On Saturday, May 28, 2022 at 3:21:33 PM UTC-7, Ross A. Finlayson wrote:
> On Friday, May 27, 2022 at 4:53:41 PM UTC-7, mitchr...@gmail.com wrote:
> > On Friday, May 27, 2022 at 11:21:08 AM UTC-7, sergi o wrote:
> > > On 5/27/2022 12:38 PM, mitchr...@gmail.com wrote:
> > > > On Friday, May 27, 2022 at 10:23:29 AM UTC-7, Ross A. Finlayson wrote:
> > > >> On Thursday, May 26, 2022 at 2:17:50 PM UTC-7, sergi o wrote:
> > > >>> On 5/26/2022 3:47 PM, mitchr...@gmail.com wrote:
> > > >>>> On Thursday, May 26, 2022 at 1:37:42 PM UTC-7, Chris M. Thomasson wrote:
> > > >>>>> On 5/26/2022 1:25 PM, Michael Moroney wrote:
> > > >>>>>> On 5/25/2022 11:49 PM, Chris M. Thomasson wrote:
> > > >>>>>>> On 5/25/2022 7:21 PM, Dan joyce wrote:
> > > >>>>>>>> On Monday, May 23, 2022 at 7:11:22 PM UTC-4, Chris M. Thomasson wrote:
> > > >>>>>>>>> On 5/23/2022 1:25 PM, Michael Moroney wrote:
> > > >>>>>>>>>> Just because in theory an infinite number of steps is required doesn't
> > > >>>>>>>>>> mean the limit cannot be reached. Consider Zeno's Paradox where
> > > >>>>>>>>>> Achilles
> > > >>>>>>>>>> races a tortoise with a head start. Each time Achilles reaches a point
> > > >>>>>>>>>> where the tortoise was, the tortoise advances somewhat. When Achilles
> > > >>>>>>>>>> reaches that point, the tortoise advances more. And so on for an
> > > >>>>>>>>>> infinite number of steps. Yet Achilles catches up to the tortoise and
> > > >>>>>>>>>> passes it and wins the race, despite taking an infinite number of
> > > >>>>>>>>>> steps
> > > >>>>>>>>>> to catch up to the tortoise.
> > > >>>>>>>>> If Achilles strictly plays by the tortoises rules on a step-by-step
> > > >>>>>>>>> basis:
> > > >>>>>>
> > > >>>>>> What are "the tortoise's rules"? The only rules are the tortoise gets a
> > > >>>>>> head start and both it and Achilles run as fast as they can to the
> > > >>>>>> finish line, and whoever does so first, wins.
> > > >>>>>>>>>
> > > >>>>>>>>> step 1: tortoise moves one meter; Achilles moves one meter. The
> > > >>>>>>>>> tortoise
> > > >>>>>>>>> is ahead because of the head start.
> > > >>>>>>>>>
> > > >>>>>>>>> step 2: tortoise moves one meter; Achilles moves one meter. The
> > > >>>>>>>>> tortoise
> > > >>>>>>>>> is still ahead because of the head start.
> > > >>>>>>>>>
> > > >>>>>>>>> on and on. The turtle will cross the finish line before Achilles.
> > > >>>>>>>>
> > > >>>>>>>> The turtle will never cross the finish line but will always be ahead
> > > >>>>>>>> of Achilles.
> > > >>>>>>>
> > > >>>>>>> Yes. True. It gets infinitely closer and closer to the finish line.
> > > >>>>>>
> > > >>>>>> That is not Zeno's Paradox. The tortoise gets a head start, at point
> > > >>>>>> A[1]. The race starts. When Achilles reaches A[1], the tortoise has
> > > >>>>>> moved ahead somewhat, to what we call A[2]. When Achilles reaches A[2].
> > > >>>>>> the tortoise has reached A[3], at A[3] the tortoise is at A[4] and so
> > > >>>>>> forth.
> > > >>>>>>
> > > >>>>>> Since Achilles is faster than the tortoise, the distances A[1], A[2],
> > > >>>>>> [A3], ... get smaller and smaller, since the time it takes Achilles to
> > > >>>>>> run from the start to A[1] equals the time it takes the slower tortoise
> > > >>>>>> to run from A[1] to A[2], and so on.
> > > >>>>>>
> > > >>>>>> The paradox is, no matter how big n gets, A[n] (Achilles' position) is
> > > >>>>>> always behind A[n+1] (the tortoise's position), even as n approaches
> > > >>>>>> infinity. So Achilles can never beat the tortoise, right? But, as long
> > > >>>>>> as the head start isn't _too_ large, in real life, Achilles passes the
> > > >>>>>> tortoise and wins, just as you'd expect. So what's wrong with this?
> > > >>>>>>
> > > >>>>>> As I said, just because there's an infinite limit, it doesn't mean the
> > > >>>>>> limit is absolute. In this case, the total time passed also reaches a
> > > >>>>>> limit (at n=infinity) but that time limit isn't infinite, so what
> > > >>>>>> happens after the "limit" on time passes? As always, time marches on...
> > > >>>>>> At that point Achilles passes the tortoise and remains ahead for the
> > > >>>>>> rest of the race, and the infinite series no longer applies.
> > > >>>>>>>
> > > >>>>>>>
> > > >>>>>>>>> Now, if Achilles tells the tortoise to f-off and just starts
> > > >>>>>>>>> running, he
> > > >>>>>>>>> will quickly pass the tortoise...
> > > >>>>>>
> > > >>>>>> In real life, yes, but in Zeno's Paradox, no.
> > > >>>>>>>>>
> > > >>>>>>>>> ;^)
> > > >>>>>>>
> > > >>>>>>
> > > >>>>>> Extra credit: Given the speeds of Achilles S1 and the tortoise S2
> > > >>>>>> (S1>S2), and the head start distance A1, how long does it take for
> > > >>>>>> Achilles to pass the tortoise? :-)
> > > >>>>>>
> > > >>>>> I did some equations on this a while back:
> > > >>>>>
> > > >>>>> https://groups.google.com/g/sci.math/c/UKBgW2IOZkI/m/6tr-_qY-3DgJ
> > > >>>>>
> > > >>>>> Here are my comments:
> > > >>>>>
> > > >>>>> Iirc, scale was speed:
> > > >>>>> ____________________________
> > > >>>>> [...]
> > > >>>>> Ahhhh, now this is a direct formula:
> > > >>>>>
> > > >>>>> n = iteration count
> > > >>>>> d = distance
> > > >>>>> s = scale
> > > >>>>>
> > > >>>>> r_[n] = (d / s^n) * (s^n - (s-1)^n)
> > > >>>>>
> > > >>>>>
> > > >>>>> just might work for finding the total distance
> > > >>>>> traveled at a given iteration count of the following
> > > >>>>> iterated equation:
> > > >>>>>
> > > >>>>> r_[n+1] = r_[n] + (d - r_[n]) / s
> > > >>>>>
> > > >>>>>
> > > >>>>>
> > > >>>>> Here is the sequence for d = 10 and s = 4 using the
> > > >>>>> iterative formula:
> > > >>>>> __________________________________
> > > >>>>> r_[0] = 0
> > > >>>>> r_[1] = 0 + (10 - 0) / 4 = 2.5
> > > >>>>> r_[2] = 2.5 + (10 - 2.5) / 4 = 4.375
> > > >>>>> r_[3] = 4.375 + (10 - 4.375) / 4 = 5.78125
> > > >>>>> r_[4] = 5.78125 + (10 - 5.78125) / 4 = 6.8359375
> > > >>>>> __________________________________
> > > >>>>>
> > > >>>>>
> > > >>>>> And here is the sequence for d = 10 and s = 4 using
> > > >>>>> the direct formula:
> > > >>>>> __________________________________
> > > >>>>> r_[0] = 10 / 1 * 0 = 0
> > > >>>>> r_[1] = 10 / 4 * 1 = 2.5
> > > >>>>> r_[2] = 10 / 16 * 7 = 4.375
> > > >>>>> r_[3] = 10 / 64 * 37 = 5.78125
> > > >>>>> r_[4] = 10 / 256 * 175 = 6.8359375
> > > >>>>> __________________________________
> > > >>>>>
> > > >>>>>
> > > >>>>> As you can see, they are identical!
> > > >>>>>
> > > >>>>> Humm...
> > > >>>>> ____________________________
> > > >>>>>
> > > >>>>>
> > > >>>>> Here is another post:
> > > >>>>>
> > > >>>>> https://groups.google.com/g/sci.math/c/UKBgW2IOZkI/m/ysjxQWu9URMJ
> > > >>>>> ____________________________
> > > >>>>> I think I found a way to find the handicap of a
> > > >>>>> runner in an infinite race on a finite track...
> > > >>>>>
> > > >>>>> How about something like:
> > > >>>>>
> > > >>>>>
> > > >>>>> Let:
> > > >>>>>
> > > >>>>> d = total distance in track
> > > >>>>> s = scale, which relates to speed
> > > >>>>> n = integer iteration count, which relates to time
> > > >>>>> r_h = a runners starting handicap
> > > >>>>>
> > > >>>>>
> > > >>>>>
> > > >>>>> Here is the iterative equation for finding the
> > > >>>>> distance a runner is down the track that I posted
> > > >>>>> up thread:
> > > >>>>>
> > > >>>>> r_[n + 1] = r_[n] + (d - r_[n]) / s
> > > >>>>>
> > > >>>>>
> > > >>>>> The handicap of the runner is equal to r_[0]
> > > >>>>> because n = 0 is the starting position of every
> > > >>>>> runner.
> > > >>>>>
> > > >>>>> The goal is to find the handicap of a runner with
> > > >>>>> a given distance, iteration count, total distance
> > > >>>>> of the track, and a scale or speed. AFAICT, the
> > > >>>>> following formula solves for the handicap of a
> > > >>>>> runner using that information:
> > > >>>>>
> > > >>>>>
> > > >>>>> r_h = ((s-1) / s)^(-n) * ( (d * (s-1)^n * s^(-n) - d + r)
> > > >>>>>
> > > >>>>>
> > > >>>>>
> > > >>>>> Here is output of a racer using the iterative equation
> > > >>>>> with the following attributes:
> > > >>>>>
> > > >>>>> d = 10
> > > >>>>> s = 4
> > > >>>>> r_h = 6.8
> > > >>>>> _______________________________________
> > > >>>>> r_[0] = 6.8
> > > >>>>> r_[1] = 6.8 + (10 - 6.8) / 4 = 7.6
> > > >>>>> r_[2] = 7.6 + (10 - 7.6) / 4 = 8.2
> > > >>>>> r_[3] = 8.2 + (10 - 8.2) / 4 = 8.65
> > > >>>>> r_[4] = 8.65 + (10 - 8.65) / 4 = 8.9875
> > > >>>>> _______________________________________
> > > >>>>>
> > > >>>>>
> > > >>>>>
> > > >>>>> As we can see this runner has a head start of 6.8 out
> > > >>>>> of 10. Also, in the third frame, the runner r_[2] has
> > > >>>>> traveled 8.2 out of a possible 10.0.
> > > >>>>>
> > > >>>>> Given that information alone, we can plug it all into
> > > >>>>> the formula for finding the handicap, and get:
> > > >>>>>
> > > >>>>>
> > > >>>>> r_h = ((4-1) / 4)^(-2) * ((10 * (4-1)^2 * 4^(-2) - 10 + 8.2) = 6.8
> > > >>>>>
> > > >>>>>
> > > >>>>>
> > > >>>>> Bingo! We now know that the handicap for the runner
> > > >>>>> is 6.8 at n = 0 by information reaped in a later moment
> > > >>>>> in time when n = 2... Three frames later.
> > > >>>>>
> > > >>>>>
> > > >>>>> Is this Kosher?!?!
> > > >>>>>
> > > >>>>>
> > > >>>>>
> > > >>>>> :^o
> > > >>>>>
> > > >>>>> ____________________________
> > > >>>>
> > > >>>> If you add zero to .999 repeating you still get .999 repeating.
> > > >>>> Add the infinitely small and you get 1 instead.
> > > >>> .999 repeating = 1.000 repeating anyway
> > > >> Mitch, for that ".999... is add infinitesimal", just first
> > > >> have it that "1 minus infinitesimal, is, .999..., lesser".
> > > >
> > > > .999 is lesser than one by the infinitely small not zero.
> > > >
> > > > Mitchell Raemsch
> > > if you add an infinitesimal to 1 you get 1.000... repeating
> > No. You get above 1 by the infinitely small.
> >
> > Mitchell Raemsch
> Zero, ....

