The Non Transcendental, Exact Value of π and the Squaring of the Circle

[C.B.]

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?

[Arthur]

> 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.

[C.B.]

> You feel insulted. Ok

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

Read your criticisms against the derivation.

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.

I said it is a similar triangle.
Currently you construct a KT with 1; √φ and φ

> 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.

you still do not understand.
I place b and b^2 in a frame where the value of b is solvable and this frame is the Pythagoras theorem. If †he solution were to be obtained with a sum then it would have been a sum.
The h=1 ….again, is to keep the nominal value of b. If I take any other then the original premise 4b=π isn’t valid anymore.
Watch out, because it is kindergarten math. Ok? Put some effort.

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.

Here we go again. You can even put the Donald Duck in the equation, you have just to explain WHY. Why would you do that, use b and b^3?
1; b; b^2 is in fact b^0; b; b^2
Three different values for b that I can put in a triangle and solve by Pythagoras to obtain the value of b. And b^0 keeps the nominal value of b.
Why shouldn’t I do that?
Actually we have here a serie of b values as you can see in the attached figure:
1; b; b^2; b^3; b^4; b^5 etc…..
You can take any group of three in a row and simplify them to their original value 1; b; b^2 and operate.
No mystery.

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.

You said something about “two hypothesis” referring to 4b=π and Pythagoras.
And you’re questioning it because you don’t pay enough attention to understand.
Of course the choice is not neutral! I have to solve the value of b. If you want to know how much is it 3 plus 2 then you choose 3 and 2 to make a sum: 3+2=5. Of course is not neutral, it is about 3 and 2.
So, I don’t really understand your objection.

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.

Pythagoras has not directly to do with the original problem because the original problem is already solved when I come to Pythagoras. The theorem’s use is just to find out the numerical value of b.
Otherwise I wouldn’t know why I should use b and 2b^2 and you don’t explain why would you do that either.
These are not arguments at all. You have to give a reason for it. But you didn’t yet.

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.

Again the same song here. You’re riding circles ignoring my prior explanations.
The choice of b^2 happens basically because it fits the theorem together with b and 1.
And here you’re incurring in gross mistakes again.
The sign ^2 by b^2 doesn’t mean automatically it is an area. b^2 is just the value of that area.
If you take the sides b and b^2 and multiply them you obtain: b^2 x b = b^3. It is the area of the rectangle b; b^2. But, according to you, it should be considered a Volume because of the ^3.
It is just the value of the area as it is b^2 the value of the square’s area. And being just the value I can use it as a linear value to obtain the numerical value of b over Pythagoras..
And the equation is invariant. You have just to multiply all three terms by the same factor, which is the same as living the equation as it is.
You^re not reading my answers.

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.

It doesn’t change the nominal value. Otherwise you could explain why but you don’t, you just shoot around with unbacked claims, which I won’t longer answer.
the relation 4b=π is not influenced by Pythagoras. the theorem just brings the numerical value to light.
Read again the nonsense you write, mixing up nominal and numerical value.

This numerical value is entirely dictated by the equation b^4 + b^2 = 1

.

What does it mean again?
Dictated?
These are values that work in the frame of the theorem to solve the numerical value.
That’s the way it works. What’s your problem with it.?

If you had chosen a different triangle, you would have a different value.

Brilliant!
But why should I have had done something like that?

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.

No fellow.
It is getting boring because I’m explaining it for the third time now.
It comes from π because of 4b=π and the triangle does not change anything about it. You put the different values of b and the theorem works out for you the real value of b, without touching that value, otherwise the theorem would be useless as a theorem. Simple.

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

Ok. Your claim. And that claim alone doesn’t mean anything meaningful.

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.

Ok, you’re still in replay modus.
As I told you already up there, the ^2 of b^2 doesn’t make it to an area!!! It is just a value that appears as an area and as a length as well.
The area of b^2 x b would be = b^3. But, according to you we can’t use it as an area because b^3 is a Volume now. See where you are failing?
These are just values we can use for what we need. And it doesn’t change any “nature”. Do You really think the theorem can differentiate btw area, length or volume when you write a number or a symbol? Really?.
Are we talking VooDoo here?
And once again, the equation is invariant, as long as you multiply all three terms by the same factor. What you didn’t because you’re so desperate to prove your point here.

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.

Your equation is blatantly wrong, thats why it is no longer true.If you want to double the values then you should write:

(2b^2)^2+(2b)^2— 2^2 = 0 (you have to double the h as well !!!)
Which yields:
4b^4 + 4b^2 — 4 = 0
which is the same as:

b^4+ b^2 — 1= 0

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.

As you have seen it is scale independent and the choice of h=1 is not arbitrary. Should I write for the fifth time WHY?

