Proportions in this blog and see how much more direct the rule is by framing information as cause and effect.

Briefly, Sheldon’s 1886 statement of the procedure:

“The solution of every example in proportion proceeds on the assumption that effects are in the same ratio as the causes that produce them. Every proportion is the comparison of two causes and two effects. In the method known as Cause and Effect, the causes form one ratio, and the effects the other. The first cause and the first effect are antecedents; the second cause and second effect consequents.”

Notice the simplicity of identifying cause and effect and then the causes forming one ratio and the effects the other. The words” antecedents” and “consequents” could be updated to 1st and 3rd term and 2nd and 4th term, respectively.

Taking the above compound problem the 1st causes are 12 men, 15 days 12 hours a day and the 1st effect is to build the wall 30 feet long, 6 feet high, and 3 feet thick. The 2nd causes are 30 men working 8 hours a day and the 2nd effect is to build a wall 300 feet long, 8 feet high, and 6 feet thick. You are to find “…in what time…” which is a 2nd cause. There is a shortcut that can be used but let me show you – in what I call slow-motion-math – one way to make sure the terms get placed correctly. I typically use the labels and then replace it with the values (for a lot of different types of problems, not just compound proportions). The proportion following Sheldon’s procedure is:

Causes                     Effects

1st       men, days, hours         length, height, thickness

2nd       men, x, hours               length, height, thickness

I used “x” for days in the second cause. If the numbers are substituted, we have:

12•15•12 = 30•6•3
30•x•8     300•8•6

Cross-multiply and divide, solving for x and the answer is 240.

Again, a detailed description of the “cause and effect” is in Sheldon’s Compound Proportions in this blog.

The essence of this posting is to demonstrate the importance of a well thought-out procedure expressed in easily understood language. If you are an instructor, you likely have done this kind of “simplifying” of the algorithm because as stated in the text, it seemed too fussy for students to follow. Not every algorithm can be simplified but I believe it’s an instructor’s responsibility to make math more accessible to students by removing the fog of awkwardly phrased rules and algorithms. Give it a try.

Posted in algebra, basic math operations, Historical Math, math instruction, mathematics, proportion, Proportions, remedial/developmental math | Tagged: , , , , | Leave a Comment »

Commentary: Algebra – yes or no?

Posted by mark schwartz on August 17, 2016

Commentary: Algebra – yes or no?

Andrew Hacker is proposing not teaching algebra but rather teaching math in a real-world context. He has created a conversation about the utility of algebra and proposes in his book “the math myth” that algebra is a cause of the loss of talent because many students can’t get past the algebra filter. I propose that it’s not the algebra content but rather the bad teaching of algebra. Consider this.

What does a student see when presented with 2(3x ̶ 1) = 10? As they have been taught, they will start rummaging through the rules for solving equations – order of operations, distributive law, operations with signed numbers, basic addition and subtraction – and anything else needed to solve for “x”.

This is sad. Algebra is an aid to help us see patterns and relationships and the equation presented above contains those things, yet that’s not what students have been taught to see. Poor teaching of algebra is the issue, not the algebra content itself. The argument about when will students use algebra in the future or in their daily lives can be asked also of history and poetry. If the premise is utility in the future, then many topics need not be taught at all. I’m going to shy away from the philosophical framework of the purpose of education, outside of stating it’s a way of introducing people to the social, economic, religious, literate and cultural concepts in which they will live their lives.

As humans, we tend to learn not individual facts but rather how these facts aggregate to patterns and relationships. Using math as the example, students in elementary school learn arithmetic – the four basic operations and maybe a few extended procedures, for example what is called long division. But what is really being learned? Are there patterns and relationships from which the four basic operations emerge?