mitchr...@gmail.com wrote:

>> > > if you add an infinitesimal to 1 you get 1.000... repeating
>> > No. You get above 1 by the infinitely small.
>> >
>> > Mitchell Raemsch
>> Zero, ....
>
> What do you believe about zero math. Add it to .999 repeating and you
> get .999 repeating don't you?

nonsense, this is incorrect.

On Saturday, May 28, 2022 at 4:03:35 PM UTC-7, Atom Hatoyama wrote:
> mitchr...@gmail.com wrote:
>
> >> > > if you add an infinitesimal to 1 you get 1.000... repeating
> >> > No. You get above 1 by the infinitely small.
> >> >
> >> > Mitchell Raemsch
> >> Zero, ....
> >
> > What do you believe about zero math. Add it to .999 repeating and you
> > get .999 repeating don't you?
> nonsense, this is incorrect.

Then what would it add to you moron?
add zero to any quantity and that quantity stays.

Mitchell Raemsch

On 5/28/2022 10:14 PM, mitchr...@gmail.com wrote:
> On Saturday, May 28, 2022 at 4:03:35 PM UTC-7, Atom Hatoyama wrote:
>> mitchr...@gmail.com wrote:
>>
>>>>>> if you add an infinitesimal to 1 you get 1.000... repeating
>>>>> No. You get above 1 by the infinitely small.
>>>>>
>>>>> Mitchell Raemsch
>>>> Zero, ....
>>>
>>> What do you believe about zero math. Add it to .999 repeating and you
>>> get .999 repeating don't you?
>> nonsense, this is incorrect.
>
> Then what would it add to you moron?
> add zero to any quantity and that quantity stays.
>
Exactly, Roy! Add 0 to 0.999... and you get 1! That's because 0.999...=1.

You can also write 1+0=0.999... or 1+0=1, or 0.999...+0=0.999...

All of those are correct.

On 5/27/2022 6:26 PM, Mandy Ugaki wrote:

> "Ukrainian President Volodymyr Zelensky, foreground, and Polish President
> Andrzej Duda pictured in Kiev, Ukraine on May 22, 2022. © AP / Efrem
> Lukatsky"
> https://www.rt.com/russia/556220-ukraine-poland-merger-yanukovich/
>
> "Duda", what on earth *nazi_name* is that, Duda??
>
>
How is "Duda" a "Nazi name", nymshifter? Now if you want a Nazi name,
there's Putin. Now that's a *real* Nazi name, Putin.
Or just Putler.

