On 06-Nov-21 2:00 pm, Sylvia Else wrote:
> For any odd number x, there exists another odd number y, such that the
> product of x and y expressed in binary is a contiguous sequence of 1s,
> though I'll express them in hex.
>
> For example
>     3 * 1 =       3
>     5 * 3 =       f
>     7 * 9 =      3f
>     9 * 7 =      3f
>     11 * 93 =   3ff
>     13 * 315 =  fff
>
>     97 * 816785180559426160758185055 = FFFFFFFFFFFFFFFFFFFFFFFF
>
> Sylvia.

Well that was interesting. Who would have thought that a quite trivial
statement could generate so much heat.

Look out for my next numerological missive - squaring the circle. But
you'll have to wait a bit.

Sylvia.

Sylvia Else wrote:

> On 06-Nov-21 2:00 pm, Sylvia Else wrote:
>> For any odd number x, there exists another odd number y, such that the
>> product of x and y expressed in binary is a contiguous sequence of 1s,
>> though I'll express them in hex. For example
>>     3 * 1 =       3 5 * 3 =       f 7 * 9 =      3f 9 * 7 =     
>>     3f 11 * 93 =   3ff 13 * 315 =  fff
>> 97 * 816785180559426160758185055 = FFFFFFFFFFFFFFFFFFFFFFFF Sylvia.
> Well that was interesting. Who would have thought that a quite trivial
> statement could generate so much heat.
> Look out for my next numerological missive - squaring the circle. But
> you'll have to wait a bit.

xor the result to get the zeros.

On Thursday, November 11, 2021 at 7:45:57 PM UTC-3, Sylvia Else wrote:
> On 06-Nov-21 2:00 pm, Sylvia Else wrote:
> > For any odd number x, there exists another odd number y, such that the
> > product of x and y expressed in binary is a contiguous sequence of 1s,
> > though I'll express them in hex.
> >
> > For example
> > 3 * 1 = 3
> > 5 * 3 = f
> > 7 * 9 = 3f
> > 9 * 7 = 3f
> > 11 * 93 = 3ff
> > 13 * 315 = fff
> >
> > 97 * 816785180559426160758185055 = FFFFFFFFFFFFFFFFFFFFFFFF
> >
> > Sylvia.
> Well that was interesting. Who would have thought that a quite trivial
> statement could generate so much heat.
>
> Look out for my next numerological missive - squaring the circle. But
> you'll have to wait a bit.
>
> Sylvia.

Why?

Just post:

Prove that you can EXACTLY build an square with sides L that verifies the same area as a circle with radius R:

L² = π R²

The pair (L, R) has to verify π EXACTLY (for any arbitrarily large number of decimals).

On Thursday, November 11, 2021 at 7:03:11 PM UTC-5, Richard Hertz wrote:

> Prove that you can EXACTLY build an square with sides L that verifies the same area as a circle with radius R:
> L² = π R²
> The pair (L, R) has to verify π EXACTLY (for any arbitrarily large number of decimals).

A known and solved problem.

https://en.wikipedia.org/wiki/Tarski%27s_circle-squaring_problem

See also: https://en.wikipedia.org/wiki/Squaring_the_circle

On Thursday, November 11, 2021 at 9:58:46 PM UTC-3, rotchm wrote:

<snip>

> > Prove that you can EXACTLY build an square with sides L that verifies the same area as a circle with radius R:
> > L² = π R²
> > The pair (L, R) has to verify π EXACTLY (for any arbitrarily large number of decimals).

> A known and solved problem.
>
> https://en.wikipedia.org/wiki/Tarski%27s_circle-squaring_problem
>
> See also: https://en.wikipedia.org/wiki/Squaring_the_circle

It has to verify π EXACTLY, for any arbitrarily K number of decimals. I think that such task is impossible to be achieved.

Your reference has not a finite solution. You have to use imaginary squares, with increasingly small sides, and truncate the
calculation to P digits.

