On Monday, 18 July 2022 at 17:53:39 UTC+3, WM wrote:
> William schrieb am Montag, 18. Juli 2022 um 16:33:18 UTC+2:
> > On Monday, July 18, 2022 at 10:17:11 AM UTC-3, WM wrote:
> > > William schrieb am Sonntag, 17. Juli 2022 um 22:54:16 UTC+2:
> > > > On Sunday, July 17, 2022 at 3:28:42 PM UTC-3, WM wrote:
> > > > > William schrieb am Sonntag, 17. Juli 2022 um 17:08:29 UTC+2:
> > > > > > On Sunday, July 17, 2022 at 11:42:23 AM UTC-3, WM wrote:
> > > > > > > William schrieb am Samstag, 16. Juli 2022 um 18:16:45 UTC+2:
> > > > > > > > On Saturday, July 16, 2022 at 12:56:46 PM UTC-3, WM wrote:
> > > > > > >
> > > > > > > > > > A trivial induction shows that each element (without exception) of |N has finitely many predecessors and infinitely many successors.
> > > > > > > > > But according to Cantor all natural numbers with no successors
> > > > > > > > Piffle, there is no such thing as a natural number with no successors.
> > > > > > > Therefore not all natural numbers are individually definable. The successors
> > > > > > each (without exception) has finitely many predecessors.
> > > > > Are all successors in the bijection with Q such that none remains?
> > > > Yes. Simple induction shows that each successor (without exception) is in the bijection.
> > > 5 lines above there was no such thing as a natural number with no successors.
> > Correct.
> > >Now they are in the bijection without successors?
> > Nope, "none remains" does not imply there is an element in the bijection without a successor.
> It implies that all elements are in the bijection and these elements have no successors outside of the bijection.
> > There is no last element. Simple induction shows that 1. each element (without exception) is in the bijection
> So we can take them (the elements of the bijction) out, and nothing will remain.
> > and 2. each element (without exception) has a successor.
> Even infinitely many. So we can take them (the elements) out, and infinitely many will remain.
>
> Contradiciton.

One of Cantor’s delusions is that ∞ + ∞ = ∞. In words, if one “infinite set” is added to another, the resulting set is infinite. This becomes problematic if the difference ∞ − ∞ is considered. Therefore, it is expressly forbidden by a formal decree issued by my intellectual inferiors of mainstream mathematics academia. There are so many problems with Cantor’s nonsensical theories and I don’t intend to address all of them here, save the blunder that the difference ∞ − ∞ is “undefined”.

Proof that ∞ − ∞ = 0 which is in contradiction to Cantor’s claims:

(1-x^(n+1))/(1-x) = 1/(1-x) -(x^(n+1))/(1-x) = 1 + x + x^2 + x^3 + ...+ x^(n-1) + x^n + [x^(n+1)+x^(n+2)+...] - [x^(n+1)+x^(n+2)+...] [ID]

where

(x^(n+1))/(1-x) = - [x^(n+1)+x^(n+2)+...]

and

(1-x^(n+1))/(1-x) = 1 + x + x^2 + x^3 + ...+ x^(n-1) + x^n

because

1/(1-x) = 1 + x + x^2 + x^3 + ...+ x^(n-1) + x^n + [x^(n+1)+x^(n+2)+....]

Now the sets {x^(n+1); x^(n+2); ...} and {- x^(n+1); - x^(n+2); ...} possess the same cardinality given that they are described by exactly the same elements and it’s clear the only difference is in sign, that is, the elements of {- x^(n+1); - x^(n+2); ...} are all negative. Therefore in order for [ID] to remain true, it must be the case that [x^(n+1)+x^(n+2)+...] - [x^(n+1)+x^(n+2)+...] = 0, that is, ∞ − ∞ = 0 must be true also which is in contradiction to Cantor’s decree!

What else can one expect? Chuckle. Cantor was placed in a mental asylum for a reason - he lost his marbles!

>
> Regards, WM

On Monday, July 18, 2022 at 4:18:10 PM UTC+2, WM wrote:
> Fritz Feldhase schrieb am Montag, 18. Juli 2022 um 16:04:28 UTC+2:
> > On Monday, July 18, 2022 at 3:38:40 PM UTC+2, WM wrote:
> >
> > A matrix can be "filled" with integer fractions
> >
> Yes, of course, but not an infinite matrix.

I meant an infinite matrix, dumbo.

Even a subset suffices.

Consider the matrix (d_n,m) defined with d_n,m = (2^n * 3^m)/2 (for all n, m e IN).
> It is not achieved unless every [element] can be checked [manually].