On 5/28/2022 5:45 PM, mitchr...@gmail.com wrote:
> On Saturday, May 28, 2022 at 3:21:33 PM UTC-7, Ross A. Finlayson wrote:
>> On Friday, May 27, 2022 at 4:53:41 PM UTC-7, mitchr...@gmail.com wrote:
>>> On Friday, May 27, 2022 at 11:21:08 AM UTC-7, sergi o wrote:
>>>> On 5/27/2022 12:38 PM, mitchr...@gmail.com wrote:
>>>>> On Friday, May 27, 2022 at 10:23:29 AM UTC-7, Ross A. Finlayson wrote:
>>>>>> On Thursday, May 26, 2022 at 2:17:50 PM UTC-7, sergi o wrote:
>>>>>>> On 5/26/2022 3:47 PM, mitchr...@gmail.com wrote:
>>>>>>>> On Thursday, May 26, 2022 at 1:37:42 PM UTC-7, Chris M. Thomasson wrote:
>>>>>>>>> On 5/26/2022 1:25 PM, Michael Moroney wrote:
>>>>>>>>>> On 5/25/2022 11:49 PM, Chris M. Thomasson wrote:
>>>>>>>>>>> On 5/25/2022 7:21 PM, Dan joyce wrote:
>>>>>>>>>>>> On Monday, May 23, 2022 at 7:11:22 PM UTC-4, Chris M. Thomasson wrote:
>>>>>>>>>>>>> On 5/23/2022 1:25 PM, Michael Moroney wrote:
>>>>>>>>>>>>>> Just because in theory an infinite number of steps is required doesn't
>>>>>>>>>>>>>> mean the limit cannot be reached. Consider Zeno's Paradox where
>>>>>>>>>>>>>> Achilles
>>>>>>>>>>>>>> races a tortoise with a head start. Each time Achilles reaches a point
>>>>>>>>>>>>>> where the tortoise was, the tortoise advances somewhat. When Achilles
>>>>>>>>>>>>>> reaches that point, the tortoise advances more. And so on for an
>>>>>>>>>>>>>> infinite number of steps. Yet Achilles catches up to the tortoise and
>>>>>>>>>>>>>> passes it and wins the race, despite taking an infinite number of
>>>>>>>>>>>>>> steps
>>>>>>>>>>>>>> to catch up to the tortoise.
>>>>>>>>>>>>> If Achilles strictly plays by the tortoises rules on a step-by-step
>>>>>>>>>>>>> basis:
>>>>>>>>>>
>>>>>>>>>> What are "the tortoise's rules"? The only rules are the tortoise gets a
>>>>>>>>>> head start and both it and Achilles run as fast as they can to the
>>>>>>>>>> finish line, and whoever does so first, wins.
>>>>>>>>>>>>>
>>>>>>>>>>>>> step 1: tortoise moves one meter; Achilles moves one meter. The
>>>>>>>>>>>>> tortoise
>>>>>>>>>>>>> is ahead because of the head start.
>>>>>>>>>>>>>
>>>>>>>>>>>>> step 2: tortoise moves one meter; Achilles moves one meter. The
>>>>>>>>>>>>> tortoise
>>>>>>>>>>>>> is still ahead because of the head start.
>>>>>>>>>>>>>
>>>>>>>>>>>>> on and on. The turtle will cross the finish line before Achilles.
>>>>>>>>>>>>
>>>>>>>>>>>> The turtle will never cross the finish line but will always be ahead
>>>>>>>>>>>> of Achilles.
>>>>>>>>>>>
>>>>>>>>>>> Yes. True. It gets infinitely closer and closer to the finish line.
>>>>>>>>>>
>>>>>>>>>> That is not Zeno's Paradox. The tortoise gets a head start, at point
>>>>>>>>>> A[1]. The race starts. When Achilles reaches A[1], the tortoise has
>>>>>>>>>> moved ahead somewhat, to what we call A[2]. When Achilles reaches A[2].
>>>>>>>>>> the tortoise has reached A[3], at A[3] the tortoise is at A[4] and so
>>>>>>>>>> forth.
>>>>>>>>>>
>>>>>>>>>> Since Achilles is faster than the tortoise, the distances A[1], A[2],
>>>>>>>>>> [A3], ... get smaller and smaller, since the time it takes Achilles to
>>>>>>>>>> run from the start to A[1] equals the time it takes the slower tortoise
>>>>>>>>>> to run from A[1] to A[2], and so on.
>>>>>>>>>>
>>>>>>>>>> The paradox is, no matter how big n gets, A[n] (Achilles' position) is
>>>>>>>>>> always behind A[n+1] (the tortoise's position), even as n approaches
>>>>>>>>>> infinity. So Achilles can never beat the tortoise, right? But, as long
>>>>>>>>>> as the head start isn't _too_ large, in real life, Achilles passes the
>>>>>>>>>> tortoise and wins, just as you'd expect. So what's wrong with this?
>>>>>>>>>>
>>>>>>>>>> As I said, just because there's an infinite limit, it doesn't mean the
>>>>>>>>>> limit is absolute. In this case, the total time passed also reaches a
>>>>>>>>>> limit (at n=infinity) but that time limit isn't infinite, so what
>>>>>>>>>> happens after the "limit" on time passes? As always, time marches on...
>>>>>>>>>> At that point Achilles passes the tortoise and remains ahead for the
>>>>>>>>>> rest of the race, and the infinite series no longer applies.
>>>>>>>>>>>
>>>>>>>>>>>
>>>>>>>>>>>>> Now, if Achilles tells the tortoise to f-off and just starts
>>>>>>>>>>>>> running, he
>>>>>>>>>>>>> will quickly pass the tortoise...
>>>>>>>>>>
>>>>>>>>>> In real life, yes, but in Zeno's Paradox, no.
>>>>>>>>>>>>>
>>>>>>>>>>>>> ;^)
>>>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>> Extra credit: Given the speeds of Achilles S1 and the tortoise S2
>>>>>>>>>> (S1>S2), and the head start distance A1, how long does it take for
>>>>>>>>>> Achilles to pass the tortoise? :-)
>>>>>>>>>>
>>>>>>>>> I did some equations on this a while back:
>>>>>>>>>
>>>>>>>>> https://groups.google.com/g/sci.math/c/UKBgW2IOZkI/m/6tr-_qY-3DgJ
>>>>>>>>>
>>>>>>>>> Here are my comments:
>>>>>>>>>
>>>>>>>>> Iirc, scale was speed:
>>>>>>>>> ____________________________
>>>>>>>>> [...]
>>>>>>>>> Ahhhh, now this is a direct formula:
>>>>>>>>>
>>>>>>>>> n = iteration count
>>>>>>>>> d = distance
>>>>>>>>> s = scale
>>>>>>>>>
>>>>>>>>> r_[n] = (d / s^n) * (s^n - (s-1)^n)
>>>>>>>>>
>>>>>>>>>
>>>>>>>>> just might work for finding the total distance
>>>>>>>>> traveled at a given iteration count of the following
>>>>>>>>> iterated equation:
>>>>>>>>>
>>>>>>>>> r_[n+1] = r_[n] + (d - r_[n]) / s
>>>>>>>>>
>>>>>>>>>
>>>>>>>>>
>>>>>>>>> Here is the sequence for d = 10 and s = 4 using the
>>>>>>>>> iterative formula:
>>>>>>>>> __________________________________
>>>>>>>>> r_[0] = 0
>>>>>>>>> r_[1] = 0 + (10 - 0) / 4 = 2.5
>>>>>>>>> r_[2] = 2.5 + (10 - 2.5) / 4 = 4.375
>>>>>>>>> r_[3] = 4.375 + (10 - 4.375) / 4 = 5.78125
>>>>>>>>> r_[4] = 5.78125 + (10 - 5.78125) / 4 = 6.8359375
>>>>>>>>> __________________________________
>>>>>>>>>
>>>>>>>>>
>>>>>>>>> And here is the sequence for d = 10 and s = 4 using
>>>>>>>>> the direct formula:
>>>>>>>>> __________________________________
>>>>>>>>> r_[0] = 10 / 1 * 0 = 0
>>>>>>>>> r_[1] = 10 / 4 * 1 = 2.5
>>>>>>>>> r_[2] = 10 / 16 * 7 = 4.375
>>>>>>>>> r_[3] = 10 / 64 * 37 = 5.78125
>>>>>>>>> r_[4] = 10 / 256 * 175 = 6.8359375
>>>>>>>>> __________________________________
>>>>>>>>>
>>>>>>>>>
>>>>>>>>> As you can see, they are identical!
>>>>>>>>>
>>>>>>>>> Humm...
>>>>>>>>> ____________________________
>>>>>>>>>
>>>>>>>>>
>>>>>>>>> Here is another post:
>>>>>>>>>
>>>>>>>>> https://groups.google.com/g/sci.math/c/UKBgW2IOZkI/m/ysjxQWu9URMJ
>>>>>>>>> ____________________________
>>>>>>>>> I think I found a way to find the handicap of a
>>>>>>>>> runner in an infinite race on a finite track...
>>>>>>>>>
>>>>>>>>> How about something like:
>>>>>>>>>
>>>>>>>>>
>>>>>>>>> Let:
>>>>>>>>>
>>>>>>>>> d = total distance in track
>>>>>>>>> s = scale, which relates to speed
>>>>>>>>> n = integer iteration count, which relates to time
>>>>>>>>> r_h = a runners starting handicap
>>>>>>>>>
>>>>>>>>>
>>>>>>>>>
>>>>>>>>> Here is the iterative equation for finding the
>>>>>>>>> distance a runner is down the track that I posted
>>>>>>>>> up thread:
>>>>>>>>>
>>>>>>>>> r_[n + 1] = r_[n] + (d - r_[n]) / s
>>>>>>>>>
>>>>>>>>>
>>>>>>>>> The handicap of the runner is equal to r_[0]
>>>>>>>>> because n = 0 is the starting position of every
>>>>>>>>> runner.
>>>>>>>>>
>>>>>>>>> The goal is to find the handicap of a runner with
>>>>>>>>> a given distance, iteration count, total distance
>>>>>>>>> of the track, and a scale or speed. AFAICT, the
>>>>>>>>> following formula solves for the handicap of a
>>>>>>>>> runner using that information:
>>>>>>>>>
>>>>>>>>>
>>>>>>>>> r_h = ((s-1) / s)^(-n) * ( (d * (s-1)^n * s^(-n) - d + r)
>>>>>>>>>
>>>>>>>>>
>>>>>>>>>
>>>>>>>>> Here is output of a racer using the iterative equation
>>>>>>>>> with the following attributes:
>>>>>>>>>
>>>>>>>>> d = 10
>>>>>>>>> s = 4
>>>>>>>>> r_h = 6.8
>>>>>>>>> _______________________________________
>>>>>>>>> r_[0] = 6.8
>>>>>>>>> r_[1] = 6.8 + (10 - 6.8) / 4 = 7.6
>>>>>>>>> r_[2] = 7.6 + (10 - 7.6) / 4 = 8.2
>>>>>>>>> r_[3] = 8.2 + (10 - 8.2) / 4 = 8.65
>>>>>>>>> r_[4] = 8.65 + (10 - 8.65) / 4 = 8.9875
>>>>>>>>> _______________________________________
>>>>>>>>>
>>>>>>>>>
>>>>>>>>>
>>>>>>>>> As we can see this runner has a head start of 6.8 out
>>>>>>>>> of 10. Also, in the third frame, the runner r_[2] has
>>>>>>>>> traveled 8.2 out of a possible 10.0.
>>>>>>>>>
>>>>>>>>> Given that information alone, we can plug it all into
>>>>>>>>> the formula for finding the handicap, and get:
>>>>>>>>>
>>>>>>>>>
>>>>>>>>> r_h = ((4-1) / 4)^(-2) * ((10 * (4-1)^2 * 4^(-2) - 10 + 8.2) = 6.8
>>>>>>>>>
>>>>>>>>>
>>>>>>>>>
>>>>>>>>> Bingo! We now know that the handicap for the runner
>>>>>>>>> is 6.8 at n = 0 by information reaped in a later moment
>>>>>>>>> in time when n = 2... Three frames later.
>>>>>>>>>
>>>>>>>>>
>>>>>>>>> Is this Kosher?!?!
>>>>>>>>>
>>>>>>>>>
>>>>>>>>>
>>>>>>>>> :^o
>>>>>>>>>
>>>>>>>>> ____________________________
>>>>>>>>
>>>>>>>> If you add zero to .999 repeating you still get .999 repeating.
>>>>>>>> Add the infinitely small and you get 1 instead.
>>>>>>> .999 repeating = 1.000 repeating anyway
>>>>>> Mitch, for that ".999... is add infinitesimal", just first
>>>>>> have it that "1 minus infinitesimal, is, .999..., lesser".
>>>>>
>>>>> .999 is lesser than one by the infinitely small not zero.
>>>>>
>>>>> Mitchell Raemsch
>>>> if you add an infinitesimal to 1 you get 1.000... repeating
>>> No. You get above 1 by the infinitely small.
>>>
>>> Mitchell Raemsch
>> Zero, ....
>
> What do you believe about zero math. Add it to .999 repeating and
> you get .999 repeating don't you?

