Discussion:
Isn't infinity (in mathematics) actually a flawed concept?
bassam king karzeddin
2017-04-15 12:35:42 UTC
Actually this only concept in mathematics needs few minutes to verify its legality of being considered as any real good and useful concept or just a mere big fallacy that is more than meaningless and a kind of madness, wonder!

It is also stranger how it led the mathematicians to built on it so many huge volumes of baseless mathematics that is good enough for little carpentry works

And even much stranger that they had generated so many types of more infinities to the science of mathematics.

But the oddest thing is that mainstream common mathematicians never realize that such concepts are so meaningless the same way their results that they obtain relying on that fake concept

And naturally such concept is indeed a real paradise for any jugglers to create so many meaningless games and shamelessly consider it as a real science

But the fact that any theorem or formula or result that is associated with such silly concept as infinity, is also so silly but may be good for entertainment

Since mainly, no definite rules with it, nor any real existence (being unreal: by its own definition)

So, this flowed concept would make it very easy in the near future to make mathematics get doubled and tripled and even grows indefinitely in the fake unreal direction to its collapsed limits, for sure

However, many refutations of such meaningless concept was provided in my posts, beside many others

Regards
Bassam King Karzeddin
15 th, April, 2017
b***@gmail.com
2017-04-15 15:25:06 UTC
So sqrt(2) = 1.41421356237... + e, where e is an infinitessimal e <> 0,
and where we have e^2 = 0, i.e. it is nilsquare.

An Invitation to Smooth Infinitesimal Analysis John L. Bell
http://publish.uwo.ca/~jbell/invitation%20to%20SIA.pdf

Or what property does your difference between sqrt(2) and 1.41421356237...
have? There must be a difference when sqrt(2) <> 1.41421356237...

e = sqrt(2) - 1.41421356237...

e <> 0

What property does your infinitessimal e have? Is it nilsquare?
Or does it have some other property?
Post by bassam king karzeddin
Actually this only concept in mathematics needs few minutes to verify its legality of being considered as any real good and useful concept or just a mere big fallacy that is more than meaningless and a kind of madness, wonder!
It is also stranger how it led the mathematicians to built on it so many huge volumes of baseless mathematics that is good enough for little carpentry works
And even much stranger that they had generated so many types of more infinities to the science of mathematics.
But the oddest thing is that mainstream common mathematicians never realize that such concepts are so meaningless the same way their results that they obtain relying on that fake concept
And naturally such concept is indeed a real paradise for any jugglers to create so many meaningless games and shamelessly consider it as a real science
But the fact that any theorem or formula or result that is associated with such silly concept as infinity, is also so silly but may be good for entertainment
Since mainly, no definite rules with it, nor any real existence (being unreal: by its own definition)
So, this flowed concept would make it very easy in the near future to make mathematics get doubled and tripled and even grows indefinitely in the fake unreal direction to its collapsed limits, for sure
However, many refutations of such meaningless concept was provided in my posts, beside many others
Regards
Bassam King Karzeddin
15 th, April, 2017
WM
2017-04-15 15:53:36 UTC
Post by b***@gmail.com
So sqrt(2) = 1.41421356237... + e,
e depends on how many digits you use. But it can only be estimated, not calculated, because sqrt(2) has no decimal representation.

Example: In sqrt(2) = 1.4, e > 0.01421356237.

Regards, WM
b***@gmail.com
2017-04-15 15:56:15 UTC
Well D = 1.41421356237... is a limes its the same as your arithmo-
geometric who knows what figure:

1
1 2
1 2 3
1 2 3 4
...

D is the limes of all chop chops D_n, i.e. D_1 = 1.4, D_2 =1.41, ..
etc.. I didnt ask for sqrt(2) <> D_n, everybody knows that

sqrt(2) <> D_n, because sqrt(2) is irrational.
Post by WM
Post by b***@gmail.com
So sqrt(2) = 1.41421356237... + e,
e depends on how many digits you use. But it can only be estimated, not calculated, because sqrt(2) has no decimal representation.
Example: In sqrt(2) = 1.4, e > 0.01421356237.
Regards, WM
b***@gmail.com
2017-04-15 15:58:51 UTC
So whats the difference between D = lim_n->oo D_n where
D_n = floor(sqrt(2)*10^n)/10^n and sqrt(2):