I'm sorry, but _only_ Chuck Norris can do this.

That's the reason why _mathematicians_ (in contrast to cranks) are satisfied with proofs already.

You really want to claim that the elements of the matrix (d_n,m) defined with d_n,m = (2^n * 3^m)/1 (for all n, m e IN)) are not just the integer fractions in {(2^n * 3^m)/1 : n, m e IN}?

On Monday, July 18, 2022 at 11:53:39 AM UTC-3, WM wrote:
> So we can take them (the elements) out, and infinitely many will remain.

Nope. only if you use a stepwise process, "infinitely many will remain" at every step. If you use, e.g., induction no element will remain.

--
William Hughes

On 7/18/2022 9:22 AM, WM wrote:
> Gus Gassmann schrieb am Montag, 18. Juli 2022 um 16:16:00 UTC+2:
>> On Saturday, 16 July 2022 at 11:20:26 UTC-3, William wrote:
>> [...]
>>>> XOOOO...
>>>> XOOOO...
>>>> XOOOO...
>>>> XOOOO...
>>>> XOOOO...
>>>> ...
>>> Starting with this matrix does not work. So it is completely irrelevant to the question of whether M exists.
>>
>> I would argue that one can make it work
>
> Nonsense. As JG has remarked completely correct, in every line there are far more fractions than integer fractions.
>
> Regarsd, WM

It is a pile of steaming poo, WM. no matter how you swapparoo your piles, it will always stink.

an endorsement by JG of your math, underscores your intent of misleading people with bad math, as that is what he does.

borrowed from Dan's post;

Even at his advanced age (60+?), John Gabriel is STILL struggling with basic, elementary-school arithmetic. As he has repeatedly posted here:

"There are no points on a line."
--April 12, 2021

"Pi is NOT a number of ANY kind!"
--July 10, 2020

"1/2 not equal to 2/4"
--October 22, 2017

“1/3 does NOT mean 1 divided by 3 and never has meant that”
-- February 8, 2015

"3 =< 4 is nonsense.”
--October 28, 2017

"Zero is not a number."
-- Dec. 2, 2019

"0 is not required at all in mathematics, just like negative numbers."
-- Jan. 4, 2017

“There is no such thing as an empty set.”
--Oct. 4, 2019

“3 <=> 2 + 1 or 3 <=> 8 - 5, etc, are all propositions” (actually all are meaningless gibberish)
--Oct. 22, 2019

No math genius our JG, though he actually lists his job title as “mathematician” at Linkedin.com. Apparently, they do not verify your credentials.

Though really quite disturbing, interested readers should see: “About the spamming troll John Gabriel in his own words...” (lasted updated March 10,
2020) at https://groups.google.com/forum/#!msg/sci.math/PcpAzX5pDeY/1PDiSlK_BwAJ

WM brought next idea :
> Tom Bola schrieb am Montag, 18. Juli 2022 um 14:08:55 UTC+2:
>
>> But, for WM, the cardinalities
>> c_a = |{[0, 1, 2, 3, ...]}|,
>> c_b = |{[1, 2, 3, ...]}|
>> are equal in it's size which is aleph_0_WM,
>> with aleph_0_WM being a quantity of fixed
>> size in the same way as any n in IN is,
>> hence, for WM is c_a = c_b - 1.
>
> Nonsense. |ℕ| is an invariable constant
>> {1, 2, 3, ...}| = |ℕ|
>> {0, 1, 2, 3, ...}| = |ℕ| + 1
>> {2, 4, 6, ...}| = |ℕ|/2 ℵo denotes actual infinity, not fixed
>> ℕ| = ℵo
>> ℕ| = n + ℵo
>> ℕ|/2 = ℵo
>
>> It's useless to repeat one's arguments each day, year in, year out.
>
> I repeated it because you have not understood it yet - after 5 years.