*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.

If you’re not jealous then there is some other problem bothering you that is preventing you from seeing evident facts and overlooking my comments explaining them.

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

This is all nonsense. I’m not making φ a premise nor the KT. -It is a result and I have no control over the result. That’s all in your head what you claim to be.

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.

By now you know that it is better for you to open such an account.

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.

The only one circular is you Arthur!
I did explain it to you already many times over. But you’re not listening, except the voices in your head.
You know already why b and b^2 and 1 are significant to compute the value of b.
Read the comments again, there is your answer many times written.

[Arthur]

> Read your criticisms against the derivation.

There is no personal attack. Why should I have to do that?

When and where i used those terms A series of personal attacks ("BS", "Ignorant", "moron", "stupidology", "jealousy")

I dont' need that.

So anyone who disagrees with you should automatically be insulted, and it's perfectly acceptable to attack them personally?


I'll be brief, because we're going in circles.

You say that the Kepler triangle "appears as a result" and that you have "no control over it." Yet you are the one who chooses to construct a triangle with sides b and b^2, and hypotenuse 1. That choice is yours. It is not dictated by the definition of a square with perimeter π.

The question I have been asking from the start is very simple:

Why must b and b^2 be the sides of a right triangle with hypotenuse 1 ?

You answer: "Because I can." But "can" does not mean "must." A mathematical proof does not work with "I can." It works with "I deduce."

You also say:

"If I take another hypotenuse, 4b=π is no longer valid."

That is precisely the problem! You impose the hypotenuse 1 so that the relation 4b = remains true. But that means you choose the scale so that your calculation works out. That is not a proof; it is an arbitrary adjustment.

You also say:

"b^2 is not an area, it's just a value."

In pure mathematics, that is true: a number is a number. But you are giving a geometric proof. In geometry, however, the nature of quantities matters. If you treat an area as if it were a length, you change the nature of the object. The fact that your equation is not scale-invariant (unless you also change the hypotenuse, which you do) proves that your construction depends on the chosen scale. A true geometric property is scale-independent. Yours is not.

Finally, you say:

"I do not make ϕ a premise."

But you do. You choose a Kepler triangle, which has the unique solution b = 1 /√ϕ. So you choose ϕ by choosing that triangle. It is not a result; it is a condition you impose.

To summarize one last time:

What you do ---> What you should do

You choose a Kepler triangle --> You prove that this triangle follows from the square of perimeter π
You solve b = 1/√ϕ --> You deduce the value of b from π alone
You substitute into 4b=π --> You obtain a value of π with no external condition

As long as you have not answered this single question:

Where does the Kepler triangle come from in your proof?

Your "proof" remains an arbitrary construction, not a demonstration.

My advice remains the same:

Post your demonstration on a mathematics forum (Math Stack Exchange, mathforums.com, or similar). Mathematicians will give you an objective opinion. If I am wrong, they will correct me. But I am convinced they will tell you the same thing.

I will stop here. I will not respond to insults or repetitions. The discussion is closed on the substance.

[C.B.]

Bildschirmfoto 2025-12-22 um 22.42.59.png (82.54 KiB) Viewed 297 times

It is not a geometric demonstration.
And the procedure is simple.
I can take the value of the Area b^2 and inscribe it on the side of the square as a length, draw the hypotenuse = 1, simply because we are working with the base circle with perimeter π and solve with Pythagoras to obtain the value of b.
And there is nothing you can say to invalidate it.
The only ting you have proven with your bunch of nonsensical objections is that you are not so versed in mathematics as you candidly believe.
Only one of many cases: One of your terrible conceptual mistakes regarding the invariance of b^4+b^2=1 was to multiply by a factor only the left side of the equation when you have to do it on both sides. Got it?

If you don’t answer you do me a favor.
So please do me that favor.

[Arthur]

On units and dimensions

You wrote: "A VALUE has no unit, moron!"

I wrote : Here, you are confusing units (metres, feet) with dimensions (length, area). 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.
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.


Detailed explanation

1. The difference between "unit" and "dimension"


This is your first confusion.

Concept --> Definition --> Example

Unit --> The standard of measurement (metre, foot, inch) --> 1 meter, 1 foot
Dimension --> The nature of the quantity being measured (length, area, volume) --> Length = L, Area = L^2, Volume = L^3

When you say "A value has no unit", you are right about units: in pure mathematics, we don't write "metres" or "feet". A number is a number.

But you are wrong about dimension: in geometry, the nature of the quantity matters. A length and an area are not interchangeable.


2. Why dimension matters in geometry

In geometry, every quantity has a dimension:

- A side has dimension L (length).

- An area has dimension L^2 (length × length).