Consider addition. The “new” concepts being presented in the common core curriculum have students thinking about numbers differently. In essence, they are taught that it is ok to rephrase the problem. They are presented with a specific procedure on how to do this but what I suspect is that they aren’t taught that they have some alternatives which also work. For example, 14 ̶ 9 can be rephrased as 5 + 9 ̶ 9, giving the answer 5. There are alternatives, which I have seen students create and use. One example is a student who realized that in the problem 14 ̶ 9 if 1 is added to both numbers, you see 15 ̶ 10, giving 5. This is a sound “theory” and if extended leads to a different conception of subtraction. This students can’t “prove” (or demonstrate) why this procedure works, but he knows it does consistently.

This type of thinking happens at the algebraic level as well. Hacker spends a lot time focusing on the idea of “rigor” and provides examples how rigor is factor which keeps students from passing algebra. Hacker misses what this student has done; Immanuel Kant stated that Math requires two things: imagination and rigorous logic and this student has employed both of these aspects, while hacker focused on only one.

I’ve strayed from the basic theme here. To repeat, it’s not algebra it’s the way it is taught. What’s the purpose of teaching math beyond the basics of arithmetic? Simplistically, our society needs to identify students with math capability who really will be needed as our society develops and grows. Mathematical modeling has become a core element across a wide spectrum of our lives.

Beyond this identification purpose, math allows us to examine a cluster of information and toss out the distractors, those elements that have no serious relationship to the core relationships in that set of information. This is what happens in what are called application problems. This idea alone is a very essential life value. Again, given a flurry of information, we tend to look for the essential pattern and relationship. This happens when driving. There’s a heap of stuff we see and hear but we pay attention only to those things that help us drive safely. Is this algebra? I believe some of it is.

Allow me the license of calling algebraic math something that we do automatically at incredible speed. The stuff we see in texts and the formulae and procedures are nothing more than slowed-down and recorded versions of what we do automatically. For example, I will start a math class by asking if there are any softball, baseball or basketball players and usually there’s more than one in the class. I ask one of them to stand and I toss them an eraser. They catch it and throw it back. I then ask if there was any math involved. As the discussion develops, it turns out that the class identifies a bunch of quantitative judgements involved. For example, how much speed must be used to get the eraser from the thrower to the catcher; how is the trajectory of the eraser tracked; how does the catcher determine where to place a hand to intercept the eraser’s path; how much energy must be applied to grip and hold the incoming eraser? There are other issues related to these core activities but the overall effect is that students start realizing that they employ “math” all the time. What’s more, they offer many more examples of activities that involve making quantitative judgements – driving, even walking up and down steps.

So, in my judgement, algebra is a tool we use to help us identify, connect and summarize quantifiable relationships and if taught this way, I tend to believe people would actually enjoy it.

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An 8th Grade Final Exam: Salina , KS – 1895

Posted by mark schwartz on August 16, 2016

Introduction

In the story below is an 8th Grade Final Exam given in Salina, KS in 1895. After you look at the problems, I’ve posed some questions and commented on a few things.

The Story

Here are the problems.

Name and define the Fundamental Rules of Arithmetic.

  1.  Name and define the Fundamental Rules of Arithmetic.
  2. A wagon box is 2 ft. Deep, 10 feet long, and 3 ft. Wide. How many bushels of wheat will it hold?
  3.  If a load of wheat weighs 3,942 lbs., what is it worth at 50cts/bushel, deducting 1,050 lbs. for tare?
  4. District No 33 has a valuation of $35,000.. What is the necessary levy to carry on a school seven months at $50 per month, and have $104 for incidentals
  5. Find the cost of 6,720 lbs. Coal at $6.00 per ton.
  6. Find the interest of $512.60 for 8 months and 18 days at 7 percent.
  7. What is the cost of 40 boards 12 inches wide and 16 ft. long at $20 per metre?
  8. Find bank discount on $300 for 90 days (no grace) at 10 percent.
  9. What is the cost of a square farm at $15 per acre, the distance of which is 640 rods

That’s it; just 9 questions. I guess the belief was that a student can demonstrate what they know in 9 questions, rather than 20 or so as we seem to do in today’s examination mode. In essence, either they learned it or they didn’t.

