Snipet from https://sourceforge.net/projects/cscall/files/MisFiles/RealNumber-en.txt/=download
=20=84=9A, x-a=E2=88=88=E2=84=9A
...
Real Nunmber(=E2=84=9D)::=3D {x| x is represented by n-ary <fixed_point_number>, the digits may be infinitely long }
=20
Note: This definition implies that repeating decimals are
irrational number. Let's list a common magic proof in the way as a
brief explanation: (1) x=3D 0.999...
(2) 10x=3D 9+x // 10x=3D 9.999...
(3) 9x=3D9 =20
(4) x=3D1
Ans: There is no axiom or theorem to prove (1) =3D> (2).
=20
Note: If the steps of converting a number x to
<fixed_point_number> is not finite, x is not a ratio of two integers,
because the following statement is always true: =E2=88=80x,a=E2=88=88=E2=
=20
---End of quote
=20
On Tue, 26 Mar 2024 22:51:40 +0800<nothing relating to C++>
wij <wyniijj5@gmail.com> wrote:
I don't know what you meant to say,
On 3/26/2024 7:51 AM, wij wrote:
[...]
=20
Repeating decimals are rational, say
=20
0.142857 142857 142857
=20
That is just 1 / 7 represented in base 10.
=20
Now, think of using a TRNG to create each digit...
=20
That would be, irrational... ;^)
On 3/26/2024 7:51 AM, wij wrote:
[...]
Repeating decimals are rational, say
0.142857 142857 142857
That is just 1 / 7 represented in base 10.
Now, think of using a TRNG to create each digit...
That would be, irrational... ;^)
26.03.2024 22:13 Chris M. Thomasson kirjutas:
On 3/26/2024 7:51 AM, wij wrote:
[...]
Repeating decimals are rational, say
0.142857 142857 142857
That is just 1 / 7 represented in base 10.
Now, think of using a TRNG to create each digit...
That would be, irrational... ;^)
Any number represented by stored digits on Earth has finite number of
digits (because Earth is finite) and therefore is rational, regardless
of how the digits are generated.
If you want to represent irrational numbers you need to use some other encoding schema, e.g. "sqrt(2)" (8 bytes, voila!).
On 3/26/2024 2:51 PM, Paavo Helde wrote:
26.03.2024 22:13 Chris M. Thomasson kirjutas:
On 3/26/2024 7:51 AM, wij wrote:
[...]
Repeating decimals are rational, say
0.142857 142857 142857
That is just 1 / 7 represented in base 10.
Now, think of using a TRNG to create each digit...
That would be, irrational... ;^)
Any number represented by stored digits on Earth has finite number of
digits (because Earth is finite) and therefore is rational, regardless
of how the digits are generated.
Even with a TRNG?
If you want to represent irrational numbers you need to use some other
encoding schema, e.g. "sqrt(2)" (8 bytes, voila!).
pi? ;^D
On 3/26/2024 7:51 AM, wij wrote:
[...]
Repeating decimals are rational, say
0.142857 142857 142857
That is just 1 / 7 represented in base 10.
Now, think of using a TRNG to create each digit...
That would be, irrational... ;^)
On Tue, 2024-03-26 at 13:13 -0700, Chris M. Thomasson wrote:
On 3/26/2024 7:51 AM, wij wrote:
[...]
Repeating decimals are rational, say
0.142857 142857 142857
That is just 1 / 7 represented in base 10.
Now, think of using a TRNG to create each digit...
That would be, irrational... ;^)
Just repeat the pattern infinitely, then it is irrational.
As said "∀x,a∈ℚ, x-a∈ℚ", if the subtraction a= 142857/10^(6*i) cannot terminate, 1/7 != 0.(142857)
In mathematics, a notion like pi actually defines an irrational number
by fixing its properties. For some such irrational numbers it is
possible to give an infinite algorithm which produces the sequence of
its digits. The funny thing is that after fixing the number there is no randomness any more, so e.g. a machine computing subsequent digits of pi would make a pretty poor RNG ;-)
On 26/03/2024 22:43, wij wrote:subtraction a=3D 142857/10^(6*i)
On Tue, 2024-03-26 at 13:13 -0700, Chris M. Thomasson wrote:=20
On 3/26/2024 7:51 AM, wij wrote:=20
[...]