On 5/28/2022 9:14 PM, mitchr...@gmail.com wrote:
> On Saturday, May 28, 2022 at 4:03:35 PM UTC-7, Atom Hatoyama wrote:
>> mitchr...@gmail.com wrote:
>>
>>>>>> if you add an infinitesimal to 1 you get 1.000... repeating
>>>>> No. You get above 1 by the infinitely small.
>>>>>
>>>>> Mitchell Raemsch
>>>> Zero, ....
>>>
>>> What do you believe about zero math. Add it to .999 repeating and you
>>> get .999 repeating don't you?
>> nonsense, this is incorrect.
>
> Then what would it add to you moron?
> add zero to any quantity and that quantity stays.
>
> Mitchell Raemsch

what would your Roy Masters say ?

On Saturday, May 28, 2022 at 8:35:50 PM UTC-7, Michael Moroney wrote:
> On 5/28/2022 10:14 PM, mitchr...@gmail.com wrote:
> > On Saturday, May 28, 2022 at 4:03:35 PM UTC-7, Atom Hatoyama wrote:
> >> mitchr...@gmail.com wrote:
> >>
> >>>>>> if you add an infinitesimal to 1 you get 1.000... repeating
> >>>>> No. You get above 1 by the infinitely small.
> >>>>>
> >>>>> Mitchell Raemsch
> >>>> Zero, ....
> >>>
> >>> What do you believe about zero math. Add it to .999 repeating and you
> >>> get .999 repeating don't you?
> >> nonsense, this is incorrect.
> >
> > Then what would it add to you moron?
> > add zero to any quantity and that quantity stays.
> >
> Exactly, Roy! Add 0 to 0.999... and you get 1! That's because 0.999...=1.
>

Since when? Add zero to .999 repeating and you get .999 repeating.
Prove otherwise. You're the biggest moron now for that claim.

Mitchell Raemsch

On 5/29/2022 1:12 PM, mitchr...@gmail.com wrote:
> On Saturday, May 28, 2022 at 8:35:50 PM UTC-7, Michael Moroney wrote:
>> On 5/28/2022 10:14 PM, mitchr...@gmail.com wrote:
>>> On Saturday, May 28, 2022 at 4:03:35 PM UTC-7, Atom Hatoyama wrote:
>>>> mitchr...@gmail.com wrote:
>>>>
>>>>>>>> if you add an infinitesimal to 1 you get 1.000... repeating
>>>>>>> No. You get above 1 by the infinitely small.
>>>>>>>
>>>>>>> Mitchell Raemsch
>>>>>> Zero, ....
>>>>>
>>>>> What do you believe about zero math. Add it to .999 repeating and you
>>>>> get .999 repeating don't you?
>>>> nonsense, this is incorrect.
>>>
>>> Then what would it add to you moron?
>>> add zero to any quantity and that quantity stays.
>>>
>> Exactly, Roy! Add 0 to 0.999... and you get 1! That's because 0.999...=1.
>>
>
> Since when?

Since always, Roy. Since 0.999... = 1, in any valid mathematical
expression I can replace 1 with 0.999... and vice versa. Therefore,
there are 4 variations of the expression 1+0=1, each of them valid.
1+0=1
1+0=0.999...
0.999...+0=1
0.999...+0=0.999...

> Add zero to .999 repeating and you get .999 repeating.

Which is also known as "1".

> Prove otherwise.

It is your responsibility to prove your own claims.

The proofs that 0.999...=1 have been shown to you many, many, many
times. You just ignore them. But if you are actually interested, Google
has kept a record, so just look.

> You're the biggest moron now for that claim.

Hey, I'm not the one who is so stupid as to insist on something which
has been shown to be wrong many times.
>
> Mitchell Raemsch

On Sunday, May 29, 2022 at 11:14:40 AM UTC-7, Michael Moroney wrote:
> On 5/29/2022 1:12 PM, mitchr...@gmail.com wrote:
> > On Saturday, May 28, 2022 at 8:35:50 PM UTC-7, Michael Moroney wrote:
> >> On 5/28/2022 10:14 PM, mitchr...@gmail.com wrote:
> >>> On Saturday, May 28, 2022 at 4:03:35 PM UTC-7, Atom Hatoyama wrote:
> >>>> mitchr...@gmail.com wrote:
> >>>>
> >>>>>>>> if you add an infinitesimal to 1 you get 1.000... repeating
> >>>>>>> No. You get above 1 by the infinitely small.
> >>>>>>>
> >>>>>>> Mitchell Raemsch
> >>>>>> Zero, ....
> >>>>>
> >>>>> What do you believe about zero math. Add it to .999 repeating and you
> >>>>> get .999 repeating don't you?
> >>>> nonsense, this is incorrect.
> >>>
> >>> Then what would it add to you moron?
> >>> add zero to any quantity and that quantity stays.
> >>>
> >> Exactly, Roy! Add 0 to 0.999... and you get 1! That's because 0.999...=1.
> >>
> >
> > Since when?
> Since always, Roy. Since 0.999... = 1, in any valid mathematical

What is your proof that .999 repeating added to by zero is more
than .999 repeating? You need to add an infinitesimal to reach
the first integer.

On 5/29/2022 1:21 PM, mitchr...@gmail.com wrote:
> On Sunday, May 29, 2022 at 11:14:40 AM UTC-7, Michael Moroney wrote:
>> On 5/29/2022 1:12 PM, mitchr...@gmail.com wrote:
>>> On Saturday, May 28, 2022 at 8:35:50 PM UTC-7, Michael Moroney wrote:
>>>> On 5/28/2022 10:14 PM, mitchr...@gmail.com wrote:
>>>>> On Saturday, May 28, 2022 at 4:03:35 PM UTC-7, Atom Hatoyama wrote:
>>>>>> mitchr...@gmail.com wrote:
>>>>>>
>>>>>>>>>> if you add an infinitesimal to 1 you get 1.000... repeating
>>>>>>>>> No. You get above 1 by the infinitely small.
>>>>>>>>>
>>>>>>>>> Mitchell Raemsch
>>>>>>>> Zero, ....
>>>>>>>
>>>>>>> What do you believe about zero math. Add it to .999 repeating and you
>>>>>>> get .999 repeating don't you?
>>>>>> nonsense, this is incorrect.
>>>>>
>>>>> Then what would it add to you moron?
>>>>> add zero to any quantity and that quantity stays.
>>>>>
>>>> Exactly, Roy! Add 0 to 0.999... and you get 1! That's because 0.999...=1.
>>>>
>>>
>>> Since when?
>> Since always, Roy. Since 0.999... = 1, in any valid mathematical
>
> What is your proof that .999 repeating added to by zero is more
> than .999 repeating? You need to add an infinitesimal to reach
> the first integer.
>

how big an infinitesimal should be added ?

