Question 7

7. Use Geometric Algebra to show that a2-b2=(a+b)(a-b). You can use the distributive rule in your demonstration. Do NOT use negative numbers.

First, let's prove it algebraically:

a² – b² = (a-b)(a+b)

a² – b² = a² – ab + ab – b²

a² – b² = a² – b²

Now, that I've seen that, let's try to prove it using geometric algebra. Let's first start with a square, with sides a.

Then, on one side, let's add b, and on the other let's go backwards b.

Once, this has been done, let's look at the part where I actually want to find the area (the orange part):

Why do I only look at this part? Because the bottom purple part has been taken out, we want to look at the side (a-b) only and not a, as a whole. Therefore, the bottom purple rectangle is no longer important. If this is the case, why is the rectangle on the right important? Because we want to look at a + b , and the rectangle on the right along with a is representative of this.

Now, let's remove the bottom part, and concentrate on what we really want to look at:

The top and bottom side is now: a + b, and the left and right side is now representative of a - b. But how is their area a2 - b2?

To make this easier, Gabe showed me how to move the rectangle with top side b around, and put it on the bottom.
The end result of the move should look like this:

Now that I've moved it, we can see that there is now a whole side of a on the left and on the top. But, wait, how did this work out? Well, when I moved it, I took the extra part of b from a + b away, so it's like a + b - b, and now we have only a on the top. Then, by moving this part of b to the part of a that I took it out of, I replaced the missing b on one side - in a way, I did a - b + b. 

Now I have a whole a on both the top side and the left side, I know that if the bottom and right side were also filled, it would be like the original square that I had way at the beginning, which has an area of::

a x a = a²

But, this square is not "whole" it has a hole that's in the shape of square in the middle with sides b. How do I know that the sides are b? Well, from the move and from knowledge of before. Before the move, the now-bottom side has a length of a - b. So after the move, the bottom side should still have a length of a - b. If  the top is length a, and the bottom is length a - b, then the difference between them is a length of b, which I have labeled. I know the left side of the small square is also length b because before I moved it, the length was "b", so after the move, the length should remain the same. Therefore, I know that the hole on the corner of square a is a small square made from two sides of b.

So, to find the area of the orange part only, I have to take the area of the large square a, and take out the area of a small square b. I already know the area of the large square a to be a², so I only need to find the area of the small square b.

The area of the small square b is:

b x b = b²

Therefore, the area of the orange rectangles which is representative of (a-b)(a+b) combined is:

a² – b² 

So, if the area of the orange rectangles is representative of  (a-b)(a+b) then:

(a-b)(a+b) = a² – b²