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tech / sci.math / Re: Repeating decimal is irrational

SubjectAuthor
* Repeating decimal is irrationalwij
+* Re: Repeating decimal is irrationalMike Terry
|`- Re: Repeating decimal is irrationalwij
+- Re: Repeating decimal is irrationalFromTheRafters
+* Re: Repeating decimal is irrationalChris M. Thomasson
|`- Re: Repeating decimal is irrationalwij
+- Re: Repeating decimal is irrationalsergi o
+* Re: Repeating decimal is irrationalTimothy Golden
|`- Re: Repeating decimal is irrationalsergi o
+- Re: Repeating decimal is irrationalSeñor Dingus
+- Re: Repeating decimal is irrationalTimothy Golden
+- Re: Repeating decimal is irrationalzelos...@gmail.com
`- Re: Repeating decimal is irrationalzelos...@gmail.com

1
Repeating decimal is irrational

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Subject: Repeating decimal is irrational
From: wynii...@gmail.com (wij)
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 by: wij - Sat, 21 May 2022 00:54 UTC

The following proofs P1,P2 and P3 prove that repeating decimal is irrational.
For brevity, repeating decimal is to be represented by "0.333...":

[P1] From long division, we can have the true equality:

1/3=0.333... + nonzero_remainder

This should be the basic instance where the repeating decimal problem is from.
The infinite steps of the long division method in the conversion process that
generates the infinitely long "0.333..." must contain a nonzero remainder
proves the above equality. Therefore, 1/3 ≠ 0.333..., i.e. p/q and the generated
repeating decimal are not equal.

[P2] Let intervals A=[0,1/3), B=[1/3,1]. (the 1/3=q can be any real number)
Firstly, the "0.333..." does exist (density property demands so), it locates at near the open end of [0,1/3). Therefore, 0.333...∈A is true. The answers "0.333...∉A" or "0.333...=1/3 ∈B" yields logical absurdity.
If q is rational, then q∈B and repeating_decimal(q)∈A. These two are in different intervals.

[P3] Let x∈ℝ, x>0, and function A is defined as: A(0)=0, A(n)=(A(n-1)+x)/2
As n increases, "A(n)< A(n+1) < x" always holds, even n counts to infinity since density property is forever true. Density property demands there ALWAYS exists a different number between A(n) and x, otherwise it is broken.
A(n) eventually is the 0.333... (when x=1/3), and A(n)≠1/3. This applies to all
repeating decimals (and more).

Note: https://sourceforge.net/projects/cscall/files/MisFiles/NumberView-en.txt/download contains more basic materials of these proofs.

Re: Repeating decimal is irrational

<k66dnd0lGISA0RX_nZ2dnUU7-YvNnZ2d@brightview.co.uk>

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Subject: Re: Repeating decimal is irrational
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From: news.dea...@darjeeling.plus.com (Mike Terry)
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 by: Mike Terry - Sat, 21 May 2022 02:19 UTC

On 21/05/2022 01:54, wij wrote:
> The following proofs P1,P2 and P3 prove that repeating decimal is irrational.
> For brevity, repeating decimal is to be represented by "0.333...":
>
> [P1] From long division, we can have the true equality:
>
> 1/3=0.333... + nonzero_remainder

No, that's not right. What long division shows is stuff like:

1/3=0.333 + nonzero_remainder
1/3=0.3333333 + nonzero_remainder

etc. where there are only finitely many 3s in the decimal expansion. You CAN'T ACTUALLY PERFORM
infinitely many division operations to get what you said.

If we have the nonterminating expansion, we have
1/3 = 0.333333...
(exactly).

>
> This should be the basic instance where the repeating decimal problem is from.
> The infinite steps of the long division method in the conversion process that
> generates the infinitely long "0.333..." must contain a nonzero remainder
> proves the above equality. Therefore, 1/3 ≠ 0.333..., i.e. p/q and the generated
> repeating decimal are not equal.

