Drexel dragonThe Math ForumDonate to the Math Forum

Ask Dr. Math - Questions and Answers from our Archives
_____________________________________________
Associated Topics || Dr. Math Home || Search Dr. Math
_____________________________________________

Product of Negative Numbers


Date: 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/   
    
Associated Topics:
College Analysis
High School Analysis
High School Negative Numbers
Middle School Negative Numbers

Search the Dr. Math Library:


Find items containing (put spaces between keywords):
 
Click only once for faster results:

[ Choose "whole words" when searching for a word like age.]

all keywords, in any order at least one, that exact phrase
parts of words whole words

Submit your own question to Dr. Math

[Privacy Policy] [Terms of Use]

_____________________________________
Math Forum Home || Math Library || Quick Reference || Math Forum Search
_____________________________________

Ask Dr. MathTM
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/