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 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.

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Archive for the ‘math instrution’ Category

Counting Sheep

Posted by mark schwartz on May 26, 2016


This was presented in a course titled Contemporary Math. The course was designed for students whose major didn’t require math beyond the basics. It was to familiarize students with a wide array of math ideas, but none in depth. This article is how students were introduced to different base number systems and operations in bases other than ten. Students were remarkably novel in their response to the idea in this article, modifying it to fit other circumstances. Some of their ideas are included. The following story is a handout to students, only up to the point where the exercise was extended. The class worked in groups.

The Story:

Below are shown 5 holding areas each individually fenced in, labelled A through E, from right to left. They are for the farmer’s sheep, which are in the East Pasture. The areas are labeled A through E, from right to left because when the sheep come to the holding areas, they enter from right to left, starting with A. The fence around each area is low enough so that the sheep can get in and out by jumping over the fence.

E        D        C        B        A            East  Pasture

South Pasture

So, a sheep that wants to get into area A or move from area to area must jump over the fence. Sheep move only from right to left between areas – A to B to C, etc. –  and only one sheep per area is allowed. The farmer may change his mind later and allow more than one sheep per area.

But these are not ordinary sheep! They are specially bred to have amazing jumping power. If an area is empty, a sheep will simply jump in. However, if they see a sheep already in the next area, they will jump over that sheep to land in the next empty area. It gets sillier! If a sheep coming in sees that there is a sheep in area A and also in area B, the incoming sheep will jump over both of them to land in area C, the next empty area.  An incoming sheep jumps over all filled areas to the next empty area.  AND THEN, any sheep that gets jumped over leaves the area by jumping into the South Pasture.  They get a little weird when other sheep fly over their head!

Now, I wasn’t allowed to bring pictures of the sheep because they are so special, so you can use anything you want to represent your sheep (a coin, a wad of paper, a stick-figure sheep, a pen, etc.). So, see what happens when your sheep come in from the East Pasture. Let’s start with 5 of them. Although 5 sheep entered, if you do it as the sheep would, sheep would be only in areas A and C when you’re done. If this doesn’t happen, call me over. Once everyone has accomplished this, we’ll play with a different number of sheep.

Extending the Concept: After playing with 5 sheep, the class was asked to extend the idea by finding out which areas would be filled if 6 sheep came in from the East Pasture. Once each group got an answer and all groups agreed, the results for 5 and 6 sheep were “tabled” on the board and then this was extended to 7, 8, 9, and 10 sheep, which was also recorded.  There was a lot of interaction within and between groups, which is a desirable outcome of this exercise. Here’s what the class saw once all of them finished the exercise and there was agreement between groups about the correct arrangement of sheep.

# sheep     E     D     C     B     A

5            0      0      1     0      1

6            0      0      1     1      0

7            0      0      1     1      1

8            0      1      0     0      0

9            0      1      0     0      1

10          0      1      0     1      0

I then asked that everyone let their sheep sleep and then asked: “Without actually doing it can you predict what the arrangement would be for a dozen sheep?” I walked around the classroom and listened to the within-group discussions. and prompted them to let their sheep sleep. Every group eventually concluded correctly that only areas C and D would be occupied. We talked about how they figured it out.

In most classes, no one in the class was aware that what they were actually doing was counting in binary. In some classes there were tech and computer savvy students, who were aware of the binary system (base 2). I asked them not to shout it out or share it with those in their group. This activity led to a discussion about the binary system and its use in computers. At this time, the discussion was extended to seeing how the base 2 and base 10 systems functioned similarly by demonstrating the place-value system. The class was then asked to wake up their sheep and do the following:  the condition for the sheep was changed to allow three sheep to be in an area at the same time and the class was asked to find the arrangement in the areas if 19 sheep came in from the East Pasture. The result was …

# sheep     E     D     C     B     A

19          0     0      1      0      3

This was discussed in detail and I asked them to identify in what base they had counted and yes, they figured it out. Then the question asked by one of the students (it always comes up at some point in this exercise) was: do all bases work the same, even if a dozen sheep were allowed in each area?  We then had another fun discussion. Then someone asked (but we didn’t discuss it) how it would work if a different number of sheep were allowed in each area – what a great question!