e = sqrt(2) - D

e <> 0

Is this the case for the reals or not. And if it is the case,
what is this e exactly? Do we have e^2 = 0?
Post by b***@gmail.com
Well D = 1.41421356237... is a limes its the same as your arithmo-
1
1 2
1 2 3
1 2 3 4
...
D is the limes of all chop chops D_n, i.e. D_1 = 1.4, D_2 =1.41, ..
etc.. I didnt ask for sqrt(2) <> D_n, everybody knows that
sqrt(2) <> D_n, because sqrt(2) is irrational.
Post by WM
Post by b***@gmail.com
So sqrt(2) = 1.41421356237... + e,
e depends on how many digits you use. But it can only be estimated, not calculated, because sqrt(2) has no decimal representation.
Example: In sqrt(2) = 1.4, e > 0.01421356237.
Regards, WM
WM
2017-04-15 16:17:29 UTC
Post by b***@gmail.com
So whats the difference between D = lim_n->oo D_n
The limit does exist, but it is not realized or reached by a digit sequence.

where
Post by b***@gmail.com
e = sqrt(2) - D
e <> 0
e = 0. Whether you call the limit sqrt(2) or D does not change anything. The limit is a landmark, not expressible by digits, that remains untouched.
Post by b***@gmail.com
Is this the case for the reals or not. And if it is the case,
what is this e exactly? Do we have e^2 = 0?
e^2 = 0 ==> e = 0.

Regards, WM
b***@gmail.com
2017-04-15 16:20:23 UTC
Post by WM
Whether you call the limit sqrt(2) or D does not change anything.
Can you rigorously proof, that D = sqrt(2), i.e. e = 0?
b***@gmail.com
2017-04-15 16:28:00 UTC
sqrt(2) is not given as a decimal representation here, you
could also imagine that you have rational numbers Q,

and the polynoms Q[X], and then you reduce them via X^2-2=0.
For example to compute:

(1 + sqrt(2))^3

You compute:

(1 + X)^3 = 1 + 3*X + 3*X^2 + X^3

And now you reduce via X^2-2=0 respectively X^2=2, and you get:

= 1 + 3*X + 3*2 + 2*X

= 7 + 5*X

Now you can write sqrt(2) for X again:

= 7 + 5*sqrt(2)

Rational numbers Q, polynoms Q[X] (as syntactic objects) and
the reduction is all pretty finite, no infinity involve.

So a countable language is enought. Check back the result:

7 + 5 sqrt(2)
https://www.wolframalpha.com/input/?i=%281%2Bsqrt%282%29%29^3
Post by b***@gmail.com
Post by WM
Whether you call the limit sqrt(2) or D does not change anything.
Can you rigorously proof, that D = sqrt(2), i.e. e = 0?
WM
2017-04-15 16:35:35 UTC
Post by b***@gmail.com
sqrt(2) is not given as a decimal representation here,
neither is D. D is a limit like sqrt(2). Since D is the limit of the decimal series approximating sqrt(2), D = sqrt(2).

And yes, no infinity is involved. x^2 = D^2 = 2 is enough.

For the first n terms of the decimal sequence we find: they are less than D. But if we choose any real number smaller than D, then we can find a finite term of the decimal sequence which surpasses this smaller number.

Regards, WM

Regards, WM
b***@gmail.com
2017-04-15 20:08:27 UTC
Well D is the decimal representation, 1.41421356237... is just a
shorthand for D = lim n->oo D_n. And its an infinite sequence.

D and sqrt(2) are not "constructed" the same way. sqrt(2) is first
there, and then from sqrt(2) I have defined D_n = floor(sqrt(2)*10^n)/10^n,

and then from D_n I have defined D = lim n->oo D_n. So there was
a process involved going from sqrt(2) to D. And this proces was a

sqrt(2) --> (D_n) --> D

super task, and now we have D = sqrt(2), right? Yes or No?
Post by WM
neither is D. D is a limit like sqrt(2). Since D is the limit of the decimal series approximating sqrt(2), D = sqrt(2).
WM
2017-04-16 18:06:12 UTC
Post by b***@gmail.com
Well D is the decimal representation
No.
Post by b***@gmail.com
, 1.41421356237... is just a
shorthand for D = lim n->oo D_n. And its an infinite sequence.
No. No sequence can define a number because every digit provably fails. You believe that infinity does change this fact?
Post by b***@gmail.com
D and sqrt(2) are not "constructed" the same way.
But both can be approximated step by step.
Post by b***@gmail.com
sqrt(2) is first
there, and then from sqrt(2) I have defined D_n = floor(sqrt(2)*10^n)/10^n,
and then from D_n I have defined D = lim n->oo D_n. So there was
a process involved going from sqrt(2) to D. And this proces was a
sqrt(2) --> (D_n) --> D
super task, and now we have D = sqrt(2), right? Yes or No?
Yes. But D is not expressible by digits.