- A volume has dimension L^3.

When we write a geometric equation, all terms must have the same dimension.

Example: the Pythagorean theorem

a^2 + b^2 = c^2

- a, b, c are lengths (dimension L).
- a^2, b^2, c^2 are areas (dimension L^2).

The equation is homogeneous: all terms have dimension L^2.

This is correct.

3. What you are doing

You write:

(b^2)^2 + b^2 = 1

Here:

- b is the side of the square → dimension L.
- b^2 is the area of the square → dimension L^2.
- But you use b^2 as a length in the triangle.


Let's analyse the dimensions:


Term --> What it represents --> Dimension

b^2 ------> A side of the triangle ---> L^2 (if it's an area) or (if it's a length)
(b^2)^2 --> The square of that side -> L^4 or L^2
b^2 ------> The other side ---------> L^2 or L
1 ---------> The hypotenuse --------> L

The terms do not have the same dimension:

- If b^2 is an area (L^2), then (b^2)^2 has dimension L^4 , and the equation mixes L^4, L^2, and L.

- If b^2 is a length (L), then b (the side of the square) is also a length, but b^2 is the area of the square — there is a contradiction.

The equation is not homogeneous. It mixes quantities of different natures.

4. Why this is a problem: scale invariance

A geometric equation must be scale-invariant. This means: if I double all lengths, the equation must remain true.

Take your equation:

b^4 + b^2 = 1

Double all lengths:

- b -> 2b
- b^2 -> (2b)^2 = 4b^2 (the area is multiplied by 4)
- The hypotenuse 1 -> 2


The new Pythagorean equation must be:

(4b^2)^2 + (2b)^2 = (2)^2

Calculate:

16b^4 + 4b^2 = 4

Divide by 4:

4b^4 + b^2 = 1

This is not the same equation as b^4 + b^2 = 1. The equation has changed.

Why?

Because b^2 is treated as a length, but it is actually an area. When you change scale, an area and a length do not transform the same way.

- A length is multiplied by 2.
- An area is multiplied by 4.

By mixing the two in the same equation, you get an equation that depends on the chosen scale. This is not a geometric property.

5. The mistake you made

When you responded to this objection, you wrote:

"If you double the values, the hypotenuse is no longer 1. If the hypotenuse is 2, b remains 0.786."

Then you wrote:

"(2b^2)^2 + (2b)^2 − 2^2 = 0 -> 4b^4 + 4b^2 − 4 = 0 -> b^4 + b^2 − 1 = 0"

You corrected the equation by doubling the hypotenuse. But you missed the point:

You showed that the equation is invariant if you double everything (sides and hypotenuse).
But this does not solve the dimension problem. The equation b^4 + b^2 = 1 remains a mixture of quantities of different natures.

Scale invariance is a necessary condition but not a sufficient one. Even if the equation is invariant (after correction), it remains geometrically incorrect because it mixes dimensions.

6. The analogy to understand

Imagine a rectangle:

Length = 3 metres
Width = 2 metres
Area = 6 m^2

The area is 6. But that does not mean the rectangle has a side of 6. The area is a surface, not a length.

If I say: "The area is 6, so I can build a triangle with a side of 6", I am making a dimension error.

This is exactly what you are doing: you take the area b^2 (a surface) and use it as a length in a triangle. This is geometrically incorrect.

7. Summary

Point --> Explanation

Unit vs Dimension ---> The unit (metre, foot) is arbitrary. The dimension (length, area) is a property of the geometric object.
Mixing dimensions ---> The equation b^4 + b^2 = 1 mixes terms that do not have the same dimension.
Scale invariance ---> A geometric equation must be scale-invariant. Yours is not (unless you multiply everything, but that doesn't solve the dimension problem).
Consequence --> The construction is not geometrically valid. It relies on a confusion between area and length.


8. The question you have never answered

"If b is a length and b^2 is an area, how can you add them in the same equation without changing scale? And how do you justify that the area of the square suddenly becomes a length in the triangle?"

There is no possible answer to this question, because your construction is geometrically incorrect from the start.

This is why your proof is invalid. It is not a question of units (metres, feet), it is a question of dimension (length vs area). And this is a fundamental problem you have never resolved.

[C.B.]

Sorry fellow but you’re writing such a pile of BS again. I’m not going to refute it once again.

You’re simply not listening and keep regurgitating your own fantasised rebuttals and don’t even know how to scale the equation!