Well, you must not be explaining it very well then.

sergi o schrieb:

> On 7/18/2022 9:22 AM, WM wrote:
>> Gus Gassmann schrieb am Montag, 18. Juli 2022 um 16:16:00 UTC+2:
>>> On Saturday, 16 July 2022 at 11:20:26 UTC-3, William wrote:
>>> [...]
>>>>> XOOOO...
>>>>> XOOOO...
>>>>> XOOOO...
>>>>> XOOOO...
>>>>> XOOOO...
>>>>> ...
>>>> Starting with this matrix does not work. So it is completely irrelevant to the question of whether M exists.
>>>
>>> I would argue that one can make it work
>>
>> Nonsense. As JG has remarked completely correct, in every line there are far more fractions than integer fractions.
>>
>> Regarsd, WM
>
>
>
> It is a pile of steaming poo, WM. no matter how you swapparoo your piles, it will always stink.
>
> an endorsement by JG of your math, underscores your intent of misleading people with bad math, as that is what he does.
>
>
>
> borrowed from Dan's post;
>
> Even at his advanced age (60+?), John Gabriel is STILL struggling with basic, elementary-school arithmetic. As he has repeatedly posted here:
>
> "There are no points on a line."
> --April 12, 2021
>
> "Pi is NOT a number of ANY kind!"
> --July 10, 2020
>
> "1/2 not equal to 2/4"
> --October 22, 2017
>
> “1/3 does NOT mean 1 divided by 3 and never has meant that”
> -- February 8, 2015
>
> "3 =< 4 is nonsense.”
> --October 28, 2017
>
> "Zero is not a number."
> -- Dec. 2, 2019
>
> "0 is not required at all in mathematics, just like negative numbers."
> -- Jan. 4, 2017
>
> “There is no such thing as an empty set.”
> --Oct. 4, 2019
>
> “3 <=> 2 + 1 or 3 <=> 8 - 5, etc, are all propositions” (actually all are meaningless gibberish)
> --Oct. 22, 2019
>
> No math genius our JG, though he actually lists his job title as “mathematician” at Linkedin.com. Apparently, they do not verify your credentials.
>
> Though really quite disturbing, interested readers should see: “About the spamming troll John Gabriel in his own words...” (lasted updated March 10,
> 2020) at https://groups.google.com/forum/#!msg/sci.math/PcpAzX5pDeY/1PDiSlK_BwAJ

Today:

|IN| is an invariable constant
|{1, 2, 3, ...}| = | IN |
|{0, 1, 2, 3, ...}| = | IN | + 1
|{2, 4, 6, ...}| = | IN |/2

On Monday, 18 July 2022 at 18:15:43 UTC+3, Eram semper recta wrote:
> On Monday, 18 July 2022 at 17:53:39 UTC+3, WM wrote:
> > William schrieb am Montag, 18. Juli 2022 um 16:33:18 UTC+2:
> > > On Monday, July 18, 2022 at 10:17:11 AM UTC-3, WM wrote:
> > > > William schrieb am Sonntag, 17. Juli 2022 um 22:54:16 UTC+2:
> > > > > On Sunday, July 17, 2022 at 3:28:42 PM UTC-3, WM wrote:
> > > > > > William schrieb am Sonntag, 17. Juli 2022 um 17:08:29 UTC+2:
> > > > > > > On Sunday, July 17, 2022 at 11:42:23 AM UTC-3, WM wrote:
> > > > > > > > William schrieb am Samstag, 16. Juli 2022 um 18:16:45 UTC+2:
> > > > > > > > > On Saturday, July 16, 2022 at 12:56:46 PM UTC-3, WM wrote:
> > > > > > > >
> > > > > > > > > > > A trivial induction shows that each element (without exception) of |N has finitely many predecessors and infinitely many successors.
> > > > > > > > > > But according to Cantor all natural numbers with no successors
> > > > > > > > > Piffle, there is no such thing as a natural number with no successors.
> > > > > > > > Therefore not all natural numbers are individually definable. The successors
> > > > > > > each (without exception) has finitely many predecessors.
> > > > > > Are all successors in the bijection with Q such that none remains?
> > > > > Yes. Simple induction shows that each successor (without exception) is in the bijection.
> > > > 5 lines above there was no such thing as a natural number with no successors.
> > > Correct.
> > > >Now they are in the bijection without successors?
> > > Nope, "none remains" does not imply there is an element in the bijection without a successor.
> > It implies that all elements are in the bijection and these elements have no successors outside of the bijection.
> > > There is no last element. Simple induction shows that 1. each element (without exception) is in the bijection
> > So we can take them (the elements of the bijction) out, and nothing will remain.
> > > and 2. each element (without exception) has a successor.
> > Even infinitely many. So we can take them (the elements) out, and infinitely many will remain.
> >
> > Contradiciton.
> One of Cantor’s delusions is that ∞ + ∞ = ∞. In words, if one “infinite set” is added to another, the resulting set is infinite. This becomes problematic if the difference ∞ − ∞ is considered. Therefore, it is expressly forbidden by a formal decree issued by my intellectual inferiors of mainstream mathematics academia. There are so many problems with Cantor’s nonsensical theories and I don’t intend to address all of them here, save the blunder that the difference ∞ − ∞ is “undefined”.
>
> Proof that ∞ − ∞ = 0 which is in contradiction to Cantor’s claims:
>
> (1-x^(n+1))/(1-x) = 1/(1-x) -(x^(n+1))/(1-x) = 1 + x + x^2 + x^3 + ....+ x^(n-1) + x^n + [x^(n+1)+x^(n+2)+...] - [x^(n+1)+x^(n+2)+...] [ID]
>
> where
>
> (x^(n+1))/(1-x) = - [x^(n+1)+x^(n+2)+...]
>
> and
>
> (1-x^(n+1))/(1-x) = 1 + x + x^2 + x^3 + ...+ x^(n-1) + x^n
>
> because
>
> 1/(1-x) = 1 + x + x^2 + x^3 + ...+ x^(n-1) + x^n + [x^(n+1)+x^(n+2)+...]
>
> Now the sets {x^(n+1); x^(n+2); ...} and {- x^(n+1); - x^(n+2); ...} possess the same cardinality given that they are described by exactly the same elements and it’s clear the only difference is in sign, that is, the elements of {- x^(n+1); - x^(n+2); ...} are all negative. Therefore in order for [ID] to remain true, it must be the case that [x^(n+1)+x^(n+2)+...] - [x^(n+1)+x^(n+2)+...] = 0, that is, ∞ − ∞ = 0 must be true also which is in contradiction to Cantor’s decree!
>
> What else can one expect? Chuckle. Cantor was placed in a mental asylum for a reason - he lost his marbles!