On 5/29/2022 2:21 PM, mitchr...@gmail.com wrote:
> On Sunday, May 29, 2022 at 11:14:40 AM UTC-7, Michael Moroney wrote:
>> On 5/29/2022 1:12 PM, mitchr...@gmail.com wrote:
>>> On Saturday, May 28, 2022 at 8:35:50 PM UTC-7, Michael Moroney wrote:
>>>> On 5/28/2022 10:14 PM, mitchr...@gmail.com wrote:
>>>>> On Saturday, May 28, 2022 at 4:03:35 PM UTC-7, Atom Hatoyama wrote:
>>>>>> mitchr...@gmail.com wrote:
>>>>>>
>>>>>>>>>> if you add an infinitesimal to 1 you get 1.000... repeating
>>>>>>>>> No. You get above 1 by the infinitely small.
>>>>>>>>>
>>>>>>>>> Mitchell Raemsch
>>>>>>>> Zero, ....
>>>>>>>
>>>>>>> What do you believe about zero math. Add it to .999 repeating and you
>>>>>>> get .999 repeating don't you?
>>>>>> nonsense, this is incorrect.
>>>>>
>>>>> Then what would it add to you moron?
>>>>> add zero to any quantity and that quantity stays.
>>>>>
>>>> Exactly, Roy! Add 0 to 0.999... and you get 1! That's because 0.999...=1.
>>>>
>>>
>>> Since when?
>> Since always, Roy. Since 0.999... = 1, in any valid mathematical
>
> What is your proof that .999 repeating added to by zero is more
> than .999 repeating?

It's not, Roy. 0.999... + 0 = 1, because 1) anything added to 0 is
itself; and 2) 0.999... = 1.

And, of course, you *still* haven't provided your proof that 0.999... is
less than 1 by "an infinitesimal". Remember, extraordinary claims
require extraordinary evidence.

> You need to add an infinitesimal to reach
> the first integer.

We are already at "the first integer". (you mean the first natural
number, of course, since 0, -1, -2, -999, -123456789 are all integers
"before" 1, of course. Why not just say 1?)
>

On Sunday, May 29, 2022 at 11:30:27 AM UTC-7, sergi o wrote:
> On 5/29/2022 1:21 PM, mitchr...@gmail.com wrote:
> > On Sunday, May 29, 2022 at 11:14:40 AM UTC-7, Michael Moroney wrote:
> >> On 5/29/2022 1:12 PM, mitchr...@gmail.com wrote:
> >>> On Saturday, May 28, 2022 at 8:35:50 PM UTC-7, Michael Moroney wrote:
> >>>> On 5/28/2022 10:14 PM, mitchr...@gmail.com wrote:
> >>>>> On Saturday, May 28, 2022 at 4:03:35 PM UTC-7, Atom Hatoyama wrote:
> >>>>>> mitchr...@gmail.com wrote:
> >>>>>>
> >>>>>>>>>> if you add an infinitesimal to 1 you get 1.000... repeating
> >>>>>>>>> No. You get above 1 by the infinitely small.
> >>>>>>>>>
> >>>>>>>>> Mitchell Raemsch
> >>>>>>>> Zero, ....
> >>>>>>>
> >>>>>>> What do you believe about zero math. Add it to .999 repeating and you
> >>>>>>> get .999 repeating don't you?
> >>>>>> nonsense, this is incorrect.
> >>>>>
> >>>>> Then what would it add to you moron?
> >>>>> add zero to any quantity and that quantity stays.
> >>>>>
> >>>> Exactly, Roy! Add 0 to 0.999... and you get 1! That's because 0.999...=1.
> >>>>
> >>>
> >>> Since when?
> >> Since always, Roy. Since 0.999... = 1, in any valid mathematical
> >
> > What is your proof that .999 repeating added to by zero is more
> > than .999 repeating? You need to add an infinitesimal to reach
> > the first integer.
> >
> how big an infinitesimal should be added ?

The fundamental or 1 divided by infinity...

On Sunday, May 29, 2022 at 11:42:26 AM UTC-7, Michael Moroney wrote:
> On 5/29/2022 2:21 PM, mitchr...@gmail.com wrote:
> > On Sunday, May 29, 2022 at 11:14:40 AM UTC-7, Michael Moroney wrote:
> >> On 5/29/2022 1:12 PM, mitchr...@gmail.com wrote:
> >>> On Saturday, May 28, 2022 at 8:35:50 PM UTC-7, Michael Moroney wrote:
> >>>> On 5/28/2022 10:14 PM, mitchr...@gmail.com wrote:
> >>>>> On Saturday, May 28, 2022 at 4:03:35 PM UTC-7, Atom Hatoyama wrote:
> >>>>>> mitchr...@gmail.com wrote:
> >>>>>>
> >>>>>>>>>> if you add an infinitesimal to 1 you get 1.000... repeating
> >>>>>>>>> No. You get above 1 by the infinitely small.
> >>>>>>>>>
> >>>>>>>>> Mitchell Raemsch
> >>>>>>>> Zero, ....
> >>>>>>>
> >>>>>>> What do you believe about zero math. Add it to .999 repeating and you
> >>>>>>> get .999 repeating don't you?
> >>>>>> nonsense, this is incorrect.
> >>>>>
> >>>>> Then what would it add to you moron?
> >>>>> add zero to any quantity and that quantity stays.
> >>>>>
> >>>> Exactly, Roy! Add 0 to 0.999... and you get 1! That's because 0.999...=1.
> >>>>
> >>>
> >>> Since when?
> >> Since always, Roy. Since 0.999... = 1, in any valid mathematical
> >
> > What is your proof that .999 repeating added to by zero is more
> > than .999 repeating?
> It's not, Roy. 0.999... + 0 = 1, because 1) anything added to 0 is
> itself; and 2) 0.999... = 1.

No. Add zero to any quantity and that quantity stays the same.
Add first fundamental quantity and there is a difference.
The difference shows .999 repeating and 1 are not the same
quantity. The equality they share has an infinitesimal
difference not zero. They balance on the algebraic scale.

Mitchell Raemsch

On 5/29/2022 3:23 PM, mitchr...@gmail.com wrote:
> On Sunday, May 29, 2022 at 11:42:26 AM UTC-7, Michael Moroney wrote:
>> On 5/29/2022 2:21 PM, mitchr...@gmail.com wrote:
>>> On Sunday, May 29, 2022 at 11:14:40 AM UTC-7, Michael Moroney wrote:
>>>> On 5/29/2022 1:12 PM, mitchr...@gmail.com wrote:
>>>>> On Saturday, May 28, 2022 at 8:35:50 PM UTC-7, Michael Moroney wrote:
>>>>>> On 5/28/2022 10:14 PM, mitchr...@gmail.com wrote:
>>>>>>> On Saturday, May 28, 2022 at 4:03:35 PM UTC-7, Atom Hatoyama wrote:
>>>>>>>> mitchr...@gmail.com wrote:
>>>>>>>>
>>>>>>>>>>>> if you add an infinitesimal to 1 you get 1.000... repeating
>>>>>>>>>>> No. You get above 1 by the infinitely small.
>>>>>>>>>>>
>>>>>>>>>>> Mitchell Raemsch
>>>>>>>>>> Zero, ....
>>>>>>>>>
>>>>>>>>> What do you believe about zero math. Add it to .999 repeating and you
>>>>>>>>> get .999 repeating don't you?
>>>>>>>> nonsense, this is incorrect.
>>>>>>>
>>>>>>> Then what would it add to you moron?
>>>>>>> add zero to any quantity and that quantity stays.
>>>>>>>
>>>>>> Exactly, Roy! Add 0 to 0.999... and you get 1! That's because 0.999...=1.
>>>>>>
>>>>>
>>>>> Since when?
>>>> Since always, Roy. Since 0.999... = 1, in any valid mathematical
>>>
>>> What is your proof that .999 repeating added to by zero is more
>>> than .999 repeating?
>> It's not, Roy. 0.999... + 0 = 1, because 1) anything added to 0 is
>> itself; and 2) 0.999... = 1.
>
> No. Add zero to any quantity and that quantity stays the same.