Besides b^2 can be a length as well as an area, as I showed it to you in the uploaded diagrams you didn’t even look at them or you didn’t understand them.
b^2 is an area in a Square of side b but it is also a length in a Square of area b^4.
These are just values

And now you scaling of the equation

Double all lengths:
- b -> 2b
- b^2 -> (2b)^2 = 4b^2 (the area is multiplied by 4)
- The hypotenuse 1 -> 2- The hypotenuse 1 -> 2

Pay attention because it is not an error it is a horror:

b^2 -> (2b)^2 = 4b^2 (the area is multiplied by 4)


The length here is not b but b^2
So, if you want to double that you have to multiply b^2 with 2 and not b alone and then square (2b)

2(b^2)

What every kindergarten kid would do is to divide again every term by 2 and see if he gets the original equation back again.

You didn’t. Otherwise you would have seen that you don’t get the original equation back again, because you have a b^2 too much.

What you’re displaying here is embarrassing.

[Arthur]

You write: 'The length here is b^2, so to double it, you multiply b^2 by 2.' That is correct if you are doubling that specific length. But when you double all lengths, you also double b and the hypotenuse. You forgot to double them.

The correct equation is: (2b^2)^2 + (2b)^2 = 2^2 , which simplifies to b^4 + b^2 = 1 . The equation is scale-invariant.

Your mistake is confusing 'doubling a length' with 'doubling the scale'. These are two different operations. You did the first, but pretended to do the second.

However, even with this correction, two fundamental problems remain :

You still haven't justified why b and b^2 must form a Kepler triangle. You simply assert it. Nothing in the definition of a square with perimeter
π implies that its side and its area must be the legs of a right triangle with hypotenuse 1. This is an arbitrary condition you impose, not a consequence of the premises. And there is no "Because i can"

You are mixing dimensions. In geometry, b is a length (dimension L), and b^2 is an area (dimension L^2). By using b^2 as a length in Pythagoras' theorem, you are treating an area as if it were a length. This is geometrically invalid unless you change the scale or provide a specific construction (which you have not done). The Pythagorean theorem applies to lengths, not to a mixture of lengths and areas. As i explained before.

Until you address these two points — especially the second one — your construction remains geometrically incorrect, and your proof remains invalid.

[Arthur]

Found a new way to calculate the value of pi :

- Let π be the perimeter of a square with side length b.
- Let a Kepler triangle with hypothenuse 1, so height h and base h^2 , so h = hypothenuse / √ɸ = 1 / √ɸ, base = hypothenuse / ɸ = (1 / √ɸ)^2 = 1/ɸ
- Let b = h

So the value of pi is 4 * 1 / √ɸ = 4 / √ɸ = 3.144...

Question :

How can b = h ?

Answer : Because i can.

[C.B.]

You write: 'The length here is b^2, so to double it, you multiply b^2 by 2.' That is correct if you are doubling that specific length. But when you double all lengths, you also double b and the hypotenuse. You forgot to double them.

I was just pointing out your gross mistake by doubling b^2.

The correct equation is: (2b^2)^2 + (2b)^2 = 2^2 , which simplifies to b^4 + b^2 = 1 . The equation is scale-invariant.

Yeah. Now it is correct.

Your mistake is confusing 'doubling a length' with 'doubling the scale'. These are two different operations. You did the first, but pretended to do the second.

BS.

However, even with this correction, two fundamental problems remain :

Oh really?

You still haven't justified why b and b^2 must form a Kepler triangle. You simply assert it.

This is just invented by you. I never claim that.

Nothing in the definition of a square with perimeter
π implies that its side and its area must be the legs of a right triangle with hypotenuse 1.

But they are the legs of a right triangle if I place them that way.
That’s the thrill in Geometry, you have to put the elements together that can solve your problem. And I did it.

This is an arbitrary condition you impose, not a consequence of the premises. And there is no "Because i can"

Yes. It is exactly because I can and do not contradict any rational parameter. And yes, it is a consequence of the premises: A square of side b.
Different from you, that put things the way you want just to justify your baseless criticism. regardless geometric parameters.

You are mixing dimensions. In geometry, b is a length (dimension L), and b^2 is an area (dimension L^2). By using b^2 as a length in Pythagoras' theorem, you are treating an area as if it were a length. This is geometrically invalid unless you change the scale or provide a specific construction (which you have not done). The Pythagorean theorem applies to lengths, not to a mixture of lengths and areas. As i explained before.

Ok. for the sixth time now.
For Pythagoras I use VALUES and not lengths or areas. And, because Pythagoras needs these VALUES to be the lengths of the triangle’s legs I use them there as lengths. It is no problem whatsoever because we’re looking for a relationship btw VALUES. So, I’m looking for it. What you’re looking for I can’t even figure it.

Until you address these two points — especially the second one — your construction remains geometrically incorrect, and your proof remains invalid.

Yeah. You would like that. But with your nonsensical objections you’re biting on granite so far.
Sorry John.!