No. You made a mistake right at the beginning. [P1 is just wrong. You have had several other
threads where you've misunderstood limits and decimals and people have explained, so no point going
through it all again.]

>
> [P2] Let intervals A=[0,1/3), B=[1/3,1]. (the 1/3=q can be any real number)
> Firstly, the "0.333..." does exist (density property demands so),

Density property does not "demand" this, although of course the number exists as a real number. No
point in explaining the density property again, as you've had several previous threads about this.

> it locates at near the open end of [0,1/3).

No, 0.333... = 1/3 is in B. Lol - "near the open end of [0,1/3)"? How near, exactly? [Rhetorical]

> Therefore, 0.333...∈A is true. The answers "0.333...∉A" or "0.333...=1/3 ∈B" yields logical absurdity.

No they don't. You just don't understand limits and decimals despite all the previous threads.
[When previously asked what exactly these logical absurdities were, you always either just make more
mistakes, or sometimes agree that there /is/ no absurdity, but then you're back a few months later
saying exactly the same, so there's no point going over it all again.]

> If q is rational, then q∈B and repeating_decimal(q)∈A. These two are in different intervals.
>
> [P3] Let x∈ℝ, x>0, and function A is defined as: A(0)=0, A(n)=(A(n-1)+x)/2
> As n increases, "A(n)< A(n+1) < x" always holds, even n counts to infinity since density property is forever true. Density property demands there ALWAYS exists a different number between A(n) and x, otherwise it is broken.
> A(n) eventually is the 0.333... (when x=1/3), and A(n)≠1/3. This applies to all
> repeating decimals (and more).

Oh, the density argument again. In a previous thread you actually agreed that there was no breaking
of the density property. No point going over it all again. (Briefly, remember that your error was
going from

"A(n)< A(n+1) < x" always holds

[which is correct] to

"A(oo) < ??? < x" always holds

which is nonsense, as n NEVER TAKES A VALUE oo.)

Regards,
Mike.

Re: Repeating decimal is irrational

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Subject: Re: Repeating decimal is irrational
From: wynii...@gmail.com (wij)
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 by: wij - Sat, 21 May 2022 07:33 UTC

On Saturday, 21 May 2022 at 10:19:24 UTC+8, Mike Terry wrote:
> On 21/05/2022 01:54, wij wrote:
> > The following proofs P1,P2 and P3 prove that repeating decimal is irrational.
> > For brevity, repeating decimal is to be represented by "0.333...":
> >
> > [P1] From long division, we can have the true equality:
> >
> > 1/3=0.333... + nonzero_remainder
> No, that's not right. What long division shows is stuff like:
>
> 1/3=0.333 + nonzero_remainder
> 1/3=0.3333333 + nonzero_remainder
>
> etc. where there are only finitely many 3s in the decimal expansion. You CAN'T ACTUALLY PERFORM
> infinitely many division operations to get what you said.
>
> If we have the nonterminating expansion, we have
> 1/3 = 0.333333...
> (exactly).
> >
> > This should be the basic instance where the repeating decimal problem is from.
> > The infinite steps of the long division method in the conversion process that
> > generates the infinitely long "0.333..." must contain a nonzero remainder
> > proves the above equality. Therefore, 1/3 ≠ 0.333..., i.e. p/q and the generated
> > repeating decimal are not equal.
> No. You made a mistake right at the beginning. [P1 is just wrong. You have had several other
> threads where you've misunderstood limits and decimals and people have explained, so no point going
> through it all again.]

1/3= Quo + Rem
1/3= 0 + 1/3
1/3= 0.3 + Rem // move 3/10 from Rem, add to Quo
1/3= 0.33 + Rem // move 3/100 from Rem , add to Quo
....
1/3= 0.333...+ nonzero_remainder // always holds, because long division is an endless algorithm. And note
// this fact already proves 1/3 cannot be expressed in decimal.

