Why a Negative Times a Negative is a Positive A discussion from the NCTM-L Mailing List How do you teach students that a negative times a negative is a positive? ~~~~~ A good example illustrating why a negative times a negative is a positive comes from thinking about paying off a debt. Suppose I have to pay $5 a week. In three weeks I will be down (-) $15, that is -5*3=15; however, three weeks ago I had $15 more than I have now----(-5)*(-3)=+15. ~~~~~ I use the following analogy: A train traveling east is +; a train traveling west is - The future is +; the past is - First, check that the students have mastered distance = speed x time. Example: how far has a car traveled which has traveled at 60 km per hour for 2 hours? Draw a number line: -15 -10 -5 0 +5 +10 +15 - -|--------|------|-----|------|------|-------|- + For each of the problems, below, the train is currently at 0. 1. A train is traveling at 5 kph to the east. Where will it be in 3 hours? +5 x +3 = +15 2. A train is traveling at 5 kph to the west. Where will it be in 3 hours? -5 x +3 = -15 3. A train is traveling at 5 kph to the east. Where was it 3 hours ago? +5 x -3 = -15 And, finally, (ta dah!) 4. A train is traveling at 5 kph to the west. Where was it 3 hours ago? -5 x -3 = +15 While I am on the subject, here is how I teach such monsters as: -3(a + 2) - 2(4 - 6a) One simple rule (which is developed when the students learn integer subtraction) - to subtract, change subtraction to addition and the term being subtracted to its opposite. Then add. The setting out looks like this: -3(a + 2) - 2(4 - 6a) = -3(a + 2) + -2(4 + -6a) = -3a + -6 + -8 + +12a = 9a + -14 An interesting aside: I accept this as correct, but tell students that the usual form is to change it back into a subtraction, ie, 9a - 14. But I don't insist on it, rather I wait until the students do this naturally. 90% do so within a year, the other 10% still leave their answers as 9a + -14 right up to the end of year 12. Interesting, eh? That same 10% will still change subtracting to adding the negative, even when they are simplifying a complex expression after using the chain rule for differential calculus. Mind you, they tend to be the C and D students, and not the A and B students. The hardest expressions of this type to teach are of this type: 4(a + 2) - (3a - 4) I have the students write this as: 4(a + 2) + -1(3a + -4) and proceed as before. This doesn't seem to be as easily accepted by the students as the previous example. ~~~~~ In the book: I. Gelfand and A. Shen, Algebra, Birkhauser, 2nd Printing, 1995, ISBN 3-7643-3677-3, $16.50, 149 pp. sections 13, 14 (p. 17-20) are devoted to the addition and the multiplication of negative numbers. The statement: (-a)*(-b) = a*b is not easy to prove at the school level. it is tied up with some of the basic results related to *abstract algebra*. I have deliberately avoided numbering the rule in any systematic order. At the basis is the (1) distributive rule for multiplication with respect to addition. Namely: a*(b+c) = a*b + a*c and (b+c)*a = b*a + c*a Of course, one may reduce these two down to a single one by using the (2) commutative rule for multiplication Namely: a*b = b*a. It should be mentioned that these rules are in fact quite special. For example, if we examine the commutative rule for multiplication in terms of the traditional algorithm, we have the following example: 36 23 x 23 36 -------------------------------------- 108 138 72 69 -------------------------------------- 828 828 Thus, we have the same answer even though the calculations take on quite different appearance. Needless to say, the above difference is not visible if the calculations are performed by using a pocket calculator. At the end of this post, there will be a different kind of example to illustrate the fact that distributive rule fails. The reason that (-a)*(-b) = a*b is our wish to retain the usual rules of arithmetic that hold for the natural numbers 0, 1, 2, .... NOTE: I believe that it is customary to use *natural numbers* in the school texts to denote the collection: 1, 2, 3, ... Actually, this can cause students to ask: Where did *0* come from when we are asked to write down the number that comes after 9? The answer here is that in the decimal system, we build up the writing of numbers by means of the ten digits: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, together with certain rules. As such, 0 is really quite indispensable. The development of the number zero in the history of mathematics took a long, long time. If one does the calculation using counting rods, one may just leave an empty space. However, if one wishes to keep a record of the calculation and avoid ambiguity, then a symbol