Diophantus of Alexandria vs Geometric Algebra

Nothing earth shattering here. Just a funny coincidence.

According to Wikipedia,


(a^2+b^2)(p^2+q^2) = (ap+bq)^2+(aq-bp)^2

Wikipedia attributes this to Diophantus of Alexandria. That is interesting. Alexandria is a remarkable place. Africa. Egypt. Greece. Anyways. It also says this is called the Brahmagupta-Fibonacci identity. Even more fascinating. India, Italy, North Africa.. anyways.

Why does that look familiar?

Well, if you say that (a,b) is a x-y coordinate on a Cartesian grid, and also say that (p,q) is another x-y coordinate on a Cartesian grid… then.

(a,b) is (x_1,y_1)

(p,q) is (x_2,y_2)

Then (aq-bp) is (x_1y_2-x_2y_1)

But that is the “Wedge Product” from the old alternative branch of geometry that goes back through Clifford, Grassman, Hamilton, etc. Basically it’s the area of a parallelogram formed by the two vectors with tails at origin, and heads at x_1,y_1 and x_2,y_2. Let’s call them vector v_1 and vector v_2.


wedge product = parallelogram area

That is pretty interesting… what is wedge product doing in this ancient formula?

And what about (ap+bq)?

(a,b) is (x_1,y_1)

(p,q) is (x_2,y_2)

Then (ap+bq) is (x_1x_2+y_1y_2)

or… the “Dot Product”, which is from Vector algebra and has a fascinating history all it’s own. But it too has an echo in this ancient Brahmagupta-Fibonacci formula.

There is a way to interpret the Dot Product as the area of a rectangle formed by dropping a perpendicular from one vector onto another. This is called called “projection” of one vector onto another. The funny thing is that you can do v1 onto v2, and it’s the same as v2 onto v1. At any rate, a picture of one version of this is below.


v1 dot v2 = area of a special rectangle


Now the modern version of the old Clifford school of algebra is called today Geometric Algebra, and there is something called a Geometric Product that is valid for “Multivectors” in certain dimensions.As described in Jaap Suter’s “Geometric Algebra Primer”, it is basically this:

 v_1 \cdot v_2 + v_1 \wedge v_2

Wow! The right-hand side of Fibonacci-Brahmagupta is


This could be re-written, using our coordinates for a p b q, like so:

(v_1 \cdot v_2)^2 + (v_1 \wedge v_2)^2

Ain’t that weird? The terms of the Geometric Product, when Squared, form the right-hand side of Brahmagupta-Fibonacci.

What about the lefthand side of Brahmagupta-Fibonacci?


Well, that can be re-written, again using our a b p q as coordinates, as above…


Wait, what? That’s just the Square of the hypoteneuse of each triangles… multiplied together. The square of the hypoteneuse could also be called the Quadrances of the vector (to borrow a term from Wildberger’s Rational Trig, perhaps incorrectly…).


Yet another way to think about the square of the Hypoteneuse is to say that it is the Vector dot-product with itself. So v_1 \cdot v_1 = Quadrance of v_1, and v_2 \cdot v_2 = Quadrance of v_2. So. Rewriting everything, we can get this:

(v_1 \cdot v_1)(v_2 \cdot v_2) = (v_1 \cdot v_2)^2 + (v_1 \wedge v_2)^2

Or, in other words….

d3multiplied by d2


equals the square of

d4added to the square of  d1

Is that interesting? I thought so.

In numbers, if a,b is 4,3 and p,q is 4,3 then… rewriting a few times…

(5^2)(5^2) = ((4,3) \cdot (3,4))^2 + ((4,3) \wedge (3,4))^2

25*25 = (4*3+3*4)^2 + (4*4-3*3)^2

25^2 = 24^2 + 7^2

625 = 576 + 49

Seems legit?

But what else is this connected to?

Well… it is saying something about Volumes in Four Dimensions… if you build your volumes off of two vectors.

Why? First, imagine how two one-dimensional objects, multiplied together, give a 2 dimensional object. An area.

Now imagine when a two-dimensional object is multiplied by a one-dimensional object. You can get a cube… or a parallelapiped, or whatnot.

Now imagine a 2 dimensional object, multiplied by another 2 dimensional object. Split the second 2-dimensional object into pieces, two one dimensional objects that are its factors. For example if it’s a square, consider two of it’s adjacent sides. Now, multiply the first two-dimensional object by one of those one-dimensional objects. That results in a three-dimensional object.

Now multiply the three dimensional object by another one-dimensional object. It can be imagined that this involves extending the 3dimensional object into the fourth dimension. Or, if you think of the fourth dimension as time, it could be considered as extending the 3 dimensional object into a segment of time… in other words imagine that it blinks into existence, and then after a few minutes, blinks out of existence. The duration of that time is the ‘one dimensional object’ of time by which the 3d object was extended.

That’s a bit abstract though. Is there nothing simpler it’s connected to? Well… if so, it will have to be another story for another time. I’m too sleepy to continue.


About donbright

don bright http://github.com/donbright
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