Let's see your claim: When does "1/3=0.333..." holds if Rem is missing ?
1/3 ≠ 0.3
1/3 ≠ 0.33
....
1/3 ≠ 0.333... // never holds (otherwise, provide an instance)
As already indicated, long division algorithm is an infinite process that never
terminate, the same as the density property is meant to be indefinitely always true.

> > [P2] Let intervals A=[0,1/3), B=[1/3,1]. (the 1/3=q can be any real number)
> > Firstly, the "0.333..." does exist (density property demands so),
> Density property does not "demand" this, although of course the number exists as a real number. No
> point in explaining the density property again, as you've had several previous threads about this.
> > it locates at near the open end of [0,1/3).
> No, 0.333... = 1/3 is in B. Lol - "near the open end of [0,1/3)"? How near, exactly? [Rhetorical]
> > Therefore, 0.333...∈A is true. The answers "0.333...∉A" or "0.333...=1/3 ∈B" yields logical absurdity.
> No they don't. You just don't understand limits and decimals despite all the previous threads.
> [When previously asked what exactly these logical absurdities were, you always either just make more
> mistakes, or sometimes agree that there /is/ no absurdity, but then you're back a few months later
> saying exactly the same, so there's no point going over it all again.]
> > If q is rational, then q∈B and repeating_decimal(q)∈A. These two are in different intervals.

A=[0,1/3), B=[1/3,1].
x=0.333... ∈ B, 0.333...∉A ? ..... I don't feel the need to argue this obvious stuff.

> > [P3] Let x∈ℝ, x>0, and function A is defined as: A(0)=0, A(n)=(A(n-1)+x)/2
> > As n increases, "A(n)< A(n+1) < x" always holds, even n counts to infinity since density property is forever true. Density property demands there ALWAYS exists a different number between A(n) and x, otherwise it is broken..
> > A(n) eventually is the 0.333... (when x=1/3), and A(n)≠1/3. This applies to all
> > repeating decimals (and more).
> Oh, the density argument again. In a previous thread you actually agreed that there was no breaking
> of the density property. No point going over it all again. (Briefly, remember that your error was
> going from
>
> "A(n)< A(n+1) < x" always holds
>
> [which is correct] to
>
> "A(oo) < ??? < x" always holds
>
> which is nonsense, as n NEVER TAKES A VALUE oo.)
You used to assume/read things strangely (like PO?)

As said, density property is meant to be indefinitely always true (there ALWAYS
exists a different number between two different numbers).
Density property demands that A(n)< A(n+1) < x" always holds (x is unreachable.
UNREACHABLE in the example above means: A(n), if x is 1/3, is eventually some
0.333... and can never be exactly 1/3).
Otherwise, you need to specify some exceptional condition/rule to differ.

Re: Repeating decimal is irrational

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Subject: Re: Repeating decimal is irrational
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 by: FromTheRafters - Sat, 21 May 2022 09:17 UTC

wij brought next idea :
> The following proofs P1,P2 and P3 prove that repeating decimal is irrational.
> For brevity, repeating decimal is to be represented by "0.333...":
>
> [P1] From long division, we can have the true equality:
>
> 1/3=0.333... + nonzero_remainder
>
> This should be the basic instance where the repeating decimal problem is
> from. The infinite steps of the long division method in the conversion
> process that generates the infinitely long "0.333..." must contain a nonzero
> remainder proves the above equality. Therefore, 1/3 ≠ 0.333..., i.e. p/q and
> the generated repeating decimal are not equal.

The CDE (Continued Decimal Expansion) of 1/3 is a *representation* of a
number, not a number itself. There is no need to finish such a
supertask for something that just represents a number.

What is the CFE (Continued Fractional Expansion) for one third? How is
it different from the CFE of Phi,Pi, or e for instance?