Regards, WM
w***@gmail.com
2017-04-16 11:53:37 UTC
Post by WM
Post by b***@gmail.com
So whats the difference between D = lim_n->oo D_n
The limit does exist, but it is not realized or reached by a digit sequence.
Of course it depends on what you mean by a "digit sequence"

take sqrt(2) = 1.414...

i: The "standard" interpretation is that the digit sequences is an actual infinite set. (in this case constructable, but that is not needed). It is assumed that all digits are fixed. In this case the limit is determined by
the sequence.

2. WM denies the existence of actual infinite sets. He views the infinite sequence as a potentially infinite set, represented by a finite function, f, on the potentially infinite set of integers. In this case the limit is not determined by the digit sequence, but both the limit and the digit sequence are determined by f. WM is fond of saying that the digit sequence does not define the limit but as any sequence must be associated with an f, which does define the limit, this seems rather pedantic
--
William Hughes
WM
2017-04-16 17:56:27 UTC
Post by w***@gmail.com
Post by WM
Post by b***@gmail.com
So whats the difference between D = lim_n->oo D_n
The limit does exist, but it is not realized or reached by a digit sequence.
Of course it depends on what you mean by a "digit sequence"
take sqrt(2) = 1.414...
i: The "standard" interpretation is that the digit sequences is an actual infinite set. (in this case constructable, but that is not needed). It is assumed that all digits are fixed. In this case the limit is determined by
the sequence.
That is counterfactual belief. For every digit we can prove that it does not determine a number because every digit is at a finite position while infinitely many follow. How can you suppress that fact?
Post by w***@gmail.com
2. WM denies the existence of actual infinite sets. He views the infinite sequence as a potentially infinite set, represented by a finite function, f, on the potentially infinite set of integers. In this case the limit is not determined by the digit sequence, but both the limit and the digit sequence are determined by f.
Yes, you have understood it. Is there a counter *argument*?
Post by w***@gmail.com
WM is fond of saying that the digit sequence does not define the limit but as any sequence must be associated with an f, which does define the limit, this seems rather pedantic
No. There are said to be uncountable many sequences, but there are not uncountably many f.

Regards, WM
pirx42
2017-04-16 18:07:21 UTC
Post by WM
Post by w***@gmail.com
Post by WM
Post by b***@gmail.com
So whats the difference between D = lim_n->oo D_n
The limit does exist, but it is not realized or reached by a digit sequence.
Of course it depends on what you mean by a "digit sequence"
take sqrt(2) = 1.414...
i: The "standard" interpretation is that the digit sequences is an actual infinite set. (in this case constructable, but that is not needed). It is assumed that all digits are fixed. In this case the limit is determined by
the sequence.
That is counterfactual belief. For every digit we can prove that it does not determine a number because every digit is at a finite position while infinitely many follow. How can you suppress that fact?
Post by w***@gmail.com
2. WM denies the existence of actual infinite sets. He views the infinite sequence as a potentially infinite set, represented by a finite function, f, on the potentially infinite set of integers. In this case the limit is not determined by the digit sequence, but both the limit and the digit sequence are determined by f.
Yes, you have understood it. Is there a counter *argument*?
Post by w***@gmail.com
WM is fond of saying that the digit sequence does not define the limit but as any sequence must be associated with an f, which does define the limit, this seems rather pedantic
No. There are said to be uncountable many sequences, but there are not uncountably many f.
Regards, WM
ach

bassam king karzeddin
2017-04-15 16:36:32 UTC
Post by b***@gmail.com
So whats the difference between D = lim_n->oo D_n where
e = sqrt(2) - D
e <> 0
Is this the case for the reals or not. And if it is the case,
what is this e exactly? Do we have e^2 = 0?
Post by b***@gmail.com
Well D = 1.41421356237... is a limes its the same as your arithmo-
1
1 2
1 2 3
1 2 3 4
...
D is the limes of all chop chops D_n, i.e. D_1 = 1.4, D_2 =1.41, ..
etc.. I didnt ask for sqrt(2) <> D_n, everybody knows that
sqrt(2) <> D_n, because sqrt(2) is irrational.
Post by WM
Post by b***@gmail.com
So sqrt(2) = 1.41421356237... + e,
e depends on how many digits you use. But it can only be estimated, not calculated, because sqrt(2) has no decimal representation.
Example: In sqrt(2) = 1.4, e > 0.01421356237.
Regards, WM
Burr...