Let’s consider these problems one at a time in the context of what an 8th grader had to know. It is most likely that the information they had to know to answer these problems was, at some point, presented and discussed in class. Basically, it’s a set of memorized information. Think about this in the pedagogy at that time compared to what students today are expected to know. What they needed to know then is in the context of their daily lives and the business of the day.

In question 1 students had to know the Fundamental Rules of Arithmetic. Do we teach this today? Would it be of value for students to know? What are they? They are the four basic operations – addition, subtraction, multiplication, division which we teach but don’t identify as the fundamental rules.

In question 2 students have to be able to calculate volume but the critical thing they have to know – because it’s not given in the problem – is the volume of a bushel of wheat. Quite likely, since this is an example in Kansas, knowing the volume of a bushel is a very handy piece of agricultural information. As it turns out, there are two possibilities and we have to assume that the teacher at that time made it clear what was being asked in the question. A bushel of wheat has a volume of about 1.2445 cubic feet. There is also a heaped bushel, which is 27.8% (sometimes 25%) larger than a regular bushel. The regular bushel is also called struck measure to indicate that the bushels have been struck, or leveled, rather than heaped. And by the way, I didn’t know any of this and had to look it up. I didn’t grow up in an agricultural area. One more thing – given 4 decimal places in the volume and the 27.8% (sometimes 25%), we have to again assume that the teacher gave precise directions on how to handle these two values, most likely – my guess – by rounding 1.2445 to 1.

Question 3 looks like a straight forward calculation problem. The only possible issues are students (1) knowing how to handle the 1050 lbs. – do I subtract the 1050 before or after calculating the worth or (2) mishandling the decimal point in the 50cts.

Question 4 is again a straight forward calculation problem. What got my attention here is carrying on school for 7 months. It made me wonder if school was a 7 month period or if they simply used 7 because it fit the other numbers well for the calculation? It seems plausible that there could be a 7 month school year because it’s an agricultural area and families worked together to get the farm work done. Just a thought.

In question 5, students need to know how many pounds in a ton, which we assume was talked about in class at some point.

Question 6 seems to be a rather sophisticated problem for the 8th grade, but again in the context of life at that time, it seems reasonable that an 8th grader might be involved in the family’s business and would use this kind of calculation. I suspect it was a simple plug-these-values into the formula they learned. At that time, rote knowledge was highly prized. Note several things: how to use percent in decimal form and also realize that this was all paper and pencil calculations; no calculators then.

Question 7 has inches, feet, and metre as measurements so the student is being tested on measurement conversions. Once all the conversions are done, it again becomes a straight forward calculation. The issue, since 1 meter is equal to 3 feet and 3.37 inches, is what decimal value was used or were they taught to use just 39 inches. But again, it’s something that they presumably were presented and were expected to know. It’s sort of cute to use 12 inches, simplifying the calculation.

Question 8 is like question 6 in the sense that it’s sophisticated for the 8th grade, but again something very useful if a 14-15 year old was helping out with a family business. I again suspect this was a simple plug-these-values into the formula they learned. And as in question 6, students had to know how to write 10 percent as a decimal and further, again, there were no calculators then; all pencil and paper calculation. And what, exactly, is a bank discount?

Question 9 presents conversion issues between acre and rod – common, useful agricultural measures at that time which I again will presume were covered in class and students had to know. But they still had to do the calculation with paper and pencil. Technically and precisely both acre and rod had decimal values but I suspect it’s possible it was rounded off when talking about it and when doing calculation. The rod was a measure of 5 and a half feet, so the students had to again know how to handle decimal calculation if that was the value used. It’s interesting that they noted a “square” farm, since acre is a measure of area and the farm could be any shape. Perhaps it was just the teacher tinkering with the students, as some teachers – even today – are wont to do.