I think that, roughly, if N = 10^P, being P an integer, then the area W of each piece is W ≈ 1.77. 10^(-P/2). So, for N = 10E16 pieces,
the area of each piece is roughly W ≈ 1.77. 10^-8, in the case of the radius R = 1. And you still have to truncate to K decimal digits.

If N is the number of squares with area W, it has to be proven that:

1) L = W.√N BE an integer, which is necessary for such imaginary construction of an square with area L x L using N pieces with area W².
This condition stresses the choices of W values, which have to be FINITE. There is the need to truncate W with K digits.

2) If adopt R =1, and L = W.√N (√N is an integer and W has K decimal digits), prove that the error π - N.W² < 10E(-K) for any set (N, W, K).

Examples of sets (N, W, K):

(10000; 0.01772; 5)
(65536; 0.00692364; 8)
(1000000; 0.017724538; 10)
(130233744; 0.000155314918; 12)
(2242780164; 0.00003742670406; 15)
(10000000000; 0.000017724538509055; 18)

Excerpt: ------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

Tarski's circle-squaring problem

Tarski's circle-squaring problem is the challenge, posed by Alfred Tarski in 1925, to take a disc in the plane, cut it into FINITELY many
pieces, and reassemble the pieces so as to get a square of equal area. This was proven to be possible by Miklós Laczkovich in 1990; the decomposition makes heavy use of the axiom of choice and is therefore non-constructive. Laczkovich estimated the number of pieces in his decomposition at roughly 10E+50. More recently, Andrew Marks and Spencer Unger (2017) gave a completely constructive solution using about 10E+200 Borel pieces.[1]

In particular, it is impossible to dissect a circle and make a square using pieces that could be cut with an idealized pair of scissors (that is, having Jordan curve boundary). The pieces used in Laczkovich's proof are non-measurable subsets.

End excerpt: ------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

Richard Hertz <hertz778@gmail.com> wrote:
> On Thursday, November 11, 2021 at 7:45:57 PM UTC-3, Sylvia Else wrote:
>> On 06-Nov-21 2:00 pm, Sylvia Else wrote:
>>> For any odd number x, there exists another odd number y, such that the
>>> product of x and y expressed in binary is a contiguous sequence of 1s,
>>> though I'll express them in hex.
>>>
>>> For example
>>> 3 * 1 = 3
>>> 5 * 3 = f
>>> 7 * 9 = 3f
>>> 9 * 7 = 3f
>>> 11 * 93 = 3ff
>>> 13 * 315 = fff
>>>
>>> 97 * 816785180559426160758185055 = FFFFFFFFFFFFFFFFFFFFFFFF
>>>
>>> Sylvia.
>> Well that was interesting. Who would have thought that a quite trivial
>> statement could generate so much heat.
>>
>> Look out for my next numerological missive - squaring the circle. But
>> you'll have to wait a bit.
>>
>> Sylvia.
>
> Why?
>
> Just post:
>
> Prove that you can EXACTLY build an square with sides L that verifies the
> same area as a circle with radius R:
>
> L² = π R²
>
> The pair (L, R) has to verify π EXACTLY (for any arbitrarily large number of decimals).
>
>

I seem to recall you were the one saying no one ever needed anything to
more than 15 digits.

--
Odd Bodkin -- maker of fine toys, tools, tables

On Friday, November 12, 2021 at 12:54:54 AM UTC-5, Richard Hertz wrote:

> It has to verify π EXACTLY, for any arbitrarily K number of decimals.. I think that such task is impossible to be achieved.

According to the references I gave you, it is possible. It has been proven. Of course to actually do it would take zillions and zillions of pieces, something we do not want to do and cannot physically do.

> Your reference has not a finite solution.

Yes it does.

"...cut it into ***finitely*** many pieces, and reassemble the pieces so as to get a square of equal area. This was proven to be possible..."

> You have to use imaginary squares, with increasingly small sides, and truncate the
> calculation to P digits.