Summary. This exercise was done typically over two class periods and explored the place-value system and the different base systems. Further, the classes played with basic operations within each base as well as conversions between bases. A final note. In one of the classes, there was an education major and she wanted to know if the basic sheep-counting exercise could be used in elementary school to introduce students to the ideas we covered in class. Her idea was to line up 5 chairs in a row and have students enter this row of chairs just as the sheep had. My answer was a definite “yes” but not to label the activity as an exploration of different base number systems. Rather, just let the children play and the concept will incubate and at some point later in their education if they study different base number systems, I strongly suspect they will have an “aha” moment.


Posted in basic math operations, math instrution, mathematics, place value systems - different bases | Leave a Comment »

Blended Factoring

Posted by mark schwartz on May 23, 2016


Factoring trinomials is a core operation in Algebra and is more than a stand-alone procedure. Consider how it is used in solving quadratic equations. Consider the basic algorithm for factoring. I’ve seen about 7 different ones, one of which I developed, which I call the “blended” method. This method is called blended because the core of it is the traditional procedure but it is blended with a set of operations which many students have found more comfortable. For this method, it is expected that students have studied and can perform identifying and factoring out a common factor from a binomial. So, a review of it, if not covered recently, should be done. Being able to do this is an essential part of the Blended Method. This became a student handout and was given to them at the time of practicing the procedure.

The Story:

When factoring a trinomial of the form ax2 + bx + c, traditionally one always looks for and factors out any common factor, referred to as a greatest common factor (GCF). Further, the method for factoring is typically presented in two steps: first, address a trinomial of the form ax2 + bx + c where a = 1, and then later, when a >1. Then there are special cases – the perfect square trinomial and the difference of two squares, which are handy but not essential. All factoring of these can be done with what is described as the blended method.  It is not necessary to find the GCF, nor to pause to recognize and recall the procedure for the special cases.

It starts with a trinomial of the form ax2 + bx + c, with a = 1 but is not limited to this. This allows for exploring why the blended method works when a = 1 and also when a > 1.

For the first demonstration of how this method works, factor x2 + 5x + 6. The key thing to notice is that the coefficient of x2 is 1. So, as in the traditional method, set up two binomials as (1x     )(1x     ) but include the coefficient of x2.  It may seem odd to include the coefficient of 1 in both binomials, but it provides for a clear transition to factoring trinomials when a>1. Have students write it in cases where the coefficient of the x2 term is 1.

Step two is to factor both ‘a’ and ‘c’, combining these factors into one set. Typically, only pairs of factors of both ‘a’ and ‘c’ are needed but using prime factoring is very helpful. In this case the factors would be 1•2•3 (again, it seems odd but include the 1). This set then has to be played with until two products add to ‘b’, the coefficient of x. For example, 1 +  2•3 = 7; 1•2 + 3 = 5, as does 1•3 + 2 = 5. Either of these latter two is acceptable. Again, this may seem trivial because ‘1’ is one of the factors but it establishes and reinforces a pattern which isn’t quite so obvious. More examples will be provided later with different sets of factors.

Place the ‘2’ and ‘3’ as the second term in each binomial, and it makes no difference which binomial. So, the factors of x2 + 5x + 6 are (1x + 2)(1x + 3). Again, the coefficient of 1 for the x term need not be written once students are comfortable with the procedure, but for now, they are to use it to demonstrate and emphasize the procedure. At this point, the next step is to examine each of the binomials for a common factor, and if any are found, factor them out. In this case, there are none, but in later examples this step will demonstrate why it is to be done. Try another one.

Factor x2 – x – 12. Step one gives (1x     )(1x     ). Step two gives several possibilities because ‘c’ is negative and one of the factors has to be negative. However, the factors don’t have to be handled as a positive and a negative; they could both be treated as positive, with the awareness that one of the products is subtracted from the other to give the value of  ─1, the coefficient of ‘b’.

Twelve has 3 sets of factors: 1•12, 2•6, and 3•4. Again, notice that prime factoring wasn’t the beginning point, but in some cases, prime factoring is recommended. Which pair can give a difference of −1? The only pair with a difference of 1 is 3 and 4, and since we’re looking for −1, the two factors are 3 and −4. Again, multiplying either of these by 1 seems trivial (and perhaps it is in this case), so the two factors are still 3 and −4¸ and the factors of x2 – x – 12 are (1x +3 )(1x −4 ). Again, look for a common factor in each binomial, factoring it out if found. In this case, there are none. More practice with a = 1 is recommended before moving to cases where a>1. This is to establish that the coefficient of ‘x’ in both binomials is the coefficient of the x2 term in the trinomial. It also provides practice in handling trinomials where ‘c’ is negative.

