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A Negative Times a NegativeDate: 06/02/98 at 21:24:45 From: El-ad Blech Subject: Integers (multiplying) Hello Dr. Math, I would like to know more about integers. My question is, why do you have to take out the negative sign when you multiply two negative numbers (i.e. -2 x -2 = 4)? Thank You, El-ad Blech
Date: 06/13/98 at 23:31:49
From: Doctor Peterson
Subject: Re: Integers (multiplying)
Hi, El-ad,
This is something a lot of people have trouble understanding. I'll
give you an explanation I find helpful.
One thing that helps is to realize that negative numbers are just an
extension of the number line, and when we try to decide how to do
operations on them, our main concern is to keep everything working the
way it does when you work with positive numbers.
Suppose I make a graph of a line; maybe it represents how far I travel
if I keep going forward at a constant speed, so the horizontal axis is
time and the vertical axis is distance (or position along a road):
D
| /
6 + X
| /
3 + X
|/
0 +-+-+---- T
0 1 2
This graph shows that I went 3 miles in the first hour, and another 3
in the second hour, so in two hours, I went forward 2 * 3 = 6 miles. I
can represent the graph by the equation:
D = 3 * T
I can continue this graph backward in time, to figure out where I
would have been at earlier times. If I do that, it should still follow
the same rules as for positive times. As I go from right to left on
the graph above, D decreases by 3 each time I decrease T by 1. It
should still do that if I continue it for negative values of T, so it
will look like this:
D
| /
6 + X
| /
3 + X
|/
------+-+-+-+-+---- T
-2-1/| 1 2
X + -3
/ |
X + -6
/ |
So for T = -2, D = 3 * (-2) = -6. That part isn't hard to believe.
All we're doing is continuing the straight line to the left. In terms
of motion, at a time before the present I was at a place behind my
starting point.
Now let's think about negative speeds. Suppose I'm driving backwards
(maybe the wrong way down a one-way street, or with my car in
reverse). Then for each hour I drive, I go -3 miles, that is, 3 miles
backward:
D
|
6 +
|
3 +
|
+-+-+---- T
|\1 2
-3 + X
| \
-6 + X
| \
This represents the equation:
D = (-3) * T
How do I continue this for negative times? Since D increases by 3 each
time T decreases by 1, it should look like this, continuing the
straight line backwards:
D
\ |
X 6+
\ |
X3+
\|
------+-+-+-+-+---- T
-2-1 |\1 2
-3+ X
| \
-6+ X
| \
But that means that at time T = -2,
D = (-3) * (-2) = 6
What all this really means is just that a negative sign on either
factor reverses the result, and reversing it twice makes it positive
again.
I hope that helps make it more visible.
-Doctor Peterson, The Math Forum
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