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 Positive
Let us begin by considering a positive value of a on the horizontal line and a positive value of b on the slant line.
Now construct a line segment from 1 on the horizontal line to b on the slant line. Construct a parallel line segment from a on 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 product ab. See the image below.
This is also the classic construction of the fourth proportional -- 1 : b = a : x leads to x = ab
Because the triangles are similar, the location of the intersection represents the product of a and b. 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 a located in the negative direction and b located in the positive direction.
The construction is done in the same way, with a on the horizontal, b on the slant line, and construction of the segments from the 1 location to the b location and then a parallel segment from the a location to the intersection point. Again the intersection point will be the product ab but now this product is located in the negative direction. So (-a)(+b) = - ab.
Again this is a direct consequence of the similar triangles.
Positive Times a Negative
Next let’s consider the construction we have just done above but begin with a located in the positive direction and b located in the negative direction.
The construction is done in the same way, with a on the horizontal, b on the slant line, and construction of the segments from the 1 location to the b location and then a parallel segment from the a location to the intersection point. Again the intersection point will be the product ab and again this product is located in the negative direction. So (+a)(-b) = - ab.
Now we are ready to move on to the case of the product of two negative numbers.
Negative Times a Negative
Again let’s consider the construction we have been doing above but begin with a located in the negative direction AND b located in the negative direction.
The construction is done in the same way, with a on the horizontal, b on the slant line, and construction of the segments from the 1 location to the b location and then a parallel segment from the a location to the intersection point. Again the intersection point will be the product ab but this product is located in the positive direction. So (-a)(-b) = ab.