That's what everyone is trying to tell you, Roy. That's why
0.999... + 0 = 1. Because anything added to 0 is the same, unchanged in
value.

Michael Moroney wrote:

>>> Exactly, Roy! Add 0 to 0.999... and you get 1! That's because
>>> 0.999...=1.
>>>
>>>
>> Since when?
>
> Since always, Roy. Since 0.999... = 1, in any valid mathematical
> expression I can replace 1 with 0.999... and vice versa. Therefore,
> there are 4 variations of the expression 1+0=1, each of them valid.
> 1+0=1 1+0=0.999...

idiot, 0.(9) never converges. You don't know what a 0.(9) stands for. This
is NOT relativity. This is my *_Divergent_Matter_of_the_Moving_Objects_*
model.

Michael Moroney wrote:

> On 5/29/2022 2:21 PM, mitchr...@gmail.com wrote:
>> What is your proof that .999 repeating added to by zero is more than
>> .999 repeating?
>
> It's not, Roy. 0.999... + 0 = 1, because 1) anything added to 0 is
> itself; and 2) 0.999... = 1.

which country would be that?

Michael Moroney wrote:

> On 5/29/2022 3:23 PM, mitchr...@gmail.com wrote:
>>> It's not, Roy. 0.999... + 0 = 1, because 1) anything added to 0 is
>>> itself; and 2) 0.999... = 1.
>>
>> No. Add zero to any quantity and that quantity stays the same.
>
> That's what everyone is trying to tell you, Roy. That's why 0.999... + 0
> = 1. Because anything added to 0 is the same, unchanged in value.

this usenet user doesn't know what a *stay_the_same* stands for in
physics. You can't pop from 0.(9) to 1.0 my friend. Strict forbidden
everywhere, especially in *Math_Modelling* and *Scientific_Computation*.

On 5/29/2022 4:42 PM, Blain Nakada wrote:
> Michael Moroney wrote:
>
>> On 5/29/2022 2:21 PM, mitchr...@gmail.com wrote:
>>> What is your proof that .999 repeating added to by zero is more than
>>> .999 repeating?
>>
>> It's not, Roy. 0.999... + 0 = 1, because 1) anything added to 0 is
>> itself; and 2) 0.999... = 1.
>
> which country would be that?

probably British, they are screwy in so many ways

On Sunday, May 29, 2022 at 2:39:02 PM UTC-7, Michael Moroney wrote:
> On 5/29/2022 3:23 PM, mitchr...@gmail.com wrote:
> > On Sunday, May 29, 2022 at 11:42:26 AM UTC-7, Michael Moroney wrote:
> >> On 5/29/2022 2:21 PM, mitchr...@gmail.com wrote:
> >>> On Sunday, May 29, 2022 at 11:14:40 AM UTC-7, Michael Moroney wrote:
> >>>> On 5/29/2022 1:12 PM, mitchr...@gmail.com wrote:
> >>>>> On Saturday, May 28, 2022 at 8:35:50 PM UTC-7, Michael Moroney wrote:
> >>>>>> On 5/28/2022 10:14 PM, mitchr...@gmail.com wrote:
> >>>>>>> On Saturday, May 28, 2022 at 4:03:35 PM UTC-7, Atom Hatoyama wrote:
> >>>>>>>> mitchr...@gmail.com wrote:
> >>>>>>>>
> >>>>>>>>>>>> if you add an infinitesimal to 1 you get 1.000... repeating
> >>>>>>>>>>> No. You get above 1 by the infinitely small.
> >>>>>>>>>>>
> >>>>>>>>>>> Mitchell Raemsch
> >>>>>>>>>> Zero, ....
> >>>>>>>>>
> >>>>>>>>> What do you believe about zero math. Add it to .999 repeating and you
> >>>>>>>>> get .999 repeating don't you?
> >>>>>>>> nonsense, this is incorrect.
> >>>>>>>
> >>>>>>> Then what would it add to you moron?
> >>>>>>> add zero to any quantity and that quantity stays.
> >>>>>>>
> >>>>>> Exactly, Roy! Add 0 to 0.999... and you get 1! That's because 0.999...=1.
> >>>>>>
> >>>>>
> >>>>> Since when?
> >>>> Since always, Roy. Since 0.999... = 1, in any valid mathematical
> >>>
> >>> What is your proof that .999 repeating added to by zero is more
> >>> than .999 repeating?
> >> It's not, Roy. 0.999... + 0 = 1, because 1) anything added to 0 is
> >> itself; and 2) 0.999... = 1.
> >
> > No. Add zero to any quantity and that quantity stays the same.
> That's what everyone is trying to tell you, Roy. That's why
> 0.999... + 0 = 1. Because anything added to 0 is the same, unchanged in
> value.

No. Add zero to any quantity and it stays the same.

On 5/29/2022 8:11 PM, mitchr...@gmail.com wrote:
> On Sunday, May 29, 2022 at 2:39:02 PM UTC-7, Michael Moroney wrote:
>> On 5/29/2022 3:23 PM, mitchr...@gmail.com wrote:
>>> On Sunday, May 29, 2022 at 11:42:26 AM UTC-7, Michael Moroney wrote:
>>>> On 5/29/2022 2:21 PM, mitchr...@gmail.com wrote:
>>>>> On Sunday, May 29, 2022 at 11:14:40 AM UTC-7, Michael Moroney wrote:
>>>>>> On 5/29/2022 1:12 PM, mitchr...@gmail.com wrote:
>>>>>>> On Saturday, May 28, 2022 at 8:35:50 PM UTC-7, Michael Moroney wrote:
>>>>>>>> On 5/28/2022 10:14 PM, mitchr...@gmail.com wrote:
>>>>>>>>> On Saturday, May 28, 2022 at 4:03:35 PM UTC-7, Atom Hatoyama wrote:
>>>>>>>>>> mitchr...@gmail.com wrote:
>>>>>>>>>>
>>>>>>>>>>>>>> if you add an infinitesimal to 1 you get 1.000... repeating
>>>>>>>>>>>>> No. You get above 1 by the infinitely small.
>>>>>>>>>>>>>
>>>>>>>>>>>>> Mitchell Raemsch
>>>>>>>>>>>> Zero, ....
>>>>>>>>>>>
>>>>>>>>>>> What do you believe about zero math. Add it to .999 repeating and you
>>>>>>>>>>> get .999 repeating don't you?
>>>>>>>>>> nonsense, this is incorrect.
>>>>>>>>>
>>>>>>>>> Then what would it add to you moron?
>>>>>>>>> add zero to any quantity and that quantity stays.
>>>>>>>>>
>>>>>>>> Exactly, Roy! Add 0 to 0.999... and you get 1! That's because 0.999...=1.
>>>>>>>>
>>>>>>>
>>>>>>> Since when?
>>>>>> Since always, Roy. Since 0.999... = 1, in any valid mathematical
>>>>>
>>>>> What is your proof that .999 repeating added to by zero is more
>>>>> than .999 repeating?
>>>> It's not, Roy. 0.999... + 0 = 1, because 1) anything added to 0 is
>>>> itself; and 2) 0.999... = 1.
>>>
>>> No. Add zero to any quantity and that quantity stays the same.
>> That's what everyone is trying to tell you, Roy. That's why
>> 0.999... + 0 = 1. Because anything added to 0 is the same, unchanged in
>> value.
>
> No. Add zero to any quantity and it stays the same.