Now try factoring 2x2 – 11x – 6. In this case, the two initial binomials are (2x     )(2x     ). Notice that the coefficient of ‘x’ in both binomials is the coefficient of the x2 term in the trinomial. The factors of ‘ac’ are 1•2•2•3. These factors have to be played with until two numbers result such that their difference is ‘b’, or – 11. The outcome is 1 and 2•2•3, or 1 and 12. The 12 has to be negative because the sum has to be – 11.  So, the two binomial factors of 2x2 – 11x – 6 are (2x + 1)(2x −12). The next step is examining each binomial for a common factor, and in this case, (2x −12) gives 2(x – 6). So, we now have (2x + 1)•2•(x – 6). The question is whether or not the factor of 2 is part of the final factoring or not. To determine this, divide this answer by the coefficient of x2 in the original trinomial. The outcome is (2x + 1)(x – 6), and these are the factors of 2x2 – 11x – 6.

Here’s a summary of the steps for the blended method. Given a trinomial of the form ax2 + bx + c, where a ≥ 1:

  1. Write 2 binomials with only a first term in each of (ax    )(ax    ), where ‘a’ is the coefficient of the x2 term. Indicate that this expression will be divided by ‘a’, giving the expression.
  2. Factor ‘a’ and ‘c’, producing a set of factors containing those of ‘a’ and ‘c’. Prime factors are not required but typically make the work easier.
  3. Use this set of factors to determine two values – “m” and “n” – which added together give the coefficient of “b”.
  4. Write the two binomials (ax & m)(ax & n), where the “&” could be either a plus or a minus.
  5. Extract any common factor in each binomial, identifying them as c1 and c2.
  6. Cancel all common factors in c1, c2 and “a”.
  7. This gives the factors of ax2 + bx + c, two binomials and in some cases a constant.



More practice is to be done with this procedure using examples from the text, but a demonstration of how to use it for the two special cases – perfect square trinomial or difference of two squares – follows.

Factor 4x2  ̶   12x + 9. First, there is no need to identify if this is a perfect square trinomial. It seems that it’s likely to be since a = 4 and c = 9, both of which are perfect squares. But for the moment, ignore that.

Step 1 then is to write (4x  )(4x   )/4 , based on step 1 of the procedure.

Step 2 gives the set of prime factors 2⦁2•3•3.

Step 3 has to give two products whose sum is 12, and that is 2⦁3 and 2⦁3. But notice that it’s a minus twelve in the trinomial so, the two products have to be   ̶ 6 and    ̶ 6.

Step 4 gives the expression (4x   ̶ 6)( 4x   ̶ 6)/4

Step 5 gives 2(2x  ̶ 3)⦁2⦁(2x  ̶ 3)/4

Applying step 6, gives the answer of (2x  ̶  3)(2x  ̶  3).

Now factor 8 x2   ̶  18. At first, it appears as though this might be the difference of two squares but it isn’t. There is a greatest common factor, and if that isn’t noticed, the question is where to start. Try the blended method.

Step 1 then is to write (8x   )(8x   )/8, based on step 1 of the procedure.

Step 2 gives 2•4•2•9.

Step 3 using 2•4•2•9 won’t give zero, the coefficient of ‘b’, so moving toward prime factoring gives 2•4•2•3•3, and this can simplify to 2•2•3 and 4•3, which is 12 and 12. Making one of them negative and adding gives zero.

Step 4 gives the expression (8x   ̶ 12) (8x   ̶ 12)/8

Step 5 gives 4(2x   ̶  3)⦁4⦁(2x   ̶  3)/8

Applying step 6, gives the answer of 2(2x   ̶  3) (2x  + 3).

Using this process seems like a rather long way to factor as opposed to learning and recognizing the special cases of a perfect square trinomial and difference of two squares …  and it may well be. However, consider learning one procedure – the blended method – or 5 procedures (a trinomial with ‘a’ equal 1, a trinomial with ‘a’ > 1, using the grouping method, a perfect square trinomial or the difference of two squares). The choice is obviously yours, but give the blended method a try. Once practiced and understood, it’s as automatic as using 4 different procedures.