Re: Repeating decimal is irrational

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From: chris.m....@gmail.com (Chris M. Thomasson)
Newsgroups: sci.math
Subject: Re: Repeating decimal is irrational
Date: Sat, 21 May 2022 12:23:43 -0700
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 by: Chris M. Thomasson - Sat, 21 May 2022 19:23 UTC

On 5/20/2022 5:54 PM, wij wrote:
> The following proofs P1,P2 and P3 prove that repeating decimal is irrational.
> For brevity, repeating decimal is to be represented by "0.333...":
>
> [P1] From long division, we can have the true equality:
>
> 1/3=0.333... + nonzero_remainder
>
> This should be the basic instance where the repeating decimal problem is from.
> The infinite steps of the long division method in the conversion process that
> generates the infinitely long "0.333..." must contain a nonzero remainder
> proves the above equality. Therefore, 1/3 ≠ 0.333..., i.e. p/q and the generated
> repeating decimal are not equal.
>
> [P2] Let intervals A=[0,1/3), B=[1/3,1]. (the 1/3=q can be any real number)
> Firstly, the "0.333..." does exist (density property demands so), it locates at near the open end of [0,1/3). Therefore, 0.333...∈A is true. The answers "0.333...∉A" or "0.333...=1/3 ∈B" yields logical absurdity.
> If q is rational, then q∈B and repeating_decimal(q)∈A. These two are in different intervals.
>
> [P3] Let x∈ℝ, x>0, and function A is defined as: A(0)=0, A(n)=(A(n-1)+x)/2
> As n increases, "A(n)< A(n+1) < x" always holds, even n counts to infinity since density property is forever true. Density property demands there ALWAYS exists a different number between A(n) and x, otherwise it is broken.
> A(n) eventually is the 0.333... (when x=1/3), and A(n)≠1/3. This applies to all
> repeating decimals (and more).
>
> Note: https://sourceforge.net/projects/cscall/files/MisFiles/NumberView-en.txt/download contains more basic materials of these proofs.

The long division hits a cycle, and is therefore, rational. Right?

Re: Repeating decimal is irrational

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Subject: Re: Repeating decimal is irrational
From: wynii...@gmail.com (wij)
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 by: wij - Sat, 21 May 2022 22:47 UTC

On Sunday, 22 May 2022 at 03:23:53 UTC+8, Chris M. Thomasson wrote:
> On 5/20/2022 5:54 PM, wij wrote:
> > The following proofs P1,P2 and P3 prove that repeating decimal is irrational.
> > For brevity, repeating decimal is to be represented by "0.333...":
> >
> > [P1] From long division, we can have the true equality:
> >
> > 1/3=0.333... + nonzero_remainder
> >
> > This should be the basic instance where the repeating decimal problem is from.
> > The infinite steps of the long division method in the conversion process that
> > generates the infinitely long "0.333..." must contain a nonzero remainder
> > proves the above equality. Therefore, 1/3 ≠ 0.333..., i.e. p/q and the generated
> > repeating decimal are not equal.
> >
> > [P2] Let intervals A=[0,1/3), B=[1/3,1]. (the 1/3=q can be any real number)
> > Firstly, the "0.333..." does exist (density property demands so), it locates at near the open end of [0,1/3). Therefore, 0.333...∈A is true. The answers "0.333...∉A" or "0.333...=1/3 ∈B" yields logical absurdity.
> > If q is rational, then q∈B and repeating_decimal(q)∈A. These two are in different intervals.
> >
> > [P3] Let x∈ℝ, x>0, and function A is defined as: A(0)=0, A(n)=(A(n-1)+x)/2
> > As n increases, "A(n)< A(n+1) < x" always holds, even n counts to infinity since density property is forever true. Density property demands there ALWAYS exists a different number between A(n) and x, otherwise it is broken..
> > A(n) eventually is the 0.333... (when x=1/3), and A(n)≠1/3. This applies to all
> > repeating decimals (and more).
> >
> > Note: https://sourceforge.net/projects/cscall/files/MisFiles/NumberView-en.txt/download contains more basic materials of these proofs.
> The long division hits a cycle, and is therefore, rational. Right?

Definition ℚ: x∈ℚ iff ∃p,q∈ℤ, x=p/q

You are all fooled by pythagoreans.