The moral of my story is that you cannot mix mathematics with Jewish fables..

Jewish and Greek fables are both interesting but it should be remembered that a fable is by definition NOT TRUE.

fable: a short story, typically with animals as characters, conveying a moral.

Mathematics does not care about morality or immorality, only about hard, cold facts. There is no Cantor's Paradise. There never was outside of Hilbert's low IQ brain.

Yes, the wool was not only pulled over your ugly eyes, but over your thick wooden heads.

You are all unbelievably stupid. There are no measures to describe your stupidity. But even though I beat your heads with a 2x4, you resist and insist on staying ignorant and stupid. Why is this so? Why do you resist truth, you incorrigible idiots? I am the voice crying out in the mainstream mathematical wilderness. Perhaps your wax-filled ears need to be steam washed.

🤣🤣🤣🤣

>
>
> >
> > Regards, WM

On Monday, 18 July 2022 at 18:43:44 UTC+3, FromTheRafters wrote:
> WM brought next idea :
> > Tom Bola schrieb am Montag, 18. Juli 2022 um 14:08:55 UTC+2:
> >
> >> But, for WM, the cardinalities
> >> c_a = |{[0, 1, 2, 3, ...]}|,
> >> c_b = |{[1, 2, 3, ...]}|
> >> are equal in it's size which is aleph_0_WM,
> >> with aleph_0_WM being a quantity of fixed
> >> size in the same way as any n in IN is,
> >> hence, for WM is c_a = c_b - 1.
> >
> > Nonsense. |ℕ| is an invariable constant
> >> {1, 2, 3, ...}| = |ℕ|
> >> {0, 1, 2, 3, ...}| = |ℕ| + 1
> >> {2, 4, 6, ...}| = |ℕ|/2 ℵo denotes actual infinity, not fixed
> >> ℕ| = ℵo
> >> ℕ| = n + ℵo
> >> ℕ|/2 = ℵo
> >
> >> It's useless to repeat one's arguments each day, year in, year out.
> >
> > I repeated it because you have not understood it yet - after 5 years.
> Well, you must not be explaining it very well then.

The problem is not with his numerous explanations, but your dense head. LMAO.

On Monday, 18 July 2022 at 18:43:36 UTC+3, knucklehead sergi gorilla o driveled:

> > Nonsense. As JG has remarked completely correct, in every line there are far more fractions than integer fractions.
> >
> > Regarsd, WM

> It is a pile of steaming poo, WM. no matter how you swapparoo your piles, it will always stink.
> an endorsement by JG of your math, underscores your intent of misleading people with bad math, as that is what he does.