One caution. Some texts include problems in the form ax3 + bx2 + cx. Notice that this is not the form

ax2 + bx + c and since it isn’t, the blended method won’t work. Given this kind of problem, the first step has to be to factor out the common term “x”, bringing the trinomial to a form that can be factored with the Blended Method.

Posted in algebra, factoring, Historical Math, math instrution, mathematics | Leave a Comment »

What is the Question to This Answer?

Posted by mark schwartz on May 11, 2016


In Greenleaf’s 1862 New Elementary Algebra, he devotes considerable time to fractions, including exercises which require considerable mastery of the operations with fractions. He introduces the section on fractions with a discussion of unit fractions, which are fractions with a 1 in the numerator. He establishes a very handy algorithm for adding them, and then posits the question “Doesn’t it seem that if given the answer one could identify the fractions that added together gave that answer?” He doesn’t answer his own question, but I found it intriguing enough to play with it and it led to an interesting relationship with another standard Algebraic operation.

The Story:

Given the answer to the addition of two unit fractions, can one find those two fractions?

For example, given the answer 5/6, what two fractions added together give this answer? For those familiar with adding fractions, the answer of 1/2 and 1/3 might be found after a little trial and error. On the other hand, finding the “question” to the answer 14/24 might take a little longer.

But there is one caveat of caution to be considered throughout this discussion, particularly if you are the one creating the answers to be used. In all cases, if the answer is one that can be reduced, use the non-reduced form first. If not, it is possible that the reduced form will give no “question” or a “question” that is correct but different from the one used to generate the answer. For example, 14/24 can be reduced to 7/12, and the answer 14/24 gives the question 1/2 + 1/12, whereas the answer 7/12 gives the question 1/3 + 1/4. So, in presenting this kind of problem, students may be asked to find both sets of answers. But continuing this caveat of caution, 20/96 can be reduced to 5/24 but 5/24 gives no answer. So, when generating the answer, check out the possible questions.

Let’s start by identifying some useful algebraic strategies that already exist and can be used to find the question to the answer.

There is a method presented in Algebra books for finding the sum of two fractions without finding a lowest common denominator. Yes, there is such a procedure and it is:

m/a + n/b = (mb + an)/ab

For purposes of this discussion, remember the caveat provided earlier. Further, since we are discussing this premised on unit fractions, the value of ‘m’ and ‘n’ in this expression would both be equal to 1, and the expression is then

Expression 1.               (a + b)/ab

This expression is linked to the quadratic equation! Take a look.

Using ax2 + bx + c = 0, where a = 1, the solution is to find two factors of the constant ‘c’ which when added together are equal to ‘b’, the coefficient of the x-term, which is exactly what is seen in Expression 1. In the quadratic, the value “b” is the numerator of Expression 1, while the value “c” is the denominator of Expression 1. Here’s how to use this.

We were given the answer of 5/6 and asked to find the two fractions that added together gave this answer. If the numerator 5 is seen as the coefficient of the x-term ‘b’ in the quadratic (the sum of two factors) and the denominator 6 is seen as the constant ‘c’ in the quadratic (the product of two factors), the quadratic equation can be stated as

x2  – 5x + 6 = 0

The x-term is negative because the values in the two fractions are all positive, and when solving the quadratic equation for the values of x, it becomes apparent that the factors of the constant term both have to be negative and therefore when added together give a negative value for the coefficient of the x-term.

Factoring this equation gives (x – 2)(x – 3) = 0 and setting both to zero and solving for x gives 2 and 3, which are the denominators of the unit fractions 1/2  and 1/3, which when added together give the answer 5/6.

So, try another one. Given the answer 8/15, what two fractions added together give this answer? Using the above concept, we have

x2  – 8x + 15 = 0

Factoring, we get (x – 3)(x – 5) = 0 and setting both to zero and solving for x gives 3 and 5, which are the denominators of the unit fractions 1/3  and 1/5, which when added together give the answer 8/15.

This type of problem may never appear in today’s texts but if it ever does, thank Mr. Greenleaf for providing an algorithm for solving it.

Posted in algebra, equations, factoring, Historical Math, math instrution, mathematics | Leave a Comment »

Are We Lying About Factoring?