Re: Repeating decimal is irrational

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 by: sergi o - Sun, 22 May 2022 04:32 UTC

On 5/20/2022 7:54 PM, wij wrote:
> The following proofs P1,P2 and P3 prove that repeating decimal is irrational.
> For brevity, repeating decimal is to be represented by "0.333...":
>
> [P1] From long division, we can have the true equality:
>
> 1/3=0.333... + nonzero_remainder
>
> This should be the basic instance where the repeating decimal problem is from.
> The infinite steps of the long division method in the conversion process that
> generates the infinitely long "0.333..." must contain a nonzero remainder
> proves the above equality. Therefore, 1/3 ≠ 0.333..., i.e. p/q and the generated
> repeating decimal are not equal.
>
> [P2] Let intervals A=[0,1/3), B=[1/3,1]. (the 1/3=q can be any real number)
> Firstly, the "0.333..." does exist (density property demands so), it locates at near the open end of [0,1/3). Therefore, 0.333...∈A is true. The answers "0.333...∉A" or "0.333...=1/3 ∈B" yields logical absurdity.
> If q is rational, then q∈B and repeating_decimal(q)∈A. These two are in different intervals.
>
> [P3] Let x∈ℝ, x>0, and function A is defined as: A(0)=0, A(n)=(A(n-1)+x)/2
> As n increases, "A(n)< A(n+1) < x" always holds, even n counts to infinity since density property is forever true. Density property demands there ALWAYS exists a different number between A(n) and x, otherwise it is broken.
> A(n) eventually is the 0.333... (when x=1/3), and A(n)≠1/3. This applies to all
> repeating decimals (and more).
>
> Note: https://sourceforge.net/projects/cscall/files/MisFiles/NumberView-en.txt/download contains more basic materials of these proofs.

Sorry but you are wrong. Repeating decimal representation of a number proves it is a rational number, and you can solve for the rational fraction.

Re: Repeating decimal is irrational

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Subject: Re: Repeating decimal is irrational
From: timbandt...@gmail.com (Timothy Golden)
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 by: Timothy Golden - Sun, 22 May 2022 16:53 UTC

On Friday, May 20, 2022 at 8:54:37 PM UTC-4, wij wrote:
> The following proofs P1,P2 and P3 prove that repeating decimal is irrational.
> For brevity, repeating decimal is to be represented by "0.333...":
>
> [P1] From long division, we can have the true equality:
>
> 1/3=0.333... + nonzero_remainder

How about 1/3 = 0.333...3
There. Now it's finished.
Numbers as two-sided entities have a familiarity that is lost with the one sided form.

The critical interpretation that I have recently found is to regard this form with the decimal point much more seriously.
It is a new representation of number beyond the purely rational form like 1/3 which includes an operator and so cannot be fundamental.
The issue of the meaning of one third is readily resolved by engaging a radix three counting system at which point
1/3 = 0.1
and so the repeating decimal is somewhat an effect of the radix system in use.
In this way the rational number can be regarded as reradixed numbers.
Under this guise their legitimacy can be challenged. This is how it goes for things which are not fundamental.

Now, back on the decimal value (as in things with decimal points rather than base ten versus base three)
we see a representation which poses no operators, but it does carry additional information to the natural value.
The meaning of the decimal place is to assign a new unity position; a secondary unity.
Removing the decimal point we witness a natural value.
Through this data structure we can now witness the continuum as natural valued... though a bit more is required.
As well sign is a structural addition to the number and can be generalized when taken this way.

The decimal point as some sort of holy separator is not born out by the digits. Each adjacent digit in a number interacts with its next digit regardless of where the decimal point goes.

