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.

Signed Numbers with Chips

Posted by mark schwartz on April 1, 2016

Signed Numbers With Chips: AN INTRODUCTION

This introduction to signed numbers is best done with students in groups, so that they can help each other, monitor each other, and teach each other.

Each group is given a “handful” of poker chips. Each chip has a plus sign written on one side and a minus sign on the other side … just use a black magic marker.

Before doing anything with the chips, draw a number line on the board, positive and negative to about ± 5. Starting at zero, indicate that if one moves from zero to positive three and then from positive three back to zero, what has happened in essence is that after adding and subtracting the same value to a given value, the result is the given value. Other examples can be drawn and each time the emphasis is on “the same number of positives and negatives added together results in zero.”

After identifying this principle, ask students to take some chips, any number of them, and make a pile on the table so that all the chips can be seen. Then ask them to remove all the zeros and see that what is left: they should have all positives or all negatives (or perhaps no chips left in the pile). Do this exercise several times, establishing that after removing all the zeros, the pile has a value.

The next step: rather than starting with a pile of chips, each group is to build a pile of chips and determine the value of the pile. Start with an empty table. There are specific rules for doing this pile building.

Write   + (+2) + (─ 3) ─ (+1­) + (+4) on the board. Before continuing, spend time pointing out several things: First, the use of the grouping symbol, in this case parentheses. The number and the sign of the number inside the parentheses indicates the number and type of chip that will either be added to or subtracted from the pile. The sign outside the parentheses indicates addition or subtraction. Adding means put it in the pile; subtracting means take it out of the pile.

Read through the problem, noting carefully the operation “sign” and the number “sign”. This problem would be read as: add a group of two positives, add a group of 3 negatives, subtract a group of 1 positive, add a group of 4 positives. This might also be voiced as: add two positives, add 3 negatives, subtract one positive, add three positives. I tend to include the word “group” as a way of emphasizing what happens when later on problems such as 3(─2) ─ 4(+1) can be voiced as: add 3 groups of 2 negatives and subtract 4 groups of 1 positive. Play with the language but carefully note that “+” and “─” can mean two things. Also point out that all the operations are to be done before removing any zeros. Removing zeros will be the last step (if students ask why, tell them it will be explained later).

Now slowly walk through this problem with the students, drawing your pile of chips on the board. You can just show the “+” and “ ̶ “ ; no need to show the chip. Your pile, as you step through the sequence would look like this:

The problem again is:   + (+2) + (─ 3) ─ (+1­) + (+4)

1st operation of adding 2 positives            +   +

2nd operation of adding 3 negatives         +   +   ─   ─   ─

3rd operation of subtracting 1 positive     +   ─   ─   ─

4th operation of adding 4 positives           +   ─   ─   ─   + +   +   +

Now that all the operations are done, what is the value of the pile? This is determined by removing all the zeros, the result being:

+   +             … or in words, 2 positives.

Note what happens after the first two operations. If 2 positives are added and 3 negatives are added, and the zeros are removed at this point, the result will be one negative in the pile. The third operation is to remove one positive … there aren’t any! That’s why there’s an emphasis at this point, to remove zeros after all operations are done.

Below are some more practice problems, crafted to always remove the zeros after all the operations … except the sixth problem, which will be used to demonstrate what to do when an operation asks that something be subtracted that isn’t in the pile. This is an important thing for students to see and demonstrates the basic rules for adding and subtracting signed numbers. The “patterns” for adding and subtracting are noted later.

After the 6th problem, make up some more or as I have found interesting, let students come to the board and make up some. Generally, 10 problems gets students comfortable (and successful) with this idea.

Write the problems, one at a time, on the board. Direct the students to do the following by adding (putting into the pile) or subtracting (taking out of the pile) positive or negative chips. Start with an empty pile. Remind them: the sign outside the parentheses tells you to add or subtract; the sign inside tells you what kind of chip ( + or ─ ) to add or subtract. If they need more chips, just ask and we’ll find some (In one case where I ran out of chips, I had students tear up paper and put “+ and ─” on the pieces).

  1.   + ( ─ 4 ) ─ ( ─ 2 ) + ( ─ 1 ) + ( + 3 ) + ( ─ 2 )     Answer:     ─   ─

2.   + ( + 5 ) + ( ─ 3 ) ─ ( + 2 ) + ( + 1 ) + ( ─ 3 )      Answer:     ─   ─

3.    + (─ 1 ) + ( ─ 4 ) + ( + 6) ─ ( ─ 3 )                     Answer:     +   +   +   +

4.   + ( + 3 ) + ( ─ 2 ) ─ ( + 1 ) + ( ─ 3 ) ─ ( ─ 4 )      Answer:    +

5.   + ( ─ 2 ) + ( + 4 ) + ( ─ 3 ) + ( + 2 ) + (─ 1)      Answer:        empty table = zero

A note about #6. Starting with an empty pile, the only way to take something away is to add something to the pile which doesn’t change its valueand that is a zero in the form of an equal number of positive and negative chips. Then the operations can be done. Let students try to work this out (usually someone figures it out). If not, show them.

