** **by

**Jim Wilson**

Consider two lines that cross and indicate 0 as the common point. From the definition of a negative number as the opposite or additive inverse of a positive number, let each line have locations in the positive and negative directions. From the multiplicative identity, 1, indicate a unit in each direction on each line..

Any point on a line has an orientation, positive or negative, determined by a location.

Positive times a PositiveLet us begin by considering a

positivevalue ofon the horizontal line and aapositivevalue ofon the slant line.bNow construct a line segment from

on the horizontal line to1on the slant line. Construct a parallel line segment frombon the horizontal lie to its intersection with the slant line. From the geometry of similar triangles, we know that the location of the intersection is the productaSee the image below.ab.This is also the classic construction of the fourth proportional --

1 : b = a : xleads tox = abBecause the triangles are similar, the location of the intersection represents the product of

aandb. In this case we have shown that the product of two positive numbers is positive.

Negative Times a Positive.Next let’s consider the construction we have just done above but begin with

alocated in thenegativedirection andblocated in thepositivedirection.The construction is done in the same way, with

aon the horizontal,bon the slant line, and construction of the segments from the1location to theblocation and then a parallel segment from thealocation to the intersection point. Again the intersection point will be the productabbut now this product is located in thenegativedirection. So(-a)(+b) = - ab.Again this is a direct consequence of the similar triangles.

Positive Times a NegativeNext let’s consider the construction we have just done above but begin with

alocated in thepositivedirection andblocated in thenegativedirection.The construction is done in the same way, with

aon the horizontal,bon the slant line, and construction of the segments from the1location to theblocation and then a parallel segment from thealocation to the intersection point. Again the intersection point will be the productaband again this product is located in thenegativedirection. So(+a)(-b) = - ab.Now we are ready to move on to the case of the product of two negative numbers.

Negative Times a NegativeAgain let’s consider the construction we have been doing above but begin with

alocated in thenegativedirection ANDblocated in thenegativedirection.The construction is done in the same way, with

aon the horizontal,bon the slant line, and construction of the segments from the1location to theblocation and then a parallel segment from thealocation to the intersection point. Again the intersection point will be the productabbut this product is located in thepositivedirection. So(-a)(-b) = ab.

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