## Geometric multiplication: an explanation

Now for an explanation/proof of that weird method of multiplication that I talked about in a previous post. You’ll recall that it’s really quite simple: to multiply a and b, draw a line from $(-a,a^2)$ to $(b,b^2)$ and see where it crosses the y-axis. But why on earth should that cross the y-axis at height $ab$? Well, I’m afraid I still don’t know how to answer the why in a way that makes intuitive sense, but we can certainly show that it is true.

The first step is to figure out the equation of the line passing through the points $(-a,a^2)$ and $(b,b^2)$. To do that, we first need the slope. Slope is “rise over run”, that is, the change in y divided by the change in x, which in this case is

$\displaystyle m = \frac{b^2 - a^2}{b + a}.$

We can simplify this, however, by noting that $b^2 - a^2$ factors as $(b+a)(b-a)$, so the $(b+a)$‘s cancel… as long as $b+a \neq 0$. In that case, we’d have $b = -a$ and we’d be trying to draw a line from a point to itself, which doesn’t make sense. This just means that we can’t use this method to compute a product like $8 \cdot (-8)$, but that’s OK since we can always compute $8 \cdot 8$ and negate the result. So, we now have

$\displaystyle m = \frac{b^2 - a^2}{b+a} = b - a.$

Now, how can we find the equation of the line? Since the slope of a line is the same everywhere, the slope between $(b,b^2)$ and any point $(x,y)$ on the line should always be equal to the slope m that we computed. We can write this requirement as

$\displaystyle \frac{y - b^2}{x - b} = m = b - a.$

This is already a perfectly good equation for the line, but let’s simplify it a bit:

$\begin{array}{rcl} \frac{y - b^2}{x - b} & = & b - a \\ y - b^2 & = & (b - a)(x - b) \\ y - b^2 & = & bx - b^2 - ax + ab \\ y & = & bx - ax + ab \end{array}$

Now that we have an equation for the line, we want to know where it crosses the y-axis. But this is simple: it crosses the y-axis when x is zero; plugging 0 into the above equation for x yields… $y = ab$!