Future Of Mankind Forum

You feel insulted. Ok
But it was you first bringing those insulting nonsensical arguments about the derivation.

Kepler Triangle

1 : √φ : φ

Any other triangle with the same property is similar to the KT.
1; b; b^2 is similar.
And it is not a condition for the derivation. It appears as a result and not a premise.
Where do you se here 4b=π, a KT?

You claim the triangle is just a "tool for calculation", not an additional hypothesis.

It is the first time I hear of the Hypothesis of Pythagoras.
What are you talking about?

But a calculation tool must be justified by the problem.

It is very well justified by the problem. Is a tool of calculation as it is +, —, ÷, x.
I have two values of b and I can construct with them a right triangle. Thats it.

You are not deriving the triangle from the square of perimeter; you are imposing it.

I’m deriving the triangle from two values of b that are compatible with the theorem of Pythagoras. The hypotenuse 1 is imposed to keep the relationship btw 4b=π. If I use any other hypotenuse then I’m not calculating 4b=π any more.
And of course I’m not deriving the triangle from the square of the perimeter, π^2. but from b and b^2 where b is 1/4 of the perimeter.
You don’t even understand the most simple terms for the derivation.

If I may illustrate: suppose I define x = π / 4 Then I arbitrarily decide that x and x^3 are the sides of a right triangle with hypothenuse 1. This would give
x^6+x^2=1 , a different value for x, and thus a different π. Would that be a valid proof? No, because the triangle is arbitrary. The same applies here.

Thats the kind of claims where I feel insulted. Because such an invented argument lacks any logic.
Of course you can put x^3 instead of x^2, you can even put Donald Duck instead but, you have to justify WHY.
In this case b is the side of the square and b^2 the Area. The values qualify to construct a right triangle and solve by Pythagoras.

The only way your derivation would be valid is if you could prove that the square of perimeter π necessarily gives rise to a right triangle with sides b and b^2 and hypothenuse 1. You have not done that. You simply assumed it.

More nonsense!
How do you bring π^2 into this derivation?
I don’t need π^2 at all to find out the value of b.
The triangle comes from the values of b and b^2 and the needed hypotenuse=1 to keep the relationship btw the terms or their nominal value.
I see you can’t grasp the basics of the derivation but put yourself on to write meter long critics.
This is insulting as well.

b is defined twice.
But it is. First, you define b = π / 4

Yeah. Thats the nominal value of b, yes.

Second, you impose that b and b^2 are the legs of a right triangle with hypothenuse 1. That second condition is not a consequence of the first; it is an independent constraint. So yes, b is being constrained by two separate conditions. That is not an insult; it is a factual observation.

My goodness!!
I do not impose anything!
I construct a computable triangle with the different values of b!
It is no constrain, as it is π/4 no constrain.
You have no idea.!
First I’m defining b and then I calculate it. Pythagoras does not change the nominal value of b, it delivers the numerical one.
What’s your problem with it?

On units and dimensions
Here, you are confusing units (metres, feet) with dimensions (length, area).

Where do I use “units”?!

In pure mathematics, numbers are dimensionless, that is true. But in geometry, a side has dimension L, and an area has dimension L^2. When you write b^2 (an area) and treat it as a length in Pythagoras' theorem, you are mixing quantities of different natures.

Nonsense!
b is the length of the side but it is the Area of the circle of diameter 1 as well. And b^2 is the area of the square.
Whatever these numbers represent they are basically numerical values and I can use those numerical values and compute them to find out another term or value. What this value finally represents it depends of what I’m looking for.
If b = 0.786 it will remain always 0.786 as a length, an area or a Volume.

The equation (b^2)^2+b^2=1 is not scale-invariant: if you double all lengths, b becomes 2b, the equation becomes 16b^4+4b^2=1, which is not the same. This proves that your construction depends on the arbitrary choice of scale which is not a property of genuine geometry.

No Arthur. that proves that you don’t have no idea whatsoever of the most basic concepts.
If you double the value of the terms the hypotenuse is not 1 anymore!
Your jealousy is blinding you even by the evident.
If you use hypotenuse =2, b is still 0.7861513777.

Convinced?

> You feel insulted. Ok

Shouldn't I? Read the list of insults again.


Kepler triangle.