They need not be squares. The shapes do not have a well-defined boundaries mathematically. But you can cut them out with scissors in such a matter you will eventually approximate to the desired accuracy. Have you read or understood the references I gave you?
No need to answer this. We are well beyond the scope of this news group And the original thread. Let's stop this here. Start a new thread if need be.

Richard Hertz wrote:

> On Thursday, November 11, 2021 at 9:58:46 PM UTC-3, rotchm wrote:
>
> <snip>
>
> It has to verify π EXACTLY, for any arbitrarily K number of decimals. I
> think that such task is impossible to be achieved.
>
> Your reference has not a finite solution. You have to use imaginary
> squares, with increasingly small sides, and truncate the calculation to
> P digits.

he doesn't know imaginary, he is an idiot.

On Friday, November 12, 2021 at 10:56:11 AM UTC-3, bodk...@gmail.com wrote:

<snip>

> I seem to recall you were the one saying no one ever needed anything to
> more than 15 digits.

You and rotchm are right. This thread has gone too far. However, I'd like to remark that using squares allow the
problem to be formulated within the limits of simple algebra (squares and circles).

I think about how the problem can be stated to fit within relativity and quantum physics fields:

************************************************************************

You have two imaginary persons orbiting Earth, in stable orbits, at heights of 3200 km and 26000 Km above the Earth.

Both are trying to solve the problem for π, squares and circles with a precision of 15 digits for π, while a third imaginary
person is observing the developments of the other two "mathenauts".

Prove that, using a circle with R = 1:

1) The "mathenaut" orbiting at height of 3200 Km finish the task faster than the other at 26000 Km, being both equally capable.

2) That the "imaginary square" built by any of them has an uncertainty lower than 10E-18 if the "imaginary matter" used to build
the squares have a "virtual mass" equal to 1, given that their relative lenght and momentum product is EXACTLY h/4π.

3) The area of the "imaginary squares", W, have a precision within 15 digits.

Addendum:

L² = N . W² ≈ π , being √N an integer and W a number with EXACTLY 15 digits.

π - N . W² < 10E-18 (remember that W has 15 digits, so W² has up to 30 valid digits).

Times t, t1 and t2 (at Earth surface, at 3200 Km and at 26000 K) are measured with a mantissa of 15 digits.

************************************************************************

Has this problem solutions that can be proven?

You can also formulate another, with an imaginary square with L = 100 Km being built at 1 radii of the Sun's center. The "plane"
that contains the center of the square is perpendicular to the Sun's radius.. Forget time and focus on squares and circles only.

On Friday, November 12, 2021 at 8:20:57 AM UTC-8, cretin Richard Hertz mused:

> I think about how the problem can be stated to fit within relativity and quantum physics fields:
>

Don't. You are way too cretin for that.

Richard Hertz wrote:

>> I seem to recall you were the one saying no one ever needed anything to
>> more than 15 digits.
>
> You and rotchm are right. This thread has gone too far. However, I'd
> like to remark that using squares allow the problem to be formulated
> within the limits of simple algebra (squares and circles).

https://www.bitchute.com/video/laqHdaOI36DC/
NEW DEATH CAMPS FOR THE UNVACCINATED...

On Friday, November 5, 2021 at 10:00:39 PM UTC-5, Sylvia Else wrote:
> For any odd number x, there exists another odd number y, such that the
> product of x and y expressed in binary is a contiguous sequence of 1s,
> though I'll express them in hex.
>
> For example
> 3 * 1 = 3
> 5 * 3 = f
> 7 * 9 = 3f
> 9 * 7 = 3f
> 11 * 93 = 3ff
> 13 * 315 = fff
>
> 97 * 816785180559426160758185055 = FFFFFFFFFFFFFFFFFFFFFFFF
>
> Sylvia.

Sorry Sylvia, Mixing bases gives me a headache, you're on your own

The girlfriend wouldn't quit smoking meth, I have been busy for a while getting her out.