0 + Roy = 0 + Masters

On Sunday, May 29, 2022 at 6:33:17 PM UTC-7, sergi o wrote:
> On 5/29/2022 8:11 PM, mitchr...@gmail.com wrote:
> > On Sunday, May 29, 2022 at 2:39:02 PM UTC-7, Michael Moroney wrote:
> >> On 5/29/2022 3:23 PM, mitchr...@gmail.com wrote:
> >>> On Sunday, May 29, 2022 at 11:42:26 AM UTC-7, Michael Moroney wrote:
> >>>> On 5/29/2022 2:21 PM, mitchr...@gmail.com wrote:
> >>>>> On Sunday, May 29, 2022 at 11:14:40 AM UTC-7, Michael Moroney wrote:
> >>>>>> On 5/29/2022 1:12 PM, mitchr...@gmail.com wrote:
> >>>>>>> On Saturday, May 28, 2022 at 8:35:50 PM UTC-7, Michael Moroney wrote:
> >>>>>>>> On 5/28/2022 10:14 PM, mitchr...@gmail.com wrote:
> >>>>>>>>> On Saturday, May 28, 2022 at 4:03:35 PM UTC-7, Atom Hatoyama wrote:
> >>>>>>>>>> mitchr...@gmail.com wrote:
> >>>>>>>>>>
> >>>>>>>>>>>>>> if you add an infinitesimal to 1 you get 1.000... repeating
> >>>>>>>>>>>>> No. You get above 1 by the infinitely small.
> >>>>>>>>>>>>>
> >>>>>>>>>>>>> Mitchell Raemsch
> >>>>>>>>>>>> Zero, ....
> >>>>>>>>>>>
> >>>>>>>>>>> What do you believe about zero math. Add it to .999 repeating and you
> >>>>>>>>>>> get .999 repeating don't you?
> >>>>>>>>>> nonsense, this is incorrect.
> >>>>>>>>>
> >>>>>>>>> Then what would it add to you moron?
> >>>>>>>>> add zero to any quantity and that quantity stays.
> >>>>>>>>>
> >>>>>>>> Exactly, Roy! Add 0 to 0.999... and you get 1! That's because 0.999...=1.
> >>>>>>>>
> >>>>>>>
> >>>>>>> Since when?
> >>>>>> Since always, Roy. Since 0.999... = 1, in any valid mathematical
> >>>>>
> >>>>> What is your proof that .999 repeating added to by zero is more
> >>>>> than .999 repeating?
> >>>> It's not, Roy. 0.999... + 0 = 1, because 1) anything added to 0 is
> >>>> itself; and 2) 0.999... = 1.
> >>>
> >>> No. Add zero to any quantity and that quantity stays the same.
> >> That's what everyone is trying to tell you, Roy. That's why
> >> 0.999... + 0 = 1. Because anything added to 0 is the same, unchanged in
> >> value.
> >
> > No. Add zero to any quantity and it stays the same.
> 0 + Roy = 0 + Masters

.... ants.

"0+0 Roy Masters ants"

Sergio, these are Mitch's ants.

Of course ants is one of the greatest contributions of this newsgroups
over several years.

The idea is there is an infinite chain of ants between zero and one,
ants at either end and ants all in between, single file, and after the
very first ant at zero is the next ant at "1 minus .999, lesser", so on
continuing through infinitely many ants at which point then there is
an end of the ants.

Here, then these are more "1 minus 1" ants than "0 + 0".

On 5/29/2022 9:11 PM, mitchr...@gmail.com wrote:
> On Sunday, May 29, 2022 at 2:39:02 PM UTC-7, Michael Moroney wrote:
>> On 5/29/2022 3:23 PM, mitchr...@gmail.com wrote:
>>> On Sunday, May 29, 2022 at 11:42:26 AM UTC-7, Michael Moroney wrote:
>>>> On 5/29/2022 2:21 PM, mitchr...@gmail.com wrote:
>>>>> On Sunday, May 29, 2022 at 11:14:40 AM UTC-7, Michael Moroney wrote:
>>>>>> On 5/29/2022 1:12 PM, mitchr...@gmail.com wrote:
>>>>>>> On Saturday, May 28, 2022 at 8:35:50 PM UTC-7, Michael Moroney wrote:
>>>>>>>> On 5/28/2022 10:14 PM, mitchr...@gmail.com wrote:
>>>>>>>>> On Saturday, May 28, 2022 at 4:03:35 PM UTC-7, Atom Hatoyama wrote:
>>>>>>>>>> mitchr...@gmail.com wrote:
>>>>>>>>>>
>>>>>>>>>>>>>> if you add an infinitesimal to 1 you get 1.000... repeating
>>>>>>>>>>>>> No. You get above 1 by the infinitely small.
>>>>>>>>>>>>>
>>>>>>>>>>>>> Mitchell Raemsch
>>>>>>>>>>>> Zero, ....
>>>>>>>>>>>
>>>>>>>>>>> What do you believe about zero math. Add it to .999 repeating and you
>>>>>>>>>>> get .999 repeating don't you?
>>>>>>>>>> nonsense, this is incorrect.
>>>>>>>>>
>>>>>>>>> Then what would it add to you moron?
>>>>>>>>> add zero to any quantity and that quantity stays.
>>>>>>>>>
>>>>>>>> Exactly, Roy! Add 0 to 0.999... and you get 1! That's because 0.999...=1.
>>>>>>>>
>>>>>>>
>>>>>>> Since when?
>>>>>> Since always, Roy. Since 0.999... = 1, in any valid mathematical
>>>>>
>>>>> What is your proof that .999 repeating added to by zero is more
>>>>> than .999 repeating?
>>>> It's not, Roy. 0.999... + 0 = 1, because 1) anything added to 0 is
>>>> itself; and 2) 0.999... = 1.
>>>
>>> No. Add zero to any quantity and that quantity stays the same.
>> That's what everyone is trying to tell you, Roy. That's why
>> 0.999... + 0 = 1. Because anything added to 0 is the same, unchanged in
>> value.
>
> No. Add zero to any quantity and it stays the same.

That's right, Roy! 1 + 0 = 0.999...!
Just like 1/2 + 0 = 2/4.

Two different representations of the same number are still the same
number. And anything + 0 is unchanged in value.

Michael Moroney wrote:

>> https://www.rt.com/russia/556220-ukraine-poland-merger-yanukovich/
>> "Duda", what on earth *nazi_name* is that, Duda??
>>
> How is "Duda" a "Nazi name", nymshifter? Now if you want a Nazi name,
> there's Putin. Now that's a *real* Nazi name, Putin.
> Or just Putler.

for the mathematicians with trouble understanding
*_The_Divergent_Matter_of_the_Moving_Objects_* model

PFIZER AIMED FOR WORLD POPULATION REDUCTION BY50%
https://www.brighteon.com/dc0885f6-fa5e-4169-a843-7e152355a76b