You said 1, b, b^2 IS NOT a Kepler triangle in your previous post.
I showed to you that it is.
Hypothenuse = 1, Height = b = 1/√ɸ, base = b^2 = 1/ɸ
The geometric progression is here with the ratio of the progression = √ɸ
Re read my previous post where i demonstrate it.


> It appears as a result and not a premise.

That is precisely the problem: it does not appear to be a result, because you are not deriving it from anything. You construct a triangle with sides b and b², and you impose the hypotenuse 1. This construction is arbitrary unless it is justified.

If I construct a triangle with sides b and b³, I’ll get a different value for b. Why is your choice the correct one ? Because b² is the area of the square? But converting an area into a length is a geometric transformation that isn’t automatic. It must be justified.


On the “Pythagorean hypothesis”

You say: “This is the first time I've heard of the Pythagorean hypothesis.”

You’re playing on words. I’m not talking about a “Pythagorean hypothesis.” I’m talking about the fact that choosing to use Pythagoras with specific values is a choice. Pythagoras is a tool, certainly, but you choose which values to plug into it. That choice is not neutral. It is that choice that I am questioning.

On the Construction of the Triangle

You say: “I have two values for b, and I can use them to construct a right triangle. That's all.”

That’s exactly what I’m saying: you can do it, but that doesn’t mean you have to, nor does it have anything to do with the original problem. You could construct a triangle using b and 2b², or using b and b + 1. All of these choices are possible. Yours is arbitrary unless it’s derived from the geometry of the square.

Regarding the comparison with x^3

You say: “Of course, I can use Donald Duck, but I have to explain WHY. Here, b² is the area of the square.”

Exactly! You justify the choice of b² because it’s the area. But an area isn’t a length. In geometry, you can’t add or use a quantity of a different nature in the Pythagorean theorem without changing the scale or using a specific construction (such as quadrature). Quadrature is possible, but it changes the scale. However, your equation b⁴ + b² = 1 is not invariant under a change of scale. This is a fundamental problem.

On the dual definition of b

You say: “I define b, then I calculate it. Pythagoras doesn’t change the nominal value.”

But it does! You use Pythagoras to determine the numerical value of b. This numerical value is entirely dictated by the equation b^4 + b^2 = 1. If you had chosen a different triangle, you would have a different value. So the numerical value of b does not come from π; it comes from the triangle you chose. And since you chose this triangle so that b = 1/ϕ, you end up with what you put into it.

That is the tautology: the value of b is determined by the constraint, not by the definition.

On Units and Dimensions

You say: “b is a length, but it’s also the area of a circle with diameter 1. No matter what the number represents, I can use it.”

In pure mathematics, yes, a number is a number. But you’re giving a geometric proof. In geometry, however, the nature of the quantities matters. The Pythagorean theorem applies to lengths. If you use an area as if it were a length, you change the nature of the object. The fact that the equation is not invariant under a change of scale proves this: a true geometric property does not depend on the unit of measurement. Here, it does depend on it. This means that your construction is not geometrically intrinsic.

On the Change of Scale

You say: “If you double the values, the hypotenuse is no longer 1. If the hypotenuse is 2, b remains 0.786.”

But that’s exactly the point! If b = 0.786 is a length, and I double all the lengths, the new side is 2b = 1.572, and the new area is (2b)² = 4b² = 2.47. The equation (2b)⁴ + (2b)² = 1 is no longer true. Yet geometry should be scale-independent. This proves that your equation is not a geometric property, but a coincidental numerical relationship that depends on the arbitrary choice of the hypotenuse = 1.

I’m not jealous, and I have nothing against the number ϕ. I’m simply pointing out that your proof is based on an arbitrary construction that isn’t derived from the original problem. You chose a Kepler triangle, you solved it, and you arrived back at what you had chosen. That’s not a proof of π; it’s a numerical curiosity.

To summarize your demonstration :

- 4b = π
- Kepler's triangle: 1, h, h², so h = height = 1/√ɸ

How do we set π equal to 4/√ɸ ?

Very simple: b = h

A few months ago, I suggested that you open an account on a math forum and post your calculations there.
I think that's the best advice i can give you.

If you want to convince me, all you need to do is answer this single question:

Can you prove, using only the definition of a square with perimeter π, that the side b and the area b² must necessarily form a right triangle with a hypotenuse of 1?

If you cannot, then your “proof” is circular.

I invite you to respond specifically to this point. I am willing to change my mind if you provide convincing proof.