Modeling Multiplying Two Negatives with Number LinesDate: 11/13/2006 at 07:36:28 From: Joan Subject: -2*-3= 6??? I'm trying to use a number line to figure out why -2 times -3 makes 6, and I can't do it. If I start at -2 and move to the right three times I wind up at 4, and if I move to the left I wind up at -8. What is the logic behind the rule of a negative times a negative makes a positive, and how do I demonstrate it on a number line? Date: 11/13/2006 at 08:28:45 From: Doctor Rick Subject: Re: -2*-3= 6??? Hi, Joan. You can find a variety of ways to think about multiplication of negatives, and also ways to *prove*, more or less formally, that a negative times a negative is positive, in the Dr. Math FAQ: Negative X Negative = Positive http://mathforum.org/dr.math/faq/faq.negxneg.html It's hard to demonstrate signed multiplication on a number line, but there is a way to do it using *two* number lines. See what you think of this: Draw two number lines that cross at the origin on both lines--the angle between the lines doesn't matter. To multiply a number x by another number y, first draw a line through 1 on the first line and x on the second line: / / / / / x / * / / / ---0--*----------- / 1 / / I can't draw that line, you'll have to do it on paper. Then draw another line, parallel to this one and passing through the number y on the first number line. / z / * / / x / * / / / ---0--*--*-------- / 1 y / / You can easily prove by similar triangles that x/1 = z/y so that z = xy That is, the position of z on the second number line is the product of x and y. Now, use this same graphical method to multiply -2 by -3: / *-2*-3=6 / / / / / / / / / / / / --+------------0----+------------- -3 / 1 / / *-2 / / Draw a line connecting 1 on the first (horizontal) number line with -2 on the second number line. Draw a line parallel to this line, and passing through -3 on the first number line. You will see that it intersects the second number line at 6, which is therefore the product of -2 and -3. - Doctor Rick, The Math Forum http://mathforum.org/dr.math/ |
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