When you look across these 9 questions, it becomes apparent that they are all what we call today “application” or real life problems. These questions also involved sophisticated calculations based on formulae they clearly had to learn. It seems that the information that 8th graders needed to know was normal for that time, yet the pencil and paper calculations they needed to do were demanding. In this day and age, would there be an outcry if students had to do similar problems, having had to memorize the formulae and had to do the calculations without a calculator? I’m not going to venture a guess.

Posted in basic math operations, Historical Math, math instruction, mathematics, percent | Tagged: , , | Leave a Comment »

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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Walk the Clock: It’s Fractions

Posted by mark schwartz on August 3, 2016

Introduction

For some reason, or perhaps reasons, fractions don’t make sense to many students. Despite the visual representations in text and/or the use of manipulatives such as Cuisenaire rods, fractions seem to remain a mystery to students. One day I asked all my basic math classes “What makes fractions so hard?” The overwhelming response focused on remembering the steps of the 4 basic operations. For them, operations with fractions seemed nothing more than trying to remember the steps to get the answer. Somehow, math instruction throughout elementary and secondary education led students to think not about what fractions mean and what they represent but rather to think about how to “solve the problem”. So, I played with an idea which seems to have provided a way for students to “see” fractions a little differently.

The Story

DON’T TELL STUDENTS THIS IS FRACTIONS! If someone asks if this is fractions, tell them it will be discussed after the activities are done. I’ve provided an idea on how to do this in the discussion section following the demonstration of the activity.

The students preferably will work in groups of 3 (or 2, depending on the size of the class). The minimum grouping is 2. Each group gets a magic marker and 12 paper plates. The students are to number the plates 1 to 12. The plates are to be placed on the ground as a clock face. This activity is best done outdoors but if not, move the desks and chairs to allow for each group to have enough space for one person to walk inside and one person to walk outside a clock face circle. If neither of these spaces are available, the plates can be cut down in size and placed on a table top. If it’s a rainy cold day and going outside is a bad idea, and if the curriculum allows and time allows, make it a “review” day and hope for sunny and warm tomorrow. This activity works indoors and on the desk top but outside is best; it’s more fun. If done on the desk top, 2 markers per group will be needed. These markers will be the “walkers” in the activity (this will be explained below).

Here’s how it works. Lay out the plates as a clock face. One of the people in the group will walk outside the circle (call this person the outsider); one person will walk inside the circle (call this person the insider); the third person will be the reader/recorder (call this person reader). Give each group a copy of the activities (a sample is below) which states what the insider, outsider and reader are to do. Once the groups have figured out who will do what, give a demonstration of what they are to do, using the 1st activity.

Using the first activity and using one group to demonstrate, note that both walkers will walk twice. Both walkers start at “12”. In each activity, the insider walks first and then the outsider. The first walk is done when the insider reaches “12”. The second walk for both starts where their first walk ended. The reader is to watch and verify that each walker takes the right number of steps (others in that group can help verify).

1st Activity: on the first walk, the insider walks 2 units while the outsider walks 1 unit. On the second walk, the insider walks 3 units while the outsider walks 1 unit. The reader will note “outsider.insider”. In this activity, the record should show 10.12.

If there is confusion about the walking and/or the recording, just repeat the first activity. When everyone’s ready, move on to the next activities.

2nd Activity: on the first walk, the “insider” walks 4 units while the “outsider” walks 1 unit. On the second walk, the “insider” walks 6 units while the “outsider” walks 1 unit. The reader should note 5.12.

3rd Activity: on the first walk, the “insider” walks 2 units while the “outsider” walks 1 unit. On the second walk, the “insider” walks 3 units while the “outsider” walks 2 unit. The reader should note 14.12. (There should be questions on how to record this. Show students “military” time.

It’s important that if more activities are to be done, don’t allow students to do it. The reason: activities provided by students may result in a very time consuming set of walks and more critically, present a new issue to handle. For example, although subtraction of fractions can be done this way, I suggest not doing it. You could get a negative answer and you might want to avoid this. Just stick with one concept at a time; adding fractions (although they may not realize it). Given this, you might want to prepare and walk through a bunch of activities and be careful that none of them take too much time, yet enough time for the students to play and enjoy it.

