Re: The Non Transcendental, Exact Value of π and the Squaring of the Circle
Posted: Wed Aug 27, 2025 7:08 pm
Ok. Good to know. Thanks.
Unsure if this will work. Let's see. If you can see the image, can you spot anything unusual?
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The Non Transcendental, Exact Value of π and the Squaring of the Circle
Page 5 of 22
Re: The Non Transcendental, Exact Value of π and the Squaring of the Circle
Posted: Sat Aug 30, 2025 11:03 am
Unsure if this will work. Let's see. If you can see the image, can you spot anything unusual?
Re: The Non Transcendental, Exact Value of π and the Squaring of the Circle
Posted: Sat Aug 30, 2025 12:31 pm
I can’t see any image…...
Re: The Non Transcendental, Exact Value of π and the Squaring of the Circle
Posted: Tue Sep 02, 2025 9:13 pm
- Screenshot_20250902-170939_Desmos.jpg (251.31 KiB) Viewed 97886 times
- Screenshot_20250902-170905_Desmos.jpg (244.64 KiB) Viewed 97886 times
Re: The Non Transcendental, Exact Value of π and the Squaring of the Circle
Posted: Tue Sep 02, 2025 9:32 pm
If I’m supposed to see a graph, I don’t.
Re: The Non Transcendental, Exact Value of π and the Squaring of the Circle
Posted: Thu Sep 04, 2025 5:16 pm
Are you able to see an image? Numbers? Anything?
Re: The Non Transcendental, Exact Value of π and the Squaring of the Circle
Posted: Thu Sep 04, 2025 6:53 pm
Only equations…..
Re: The Non Transcendental, Exact Value of π and the Squaring of the Circle
Posted: Mon Sep 08, 2025 12:21 pm
That is basically what I want you to look at 
Re: The Non Transcendental, Exact Value of π and the Squaring of the Circle
Posted: Mon Sep 08, 2025 12:23 pm
Ok. I have seen them but you should tell me something about the meaning of all that.
Re: The Non Transcendental, Exact Value of π and the Squaring of the Circle
Posted: Tue Sep 09, 2025 4:52 pm
well I figured that since Pi is Circumference divided by the diamter of a circle, I tried to somewhat "recreate" Pi using that age-old formula.
I had to re-do it through different iterations and figured out a flaw in the Pi we currently use (3.1416...).
In the other iterations I tried, I started with C as the circumference and gave it 1 as the base unit.
Afterwards I presented the diamtere as "d" and gave it a value of (1/pi)*C, to which the value is 0.31830988618 (while using the current conventional Pi).
Then I tested, and this gave me a result of 3.1416 as I would've expected.
Later on, I created "r" as the radius of the circle with a value of d/2 (that's basic arithmetic... just divide the diameter by two), and also created half the side of a square as "s" with a value of "C/8" with in mind the squaring the circle. In doing so, the value of "C" always remains the same, no matter the Pi which is being used.
As a matter of fact, if we were to do 4s/r (which would be representative of 4/sqrt(phi)), then the resulting Pi would reflect exactly the one which was chosen. In order to ensure that the proof is sound, we use the following equations for both the redius and half side of a square:
1) radius = r_1 = sqrt((r^2/s)^2-s^2) -> the result MUST BE THE SAME AS THE INITAL RADIUS
2) 1/2 side of a square = s_1 = r^2/sqrt(r^2+s^2) -> the result MUST BE THE SAME AS THE INITAL 1/2 SIDE FOR A SQUARE
then we formulate for the TRUE PI -> 4s_1/r_1 -> this will reveal discrepencies in the Pi we input in d = (1/pi)*C, and only 3.144605511029693144... will come out of this mathematical showdown as the victor, while at the same time revealing the Golden Ratio through (r_1/s_1)^2.
I made a video on this recently on Youtube. You should check it out: https://youtu.be/TT5FAvMap2o
I had to re-do it through different iterations and figured out a flaw in the Pi we currently use (3.1416...).
In the other iterations I tried, I started with C as the circumference and gave it 1 as the base unit.
Afterwards I presented the diamtere as "d" and gave it a value of (1/pi)*C, to which the value is 0.31830988618 (while using the current conventional Pi).
Then I tested, and this gave me a result of 3.1416 as I would've expected.
Later on, I created "r" as the radius of the circle with a value of d/2 (that's basic arithmetic... just divide the diameter by two), and also created half the side of a square as "s" with a value of "C/8" with in mind the squaring the circle. In doing so, the value of "C" always remains the same, no matter the Pi which is being used.
As a matter of fact, if we were to do 4s/r (which would be representative of 4/sqrt(phi)), then the resulting Pi would reflect exactly the one which was chosen. In order to ensure that the proof is sound, we use the following equations for both the redius and half side of a square:
1) radius = r_1 = sqrt((r^2/s)^2-s^2) -> the result MUST BE THE SAME AS THE INITAL RADIUS
2) 1/2 side of a square = s_1 = r^2/sqrt(r^2+s^2) -> the result MUST BE THE SAME AS THE INITAL 1/2 SIDE FOR A SQUARE
then we formulate for the TRUE PI -> 4s_1/r_1 -> this will reveal discrepencies in the Pi we input in d = (1/pi)*C, and only 3.144605511029693144... will come out of this mathematical showdown as the victor, while at the same time revealing the Golden Ratio through (r_1/s_1)^2.
I made a video on this recently on Youtube. You should check it out: https://youtu.be/TT5FAvMap2o