Posted by mark schwartz on May 9, 2016

This is a short piece. I didn’t mean to do this but I was playing with factoring and here’s the thought I had.

Using a simple example, the factors of x2 + 2x + 1 are (x + 1)(x + 1). However, what is also a set of factors of this trinomial is ( ̶  x  ̶ 1) ( ̶  x  ̶ 1). Although this latter is technically correct, it is not traditionally correct. If a student in class gave you this as an answer, would you accept it? Technically, yes; traditionally, no. You know what a computer-based delivery system would say … and that’s because it’s been programmed by traditionalists.

So, why  won’t we accept  ( ̶  x  ̶ 1) ( ̶  x  ̶ 1) as a pair of factors? We teach factoring as a prelude to solving an equation. The expression x2 + 2x + 1 becomes an equation by simply stating that the expression equals something, like 5 or y. In solving the equation using the factoring method, would it make a difference?

Given the equation x2 + 4x + 1 = – 3, adding 3 to both sides gives x2 + 4x + 4 = 0. Solving this by factoring we get (x + 2)(x + 2) = 0 and setting both to zero, x equals – 2. If we now set the factors to be (-x – 2)(-x – 2) and set both to zero, x again equals – 2. So far, it seems to work.

When wouldn’t it work? I don’t know. Just a thought to share with you …


Posted in algebra, factoring, math instrution, mathematics, remedial/developmental math | Leave a Comment »

Are We Adding Ratios (rates?) or Fractions?

Posted by mark schwartz on May 5, 2016


This writing is a discussion and speculation on why I think fractions are so hard. I say speculating, yet when this was presented in class, the speculation was transformed into reality based on what students said. The ideas in this writing were sometimes presented to a class at this level of detail but even if the discussion never reached this depth, the core idea was discussed with the class. After such a discussion, and when assignments were given, there was always at least one student who would ask “What kind of problem are you asking us to do?” You’ll understand why this question was asked after you’ve looked at this article.

The Story

One of the elements of this story is embedded in the title. For the most part, we tend to not pay attention to the difference between ratio and rate; they tend to blend together. Yet, texts will carefully distinguish between them. I bring this up now because some of my colleagues may be distressed by my using “ratio” where they would see what’s being described, technically and formally, as a ‘rate”. For purposes of this discussion, ratio will be the term used although it’s a fine distinction in the examples I use.

What seems to be somewhat confusing is the concept of adding ratios as opposed to the concept of adding fractions. But you contest, ratios are fractions. Yes, they are but adding ratios is a different procedure from adding fractions. At least it is as we present addition of fractions in the classroom.

I present this because I believe it carries the kernel of misconception about adding fractions. For example, for students who have struggled with addition of fractions, a question like 1/2 + 1/3 will commonly get the erroneous result 2/5. In essence, add the numerators and add the denominators. Most math instructors would say this is wrong. However, it is a correct answer depending on what question was asked!

For example, if there are 5 people in the room, 2 men and 3 women, and 1 of the men is wearing glasses and 1 of the women is wearing glasses, the ratio of men who wear glasses to all the men in the room is 1/2, while for women, the ratio of women who wear glasses to all the women in the room is 1/3. If asked “what is the ratio of those wearing glasses to all those in the room?” the answer is 2/5. This comes from simply doing a count or from adding 1/2 + 1/3.

So, the answer to the abstract question, “what is 1/2 + 1/3?” depends on what the 1/2 and 1/3 represent. If it represents 1/2 (1 of 2 equal parts of something) and 1/3 (1 of 3 equal parts of the same thing), then it represents a fraction and the fraction addition algorithm is to be used. If it represents 2 ratios — how many of the total has a certain characteristic and there are two of these ratios – then the ratio addition algorithm is to be used.

This difference between adding fractions and adding ratios is a critical and typically unvoiced point. It typically is unvoiced because the concept of “lowest common denominator” (LCD) pushes us past this consideration. Most LCD algorithms involve students’ first knowing prime numbers, prime factors, equivalent fractions and how to build the LCD. After this, the students then return to “addition”.

I would contend that students should ask (or be told) what circumstance is the context for the question. In most texts, the addition of ratios may be noted but not dwelled upon because the focus is on adding fractions. This, however isn’t explicitly stated, but it is the assumption on which “adding like fractions” is based. Yes, ratios are fractions but as noted, adding two different ratios is allowable but doesn’t use the same addition algorithm as the traditional algorithm for addition of fractions.