As to getting into remainders: you just got out of remainders by introducing the decimal point so going back into remainders to resolve anything doesn't seem too serious to me. Closure through division is somewhat upheld by the remainder format. The rational value abolished closure and this is how it claims to develop the continuum. This detail indicates that this discussion has far reaching consequences.
>
> This should be the basic instance where the repeating decimal problem is from.
> The infinite steps of the long division method in the conversion process that
> generates the infinitely long "0.333..." must contain a nonzero remainder
> proves the above equality. Therefore, 1/3 ≠ 0.333..., i.e. p/q and the generated
> repeating decimal are not equal.
>
> [P2] Let intervals A=[0,1/3), B=[1/3,1]. (the 1/3=q can be any real number)
> Firstly, the "0.333..." does exist (density property demands so), it locates at near the open end of [0,1/3). Therefore, 0.333...∈A is true. The answers "0.333...∉A" or "0.333...=1/3 ∈B" yields logical absurdity.
> If q is rational, then q∈B and repeating_decimal(q)∈A. These two are in different intervals.
>
> [P3] Let x∈ℝ, x>0, and function A is defined as: A(0)=0, A(n)=(A(n-1)+x)/2
> As n increases, "A(n)< A(n+1) < x" always holds, even n counts to infinity since density property is forever true. Density property demands there ALWAYS exists a different number between A(n) and x, otherwise it is broken.
> A(n) eventually is the 0.333... (when x=1/3), and A(n)≠1/3. This applies to all
> repeating decimals (and more).
>
> Note: https://sourceforge.net/projects/cscall/files/MisFiles/NumberView-en.txt/download contains more basic materials of these proofs.

Re: Repeating decimal is irrational

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 by: sergi o - Sun, 22 May 2022 18:03 UTC

On 5/22/2022 11:53 AM, Timothy Golden wrote:
> On Friday, May 20, 2022 at 8:54:37 PM UTC-4, wij wrote:
>> The following proofs P1,P2 and P3 prove that repeating decimal is irrational.
>> For brevity, repeating decimal is to be represented by "0.333...":
>>
>> [P1] From long division, we can have the true equality:
>>
>> 1/3=0.333... + nonzero_remainder
>
> How about 1/3 = 0.333...3
> There. Now it's finished.

just use base(0.333....)

0.333... base10 = 1 base(0.333....) problem solved

> Numbers as two-sided entities have a familiarity that is lost with the one sided form.
>
> The critical interpretation that I have recently found is to regard this form with the decimal point much more seriously.
> It is a new representation of number beyond the purely rational form like 1/3 which includes an operator and so cannot be fundamental.
> The issue of the meaning of one third is readily resolved by engaging a radix three counting system at which point
> 1/3 = 0.1
> and so the repeating decimal is somewhat an effect of the radix system in use.
> In this way the rational number can be regarded as reradixed numbers.
> Under this guise their legitimacy can be challenged. This is how it goes for things which are not fundamental.
>
> Now, back on the decimal value (as in things with decimal points rather than base ten versus base three)
> we see a representation which poses no operators, but it does carry additional information to the natural value.
> The meaning of the decimal place is to assign a new unity position; a secondary unity.
> Removing the decimal point we witness a natural value.
> Through this data structure we can now witness the continuum as natural valued... though a bit more is required.
> As well sign is a structural addition to the number and can be generalized when taken this way.
>
> The decimal point as some sort of holy separator is not born out by the digits. Each adjacent digit in a number interacts with its next digit regardless of where the decimal point goes.
>
> As to getting into remainders: you just got out of remainders by introducing the decimal point so going back into remainders to resolve anything doesn't seem too serious to me. Closure through division is somewhat upheld by the remainder format. The rational value abolished closure and this is how it claims to develop the continuum. This detail indicates that this discussion has far reaching consequences.
>
>>
>> This should be the basic instance where the repeating decimal problem is from.
>> The infinite steps of the long division method in the conversion process that
>> generates the infinitely long "0.333..." must contain a nonzero remainder
>> proves the above equality. Therefore, 1/3 ≠ 0.333..., i.e. p/q and the generated
>> repeating decimal are not equal.
>>
>> [P2] Let intervals A=[0,1/3), B=[1/3,1]. (the 1/3=q can be any real number)
>> Firstly, the "0.333..." does exist (density property demands so), it locates at near the open end of [0,1/3). Therefore, 0.333...∈A is true. The answers "0.333...∉A" or "0.333...=1/3 ∈B" yields logical absurdity.
>> If q is rational, then q∈B and repeating_decimal(q)∈A. These two are in different intervals.
>>
>> [P3] Let x∈ℝ, x>0, and function A is defined as: A(0)=0, A(n)=(A(n-1)+x)/2
>> As n increases, "A(n)< A(n+1) < x" always holds, even n counts to infinity since density property is forever true. Density property demands there ALWAYS exists a different number between A(n) and x, otherwise it is broken.
>> A(n) eventually is the 0.333... (when x=1/3), and A(n)≠1/3. This applies to all
>> repeating decimals (and more).
>>
>> Note: https://sourceforge.net/projects/cscall/files/MisFiles/NumberView-en.txt/download contains more basic materials of these proofs.