Had you ever asked your so selly self, what is the largest rational number that is less than sqrt(2),

Or similarly, Had you ever asked your so tiny self, what is the least rational number that is greater than sqrt(2),

Of course, it is more than trivial to see and confess openly that those (greatest, or least) rational numbers do not not exist, do they? wonder!

But why then you and the so broken mathematics insist that exist at infinity? wonder
And a Fool professional answer would say yes they exist at infinity, but infinity is not there moron (except in your toooo... tiny empty skull) and for sure
And how magically your rational decimal would turn suddenly irrational, after many billions of digits, after many billions of years, it would remain rational as long as you can count, but you can not count indefinitely for sure

Thus, you are after a clear illusion in your mind

And that does not mean in any case that sqrt(2) does not exist, or a symptom of a number as some famous idiots constantly say

Yes, sqrt(2), and sqrt(sqrt(2)), and generally [2^{2^{- n}}] do all exist exactly

for any + ve integer (n), and independently from rational numbers even that rational or decimal goes to

And guess what that positive integer (n) could be?

It can be for little explanation as large as we wish, I mean it can be as finite integer with sequence digits such that it can fill up trillions of galaxy size, where every trillion of sequence digits are stored only with one mm cube (imagine), but still rational, is not it?

And even much larger finite integers can be assumed for sure, because simply infinity IS NOT THERE EXCEPT IN EMPTY SKULLS, got it? wonder!

So, finite and infinite are not different, since you did not define the finite in order to define correctly the infinite

And may be the AP can teach you a better lesson in this regard

But, still you would not get anything out of this free lecture for sure

And it does not any matter

And the real difference of sqrt(2) and its decimal rational expansion is really unreal number, because your infinity is defined as unreal number

Regards
Bassam King Karzeddin
15th, April, 2017
WM
2017-04-15 16:16:39 UTC
Post by b***@gmail.com
Well D = 1.41421356237... is a limes its the same as your arithmo-
1
1 2
1 2 3
1 2 3 4
...
No. The limit does exist whereas I show that the figure cannot exist.

Regards, WM
2017-04-15 20:06:53 UTC
Post by bassam king karzeddin
Actually this only concept in mathematics needs few minutes to verify its legality of being considered as any real good and useful concept or just a mere big fallacy that is more than meaningless and a kind of madness, wonder!
It is also stranger how it led the mathematicians to built on it so many huge volumes of baseless mathematics that is good enough for little carpentry works
And even much stranger that they had generated so many types of more infinities to the science of mathematics.
But the oddest thing is that mainstream common mathematicians never realize that such concepts are so meaningless the same way their results that they obtain relying on that fake concept
And naturally such concept is indeed a real paradise for any jugglers to create so many meaningless games and shamelessly consider it as a real science
But the fact that any theorem or formula or result that is associated with such silly concept as infinity, is also so silly but may be good for entertainment
Since mainly, no definite rules with it, nor any real existence (being unreal: by its own definition)
So, this flowed concept would make it very easy in the near future to make mathematics get doubled and tripled and even grows indefinitely in the fake unreal direction to its collapsed limits, for sure
However, many refutations of such meaningless concept was provided in my posts, beside many others
Regards
Bassam King Karzeddin
15 th, April, 2017
The only odd thing is that you cannot see how wrong you are.
Barstool Kim Kardashian
2017-04-15 23:02:06 UTC
Shut the fuck up mathforum.org imbecile.
bassam king karzeddin
2017-04-16 06:02:33 UTC
Post by Barstool Kim Kardashian
Shut the fuck up mathforum.org imbecile.
Imbecile No. 18, I think, identified immediately from his first and hopefully the last post

And guess what kind of maths those idiots can teach you?

And guess what kind of force that makes them register just to say a half a line of nonsense?

BK
Dan Christensen
2017-04-16 15:56:25 UTC