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 ‘Historical Math’

A Short, Short Discourse on Digit Sum

Posted by mark schwartz on January 16, 2017

Introduction

My daughter and I play with numerology; it’s just play, nothing serious. We play with calendar dates, prime numbers, birthdays, etc., looking for patterns and such. Most recently she texted me that she added the digits in my wife’s birthday and continued to add them until there was a single digit, the result was 4. She wasn’t aware of digit sum nor casting-out-9s. So, I played back.

The Story

I first took a look through some of my old math texts, dating from about 1850 to 1900. I looked there because in those days, having a way to check your work was important and digit sum was popular, but not noted in all texts. What I don’t know is whether instructors may have taught it although it wasn’t in the text. I did find a few (I have about 75 old texts) that actually demonstrated how addition can be checked using digit sums. Oddly though, none of those that presented how to check addition indicated that digit sum can be used to check subtraction, multiplication and division as well. Yes, it can.

But, let’s take a look at why digit sum works. It’s based on what is called casting-out-9s. In essence, given a number – 23 – if you cast out 9s (which can be done by subtracting 9 until you have a single digit), you get 23 – 9 = 14, then 14 – 9 = 5. Notice that if you simply added the 2 and 3 in the number 23, you also get 5, so what simplifies getting a digit sum is simply add the digits repeatedly until you get a single digit. For the number 268, at first you get 16, then 7.

Why does this work? Let’s get basic. Using 23 again, this is really 2(10) + 3(1). Rewriting this in what I call ‘slow motion’ math, it becomes 1(10) + 1(10) + 3(1), then 1(9) + 1(1) + 1(9) + 1(1) + 3(1). If the ‘1(9)s’ are now ‘cast out’, the result is 1(1) + 1(1) + 3(1), giving 5(1).

This of course is not a rigid proof but rather a demonstration of casting-out-9s.

For example, 235 + 568 = 803. The digit sum for 235 is 1; the digit sum for 568 is 1, and the sum of these is 2. This equals the digit sum of 803, so it checks. I realize that in today’s technical world, this procedure isn’t likely to be taught nor used but in olden days without calculators, it seemed reasonable to check your work.

Now back to what I sent back to my daughter. I generated the digit sum for the birthdays for all the members of our family and then generated the digit sum for the sum of them and lo and behold the result was 1! Of course, the family is unity!

Told you it was a short discourse, but couldn’t resist sharing it. Can’t wait to see what she comes up with next.

 

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Yet Another Subtraction Algorithm!

Posted by mark schwartz on November 4, 2016

Introduction

I recently posted Revisiting Mr. Stoddard’s 1852 Subtraction. In that posting I modified Mr. Stoddard’s idea by introducing a procedure which allows for subtraction without borrowing. This posting modifies that modification.

The Story

I’ll use a simple subtraction example to demonstrate the procedure, but I have examined much more sophisticated problems such as 20801 ̶ 278 and the procedure is still good.

Basically, treating ‘ab’ as a 2-digit number and ‘c’ as a single digit number, in the problem “ab ̶ c”, if c > b, the answer to ‘b ̶ c’ is 10 ̶ ( c ̶ b ) and then add 1 to the 10s place value in the subtrahend. For example, 12 ̶ 8 gives 10 ̶ (8 ̶ 2), or 4, then add 1 to the 10s place value in the subtrahend, giving 1 ̶ 1 or 0, which isn’t written.

What I didn’t note clearly are two things. First, if in that example, b > c, then write down that value as the answer. Do not add 1 to the next place value in the subtrahend. However, if c > b, then the algorithm as noted is to be used. And here’s the modification – continue with this algorithm!

Here’s an example in slow-motion math. Using the problem 7234 ̶ 567 as a traditional ‘vertical’ problem, we hav

7234
–567

In the 1s column, 7 is greater than 4, so the answer is 10 ̶ ( 7 ̶ 4) which is 7. Add 1 to the 6 in the subtrahend 10s column. Then in the tens column, 7 is greater than 3, so the answer is 10 ̶ ( 7 ̶ 3), which is 6. Add 1 to the 5 in the subtrahend 100s column. Then in the 100s column, 6 is greater than 2, so the answer is 10 ̶ ( 6 ̶ 2), which is 6. Add 1 to the zero in the subtrahend 1000s column. Then in the 1000s column, 7 is greater than 1, so the answer is simply the difference of 6. The solution looks like this:

7234
– 567
6667

There are many subtraction algorithms posted in this blog and most of them focus on avoiding the need to borrow, so if you feel like trolling through the entire blog and compiling them, you might find one you like.

Posted in basic math operations, Historical Math, math instruction, mathematics, remedial/developmental math, subtraction | Tagged: , , , | 1 Comment »

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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Revisiting Mr. Stoddard’s 1852 Subtraction

Posted by mark schwartz on September 29, 2016

Introduction

In this blog is a posting Mr. Stoddard Subtracts in 1852. If you haven’t read it, you don’t need to (but of course you can!). Mr. Stoddard presents an idea in subtraction which avoided the need for “borrowing”. For some reason, I was playing with a subtraction idea and after I had written out the entire algorithm, I realized that I basically had modified Mr. Stoddard’s; thus the title.

The Story

I’ll use a simple subtraction example to demonstrate the procedure, but I have examined much more sophisticated problems such as 20801 ̶ 278 and the procedure is still good.

Basically, treating ‘ab’ as a 2-digit number and ‘c’ as a single digit number, in the problem “ab ̶ c”, if c > b, the answer to ‘b ̶ c’ is 10  ̶  ( c ̶ b ) and then add 1 to the 10s place value in the subtrahend. For example, 12 ̶ 8 gives 10 ̶ (8 ̶ 2), or 4, then add 1 to the 10s place value in the subtrahend, giving 1 ̶ 1 or 0, which isn’t written.

Here’s why it works. In essence, it could be said that borrowing has happened but it’s hidden as well as not written!

In essence, 10  ̶  ( c ̶ b ) is borrowing, but it’s hidden. The ‘10’ in the 10 ̶ (8 ̶ 2) could be said to have been borrowed from the 10s column in the minuend. Given that, that ‘10’ can be said to have been subtracted from the10s column in the minuend. It’s known that if the same value is subtracted (or added) from both the minuend and subtrahend of a subtraction problem, the answer will be the same. Thus, adding a 1 to the next place value in the subtrahend adds a value which will be subtracted.

There it is. It’s a mild modification to Mr. Stoddard, but my ‘aha’ moment with 10 ̶  ( c ̶ b ) may well have been his idea incubating all this time. Try it with other problems – like 20801 ̶ 278 and after a while it becomes as automatic as doing the problem using borrowing.

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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.

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

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.

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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 »

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.

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