"I know that most men, including those at ease with problems of the greatest complexity, can seldom accept even the simplest and most obvious truth if it be such as would oblige them to admit the falsity of conclusions which they have delighted in explaining to colleagues, which they have proudly taught to others, and which they have woven, thread by thread, into the fabric of their lives." -Tolstoy

<snip: shit>

måndag 18 juli 2022 kl. 15:34:52 UTC+2 skrev WM:
> Tom Bola schrieb am Montag, 18. Juli 2022 um 14:08:55 UTC+2:
>
> > But, for WM, the cardinalities
> > c_a = |{[0, 1, 2, 3, ...]}|,
> > c_b = |{[1, 2, 3, ...]}|
> > are equal in it's size which is aleph_0_WM,
> > with aleph_0_WM being a quantity of fixed
> > size in the same way as any n in IN is,
> > hence, for WM is c_a = c_b - 1.
> Nonsense. |ℕ| is an invariable constant

Demonstrating you are uneducated

> |{1, 2, 3, ...}| = |ℕ|
> |{0, 1, 2, 3, ...}| = |ℕ| + 1

False, it is |ℕ|

> |{2, 4, 6, ...}| = |ℕ|/2

false, it is |ℕ|

> ℵo denotes actual infinity, not fixed

It is a fixed cardinality

> |ℕ| = ℵo

By definition, correct.

> |ℕ| = n + ℵo

False

> |ℕ|/2 = ℵo

Yes

> > It's useless to repeat one's arguments each day, year in, year out.
> I repeated it because you have not understood it yet - after 5 years.

Comes from the dude failing basic cardinal arithmetic.

>
> Regards, WM
De

måndag 18 juli 2022 kl. 18:10:03 UTC+2 skrev Eram semper recta:
> On Monday, 18 July 2022 at 18:15:43 UTC+3, Eram semper recta wrote:
> > On Monday, 18 July 2022 at 17:53:39 UTC+3, WM wrote:
> > > William schrieb am Montag, 18. Juli 2022 um 16:33:18 UTC+2:
> > > > On Monday, July 18, 2022 at 10:17:11 AM UTC-3, WM wrote:
> > > > > William schrieb am Sonntag, 17. Juli 2022 um 22:54:16 UTC+2:
> > > > > > On Sunday, July 17, 2022 at 3:28:42 PM UTC-3, WM wrote:
> > > > > > > William schrieb am Sonntag, 17. Juli 2022 um 17:08:29 UTC+2:
> > > > > > > > On Sunday, July 17, 2022 at 11:42:23 AM UTC-3, WM wrote:
> > > > > > > > > William schrieb am Samstag, 16. Juli 2022 um 18:16:45 UTC+2:
> > > > > > > > > > On Saturday, July 16, 2022 at 12:56:46 PM UTC-3, WM wrote:
> > > > > > > > >
> > > > > > > > > > > > A trivial induction shows that each element (without exception) of |N has finitely many predecessors and infinitely many successors.
> > > > > > > > > > > But according to Cantor all natural numbers with no successors
> > > > > > > > > > Piffle, there is no such thing as a natural number with no successors.
> > > > > > > > > Therefore not all natural numbers are individually definable. The successors
> > > > > > > > each (without exception) has finitely many predecessors.
> > > > > > > Are all successors in the bijection with Q such that none remains?
> > > > > > Yes. Simple induction shows that each successor (without exception) is in the bijection.
> > > > > 5 lines above there was no such thing as a natural number with no successors.
> > > > Correct.
> > > > >Now they are in the bijection without successors?
> > > > Nope, "none remains" does not imply there is an element in the bijection without a successor.
> > > It implies that all elements are in the bijection and these elements have no successors outside of the bijection.
> > > > There is no last element. Simple induction shows that 1. each element (without exception) is in the bijection
> > > So we can take them (the elements of the bijction) out, and nothing will remain.
> > > > and 2. each element (without exception) has a successor.
> > > Even infinitely many. So we can take them (the elements) out, and infinitely many will remain.
> > >
> > > Contradiciton.
> > One of Cantor’s delusions is that ∞ + ∞ = ∞. In words, if one “infinite set” is added to another, the resulting set is infinite. This becomes problematic if the difference ∞ − ∞ is considered. Therefore, it is expressly forbidden by a formal decree issued by my intellectual inferiors of mainstream mathematics academia. There are so many problems with Cantor’s nonsensical theories and I don’t intend to address all of them here, save the blunder that the difference ∞ − ∞ is “undefined”.
> >
> > Proof that ∞ − ∞ = 0 which is in contradiction to Cantor’s claims:
> >
> > (1-x^(n+1))/(1-x) = 1/(1-x) -(x^(n+1))/(1-x) = 1 + x + x^2 + x^3 + ....+ x^(n-1) + x^n + [x^(n+1)+x^(n+2)+...] - [x^(n+1)+x^(n+2)+...] [ID]
> >
> > where
> >
> > (x^(n+1))/(1-x) = - [x^(n+1)+x^(n+2)+...]
> >
> > and
> >
> > (1-x^(n+1))/(1-x) = 1 + x + x^2 + x^3 + ...+ x^(n-1) + x^n
> >
> > because
> >
> > 1/(1-x) = 1 + x + x^2 + x^3 + ...+ x^(n-1) + x^n + [x^(n+1)+x^(n+2)+....]
> >
> > Now the sets {x^(n+1); x^(n+2); ...} and {- x^(n+1); - x^(n+2); ...} possess the same cardinality given that they are described by exactly the same elements and it’s clear the only difference is in sign, that is, the elements of {- x^(n+1); - x^(n+2); ...} are all negative. Therefore in order for [ID] to remain true, it must be the case that [x^(n+1)+x^(n+2)+....] - [x^(n+1)+x^(n+2)+...] = 0, that is, ∞ − ∞ = 0 must be true also which is in contradiction to Cantor’s decree!
> >
> > What else can one expect? Chuckle. Cantor was placed in a mental asylum for a reason - he lost his marbles!
> The moral of my story is that you cannot mix mathematics with Jewish fables.
>
> Jewish and Greek fables are both interesting but it should be remembered that a fable is by definition NOT TRUE.
>
> fable: a short story, typically with animals as characters, conveying a moral.
>
> Mathematics does not care about morality or immorality, only about hard, cold facts. There is no Cantor's Paradise. There never was outside of Hilbert's low IQ brain.
>
> Yes, the wool was not only pulled over your ugly eyes, but over your thick wooden heads.
>
> You are all unbelievably stupid. There are no measures to describe your stupidity. But even though I beat your heads with a 2x4, you resist and insist on staying ignorant and stupid. Why is this so? Why do you resist truth, you incorrigible idiots? I am the voice crying out in the mainstream mathematical wilderness. Perhaps your wax-filled ears need to be steam washed.
>
> 🤣🤣🤣🤣
>
> >
> >
> > >
> > > Regards, WM
my god you are are a nazi