Again, do not say anything about fractions at this point, but what has happened is that the problem 1/2 + 1/3 has been done. The record “outsider.insider” is 10.12 , or in reduced fractional notation is 5/6. Most likely, someone has noticed that the insider always has a value of 12. You sort of have to weasel your way around this and don’t yet call it a common denominator.

A Little Discussion. After these activities, you can transition to presenting fractions as you usually do. But, here’s one idea to consider in talking to students about how this activity demonstrates addition of fractions. What is seen and used but not referenced is the common denominator of 12. This explains why the insider’s walking the line twice isn’t counted twice. In the problem 2/3 + 3/4 , the denominator could be any multiple of 12 but in this case since it is 12 and you know it, don’t count it twice. Students may balk at this idea but it can be explained further. The insider always walks the line twice but always restarts the 2nd walk at “12”, while the outsider restarts the 2nd walk where the first walk ended so the insider’s walks aren’t added, rather they simply repeat.

Also not seen is the addition, but it occurs in the outside walk when the second walk starts where the first walk ended. The outside walker’s position at the end of the first walk is added to the beginning of the second walk. Please note that using this method for a problem such as 1/2 + 1/3 would give the answer 10/12, not 5/6, so clearly reducing fractions has to be addressed before this activity. Further, you might question how to get from this activity to the “rules” for addition and subtraction, but that’s not the point, although it can be seen because both fractions in this example, were converted to equivalent fractions with a denominator of 12, although in this case and others, it wouldn’t necessarily be the lowest common denominator. This again, could create a teaching moment, discussing the issue of common denominator versus lowest common denominator.

I suggest that different sets of students get a chance to walk the line. In fact, teams of students could do it; two walk and the others verify that their walking is accurate. Further, the point at which the transition from this activity to the traditional fraction work is to be made is a matter of how the class is collectively responding. In some instances, students caught on and realized that this was adding fractions. But even if they caught on, I still had them walk through all the activities. In several classes, students wanted more exercises. I think it was because it was a nice warm day. It’s a judgment call.

One more thing. Recall that the purpose of this activity is to give a visual and kinesthetic sense to the “rules” and it does seem to have a positive effect on students. When we got to the traditional rules and procedures, I heard students talking about how it “matched up” with what they were doing outside. Play with it.

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Metric to Metric Conversion: Ultimately, it’s a Proportion!

Posted by mark schwartz on July 27, 2016

Introduction

This discussion came about because one student in one class simply asked “Why does this work”? He was referring to the procedure for converting metric units to other metric units, for example, “how many centimeters are there in 10 kilometers?” He could see that the “moving the decimal point” procedure worked but he kept insisting that there must be more to it; that somehow someone had figured this out and he wanted to know how it had been figured out. I had no clear answer to this and told him (and the class) that I would think about it. What I came up with isn’t necessarily the reality of the derivation of that procedure, but it did start with something that we had already discussed in class – proportions – and he was willing to accept this as a demonstration of why it works but wasn’t about to consider it a true explanation. Loved this guy!

The Story

In Colaw and Ellwood’s 1900 School Arithmetic: Advanced Book (page 252) is a discussion of the metric system. Among other interesting things, they note that kilo hecta and deka are Greek, while deci, centi, milli are Latin. In reading through their discussion, I got to thinking about how they, as we do today, convert one metric unit to another: a 7-point scale and simply “move the decimal point” … but some of their commentary made me think about how and why this 7-point scale works.

If asked how many decimeters are in .04 kilometers, one has a variety of strategies to use. If the 7-point scale (kilo hecta, deka, unit, deci, centi, milli) is known, one can write .04 at the kilo point on the scale and then visualize moving from kilo to deci, which would give a move of 4 places to the right. If the decimal point is moved four places to the right, this shows that .04 kilos is 400 decis. Typically, students are accepting of this ‘shortcut’ because it is much more manageable than other systems. But, the question was “why does it work?”