fredag 27 maj 2022 kl. 19:38:33 UTC+2 skrev mitchr...@gmail.com:
> On Friday, May 27, 2022 at 10:23:29 AM UTC-7, Ross A. Finlayson wrote:
> > On Thursday, May 26, 2022 at 2:17:50 PM UTC-7, sergi o wrote:
> > > On 5/26/2022 3:47 PM, mitchr...@gmail.com wrote:
> > > > On Thursday, May 26, 2022 at 1:37:42 PM UTC-7, Chris M. Thomasson wrote:
> > > >> On 5/26/2022 1:25 PM, Michael Moroney wrote:
> > > >>> On 5/25/2022 11:49 PM, Chris M. Thomasson wrote:
> > > >>>> On 5/25/2022 7:21 PM, Dan joyce wrote:
> > > >>>>> On Monday, May 23, 2022 at 7:11:22 PM UTC-4, Chris M. Thomasson wrote:
> > > >>>>>> On 5/23/2022 1:25 PM, Michael Moroney wrote:
> > > >>>>>>> Just because in theory an infinite number of steps is required doesn't
> > > >>>>>>> mean the limit cannot be reached. Consider Zeno's Paradox where
> > > >>>>>>> Achilles
> > > >>>>>>> races a tortoise with a head start. Each time Achilles reaches a point
> > > >>>>>>> where the tortoise was, the tortoise advances somewhat. When Achilles
> > > >>>>>>> reaches that point, the tortoise advances more. And so on for an
> > > >>>>>>> infinite number of steps. Yet Achilles catches up to the tortoise and
> > > >>>>>>> passes it and wins the race, despite taking an infinite number of
> > > >>>>>>> steps
> > > >>>>>>> to catch up to the tortoise.
> > > >>>>>> If Achilles strictly plays by the tortoises rules on a step-by-step
> > > >>>>>> basis:
> > > >>>
> > > >>> What are "the tortoise's rules"? The only rules are the tortoise gets a
> > > >>> head start and both it and Achilles run as fast as they can to the
> > > >>> finish line, and whoever does so first, wins.
> > > >>>>>>
> > > >>>>>> step 1: tortoise moves one meter; Achilles moves one meter. The
> > > >>>>>> tortoise
> > > >>>>>> is ahead because of the head start.
> > > >>>>>>
> > > >>>>>> step 2: tortoise moves one meter; Achilles moves one meter. The
> > > >>>>>> tortoise
> > > >>>>>> is still ahead because of the head start.
> > > >>>>>>
> > > >>>>>> on and on. The turtle will cross the finish line before Achilles.
> > > >>>>>
> > > >>>>> The turtle will never cross the finish line but will always be ahead
> > > >>>>> of Achilles.
> > > >>>>
> > > >>>> Yes. True. It gets infinitely closer and closer to the finish line.
> > > >>>
> > > >>> That is not Zeno's Paradox. The tortoise gets a head start, at point
> > > >>> A[1]. The race starts. When Achilles reaches A[1], the tortoise has
> > > >>> moved ahead somewhat, to what we call A[2]. When Achilles reaches A[2].
> > > >>> the tortoise has reached A[3], at A[3] the tortoise is at A[4] and so
> > > >>> forth.
> > > >>>
> > > >>> Since Achilles is faster than the tortoise, the distances A[1], A[2],
> > > >>> [A3], ... get smaller and smaller, since the time it takes Achilles to
> > > >>> run from the start to A[1] equals the time it takes the slower tortoise
> > > >>> to run from A[1] to A[2], and so on.
> > > >>>
> > > >>> The paradox is, no matter how big n gets, A[n] (Achilles' position) is
> > > >>> always behind A[n+1] (the tortoise's position), even as n approaches
> > > >>> infinity. So Achilles can never beat the tortoise, right? But, as long
> > > >>> as the head start isn't _too_ large, in real life, Achilles passes the
> > > >>> tortoise and wins, just as you'd expect. So what's wrong with this?
> > > >>>
> > > >>> As I said, just because there's an infinite limit, it doesn't mean the
> > > >>> limit is absolute. In this case, the total time passed also reaches a
> > > >>> limit (at n=infinity) but that time limit isn't infinite, so what
> > > >>> happens after the "limit" on time passes? As always, time marches on...
> > > >>> At that point Achilles passes the tortoise and remains ahead for the
> > > >>> rest of the race, and the infinite series no longer applies.
> > > >>>>
> > > >>>>
> > > >>>>>> Now, if Achilles tells the tortoise to f-off and just starts
> > > >>>>>> running, he
> > > >>>>>> will quickly pass the tortoise...
> > > >>>
> > > >>> In real life, yes, but in Zeno's Paradox, no.
> > > >>>>>>
> > > >>>>>> ;^)
> > > >>>>
> > > >>>
> > > >>> Extra credit: Given the speeds of Achilles S1 and the tortoise S2
> > > >>> (S1>S2), and the head start distance A1, how long does it take for
> > > >>> Achilles to pass the tortoise? :-)
> > > >>>
> > > >> I did some equations on this a while back:
> > > >>
> > > >> https://groups.google.com/g/sci.math/c/UKBgW2IOZkI/m/6tr-_qY-3DgJ
> > > >>
> > > >> Here are my comments:
> > > >>
> > > >> Iirc, scale was speed:
> > > >> ____________________________
> > > >> [...]
> > > >> Ahhhh, now this is a direct formula:
> > > >>
> > > >> n = iteration count
> > > >> d = distance
> > > >> s = scale
> > > >>
> > > >> r_[n] = (d / s^n) * (s^n - (s-1)^n)
> > > >>
> > > >>
> > > >> just might work for finding the total distance
> > > >> traveled at a given iteration count of the following
> > > >> iterated equation:
> > > >>
> > > >> r_[n+1] = r_[n] + (d - r_[n]) / s
> > > >>
> > > >>
> > > >>
> > > >> Here is the sequence for d = 10 and s = 4 using the
> > > >> iterative formula:
> > > >> __________________________________
> > > >> r_[0] = 0
> > > >> r_[1] = 0 + (10 - 0) / 4 = 2.5
> > > >> r_[2] = 2.5 + (10 - 2.5) / 4 = 4.375
> > > >> r_[3] = 4.375 + (10 - 4.375) / 4 = 5.78125
> > > >> r_[4] = 5.78125 + (10 - 5.78125) / 4 = 6.8359375
> > > >> __________________________________
> > > >>
> > > >>
> > > >> And here is the sequence for d = 10 and s = 4 using
> > > >> the direct formula:
> > > >> __________________________________
> > > >> r_[0] = 10 / 1 * 0 = 0
> > > >> r_[1] = 10 / 4 * 1 = 2.5
> > > >> r_[2] = 10 / 16 * 7 = 4.375
> > > >> r_[3] = 10 / 64 * 37 = 5.78125
> > > >> r_[4] = 10 / 256 * 175 = 6.8359375
> > > >> __________________________________
> > > >>
> > > >>
> > > >> As you can see, they are identical!
> > > >>
> > > >> Humm...
> > > >> ____________________________
> > > >>
> > > >>
> > > >> Here is another post:
> > > >>
> > > >> https://groups.google.com/g/sci.math/c/UKBgW2IOZkI/m/ysjxQWu9URMJ
> > > >> ____________________________
> > > >> I think I found a way to find the handicap of a
> > > >> runner in an infinite race on a finite track...
> > > >>
> > > >> How about something like:
> > > >>
> > > >>
> > > >> Let:
> > > >>
> > > >> d = total distance in track
> > > >> s = scale, which relates to speed
> > > >> n = integer iteration count, which relates to time
> > > >> r_h = a runners starting handicap
> > > >>
> > > >>
> > > >>
> > > >> Here is the iterative equation for finding the
> > > >> distance a runner is down the track that I posted
> > > >> up thread:
> > > >>
> > > >> r_[n + 1] = r_[n] + (d - r_[n]) / s
> > > >>
> > > >>
> > > >> The handicap of the runner is equal to r_[0]
> > > >> because n = 0 is the starting position of every
> > > >> runner.
> > > >>
> > > >> The goal is to find the handicap of a runner with
> > > >> a given distance, iteration count, total distance
> > > >> of the track, and a scale or speed. AFAICT, the
> > > >> following formula solves for the handicap of a
> > > >> runner using that information:
> > > >>
> > > >>
> > > >> r_h = ((s-1) / s)^(-n) * ( (d * (s-1)^n * s^(-n) - d + r)
> > > >>
> > > >>
> > > >>
> > > >> Here is output of a racer using the iterative equation
> > > >> with the following attributes:
> > > >>
> > > >> d = 10
> > > >> s = 4
> > > >> r_h = 6.8
> > > >> _______________________________________
> > > >> r_[0] = 6.8
> > > >> r_[1] = 6.8 + (10 - 6.8) / 4 = 7.6
> > > >> r_[2] = 7.6 + (10 - 7.6) / 4 = 8.2
> > > >> r_[3] = 8.2 + (10 - 8.2) / 4 = 8.65
> > > >> r_[4] = 8.65 + (10 - 8.65) / 4 = 8.9875
> > > >> _______________________________________
> > > >>
> > > >>
> > > >>
> > > >> As we can see this runner has a head start of 6.8 out
> > > >> of 10. Also, in the third frame, the runner r_[2] has
> > > >> traveled 8.2 out of a possible 10.0.
> > > >>
> > > >> Given that information alone, we can plug it all into
> > > >> the formula for finding the handicap, and get:
> > > >>
> > > >>
> > > >> r_h = ((4-1) / 4)^(-2) * ((10 * (4-1)^2 * 4^(-2) - 10 + 8.2) = 6.8
> > > >>
> > > >>
> > > >>
> > > >> Bingo! We now know that the handicap for the runner
> > > >> is 6.8 at n = 0 by information reaped in a later moment
> > > >> in time when n = 2... Three frames later.
> > > >>
> > > >>
> > > >> Is this Kosher?!?!
> > > >>
> > > >>
> > > >>
> > > >> :^o
> > > >>
> > > >> ____________________________
> > > >
> > > > If you add zero to .999 repeating you still get .999 repeating.
> > > > Add the infinitely small and you get 1 instead.
> > > .999 repeating = 1.000 repeating anyway
> > Mitch, for that ".999... is add infinitesimal", just first
> > have it that "1 minus infinitesimal, is, .999..., lesser".
> .999 is lesser than one by the infinitely small not zero.
>
> Mitchell Raemsch
> >
> > Then though it's always that "the .999..., lesser, is
> > only on its way to zero, least or none", because there
> > are two kinds of relations: related motion and lattice
> > relations, that the field defines lattice relations while
> > the infinitesimals is only part of a "range" or "course".
> >
> > I.e., the infinitesimal changes between 1.0 and 0.0,
> > going through each .aaa... as far as it could be measured,
> > are instead of that "this .333... times 3 = .999... = 1", that
> > this "1 minus .000...1" is writing out a notation, where
> > the ...1's "sum their differences, to zero", while the numbers,
> > "round up".
> >
> > So, when someone writes ".999, ..., repeating", is mostly
> > reflecting the notion that the notation after numbers introducing
> > the "..." or over-bar or the usual way of indicating the
> > repeating part for any rational number, basically works from
> > the field of course that _all_ and _only_ rational numbers,
> > end with a repeating terminus.
> >
> > Then there's only that
> >
> > 000... <- 0
> > 000...
> >
> > 011...
> > 011... <- 1/2
> > 100...
> >
> > 111...
> > 111... <- 1
> >
> > Notice the bounds are only at the ends,
> > and each column is half 1's and half 0's.
> >
> > It's easier to reduce the discussion to [0,1] instead of
> > involving all the real numbers.
There are no infinitesimals in real numbers. The real numbers are archimedian. I have told you this