On Monday, July 18, 2022 at 4:18:10 PM UTC+2, WM wrote:
> Fritz Feldhase schrieb am Montag, 18. Juli 2022 um 16:04:28 UTC+2:
> > On Monday, July 18, 2022 at 3:38:40 PM UTC+2, WM wrote:
> >
> > A matrix can be "filled" with integer fractions
> >
> Yes, of course, but not an infinite matrix.

I meant an infinite matrix, dumbo.

________________________________________________________

Actually. even a subset of the set of all integer fractions suffices.

Consider the matrix (d_n,m) defined with d_n,m = (2^n * 3^m)/1 (for all n, m e IN).

> It is not achieved unless every [element] can be checked [manually].

I'm sorry, but _only_ Chuck Norris can do this.

That's the reason why _mathematicians_ (in contrast to cranks) are satisfied with proofs already.

You really want to claim that the elements of the matrix (d_n,m) defined with d_n,m = (2^n * 3^m)/1 (for all n, m e IN)) are not just the integer fractions in {(2^n * 3^m)/1 : n, m e IN}?

On Monday, 18 July 2022 at 11:22:23 UTC-3, WM wrote:
> Gus Gassmann schrieb am Montag, 18. Juli 2022 um 16:16:00 UTC+2:
> > On Saturday, 16 July 2022 at 11:20:26 UTC-3, William wrote:
> > [...]
> > > > XOOOO...
> > > > XOOOO...
> > > > XOOOO...
> > > > XOOOO...
> > > > XOOOO...
> > > > ...
> > > Starting with this matrix does not work. So it is completely irrelevant to the question of whether M exists.
> >
> > I would argue that one can make it work
> Nonsense.

You are an idiot, so nonsense is all you can ever spout.

As I said previously, once you (correction: anyone *BUT* you) know how the stepwise process transforms the matrices, (again: anyone other than) you can think about a limit and about a metric that makes the process convergent.

On Monday, July 18, 2022 at 3:38:40 PM UTC+2, WM wrote:

A matrix can be "filled" with integer fractions

> by [...] shuffling the [integer fractiions] in a sophisticated way [...]

No, not by "shuffling" but by "distributing" (but n o t as the result of an infinite process) them in a clever way.

Hint: A matrix is a function with 2 parameters. Hence it suffices to specify an appropriate _function_ for achieving this feat.

No "shuffling" needed, idiot!