I believe the underpinning for the move-the-decimal method is to do the problem by first converting all the units to the amount at each point on the scale that equals one unit. This by no means is a rigid mathematical derivation but rather a way of demonstrating the relationships using a previously studied math relationship, namely proportions.

The traditional 7-point scale looks like this:

Kilo                 hecta                deka                unit                  deci                 centi                milli

1000             100                  10                    1                      1/10                 1/100               1/1000

This scale shows the number of units in a named place-value. “Kilo” means 1000 units; “deci” means 1/10 of a unit, etc.. But let’s ask the question from the point of view of the unit: how many kilos would it take to make a unit? How many decis would it take to make a unit, etc.?

Here’s how the “unitized” 7-point scale would look:

Kilo                 hecta                deka                unit                  deci                 centi                milli

1/1000          1/100                 1/10                 1                      10                    100                  1000

It appears as though the scale has been reversed, and it has because we are viewing the information from the perspective of what it takes to make one unit. For example, it can be read as “1/1000 of a kilo equals 1 unit” or “10 decis equals 1 unit”, etc. The point of this is that all of the place-value names are now on the same scale and having them on the same scale permits one to establish proportions.

For example, on this scale 1/1000 of a kilo equals 10 decis because they both equal 1 unit. Another way of stating this relationship is to state that “1/1000 kilos is to 10 decis”, which is a phrase describing the first rate of two rates that would make up a proportion. What would be the second rate? The original problem was “how many decimeters are in .04 kilometers?”

In this case, being consistent with the idea in proportions that the numerators are all the same type of units and the denominators are all the same type of units, what is seen is the relationship of kilos to decis, is:

kilo  1/1000  =  .04  = 400 decis
deci    10        x

It is this proportional relationship which provides the basis for conversions from one metric unit to another, as long as the units used are those that “equate” them to 1. Students must be comfortable knowing how many ‘dekas’ it takes to make one unit (since a ‘deka’ is 10 units, it takes 1/10 of a ‘deka’ to equal a unit). This may seem counterintuitive since we typically say, for example, that a kilo is a thousand units, which is true but the focus here is with how many kilos it takes to make a unit.

Given this discussion of the two methods, it seems most likely that students would tend toward the ‘move the decimal point’ system. It doesn’t require any computation. But the point of presenting both of these it to bring out the reality that the ‘easier’ system is based on a proportional system. Just another example of the power of proportions based on an interested student’s insightful inquiry.

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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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Rephrase That Impossible Application Problem

Posted by mark schwartz on July 19, 2016

Introduction

As I was presenting a topic one day, a student said that what I was saying didn’t make any sense and could I please say it differently. My first reaction was to ask the class if it was true for them too; some agreed. It wasn’t a rude statement and I took the comment seriously and did rephrase what I said and asked if that made more sense and apparently I got it right. Then I got to thinking not only about that moment but other moments where what I was saying may not have made sense, but nobody bothered to stop me. As far as I can tell, it wasn’t the math content but the language I used to describe the content that bothered them. Thus the story that follows.

The Story

Question: Let’s say that the first city 4th of July fireworks I attended was in 2005. Since then, I attended the city fireworks every year including this year, 2016. How many fireworks have I attended?

Before considering the answer, consider if that question is the same as: how many years have I attended the city fireworks display on the fourth of July?

If your answer to the first question is 11, you’re wrong and as vague as the second question is, the answer is 11.