Re: Repeating decimal is irrational

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Subject: Re: Repeating decimal is irrational
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 by: Señor Dingus - Sun, 22 May 2022 18:06 UTC

On 5/20/2022 5:54 PM, wij wrote:
> The following proofs P1,P2 and P3 prove that repeating decimal is irrational.
> For brevity, repeating decimal is to be represented by "0.333...":

Double-idiot.

Re: Repeating decimal is irrational

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Subject: Re: Repeating decimal is irrational
From: timbandt...@gmail.com (Timothy Golden)
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 by: Timothy Golden - Sun, 22 May 2022 20:04 UTC

On Friday, May 20, 2022 at 8:54:37 PM UTC-4, wij wrote:
> The following proofs P1,P2 and P3 prove that repeating decimal is irrational.
> For brevity, repeating decimal is to be represented by "0.333...":
>
> [P1] From long division, we can have the true equality:
>
> 1/3=0.333... + nonzero_remainder

This just is wrong. Allowing the ellipses the expression is correct and the 'remainder' is exactly zero.
If anything is wrong with the expression it is out further. You are dividing two natural values and getting a continuous value.
1/3 = 0
is more accurate. Yet even this form is a complicated algorithm when done in general.

division is an inverse operator and is not fundamental.
If you believe that you can magically enter into the continuum on your first division by a natural value what happens if you do it again?
Must be quite some space you get into there, eh?

>
> This should be the basic instance where the repeating decimal problem is from.
> The infinite steps of the long division method in the conversion process that
> generates the infinitely long "0.333..." must contain a nonzero remainder
> proves the above equality. Therefore, 1/3 ≠ 0.333..., i.e. p/q and the generated
> repeating decimal are not equal.
>
> [P2] Let intervals A=[0,1/3), B=[1/3,1]. (the 1/3=q can be any real number)
> Firstly, the "0.333..." does exist (density property demands so), it locates at near the open end of [0,1/3). Therefore, 0.333...∈A is true. The answers "0.333...∉A" or "0.333...=1/3 ∈B" yields logical absurdity.
> If q is rational, then q∈B and repeating_decimal(q)∈A. These two are in different intervals.
>
> [P3] Let x∈ℝ, x>0, and function A is defined as: A(0)=0, A(n)=(A(n-1)+x)/2
> As n increases, "A(n)< A(n+1) < x" always holds, even n counts to infinity since density property is forever true. Density property demands there ALWAYS exists a different number between A(n) and x, otherwise it is broken.
> A(n) eventually is the 0.333... (when x=1/3), and A(n)≠1/3. This applies to all
> repeating decimals (and more).
>
> Note: https://sourceforge.net/projects/cscall/files/MisFiles/NumberView-en.txt/download contains more basic materials of these proofs.