William schrieb am Montag, 18. Juli 2022 um 16:33:18 UTC+2:
> On Monday, July 18, 2022 at 10:17:11 AM UTC-3, WM wrote:

> Simple induction shows that 1. each element (without exception) is in the bijection

Removing them leaves nothing.

> and 2. each element (without exception) has a successor.

Removing them leaves infinitely many succesors.

Contradiction.

Regards, WM

måndag 18 juli 2022 kl. 19:47:34 UTC+2 skrev WM:
> William schrieb am Montag, 18. Juli 2022 um 16:33:18 UTC+2:
> > On Monday, July 18, 2022 at 10:17:11 AM UTC-3, WM wrote:
>
> > Simple induction shows that 1. each element (without exception) is in the bijection
> Removing them leaves nothing.
> > and 2. each element (without exception) has a successor.
> Removing them leaves infinitely many succesors.
>
> Contradiction.
>
> Regards, WM
can you please stop being this retarded?

On Monday, July 18, 2022 at 2:47:34 PM UTC-3, WM wrote:
> William schrieb am Montag, 18. Juli 2022 um 16:33:18 UTC+2:
> > On Monday, July 18, 2022 at 10:17:11 AM UTC-3, WM wrote:
>

Let |N_b be the subset of |N of elements that are in the bijection,
Let |N_s be the subset of |N of elements that have a successor
> > Simple induction shows that 1. each element (without exception) is in the bijection

So 1. is |N_b = |N

> Removing them leaves nothing.

|N\|N_b = {}

> > and 2. each element (without exception) has a successor.

So 2. is |N_s = |N

> Removing them leaves infinitely many succesors

Nope |N\|N_s={}. Correct is to say "removing any element of |N_s leaves infinitely many successors". However "them", |N_s, is not an element of |N_s , so the fact that "removing them" leave nothing is not a contradiction.

From "removing any element of |N_s leaves infinitely many successors" we can conclude that "removing any finite subset of |N_s leaves infinitely many successors". However, "them", |N_s, is not a finite subset of |N_s , so the fact that "removing them" leave nothing is not a contradiction.

--
William Hughes

On Monday, July 18, 2022 at 7:47:34 PM UTC+2, WM wrote:

> Removing them leaves nothing.
> Removing them leaves infinitely many succesors.

See: https://en.wikipedia.org/wiki/Equivocation

Try to use mathematical notation, dumbo:

1. IN \ {n : n e IN} = {}
2. An e IN: IN \ {1, ..., n} is infinite.

Note that in mathematics the following "rule of derivation" is not valid: An e IN: Phi({1, ..., n}) implies Phi(IN). (Mückenschluss)
Hence 1. and 2. do not "form" a "contradiction" in mathematics.

On Monday, July 18, 2022 at 4:10:56 PM UTC-3, Fritz Feldhase wrote:

> Note that in mathematics the following "rule of derivation" is not valid: An e IN: Phi({1, ..., n}) implies Phi(IN).

It works if you throw in a pink elephant

An e IN: Phi({1, ..., n})

Look! Over There! A Pink Elephant!

Thus P(|N)

--
William Hughes

On Monday, July 18, 2022 at 9:23:50 PM UTC+2, William wrote:
> On Monday, July 18, 2022 at 4:10:56 PM UTC-3, Fritz Feldhase wrote:
> >
> > Note that in mathematics the following "rule of derivation" is not valid: An e IN: Phi({1, ..., n}) implies Phi(IN).
> >
> It works if you throw in a pink elephant
>
> An e IN: Phi({1, ..., n})
>
> Look! Over There! A Pink Elephant!
>
> Thus P(|N)

Holy shit! You are right: It works!

William schrieb am Montag, 18. Juli 2022 um 20:17:17 UTC+2:
> On Monday, July 18, 2022 at 2:47:34 PM UTC-3, WM wrote:
> > William schrieb am Montag, 18. Juli 2022 um 16:33:18 UTC+2:
> > > On Monday, July 18, 2022 at 10:17:11 AM UTC-3, WM wrote:
> >
> Let |N_b be the subset of |N of elements that are in the bijection,
> Let |N_s be the subset of |N of elements that have a successor
> > > Simple induction shows that 1. each element (without exception) is in the bijection

This thread again has become to slow. I open "Natural numbers and vases III"

Regards, WM

Fritz Feldhase schrieb am Montag, 18. Juli 2022 um 21:10:56 UTC+2:
> On Monday, July 18, 2022 at 7:47:34 PM UTC+2, WM wrote:
>
> > Removing them leaves nothing.
> > Removing them leaves infinitely many succesors.
> See: https://en.wikipedia.org/wiki/Equivocation

> 1. IN \ {n : n e IN} = {}
> 2. An e IN: IN \ {1, ..., n} is infinite.
>
> Note that in mathematics the following "rule of derivation" is not valid: An e IN: Phi({1, ..., n}) implies Phi(IN).
> Hence 1. and 2. do not "form" a "contradiction" in mathematics.