for zero is needed. For example: Is | ||| 13, 103, 1,003, etc? Admittedly, the inclusion of zero as a *natural number* is often not done in many of the university level text on abstract algebra texts. To show that (-2)*(-3) = 2*3, we may begin with the distributive rule applied to the following special cases: (3) a*0 = 0 = 0*a To see (3), we consider: a*(0+0) = a*0 (using 0+0 = 0) a*(0+0) = a*0 + a*0 (using the distributive rule) These two then lead to: a*0 = a*0 + a*0 (transitivity of equality). By adding the *negative of a*0 to both sides and use the (4) associative rule for addition, we have: 0 = -(a*0) = -(a*0)+[a*0 + a*0] = [-(a*0)+(a*0)] + a*0 = 0 + a*0 = a*0 We have used the basic property that (5) 0 + x = x. (left additive identity property for zero). By a similar argument exchanging left and right, we have the *mirror images* of the rules. In other words, the rules for arithmetics are left-right symmetric. At this point, we can consider the special case: (-2)*[(-3)+3] = (-2)* 0 = 0 (-2)*[(-3)+3] = (-2)*(-3)+(-2)+3 [(-2)+2]*3 = 0*3 = 0 [(-2)+2]*3 = (-2)*3 + 2*3 We may either cite the cancellation law or just repeat the argument using the associate law again: [(-2)*(-3) + (-2)*3] + 2*3 = 0 + 2*3 = 2*3 (-2)*(-3) + [(-2)*3 + 2*3] = (-2)*(-3) + 0 = (-2)*(-3) This proof can be formalized to the general case to show that (-a)*(-b) = a*b. It is not at all clear at this point whether this long winded argument will be convincing to a student. It may be one reason why the argument is typically omitted in school algebra texts. It is perhaps worth stating that the rule: -(-a) = a does not require the use of the distributive rule. It is a consequence of the associative rule and the properties of zero: [-(-a)+(-a)] + a = 0 + a = a -(-a) + [(-a)+a] = -(-a) + 0 = -(-a) The assertion that (-1)*a = -a then follows by using the distributive rule. (This may be used as an exercise.) Examples of the failure of the two-sided distributive rule abounds. They are the source of many of the mistakes made by students in schools as well as in the colleges and universities. Here are some: cos(a + b) is not the same as cos a + cos b. (a+b)^2 is not the same as a^2 + b^2. In calculus classes, the sum of two functions f and g is defined by the rule: (f+g) {x} = f {x} + g {x} Formally, we have the *left distributive rule*. However, it is rarely the case that a function f will satisfy the *right distributive rule*: f {x+y} = f {x} + f {y} Students may need frequent reminder that the notation: f {x} denotes the value of the function f at the point x and should not be confused with the *product* of f and x. In this example, the definition of the sum of functions uses the fact that we can add the values of the functions. It follows from the definition that the composition of functions satisfies also the *right distributive rule*: (f + g) o h = (f o h) + (g o h) However, the *right distributive rule* usually fails. This aspect cause quite a bit of problem in the study of *iteration of functions* which has much to do with *fractals*. Functions that satisfy the *right distributive rule* are said to be *additive*. At the level of numbers, the only such functions arise from multiplication by numbers. Namely, one may use an induction argument to show that f {1 + ... + 1} = f {1} + ... + f {1} (n summands) so that f {n} = n* f{1}. Now, from the commutative rule for multiplication, we obtain: f {n} = f{1} * n. In other words, f is the function that multiplies each number n by the number f{1}. In particular, f{0} = f{1}*0 = 0. From this then follows that: 0 = f{0} = f{(-n) + n} = f{(-n)} + f{n} so that f{(-n)} = -(f{n}). It is important to note that the additivity of a *real-valued* function does not automatically lead to the conclusion that the function must be equal to the multiplication by a real number. One needs an additional assumption that the function be *continuous*. It is, unfortunately, not possible to write down any example (!) of a real-valued function which is additive, but not continuous. This is a technical aspect of advanced mathematics that frequently lead to *backlash* against mathematics. In fact, on a graphing calculator, the concept of *continuity* makes no sense. Namely, at any stage of the display, there are only a finite number of pixels. Here is a problem I posed in my abstract algebra class for juniors and seniors. Consider the quadratic equation: 0.0001*X^2 - 10*X + 1 = 0 It is apparent, with or without graphing calculator, that X = 0.1 is an approximate solution of this quadratic equation. Question: Where is the *other* solution, approximately? As a final note: The text by Gelfand and Shen has a total of 342 problems. Rarely are the problems in the *drill* category. The problems are typically challenging. Section 44 has the title: The well-tempered clavier. Many of the other sections deal with common mistakes made by students. There is one shortcoming: There are not many historical references. I should mention: Gelfand is the oldest holder of the prestigious MacArthur Fellowship. He is one of the greatest mathematicians in the 20th century. His research not only covered a broad range of mathematics, mathematical physics, but also biology. He is also involved in mathematical education in the former Soviet Union. Han Sah, sah@math.sunysb.edu ~~~~~ Here I am again. Peter Dinin referred to not having to use manipulatives so much. I am sold on manipulatives. I do not care what age the person is. I think that if "students" can see why it works and have hands-on experiences that they will remember. I have taught grades 6-12 and have taught + and - negative numbers to all grades. I have taught for many many years just having students memorize the rules for adding, subtracting, multiplying and dividing. And that is what they did.. memorize and then forget. For the past 4 or 5 years, I have used two colored counters for all levels. Right now I teach 7th grade Algebra I, PRe-Algebra, and "average" level 7th math. I Have seen the AHA. I get the experience every time I use the counters. I never tell the students the rule. I have students who can add, subtract, multiply and divide integers and really can not quote you the rule. They just know how it works. I can see them visualizing it as they work. That is what I think it is all about. I teach them first to use the counters, then how to draw what they have and then it just sinks in and they don't have to do either of the above. I guess I have been to so many workshops, etc on manipulatives that I am sold. I know that I have said this before but Arkansas is really doing a great job training our teachers on the standards. It is hard to explain what is really going on but a class has been developed called Math Crusades. It is funded by State Dept. of Education $ plus a NSF grant plus business contributions. Teachers in Arkansas have manipulatives in their classrooms and they are being used. So much for the soap box spill..Please excuse. But I get excited seeing students discovering for themselves what is going on. I really do think that they remember it better. ~~~~~ I like "the distributive property must work" approach from an earlier post: If you believe that a(b - c) = ab - ac, then -3(5 - 7) = (-3)(5) - (-3)(7) = 6 but -3(5 - 7) = -3(-2), so -3(-2) = +6. When I taught algebra, I used a "pattern" approach: 2 x (-8) = -16, 1 x (-8) = -8, 0 x (-8) = 0, what's next? Each product is bigger by 8, so -1 x (-8) has to be 8. ~~~~~ Minus times minus is plus, The reason for this we need not discuss. --W. H. Auden Negative numbers, like many new ideas in mathematics, achieved practical acceptance long before they achieved theoretical acceptance. The Babylonians allowed negative numbers in setting up quadratic equations; the Chinese, as described in the Nine Chapters, solved systems of linear equations that often involved negative numbers. The Chinese method, by the way, would impress any modern adherent to the use of manipulatives as instructional tools. Coefficients of the unknowns and constants were represented by rods on a counting-board ... the Chinese used red rods on the counting-board to represent negative numbers, while black rods stood for positive numbers. A coefficient of zero was indicated by a blank space on the board. The conception of negative numbers as "smaller" than positive numbers has caused considerable concern for mathematicians, as well as for students of beginning algebra, over the years. An interesting argument against negative numbers was given by Antoine Arnauld (1612-1694), theologian, mathematician, and close friend of Pascal. Arnauld questioned that -1:1=1:-1 because, he said, -1 is less than +1; hence, how could a smaller be to a greater as a greater is to a smaller? ... On the whole, not many 16th- and 17th-century mathematicians felt at ease with or accepted negative numbers at all, and of course they did not recognize them as true roots of equations. There were also curious beliefs about them. Though Wallis was advanced for his times and accepted negative numbers he thought they were larger than infinity as well as less than 0. In fractional form, Arnauld's ratio becomes -1/1=1/-1. Obviously, 3/4 could never be said to be equivalent to 4/3. How could any fraction with the numerator larger than the denominator be equivalent to one with its denominator larger than its numerator? Wallis' argument went like this: Consider the sequence ...,1/5,1/4,1/3,1/2,1/1,1/0,1/-1,1/-2,1/-3,1/-4,1/-5,... . As the sequence approaches 1/0 from the left the value increases, with 1/0 being infinite. Thus for 1/-1, and so on, the denominator is smaller than 0; hence the value, while negative (i.e., less than 0) must also be larger than infinity. ~~~~~ How about another approach using balloons, weights, and a two-pan beam balance scale (the simple old kind of balance scale.... a glorified miniature see-saw). Suppose you have a bunch of helium-filled toy balloons, and each balloon exerts an upward force of exactly one dyne (small unit of force in the metric system). Suppose you also have a bunch of pebbles, each of which is pulled downward by gravity with a force of one dyne. You can balance the scale with equal numbers of pebbles on both sides, or equal numbers of balloons fastened to both sides. With the scales in balance you can add pebbles (positive = downward forces) or add balloons (upward = negative forces) and still keep the system in balance. If you add pebbles (add positive values) to one side it becomes heavier. If you take away balloons (subtract negatives)/ from a side it also becomes heavier. This might be something one could actually demonstrate in the classroom. It also leads naturally to the idea of an equation which is "balanced" in the sense that the right side balances the left. ~~~~~ Well I am glad to see that others want to use colored counters. I really am sold on them. I hope that I can explain in writing how to use them in subtraction. Did you see my earlier message on multiplication ? If not I will copy it and send it to you . Subtraction is very easy. You need to know that 1 positive (y) and 1 negative (r) make a zero to do this one. You also need to understand that subtraction means to take away. That is all the pre-knowledge that you need except of course that y means +1 and r means -1. Let's take the problem 5 - 3 that means 5 positives take away 3 positives so on the table you would have y y y y y then take away y y y and you would have left y y = +2 Now the problem -5 - -3 that means 5 negatives take away 3 negatives so on the table you would have r r r r r then take away r r r and you would have left r r =-2 On the problem 5 - -3 that means 5 positives take away 3 negatives so on the table you would put y y y y y and then you would try to take away r r r but there are no r's on the table so you would add yr which is zero. Aha--you now have l r to take away but you need 3 r's so you must add yr yr and now you have 3 r's. Now on the table you have: y y y y y yr yr yr and that is equivalent to 5. You can then take away r r r and what is left is y y y y y y y y which is +8. On the problem -5 - 3 that means 5 negatives take away 3 positives. You would do the same as the last example. r r r r r and try to take away y y y but there are no y's to take away. You would add yr yr yr which is just zero so on the table you have: r r r r r yr yr yr, then take away y y y (+3). You would have left r r r r r r r r which is -8. ~~~~~ An approach that I have adopted over the past several years is to give the students the following table, ask them to fill it in by using patterns that they see, then we generalize the multiplication rules from the quadrants. This is usually done after modeling the 3(-2) = -6 example with bingo chips : 3(-2) = (-2) + (-2 + (-2) = -6. For the most part, they seem to recognize that it is a multiplication table and will readily fill in the pos X pos Quadrant. Usually need some initial guidance in extending the vertical and horizontal factors into the negatives, but doesn't take long for the kids to fill in the table by patterns. Then we look at the quadrants and see that the pos X neg = neg relation is supported by the appropriate quadrant. The rest comes pretty readily and there doesn't seem to be too much argument about the negXneg=positve result. I think it's a pretty good activity. And is a good an argument I've seen as to convincing someone that the multiplication rules for integers are consistent with something they are familiar with -- the multiplication table. | 5 | 4 | 3 | 2 | 1 | 0 | | | | | | | ------------------------------------------------------- 5 | |20 | | | | | | | | | | | ------------------------------------------------------- 4 | | | | | 4 | | | | | | | | ------------------------------------------------------- 3 | |12 | | | | | | | | | | | ------------------------------------------------------- 2 | | | 6 | | | | | | | | | | ------------------------------------------------------- 1 | | 4 | | | | 0 | | | | | | | ------------------------------------------------------- 0 | | 0 | | | | | | | | | | | ------------------------------------------------------- | | | | | | | | | | | | | ------------------------------------------------------- | | | | | | | | | | | | | ------------------------------------------------------- ~~~~ I am coming to the conclusion that graphing should be an integral part of the Algebra I curriculum. Attempting to teach direct and indirect variation without graphs is too abstract and meaningless. Polynomials become just one more math rule. Students who have real math apathy become engaged when doing population graphs. I had a B - - student come in the other day with a projection of the future population of the Blue Whale on a poster board - she was proud of the project and explained that she had stayed up all night working on it. I am absolutely positive that she never spent all night working on one of the exercise sets. I am trying an experiment in applied math I. When it came time to introduce negative numbers, I first introduced the 4 - quadrant graph. Adding and subtracting signed numbers became a logical result of graphing lines. Furthermore, they can check their answers - if the points don't form a line, they've made a mistake which they diligently try to find. I explained that right now we are studying equations of lines so that everything we do will form a line. Gradually they are becoming convinced of the fact although at first they were convinced that the points were plotted right but the equation didn't generate a line. When they add a negative, they go down the Y axis which seems more logical then going to the left on the number line. Next week I'll give them the data and they will give me the equation. I'm holding my breath, but right now they are telling me that "this stuff is easy!" It's the first time I've been told that adding, subtracting, multiplying, and dividing signed numbers was easy stuff. No - we are not using graphing calculators since my students also need lots of basic arithmetic practice. ~~~~~ Maybe I missed it, but I haven't seen anyone mention the use of Algebra Tiles (counting buttons, or whatever) to demonstrate the multiplication of negative numbers. It's been extremely effective with my Paced Algebra and Informal Geometry students. Yellow = positive; red = negative Students will have already accepted that one red (-) and one yellow (+) equal zero and we can have many of these pairs still giving a zero value. Example: Start with ten zeros (10 -'s and 10 +'s). To model -4 * -2, we take away 4 sets of 2 -'s. You will have left 10 +'s and 2 -'s. Two of the + and -'s give zero value so we throw them out, leaving only 8 +'s. Giving a visualization of -4 * -2 = +8. If you use transparent tiles on the overhead projector and each student also works with a set, it's much easier than trying to type out on this screen. This has been very effective for me and students even go get the tiles when they forget the rules. ~~~~~ In reponse to the above: In this modelling process, *two* different interpretations have been assigned to the use of "-". One assigns a color and the other assigns the action of taking away. While this may work well in the short term, it may cause problems later. I don't think it is the kind of *habit* that one wants to instill in the minds of students. ~~~~ I attended a workshop with Barry Scully who, along with Janet Scully and Jack Lesage, wrote the book on Alge-Tiles, at least the Canadian version (_Active Learning Series_, Barrie, ON: Exclusive Educational Products, 1991). Since that time I have taken his advice to make a clear and consistent distinction between the *quality* called "negative" and the *operation* called "subtraction", "take away", "minus". I do not permit the students to use the words interchangeably, and address the concepts of "quality" and "operation" whenever I can in all subjects (e.g., a basketball has the quality of being spherical, a poem may have a dreamy quality, etc., etc. and we perform the operations of dribbling and writing on them) to reinforce this distinction. It takes several weeks and constant reminding, but it helps, but the habit that is usually already developed by grade 7-8 is questioned, if not broken. ~~~~~ I believe it is important to understand that students' confusion arises because the "-" symbol actually has *three* interpretations unlike the "+" symbol (operation of addition and sign of a number). I think the language is important and here's how I have chosen to do it: "Minus" means the binary operation of subtraction—always "Negative" is used only with numerals (the sign of the number) -5"Opposite of" is the unary operation of negation -x is read "opposite of x." You can substitute "additive inverse" for "opposite of" but I find that a little cumbersome for most students.