In both cases, which seems to be the same case, I suspect you got your answer by simple subtraction, 2016 – 2005 = 11. The thing to consider though is what exactly is being subtracted? Let’s bring this into a more manageable range, like 5 – 1. If you were to do this operation on a number line, you could put your finger on the five and move to the one, counting as you go and thus you would get 4. That four represents the number of movements from point 5 to point 1 on the number line. When you move from 5 to 4, you say “1”, in essence, “scaling” the distance between 5 and 4 as 1 unit, regardless of the actual distance. Given that it’s a number line, the distance between the points on the number line will all be the same. So, when we say 5 – 1, we are asking how many distances are there between 5 and 1. By the way, this distance analogy is similar to the idea of having 5 kittens and giving 1 away – how many kittens have you? In this case, it’s not distance, it’s kittens but conceptually it’s the same. We need not bother with scaling the number of kittens, because it’s a quantity not a distance, although some consider distance a quantity. As far as “how many years have I attended the fireworks?”, what’s being counted here is the number of years – an amount of time scaled rather than a distance. So, 2016 – 2015 is one, etc. as far as counting.

What’s the point? Remember the first question? To repeat: “Let’s say that the first city 4th of July fireworks I attended was in 2005. Since then, I attended the city fireworks every year including this year, 2016. How many fireworks have I attended?”

What is being counted here? Again, consider the number line. We’re not counting the distance between points on the number line rather were counting the number of points. The first question then has to be a subtraction plus one, which really is asking for the inclusive count.

You might say “so what?” to the difference between the first and second questions but looking at them as I did points out that there is a difference. The real issue here is the nature of asking questions in a math class. If we, as instructors, ask ambiguous questions, or questions which require students to reflect on the context of the information as well as the information in the question (and students don’t see the need to reflect on these issues) then we are, in a sense, misleading them and adding to their confusion about math. The context in this case is the words we instructors use.

I’ve seen this in questions in texts. We glibly accept the questions and answers at the end of the chapter and if some of those questions are questionable, we simply don’t assign them. But it’s not just the questions in texts. It’s how we state information, it’s how we use the language to structure questions and present concepts. The difference between the first and second question demonstrates this.

We should be attempting to be better at some precision in our questions and presentations because, like it or not, instructors are math role models for students. If we expect precision and accuracy from students, we should also expect that they can phrase good questions and it’s the instructor that establishes the idea of a well-phrased statement. It also seems it’s a critical component of being able to arrive at a correct answer to a problem. The caution to read “word” problems until you understand it is reasonable, but what if you never “understand” the problem? My thought is that students have to have license to and practice in rephrasing problems, without changing any of the relationships in the problem.

For example, when teaching percent using the percent proportion model (you can see how this is presented in the Percent Proportion posting in this blog), I point out to students that most percent problems can be rephrased. An example: A farmer sold 180 sheep, which represented 16% of all the sheep he had. How many sheep had he after the sale?

There are a lot of extra words in this problem, but only two numbers. I asked the class to rephrase this problem focusing on the relationship between the numbers. The students wrote all their attempts on the board, so that we could discuss the thinking that brought about their answer. What they produced, based on the percent proportion model that I presented to them, were two rephrased problem: First, “180 is 16% of what number of sheep?” to get the total number of sheep and then subtract the 180 he sold. The second was to see that if 180 sheep were subtracted from the total (written as x – 180), this would represent the number of sheep left and the percent left would be 84 (100 – 16). The rephrasing then would be “The number of sheep remaining is 84% of what number of sheep?” The result of this rephrased problem still is the total but again, simply subtract 180 sheep. If you want to play with it, the answer is 945. Even if the percent proportion model isn’t used, this rephrased problem is much more manageable.

But here’s how to set up both results using the percent proportion model.

180  16
 x   100      … solving, x = 1125 … total after sale = 945
x - 180  84
  x      100   … solving, x = 1125 … total after sale = 945

We practiced this rephrasing idea some more and I reminded them that they don’t have to rephrase every problem, but if the problem seems “impossible”, rephrase it.

Posted in algebra, Historical Math, math instruction, mathematics, proportion, Proportions, remedial/developmental math | Tagged: , , , | Leave a Comment »

What? That Much Percent Increase?