Note that removing all elements of a set is same as removing the set. Otherwise something is rotten in Cantor land.

Hence 1. and 2. do "form" a "contradiction" in mathematics.

Regards, WM

On Monday, July 18, 2022 at 11:32:22 PM UTC+2, WM wrote:
> Fritz Feldhase schrieb am Montag, 18. Juli 2022 um 21:10:56 UTC+2:
> > On Monday, July 18, 2022 at 7:47:34 PM UTC+2, WM wrote:
> >
> > > Removing them leaves nothing.
> > > Removing them leaves infinitely many succesors.
> > See: https://en.wikipedia.org/wiki/Equivocation
> > 1. IN \ {n : n e IN} = {}
> > 2. An e IN: IN \ {1, ..., n} is infinite.
> >
> > Note that in mathematics the following "rule of derivation" is not valid: An e IN: Phi({1, ..., n}) implies Phi(IN).
> > Hence 1. and 2. do not "form" a "contradiction" in mathematics.
> >
> Note that <bla bla>
>
> Hence 1. and 2. do "form" a "contradiction" in mathematics.

Huh?! Proof by the axiom "because I said so", again?

You know: "[WM’s] conclusions are based on the sloppiness of his notions, his inability of giving precise definitions, his fundamental misunderstanding of elementary mathematical concepts, and sometimes, as the late Dik Winter remarked [...], on nothing at all."

Du redest wieder mal Schwachsinn daher, Mückenheim.

Gus Gassmann schrieb am Montag, 18. Juli 2022 um 19:22:04 UTC+2:
> On Monday, 18 July 2022 at 11:22:23 UTC-3, WM wrote:
> > Gus Gassmann schrieb am Montag, 18. Juli 2022 um 16:16:00 UTC+2:
> > > On Saturday, 16 July 2022 at 11:20:26 UTC-3, William wrote:
> > > [...]
> > > > > XOOOO...
> > > > > XOOOO...
> > > > > XOOOO...
> > > > > XOOOO...
> > > > > XOOOO...
> > > > > ...
> > > > Starting with this matrix does not work. So it is completely irrelevant to the question of whether M exists.
> > >
> > > I would argue that one can make it work
> > Nonsense.

> As I said previously, once you (correction: anyone *BUT* you) know how the stepwise process transforms the matrices, (again: anyone other than) you can think about a limit and about a metric that makes the process convergent.

We know the process. It is given by
k = (m + n - 1)(m + n - 2)/2 + m
1/1, 1/2, 2/1, 1/3, 2/2, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5, 2/4, 3/3, 4/2, 5/1, 1/6, 2/5, 3/4, 4/3, 5/2, 6/1, ...

But we know also that never an exchange of X and O will change the relative shares or delete any O. Therefore you are wrong.

Regards, WM

On 7/18/2022 4:42 PM, WM wrote:
> Gus Gassmann schrieb am Montag, 18. Juli 2022 um 19:22:04 UTC+2:
>> On Monday, 18 July 2022 at 11:22:23 UTC-3, WM wrote:
>>> Gus Gassmann schrieb am Montag, 18. Juli 2022 um 16:16:00 UTC+2:
>>>> On Saturday, 16 July 2022 at 11:20:26 UTC-3, William wrote:
>>>> [...]

>>>>>> ...
>>>>> Starting with this matrix does not work. So it is completely irrelevant to the question of whether M exists.
>>>>
>>>> I would argue that one can make it work
>>> Nonsense.
>
>> As I said previously, once you (correction: anyone *BUT* you) know how the stepwise process transforms the matrices, (again: anyone other than) you can think about a limit and about a metric that makes the process convergent.
>
> We know the process. It is given by
> k = (m + n - 1)(m + n - 2)/2 + m
> 1/1, 1/2, 2/1, 1/3, 2/2, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5, 2/4, 3/3, 4/2, 5/1, 1/6, 2/5, 3/4, 4/3, 5/2, 6/1, ...
>
> But we know also that never an exchange of X and O will change the relative shares or delete any O.

wrong, your swapadoodlefest will not save your imagination from real math.

>
> Regards, WM