You see, I am trying to recollect what I had written in May of this year, and put off writing this book until now in August of 2023.

In the title I had written I found another proof that dimension of Space can only be 3rd dimension at tops, and cannot be 4th or higher.

So I am trying to recollect why I said that.

I know for sure I was very unhappy with making 3 arbitrary points in Space, noncollinear, form a unique plane as being a Axiom-Postulate. Very unhappy with that idea, and thus I found a derived theorem proof.

I do recall now the ellipse requiring just 3 arbitrary noncollinear points in Plane to form a unique ellipse. And that was resolved by a HourGlass type figure of 2 right triangles, smallest legs joined, one reversed and moved to one vertex point.

But I am trying to jog my memory as to how all of this proves Space can be 3rd dimension at tops.

It must have something to do with the fact that 3 arbitrary points, noncollinear in Space forms a Unique Plane, and so if Space had 4th dimension--- there then exists no way this plane can enter or intercept a 4th dimension.

So I am trying to jog my memory.

AP

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On Tuesday, August 15, 2023 at 2:22:41 PM UTC-5, Archimedes Plutonium wrote:

You see, I am trying to recollect what I had written in May of this year, and put off writing this book until now in August of 2023.

In the title I had written I found another proof that dimension of Space can only be 3rd dimension at tops, and cannot be 4th or higher.

So I am trying to recollect why I said that.

I know for sure I was very unhappy with making 3 arbitrary points in Space, noncollinear, form a unique plane as being a Axiom-Postulate. Very unhappy with that idea, and thus I found a derived theorem proof.

I do recall now the ellipse requiring just 3 arbitrary noncollinear points in Plane to form a unique ellipse. And that was resolved by a HourGlass type figure of 2 right triangles, smallest legs joined, one reversed and moved to one vertex point.

But I am trying to jog my memory as to how all of this proves Space can be 3rd dimension at tops.

It must have something to do with the fact that 3 arbitrary points, noncollinear in Space forms a Unique Plane, and so if Space had 4th dimension--- there then exists no way this plane can enter or intercept a 4th dimension.

So I am trying to jog my memory.

Apparently I had no proof of no 4th dimension and beyond, for I did not have a proper proof of theorem derived 3 arbitrary points, noncollinear determines a unique plane.

But now that I do have that proof, I can go on to prove Space has to be 3rd dimension and no higher.

And the proof of no 4th dimension space and higher is the fact and idea that how does a plane traverse or lie in this higher dimension. We can think of a 3rd dimensional space as being a packaging container full of Planes. A volume of planes makes 3rd
dimension. If we had a 4th dimension, none of our planes picks up any added dimension.

So now, in 10 Grid we have 100 points on x-axis and 100 on y-axis and 100 on z-axis, not counting zero (0,0,0). In total that would be 100^3 = 1,000,000 coordinate points or 10^6. Now how many possible arrangements are there of 3 distinct points? Is it
the horribly large number of 1,000,000 factorial? No, for that is permutation math. Here I have Combination math and we have 1,000,000*999,999*999,998/ 3*2*1 = some huge number, not as bad as 1,000,000 factorial.

Now in my earlier proof, each triangle yields a unique plane, but many different triangles are also in that same plane. And that is a stumbling block I fell on before. Each triangle does produce a unique plane for that triangle, but that plane once
formed carries many different triangles.

Now, how do I separate out all the triangles that lie within one plane in the 10 Grid? I do I extract the number of unique planes in 10 Grid?

So I have this number as the total number of unique triangles in 10 Grid= 1,000,000*999,999*999,998/ 3*2*1 reduced to 1,000,000*333,333*499,999 which is the number of unique triangles in the 10 Decimal Grid.

Now in the 10 Grid, how many triangles would each plane on average carry? If I examine the xy plane, a flat plane as representative of an average plane I am looking for an average number of triangles that forms that plane through expansion of its sides
via the parallel axiom. So here is a duplicate Combination problem. 10,000*9,999*9,998 / 3*2*1 = 10,000*3,333*4,999. And now I divide 1,000,000*333,333*499,999 by that of 10,000*3,333*4,999 to arrive at an approximate average number of distinct planes
that reside in the Space of a 10 Grid. This is approximately 100*10*10 = 10,000. Is this a surprise?? That the total number of points in 10 Grid 3D Space is 1,000,000 and the total number of planes in 10 Grid is approximately 10,000.

Does that look right?

So then, back to the question of a 4th Dimension?

AP

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Alright, when I started writing this book back in May 2023, I thought I discovered a new proof that 3rd dimension was tops and there was no 4th or beyond. I based that on the idea each Grid system had a center, such as 10 Grid was (5,5,5) and 100 Grid
was (50,50,50). Turns out that was not a proof.

But I still have a proof of no 4th dimension or higher based on this proof of 3 arbitrary noncollinear points in Space forms a unique plane. I arrived at that proof as a derived theorem using the Axiom of parallelism, by extending the sides of every
triangle in 3D Space to construct a unique plane from that triangle. In essence, I replaced Old Math's proof of 3 arbitrary noncollinear points in Space forms a unique plane. Old Math had that as a axiom itself. So their proof was not a proof at all but
a axiom.

The real proof as derived uses the parallel postulate of Euclid.

In Harold Jacobs GEOMETRY, 2nd edition, 1987, he lists the postulate 3 on page 51 saying "Postulate 3 If there are three noncollinear points, then there is exactly one plane that contains them.

On page 204, "Postulate 10 (The Parallel Postulate) Through a a point not on a line, there is exactly one line parallel to the line."

So, well, what AP did, by constructing 3 points as triangles and proving 3 arbitrary noncollinear points determines a unique plane, is AP proved it from using the parallel postulate and thus we do not need postulate 3. It is redundant.

And it is here, the parallel postulate itself denies the existence of 4th dimension or higher. So in one instance I throw out a axiom in Old Math Geometry, and use the Parallel axiom to prove it as a theorem.

Yet this same parallel postulate is going to throw out the 4th dimension and higher. And the way it does that, through consistency.

A 4th dimension or higher does not give a unique line parallel to given line from a point outside the given line.

There is a better way of stating Euclid's parallel postulat in the idea that a line parallel to another line is a perpendicular on one is a perpendicular on the other.

____________ B

____|________A

Every point on line A when drawn a perpendicular is a perpendicular on line B.

In 3rd dimension we have x-axis perpendicular to y-axis and we have z-axis perpendicular to x-axis and we have z-axis perpendicular to y-axis. We have 3 perpendicular tests. When we talk of 4th dimension we need a perpendicular to x and y and z. But no
new axis is available to be perpendicular to x,y, z. But say there was such, some would call it a hidden dimension, or a curled up dimension. But the trouble is, it would destroy the Parallel Postulate Axiom.

So the reason dimension ends at 3rd, is that higher than 3rd destroys the Parallel Postulate.

And this is a new method of proof, a proof by consistency, that when a statement is made to be proven, and if it contradicts an existing axiom, it is a false statement.

Now do not confuse this proof method as Reductio Ad Absurdum, for that is not a valid proof method. Proof by Consistency is a class of itself, and not reductio ad absurdum.

AP

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