I propose that this issue be discussed with students as part of the introduction to fractions. I believe it clarifies the issue for students who erroneously want to add the numerators and add the denominators when adding fractions. In a sense, this is a more “natural” act than adding fractions as we teach students to do, and because it is more common in everyday situations, it seems the correct procedure to follow. For example, in a classroom of 20 students (9 females and 11 males), I could ask for 3 female and 4 male volunteers. The unvoiced part – as happens in many “selection” situations – is that 3/9 and 4/11, or 7/20 were selected. We simply don’t pay attention to the denominator!

If given the abstract question 1/2 + 1/3, I would expect students to ask “Am I adding ratios or fractions?” In my experience of allowing a discussion and demonstration of this point, students have come away with a better understanding of adding fractions. The traditional addition algorithm seemed to make more sense after discussing the difference between adding ratios vs. adding fractions. The discussion actually clarified both algorithms, as the students asked very incisive questions about the difference and how they can determine which algorithm to apply.

The reason I believe it is important to allow this as a legitimate question (an instructor can always state “from now on, all abstract statements are to be understood to be adding fractions, not ratios … unless otherwise stated”) stems from the way most texts define ratio or ratio notation. Collectively, they tend to state something like “the ratio of a to b is given by the fraction notation a/b, where a is the numerator and b is the denominator, or by the colon notation a:b”. Most texts only cite the fraction notation. But again, most texts discuss and have examples and exercises for students showing how to determine and write a ratio. But, I have never seen a text present a problem of adding ratios. I contend this is done intentionally to avoid the need to identify the difference between adding ratios and adding fractions. Yet, it seems that this identification of the difference should be included, for the reasons previously stated.

And this, I propose, is one reason why students stumble through fractions. Addition of ratios is not addressed. Further, when addition of fractions is presented it is usually preceded by a lengthy discussion of factors, prime factors, greatest common factor, least common factor, like fractions, least common denominator, and finally, equivalent fractions. Wow! … all that just to add 1/2 and 1/3. I can see why students take the easy way out and apply the ratio addition algorithm!

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Exploring an 1864 Demonstration that ( – )( – ) = +

Posted by mark schwartz on May 1, 2016


When introducing (or reviewing) signed numbers to a remedial/developmental community college class, I try to identify how many in the class already have a mantra for some of the operations with signed numbers. My favorite way to check is to ask them to complete the sentence, out loud, that I’m going to say.  I make sure they’re ready and then say “a minus and a minus is a _____”. For those who voice the answer of “plus”, I pause and then ask them when. Here, I get muddled responses although that was the right answer. I also get a clue about the class because usually there are about only 5 to 10 who answer correctly. Students then take to the board to teach each other how this mantra works and this provides some more insight into how strongly this and other memorized shards of math interfere with learning and applying some core basic math resulting in blurry spots in their math experience.


In the instance of addressing multiplying two negatives to obtain a positive, many math instructors have created a variety of approaches – physical, tactile, and melodic. Of the non-physical, non-tactile, and non-melodic approaches I’ve used, Mr. Greenleaf in his 1864 New Higher Algebra (pg.25) provides a simple and direct demonstration that seems to make sense to students.

Here are his statements:

“When the quantity to be subtracted is partly or wholly, negative. Let it be required to take b  ̶  c from a.”

“Operation.  a  ̶  (b  ̶   c) = a   ̶   b + c.”

“If we take b from a, we have a   ̶   b. But, in doing this, we take away c too much, consequently the true difference will be a   ̶   b increased by c, or a   ̶   b + c.”

“Hence, the ALGEBRAIC DIFFERENCE between two quantities may be numerically greater than either quantity.”

I took license with Mr. Greenleaf’s algebraic demonstration and put it in a numerical demonstration. I didn’t go through his algebraic presentation with the class.

First, I presented to the class the statement 8  ̶   (5  ̶  3) and asked them to carefully and in slow-motion do the order of operations to get the answer. Almost everyone knew to do the work in the parenthesis first, getting 8  ̶   2, and subtracting gives 6. After everyone was satisfied that this was correct, I left this on the board and I wrote the original problem on the board again.