No .... I haven't used any illegal substances in the last half century

(Resonant Fluid Dynamic Medium) > (Einstein's Aetherless Spacetime)

(Einstein's Aetherless Spacetime) = (Mainstream's Current Road Block)

(Gravitation in a (Resonant Fluid Dynamic Medium)) = https://www.youtube.com/watch?v=qriZc1NNrak

hbar/planck length/2pi = 1.03846153846

(27 / 13) / 2 = 1.03846153846

(27 / 13) * pi = 6.52484628053

(3m)^3/((3m)^3-(1m)^3*2pi/(hbar/planck length) = 0.999992764

((4pi * (1.5^3)) / ((4pi * (1.5^3)) - ((4pi) * (0.5^3)))) / ((hbar / planck length) / (2pi)) = 0.999992764

https://photos.app.goo.gl/sfrZJ14iZN4t9LdE8

The Brachistochrone & Snell's Law & & Catenary & Cycloid

https://youtu.be/Cld0p3a43fU?t=476

Relativistic Mass & Time Dilation is ( https://en.wikipedia.org/wiki/Bernoulli%27s_principle )

On Sunday, November 28, 2021 at 11:44:35 AM UTC-6, David (Kronos Prime) Fuller wrote:
> On Friday, November 5, 2021 at 10:00:39 PM UTC-5, Sylvia Else wrote:
> > For any odd number x, there exists another odd number y, such that the
> > product of x and y expressed in binary is a contiguous sequence of 1s,
> > though I'll express them in hex.
> >
> > For example
> > 3 * 1 = 3
> > 5 * 3 = f
> > 7 * 9 = 3f
> > 9 * 7 = 3f
> > 11 * 93 = 3ff
> > 13 * 315 = fff
> >
> > 97 * 816785180559426160758185055 = FFFFFFFFFFFFFFFFFFFFFFFF
> >
> > Sylvia.
> Sorry Sylvia, Mixing bases gives me a headache, you're on your own
>
> The girlfriend wouldn't quit smoking meth, I have been busy for a while getting her out.
>
> No .... I haven't used any illegal substances in the last half century
>
>
>
> (Resonant Fluid Dynamic Medium) > (Einstein's Aetherless Spacetime)
>
> (Einstein's Aetherless Spacetime) = (Mainstream's Current Road Block)
>
> (Gravitation in a (Resonant Fluid Dynamic Medium)) = https://www.youtube.com/watch?v=qriZc1NNrak
>
> hbar/planck length/2pi = 1.03846153846
>
> (27 / 13) / 2 = 1.03846153846
>
> (27 / 13) * pi = 6.52484628053
>
> (3m)^3/((3m)^3-(1m)^3*2pi/(hbar/planck length) = 0.999992764
>
> ((4pi * (1.5^3)) / ((4pi * (1.5^3)) - ((4pi) * (0.5^3)))) / ((hbar / planck length) / (2pi)) = 0.999992764
>
https://photos.app.goo.gl/MnnEuBdq6Ag9yTMeA

>
> https://photos.app.goo.gl/sfrZJ14iZN4t9LdE8
>
>
> The Brachistochrone & Snell's Law & & Catenary & Cycloid
>
> https://youtu.be/Cld0p3a43fU?t=476
>
> Relativistic Mass & Time Dilation is ( https://en.wikipedia.org/wiki/Bernoulli%27s_principle )

On Friday, November 5, 2021 at 10:00:39 PM UTC-5, Sylvia Else wrote:
> For any odd number x, there exists another odd number y, such that the
> product of x and y expressed in binary is a contiguous sequence of 1s,
> though I'll express them in hex.
>
> For example
> 3 * 1 = 3
> 5 * 3 = f
> 7 * 9 = 3f
> 9 * 7 = 3f
> 11 * 93 = 3ff
> 13 * 315 = fff
>
> 97 * 816785180559426160758185055 = FFFFFFFFFFFFFFFFFFFFFFFF
>
> Sylvia.

FFFFFFFFFFFFFFFFFFFFFFFF < 16^24