6.       ─ ( ─ 1 ) ─ ( ─ 2 ) ─ ( ─ 3 )

In order to do the first operation, add 1 zero:     +

Now do the first operation, giving:   +

In order to do the second operation, add 2 zeroes:     +   +   ̶     +   ̶

Now do the second operation, giving:   +   +   +

In order to do the third operation, add 3 zeroes:     +   +   +   +   ̶     +   ̶   +   ̶

Now do the third operation, giving:   +   +   +   +   +   +

The answer is +6

I recommend making up more problems and have students do a total of about 10 problems before moving on – maybe even have the students make up some problems. At this point, if students want to remove zeroes as they do the problem, that’s their choice. They now know to add a zero if needed.

At this point, the “rules” for operations can be generated. After this demonstration, a pattern will be summarized. In every case, it’s important to note that the value after the operation may be positive or negative, but the value of the pile goes up or down. Have the students do the following:

First Case: start with the value of the pile at zero; meaning, no chips in the pile. What happens to the value of the pile if the operation + (+2) is done? The value of the pile goes up. So, a substitution can be made because + ( + n ) = + n ….. which means that if positives are added to a value, the value of the pile goes up.

For a second case, again start with the value of the pile at zero; meaning, no chips in the pile. What happens to the value of the pile if the operation + (─2) is done? The value of the pile goes down. So, a substitution can be made because + ( ─ n ) = ─ n ….. which means that if negatives are added to a value, the value of the pile goes down.

For a third case, start with the value of the pile at zero; meaning, no chips in the pile. What happens to the value of the pile if the operation ─ (+2) is done? Give students a moment to figure this out and if they can’t, point out that this is the same as a previous problem where you had to subtract something that wasn’t there. So, add as many zeros as needed to allow the operation to be done. The result is that the value of the pile goes down. So, a substitution can be made because ─ ( + n ) = ─ n ….. which means that if positives are subtracted from a value, the value goes down. It’s important to note that the value after the operation may be positive or negative, but the value of the pile goes down.

And a fourth case, start with the value of the pile at zero; meaning, no chips in the pile. What happens to the value of the pile if the operation ─ (─2) is done? Give students a moment to figure this out and if they can’t, point out that this is the same as a previous problem where you had to subtract something that wasn’t there. So, add as many zeros as needed to allow the operation to be done. The result is that the value of the pile goes up. So, a substitution can be made because ─ ( ─ n ) = + n ….. which means that if negatives are subtracted from a value, the value of the pile goes up.

The following table summarizes these 4 cases and shows the pattern for substituting expressions. After the table, an example demonstrates substituting. In essence, these patterns are the operations for adding, subtracting and multiplying signed numbers.

This     can be replaced with      this

+ (+n)                                               + n

+ (─n)                                              ─ n

─ (+n)                                              ─ n

─ (─n)                                              + n

Take a look at this problem:       + ( + 2 )   + ( ─ 3 ) + ( + 4 )   ─ ( ─ 5 )   ─ ( + 2 ).

Look at the substitutions that can be made using the pattern: + ( + 2 )   becomes 2,   + ( ─ 3 ) becomes ─ 3, + ( + 4 ) becomes + 4,   ─ ( ─ 5 ) becomes + 5 and   ─ ( + 2 ) becomes ─ 2.

So, + ( + 2 )   + ( ─ 3 ) + ( + 4 )   ─ ( ─ 5 )   ─ ( + 2 )  becomes 2 ─ 3 + 4 + 5 ─ 2 .

I recommend that at this point, students practice the substitution with at least 10 more problems. Prepare some in advance. Most of the problems in a text don’t have as many terms and I propose that they practice with problems having 4 or more terms.

One more note. Given 2 + ( ─ 3 ), I propose to students that this can be “read” two ways, and both are correct. It can be read as “two plus three negatives”, in which case it’s addition of two numbers with opposite signs. It can also be read as “2 plus one group of three negatives”. By putting the word “group” in the sentence, the question is; can there be more than one group of three negatives? … and the answer is yes. So, if the problem were 2 + 4( ─ 3 ) although it still looks like addition, “4 groups of 3 negatives is 12 negatives” by multiplication. So, in essence the + 4( ─ 3 ) can be replaced with ─12 consistent with the pattern above, which contains the rules for multiplying signed numbers.

The essence of this exercise is the pattern that evolves from adding and subtracting chips in a pile. It demonstrates the traditional relationships summarized in the rules in a text. But also note that the pattern which allows for substitution gives the same outcome as applying the rules for addition, subtraction, and multiplication (and you might note, also division) of signed numbers. The point is that students now have a visual and tactile basis for the rules in the text. There are several strategies for how to handle these terms after the substitutions are made, so employ the one that you have found successful.

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One Response to “Signed Numbers with Chips”

  1. sarablack said

    thanks for the excellent research!

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