Now, as I started to step through this differently, the class pointed out to me that I was doing it incorrectly; that I was cheating, that I had to do the work inside the parenthesis first, etc.  Technically, they were correct but I proposed that I would do it a piece at a time in slow motion. After much interesting discussion, they granted me permission to continue. Based on Mr. Greenleaf’s method, I first did 8  ̶  5,  noting to the class that this would be the first operation to do. They accepted this, so the result at this point is 8  ̶  5 = 3.  Are we done, I asked? Several sharp-eyed students pointed out that if the problem were done the “correct” way, the answer would be 6, not 3. Another student then pointed out that we hadn’t finished the problem! We hadn’t yet done the 8  ̶   (  ̶  3) part.

So, it was determined that when we subtracted 5 from 8, we were subtracting too much (because the real problem was to subtract less because 5  ̶  3 is less than 5), we had to account for the 3. So, how do we do this?

Again, it caught the attention of several students that the difference between the real answer of 6 and the answer here was 3, so can’t we just add the 3? A lot of back and forth about this until there seemed to a consensus that it worked.

So, after doing 8  ̶  5 and getting 3 – which was too low since we took too much away – we had to add 3 back and this +3 came from  8  ̶   (  ̶ 3).  So, can we conclude that 8  ̶   (   ̶ 3) = 8 + 3?

Again, more fun discussion and finally acceptance! A lot of the students expressed thanks for showing them why a minus and a minus is a plus and I was careful to remind them that they need to modify their mantra to say a minus times a minus is a plus. The class was quite engaged in seeing the “rule” emerge rather than simply being told it’s true. They owned it!

Posted in Category One, Historical Math, math instrution | Tagged: , , , , | Leave a Comment »

Driving the Integer Road

Posted by mark schwartz on April 27, 2016



A lot of texts these days use colored chips or other manipulatives to introduce students to the operations with signed numbers. This article goes one step further. I was using a “chip” concept but then realized that even the chip idea is somewhat abstract. So I visualized something that turned out to be effective … and fun.

The Story

Operations with signed numbers seem to escape a lot of students. This is particularly so when they are confronted with all the rules for the four basic operations. The one they seem to remember is “a minus and a minus is a plus” but they tend not to remember what operation is involved. This rule tends to get applied whenever they see two negatives in proximity to each other in a problem.

.Rather than presenting the rules to the class, they were provided a visual and kinesthetic activity which “generated” the rules. They were shown one way in which the rules were embedded in a daily activity. In essence, any rule or formula is a statement of a relationship between quantifiable activities and in our daily lives, one was found that could be used.

So looking at activities and playing with them might lead to the realization that there are quantifiable components, which ultimately may lead to or actually be the rule. It’s a matter of observation and a matter of determining which symbolizing system to use.

For example, in the following demonstration, letters and numbers will be used to indicate an action. The action will be precisely prescribed. Students will be asked to do a specified series of actions. These actions are based on something that they (well, at least most of them) do.

Here we Go …

The activity that most students do is drive. Their path is a road, but in this activity their path will be a number line. Certain conditions must be set: (1)  a car on the number line (the road) can face forward ( FF, toward the positive end of the number line) or face back ( FB, toward the negative end of the number line) and (2)  the car can move forward (MF, put it in drive) or move back (MB, put it in reverse).

A car (I use Tonka toys: small cars and trucks) can be put on the number line either facing forward or facing back. Once it’s on the number line¸ it can move either forward or back. At this point, it has to be clarified and emphasized that facing is with reference to the number line and moving is with reference to the car. I believe this to be a critical distinction because in our math symbolizing system, we use “ – “ for the action of subtraction but also the indicator for the sign of a number (negative). We use “ + “ to indicate the action of addition, but also the indicator of the sign of the number (positive). This is a confusing element for a lot of students. The notation needs to be discussed, examined, and understood. The exercise with the car makes this point very apparent, by having students begin with the FF, FB, MF, MB symbolizing system and ultimately showing how it is comparable to the use of “ + “,“ – “, and parentheses.

A demonstration of how the system works is given by drawing a number line on the board and using a picture of a car to face and move. I drew one on a piece of cardboard so that facing forward or facing back can be on opposite sides of the cardboard. The number line need only go to plus and minus five. The exercise always begins at zero. The direction to follow is stated as FFMF2. This means “face forward, move forward two”.  But combinations of facing and movement can be strung together as a “command line”. For example, FFMF2, FBMF3, FBMB1, FFMB3 would be done in sequence, beginning at zero with the second, third, and fourth command done from the previous end point.