Posted by mark schwartz on July 8, 2016

Introduction

I like coincidences. Particularly when they provide learning opportunities for my students. We had just spent time learning about percent and percent increase and decrease. The problems in the text were good but not really challenging. The coincidence was that I was reading John McPhee’s The Curve of Binding Energy (1974) and I’ll start the story with what he said on page 18.

The Story

“Thousands of miles of tubes, pipes, and other conduits were needed to create a network of flow wherein the gas could now go through a membrane, now return to try again, now go on to a new membrane, gradually advancing, in a process of separation and elimination, until what had begun as seven-tenths of one percent U-235 was more than ninety percent U-235 – fully enriched, weapons-grade uranium.”

I’d never heard this detailed an explanation of how weapons-grade uranium was made. But what really got my attention was that his statement could be a percent increase problem. I worked it out before I gave it to the class, rounding off the initial “more than ninety percent” to a manageable 90%.

Further, I decided that it would be an in-class extra credit exercise and allowed that the students had to first work within their assigned group, but once they had an answer they could discuss it with other groups.

I did this because the percent increase is 12,757% and this size percent increase would cause the groups to question what they did, even if they got that number. There were occasional answers to the problems in the text that resulted in percent increases of more than 100% but nothing quite like this. Once I gave them the problem and answered any preliminary questions and they got to work, I roamed the room listening to the strategies they came up with to do the problem.

The first issue was how to numerically express seven-tenths of one percent. One group asked if they could talk to another group to get help expressing it. So, I stopped the class and said that if they are willing to accept the following condition, they can work as a class to get the answer. The condition was that everyone in the class would get the same grade. They accepted. There was an eruption of conversation and as I roamed around, I was asked if what they got was right. I just referred them to other members of the class.

Once there was consensus on how to represent the initial percent, they simply continued with what they had learned about setting up percent increase problems. By the way, I taught a somewhat non-traditional method that doesn’t use a formula, rather it uses a somewhat modified percent proportion approach. You can look at it in this blog at Percent Proportion.

Several groups quietly called me over to show me their result, asking if they were right. Some were and some weren’t but I wouldn’t say yes or no, reminding them of the condition under which they were working. So, more talk, discussion and exchange of how to set up the problem.

It was interesting to watch the evolution of the shared work – people got up and moved around the room; some asked to and did use the white board; I heard a lot of “show me” and “why did you do that?” and “that doesn’t seem right”. But, ultimately there was class consensus on the right answer.

They did, however, insist that I walk through how I thought about it even though they got it. So, I put on what I call my “slow-motion-math” hat and gave them the following:

Ninety percent is 90/100, so the amount of increase is 90 – .7, or 89.3%. Seven-tenth of one percent is (7/10)(1/100) or 7/1000 (I did this because I saw a lot of questioning on how to express it). This in percent is .7/100 (or if you were sure, you could have just written .7 over 100). So the question can be put in a percent increase frame. First, the amount of increase is 89.3 and since it started at .7, the amount of increase relative to the beginning point can be expressed as 89.3/.7, or 893/7 (they questioned if doing this would give the same answer and we discussed this). Using the proportion statement 893/7 = x/100 gives a percent increase of 12,757%, rounded. So, doing the original problem led to some other related talk about fractions, decimals and rounding. Neat.

After all was said and done, I got questioned about this exercise because there was a sense that it was a trick question. I have noticed that when students feel uncertain about a math problem, the frequently asked question is just that. I then heard stories from the class about their prior math experiences where trick questions unfortunately were used to presumably teach them something about math, but the only learning was frustration because a lot of the tricks were beyond the bounds of what had been taught and in essence they quit. Given what I heard, I may have quit too. Somehow they concluded that math is just knowing the right tricks.

But once they were accepting that it was an interesting problem, I noted to them that as they read books, magazines, watch TV or come across “mathy” stuff, they might play with it as we did with this problem. And of course they noted to me to record the “A” for all of them.

Posted in basic math operations, Historical Math, math instruction, mathematics, percent, proportion, Proportions, remedial/developmental math | Tagged: , , , , | Leave a Comment »

 
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