Below post is a back-up of my posts in my Quora blog. I take a copy of them to here in case if something goes wrong in there.
In my
previous post about the Pythagorean Theorem, I have tried to give a different look at the subject. In summary; it was about applying the theorem in 2 dimensions on a coordinate system instead of a triangle and to show the geometrical evidence of it using one higher dimensional axis as in below sketch.
That gave us a new concept which is similar to the cross product of the vectors. In this post, I will pursue the same idea in higher dimensions than 2 and look for new perspectives.
To deal with a 3 dimensional object, I have set up my playground in a 3 dimensional coordinate system as in below sketch.
There is nothing unusual about it. I can set the lengths of x, y, z and carry on to apply Pythagorean Theorem to find the result and try to imagine its correspondence square in 4th dimension, which I already did in my previous post.
But I have noticed that it doesn’t take me anywhere. Maybe it is helpful to find insights about imagining the 4th dimension but while playing within this playground, I have noticed something else. First, I have connected x, y, z points as in below sketch and created some triangles.
You can notice that they resemble a
tethrahedron but a face of it is missing. You can see that face marked as d which is made of x, y, z points as below.
There is something very similar with our first Pythagorean Theorem application with the 1d lines. But instead of lines, this time 3 pieces of triangles are resulting in a 4th one (d). I have wondered if Pythagorean Theorem was also applicable to the areas of these triangles. I have created a simple excel sheet where I can enter different values for each edge length and can easily see the results for different values quickly. That exercise has showed me that that it works as in below equation.
a² + b² + c² = d²
That’s strange. Nobody has taught us that the Pythagorean Theorem is also applicable to the the areas of the triangles in such an arrangement. Probably that was because they didn’t find it useful or thought that the 1d application of it is all we needed. In any case, there is definitely something to pursue.
After noticing that, I have tried to apply the geometrical proof of 1d application into this one and I have quickly failed. That was most probably because taking the square of a 2 dimensional object requires to imagine a 4 dimensional one which we are not quite capable of.
After that, I have wondered, if the same logic can be taken into further steps. For example, can we apply it to volumes? But there is one problem with that. Since our scenario about 2d surfaces above had to be set in a 3d coordinate system, to apply it to the volumes, we need to set a scenario in 4d coordinate system, similarly. The problem is, we are not capable of visualizing a 4d coordinate system effectively. To overcome this problem, we need to push the limits of our imagination a bit. In below sketch, you see an illustration of a 4d coordinate system with 4 axes which each are perpendicular to the others.
In order to test that if we can apply Pythagorean Theorem to the volumes, first we need to define what we are looking for.
- 1d application of Pythagorean Theorem, was giving us the length of the hypotenuse of a right triangle by taking squares of the lengths of 2 short edges.
- In surface application of it, we have shown that if we take the sum of squares of the areas of 3 triangles in a 3 dimensional environment, we can get the square for the area of their interconnecting triangle.
- Now, using this principle, we can claim that; in a 4 dimensional coordinate system, taking the sum of the squares of the 4 tetrahedrons should give us the square of the volume of the 5th interconnecting volume of the 5-cell.
Since it is difficult to express this idea by words, I will try to explain it by sketches. But before that, since talking about higher dimensional objects’ volumes and surfaces is a very tricky subject, we need to declare our common language.
The difficulty is; while it is very clear when you talk about the length and area of a 2 dimensional object like triangle, it quickly becomes confusing after 3 dimensional objects. Because, for example; a 4 dimensional object has a surface area, surface volume and as an additional concept; a hyper volume. This is where things get mixed. In order to avoid that, people have invented clever ways to name those concepts. One of these ways is, instead of giving individual names to each deeper! volume of higher dimensional objects, we can call them simply as face and give them face numbers as shown in
Wikipedia page of the Simple.So according to this table, whenever I say;
Area of a 1 face, it will mean the length of a line,
Area of a 2 face will mean the surface area of a triangle,
Area of a 3 face will mean the volume of a tetrahedron,
Area of a 4 face will mean the hyper volume of 5-cell and so on.
And whenever I use them in a calculatio,n I will use them in that fashion: |xyzw…| number of letters will indicate the number of faces.
Deal? Ok let’s back to our subject. Here is our 4 dimensional object, a 5-cell:
Looks confusing, right? That’s expected since it is a 4d object which our minds are not familiar with. As I said, you have to push your imagination to its limits in order to visualize it. If you cannot, no problem. We can use our previous tetrahedron example as an analogue to our 5-cell in order to understand what are we dealing with.
Surface of 3d tetrahedron is covered with 2d triangles, Similarly, a 4 dimensional 5-cell is also covered with 3d tetrahedrons. So, the a, b and c triangles of below tetrahedron are analogue to the volumes of our 5-cell:
So, the a, b and c triangles in below sketch,
Are analogue to the below volumes of our 5-cell:
And the interconnecting d triangle ( |xyz| ) in below sketch,
Is analogue to interconnecting 5th |xyzw| volume below:
With this volume, the number of volumes in a 5-cell becomes 5 in total, which gives its name.
So, what we need to do is to verify if below equation is correct or not:
|xyzw|² = |oxyz|² + |oxyw|² + |oxzw|² + |oyzw|²
So, if we calculate those volumes for below values using the determinant of (A . ( B x C)) / 6 formula:
x: 2
y: 3
z: 4
w: 5
|oxyz| = 4
|oxyw| = 5
|oxzw| = 6,6666…
|oyzw| = 10
|xyzw| = 13,6178…
Which proves our formula by;
13,6178…² = 4² + 5² + 6,6666…² + 10²
I guess that, it gives us enough evidence for our case. I don’t know if this has been previously declared anywhere before. If not, I would like to call it Pythagorean Simplex Conjecture and declare it as;
In a n dimensional simplex; square of n-1 faced hypotenuse area is equal to sum of square of all other n-1 dimensional faces’ area.
I don’t have enough mathematical skills to write it as an equation. If you are interested and have an idea about how to express it, you are welcome to the comments section.
P.S. After a careful read at
Pythagorean theorem page in Wikipedia I have come to know that this has been previously thought as you can see in below picture. That means; a) there is no need to name it as a new conjecture, b) I have discovered an already thought concept once again. What can I say, it was a good exercise.
And from the page of
De Gua's theorem:
So it is possible to write it as a general equation like that.
P.S.2 As I understand, De Gua has declared this only for 3 dimensional tetrahedron and didn’t generalize it for higher dimensions. I don’t know if anybody else did it before but I could find
this paper in internet which is generalizing this idea to all dimensions and it is dated 2002. It seems like I wasn’t 300 years late but just about 14. Not that bad.