Using an arrow to indicate where the car starts and which way the car is facing, and using a ● to indicate where the car would finish, the sequence FFMF2, FBMF3, FBMB1, FFMB3 would look like this.


-5     -4     -3     -2     -1     0     1     2     3     4     5

FFMF2                                →           


-5     -4     -3     -2     -1     0     1     2     3     4     5

FBMF3                        ●                  


-5     -4     -3     -2     -1     0     1     2     3     4     5

FBMB1                         ←   


-5     -4     -3     -2     -1     0     1     2     3     4     5

FFMB3        ●                      

After having students practice a bunch of these, a shift to conventional symbols can be made. But don’t hurry the class through this because it takes a while for them to get comfortable with the idea that facing is with respect to the line and moving is with respect to the car.

For example, FF and MF are +; FB and MB are −, and here’s a substitution to show how it works (BUT THIS IS NOT TO BE SHOWN TO STUDENTS UNTIL LATER).

FFMF2,    FBMF3,    FBMB1,    FFMB3

+ ( +2 )  – ( +3)   – ( –1 )  + ( –3 )

This standard notation can be further simplified but this relationship between facing and moving and the standard notation needs to be discussed, examined and understood before proceeding.

There are a few side benefits with this. Doing the facing and movement slows them down, engages cognition, sight, hand movement, decision making, all of which might be put in the context of problem solving. Further, when they are working in groups (in all my classes, students work in groups), they learn how to accept and give questioning and correction from peers, evaluate others’ actions as well as their own, and be deliberate. It also provides a kinesthetic and visual representation that aids the transition from positive-number-only operations to operations with signed numbers.

The key outcome, however, is that the “code” of facing and moving has a parallel pattern with the traditional notation. The pattern can be generated by asking the following for each command: Ask the class: for FFMFn (“n” represents any number), in what direction do you move? For FFMFn, you move in a “forward” direction. Asking the question for FFMB, FBMF and FBMB; for FFMBn, you move in a “back” direction; for FBMFn, you move in a “back” direction and for FBMBn, you move in a “forward” direction. As the answer to each of these questions is given, write down the following pattern:

Command       direction moved

FFMFn            forward

FFMBn           back

FBMFn           back

FBMBn           forward

This pattern contains most of the rules for operations with signed numbers, but this “code” has to be translated to the traditional notation. This is done by talking to the students about the slight difference in notation, including the use of parentheses as an indicator of multiplication. We earlier established (and it’s still on the number line on the board) that forward is “+” and back is “─”. Also remind students that facing and moving are two separate events.

The traditional notation can now be substituted in the above “coded” pattern. The “command” code list is the traditional notation “action’ list and the “direction moved” code is the traditional “outcome” list.

……  code   ……                                   …  traditional notation …

Command       direction moved             action           outcome

FFMFn                forward                       + (+n)              + n

FFMBn               back                             + (─n)              ─ n

FBMFn                back                            ─ (+n)              ─ n

FBMBn                forward                       ─ (─n)             + n

For example, take the problem 2 + ( ─ 3 ) + 4 ─ ( ─ 5 ) ─ ( +2 ). The first step is to look at the problem and see if any substitutions can be done based on the patterns. In talking about substitution, stress to students that this is, in fact, multiplication.

Look at the substitutions that can be made:  2 + (─ 3) + 4 ─ (─ 5) ─ (+2):

+ (─ 3) becomes ─ 3 ….. ─ (─ 5) becomes + 5 ….. ─ (+ 2) becomes ─ 2

So, the problem now is 2 ─ 3 + 4 + 5 ─ 2.

I recommend that at this point, students practice only the substitution idea with at least 10 more problems. I spend a lot of time at this point and don’t talk about what to do with this string of numbers. Finishing this problem after the substitution can be done a variety of ways. I prefer to have this string of numbers seen as a string of signed numbers to be added, employing the commutative and associate properties as needed. Again, there are multiple approaches but the focus of this article is to get to the string of numbers, not how to handle them, so the way to handle the numbers should come from experiences which you have found successful.

Posted in algebra, basic math operations, Category One, Category Two, math instrution, remedial/developmental math, signed number operations | Tagged: | Leave a Comment »