Product of Negative NumbersDate: 02/04/97 at 18:05:04 From: Tapler, Tom Subject: multiplication of negatives Hey! Dr. Math is a great idea. I hope you can help me to formulate a SIMPLE explanation of why a negative times a negative is a positive. I would like it to be along the same lines as this explanation: (4)(5) = 4 + 4 + 4 + 4 + 4 = 20 <---- 5 additions (times) (-3)(5) = (-3) + (-3) + (-3) + (-3) + (-3) = -15 This covers pos x pos and neg x pos. When I explain neg x neg, I FEEL that I fudge when I factor out a (-1) and use the "rule" that (-1) times a number reverses that number's sign. Can you show me how to explain this to 8th graders using addition? Thanks, Tom Tapler Date: 02/07/97 at 14:35:35 From: Doctor Pete Subject: Re: multiplication of negatives Hi, This is one of the most commonly-asked questions among students who have been introduced to the concept of negative numbers. There is a rigorous explanation for the multiplicative properties of complex numbers in terms of what mathematicians call the "field axioms," but this is somewhat abstract for students who have not yet been familiarized with the concept of variables (e.g., "x"). A more "concrete" example is as follows: Suppose you film someone walking forwards at 3 ft/s. Then if you play back the film for 4 seconds, he will walk a distance of (3)(4) = 12 ft. Now, suppose he walks backwards at 3 ft/s. then playing the film for 4 seconds will show that he walked a distance of (-3)(4) = -12 ft. Now, if you play the film *backwards* for 4 seconds, the person will walk *forward* a distance of (-3)(-4) = 12 ft. It's probably not a good idea to try to develop an extension of your example to explain why the product of two negative numbers is positive because this method doesn't really work for various types of positive numbers. How do you explain something like: 2.145 x 9.2537 or Sqrt[6] x Sqrt[2] = 2 Sqrt[3] You can say that (3)(5) = 3 + 3 + 3 + 3 + 3 because of the distributive law and the multiplicative identity axiom. You are actually writing: (3)(5) = (3)(1 + 1 + 1 + 1 + 1) and distributing the 3, noting 3 x 1 = 3. But this only works because 5 is a positive integer. So in a sense, the example is motivated by the axioms, not the other way around. To explain in further detail, we describe these "field axioms" for the real numbers: Let F be a set with two operations, called addition and multiplication, which satisfy the following properties: (A1) If x is in F, and y is in F, then their sum x + y is in F. (A2) Addition is commutative: x + y = y + x for all x,y in F. (A3) Addition is associative: (x+y)+z = x+(y+z) for all x,y,z in F. (A4) F contains an element 0 such that 0 + x = x for every x in F. (A5) To every x in F corresponds an element -x in F such that x + (-x) = 0. (M1) If x is in F, and y is in F, then their product xy is in F. (M2) Multiplication is commutative: xy = yx for all x,y in F. (M3) Multiplication is associative: (xy)z = x(yz) for all x,y,z in F. (M4) F contains an element 1 not equal to 0 such that 1x = x for every x in F. (M5) If x is in F, and x not equal to 0, then there exists an element 1/x in F such that x(1/x) = 1. (D) x(y+z) = xy + xz for all x,y,z in F. If all of these axioms hold true, then we call F a *field*. R, the real numbers, is the most familiar field, but by no means is it the only one. There are finite fields (i.e., containing a finite number of elements), and there are larger fields, namely C, the field of complex numbers. From (A1-A5) one can prove: (Aa) If x + y = x + z then y = z (Ab) If x + y = x then y = 0 (Ac) If x + y = 0 then y = -x (Ad) -(-x) = x From (M1-M5) one can also prove, for x nonzero: (Ma) If xy = xz, then y = z (Mb) If xy = x, then y = 1 (Mc) If xy = 1 then y = 1/x (Md) 1/(1/x) = x Finally, from all of these, we can prove: (Fa) 0x = 0 (Fb) If x and y are nonzero, then xy is nonzero (Fc) (-x)y = -(xy) = x(-y) (Fd) (-x)(-y) = xy The proofs of these last 12 statements are not too difficult, but require some care. Note that (Fc) and (Fd) are what we are interested in. Also note that there is no mention of -1 in the above axioms, only the existence of an additive (0) and multiplicative (1) identity, and the existence of an additive (-x) and multiplicative (1/x) inverse (the latter only when x is nonzero). For the proofs of these, see Rudin, _Principles of Mathematical Analysis_. I know this is probably not quite what you were looking for, but I felt it was important to point out that it is not always the best approach to try to "stretch" a simple model to encompass more advanced concepts, especially when the model is actually a consequence of more fundamental, albeit abstract, ideas. In the long run, it would be more helpful to one's students to broaden their sense of mathematics by introducing them to many examples that point to the same result, thereby showing how everything intricately "works" together. -Doctor Pete, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2015 The Math Forum
http://mathforum.org/dr.math/