Re: Repeating decimal is irrational

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 by: zelos...@gmail.com - Mon, 23 May 2022 04:53 UTC

lördag 21 maj 2022 kl. 02:54:37 UTC+2 skrev wij:
> The following proofs P1,P2 and P3 prove that repeating decimal is irrational.
> For brevity, repeating decimal is to be represented by "0.333...":
>
> [P1] From long division, we can have the true equality:
>
> 1/3=0.333... + nonzero_remainder
>
> This should be the basic instance where the repeating decimal problem is from.
> The infinite steps of the long division method in the conversion process that
> generates the infinitely long "0.333..." must contain a nonzero remainder
> proves the above equality. Therefore, 1/3 ≠ 0.333..., i.e. p/q and the generated
> repeating decimal are not equal.
>
> [P2] Let intervals A=[0,1/3), B=[1/3,1]. (the 1/3=q can be any real number)
> Firstly, the "0.333..." does exist (density property demands so), it locates at near the open end of [0,1/3). Therefore, 0.333...∈A is true. The answers "0.333...∉A" or "0.333...=1/3 ∈B" yields logical absurdity.
> If q is rational, then q∈B and repeating_decimal(q)∈A. These two are in different intervals.
>
> [P3] Let x∈ℝ, x>0, and function A is defined as: A(0)=0, A(n)=(A(n-1)+x)/2
> As n increases, "A(n)< A(n+1) < x" always holds, even n counts to infinity since density property is forever true. Density property demands there ALWAYS exists a different number between A(n) and x, otherwise it is broken.
> A(n) eventually is the 0.333... (when x=1/3), and A(n)≠1/3. This applies to all
> repeating decimals (and more).
>
> Note: https://sourceforge.net/projects/cscall/files/MisFiles/NumberView-en.txt/download contains more basic materials of these proofs.

and wrong

https://math.stackexchange.com/questions/198810/proof-that-every-repeating-decimal-is-rational

Re: Repeating decimal is irrational

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Subject: Re: Repeating decimal is irrational
From: zelos.ma...@gmail.com (zelos...@gmail.com)
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 by: zelos...@gmail.com - Mon, 23 May 2022 04:56 UTC

lördag 21 maj 2022 kl. 02:54:37 UTC+2 skrev wij:
> The following proofs P1,P2 and P3 prove that repeating decimal is irrational.
> For brevity, repeating decimal is to be represented by "0.333...":
>
> [P1] From long division, we can have the true equality:
>
> 1/3=0.333... + nonzero_remainder
>
> This should be the basic instance where the repeating decimal problem is from.
> The infinite steps of the long division method in the conversion process that
> generates the infinitely long "0.333..." must contain a nonzero remainder
> proves the above equality. Therefore, 1/3 ≠ 0.333..., i.e. p/q and the generated
> repeating decimal are not equal.
>
> [P2] Let intervals A=[0,1/3), B=[1/3,1]. (the 1/3=q can be any real number)
> Firstly, the "0.333..." does exist (density property demands so), it locates at near the open end of [0,1/3). Therefore, 0.333...∈A is true. The answers "0.333...∉A" or "0.333...=1/3 ∈B" yields logical absurdity.
> If q is rational, then q∈B and repeating_decimal(q)∈A. These two are in different intervals.
>
> [P3] Let x∈ℝ, x>0, and function A is defined as: A(0)=0, A(n)=(A(n-1)+x)/2
> As n increases, "A(n)< A(n+1) < x" always holds, even n counts to infinity since density property is forever true. Density property demands there ALWAYS exists a different number between A(n) and x, otherwise it is broken.
> A(n) eventually is the 0.333... (when x=1/3), and A(n)≠1/3. This applies to all
> repeating decimals (and more).
>
> Note: https://sourceforge.net/projects/cscall/files/MisFiles/NumberView-en.txt/download contains more basic materials of these proofs.
your P1 is just you not understanding that long division is a method to get decimal representation. You can stop the moment you see a repeat and then know it will continue on forever and thus it is repeating, ergo proving it is rationa.

P2 your q is the only real number 1/3,

And no, you just declare it is in A without justification. Given ti is 1/3, no it is not in A

your P3 demonstrates that 0.333... is indeed 1/3, congrats you played yourself.

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