=20
Repeating decimals are rational, say
=20
0.142857 142857 142857
=20
That is just 1 / 7 represented in base 10.
=20
Now, think of using a TRNG to create each digit...
=20
That would be, irrational... ;^)
Just repeat the pattern infinitely, then it is irrational.
Nonsense.
=20
As said "=E2=88=80x,a=E2=88=88=E2=84=9A, x-a=E2=88=88=E2=84=9A", if the=
cannot terminate, 1/7 !=3D 0.(142857)=20
=20
Nonsense.
=20
Simply stating random things does not make them so.
=20
I recommend you stick to C++ in this C++ newsgroup.
=20
As for your maths, you'd do better learning some basics of the=20=20
mathematics of real numbers and rational numbers, and that being able to=
find the Unicode characters for some logic symbols does not mean you=20 understand how to write a proof.
=20
=20
On Wed, 2024-03-27 at 12:50 +0100, David Brown wrote:he subtraction a=3D 142857/10^(6*i)
On 26/03/2024 22:43, wij wrote: =20
=20=20
Just repeat the pattern infinitely, then it is irrational. =20
Nonsense.
=20
As said "=E2=88=80x,a=E2=88=88=E2=84=9A, x-a=E2=88=88=E2=84=9A", if t=
=20cannot terminate, 1/7 !=3D 0.(142857)=20
=20
Nonsense.
=20
I am surprise your math. knowledge is so low worse than teenagers.
Am Wed, 27 Mar 2024 20:12:38 +0800=20
schrieb wij <wyniijj5@gmail.com>:
=20
On Wed, 2024-03-27 at 12:50 +0100, David Brown wrote:
On 26/03/2024 22:43, wij wrote:=C2=A0=20
=20
Just repeat the pattern infinitely, then it is irrational.=C2=A0=
the subtraction a=3D 142857/10^(6*i)=20
Nonsense.
=C2=A0=20
As said "=E2=88=80x,a=E2=88=88=E2=84=9A, x-a=E2=88=88=E2=84=9A", if=
=20=20cannot terminate, 1/7 !=3D 0.(142857)=20
=C2=A0=20
Nonsense.
=C2=A0=20
I am surprise your math. knowledge is so low worse than teenagers.
Use the standard trick:
=20
x=3D0.[142857] =3D> 1,000,000*x=3D142857.[142857]
=20
subtract the first equation from the second:
=20
999,999*x=3D142857 =3D> x=3D142857/999,999=3D1/7
=20
On Wed, 2024-03-27 at 13:57 +0100, Ralf Goertz wrote:=20
Am Wed, 27 Mar 2024 20:12:38 +0800
schrieb wij <wyniijj5@gmail.com>:
=20
On Wed, 2024-03-27 at 12:50 +0100, David Brown wrote:
On 26/03/2024 22:43, wij wrote:=C2=A0=20
=20
Just repeat the pattern infinitely, then it is irrational.=C2=A0=
if the subtraction a=3D 142857/10^(6*i)=20
Nonsense.
=C2=A0=20
As said "=E2=88=80x,a=E2=88=88=E2=84=9A, x-a=E2=88=88=E2=84=9A", =
tract=20=20=20cannot terminate, 1/7 !=3D 0.(142857)=20
=C2=A0=20
Nonsense.
=C2=A0=20
I am surprise your math. knowledge is so low worse than teenagers.
Use the standard trick:
=20
x=3D0.[142857] =3D> 1,000,000*x=3D142857.[142857]
=20
subtract the first equation from the second:
=20
999,999*x=3D142857 =3D> x=3D142857/999,999=3D1/7
=20
To determine whether a number x is rational or not, we can repeatedly sub=
rational numbers a? from x. If x-a1-a2-a3-...=3D0 can be verified in fini=te
steps, then x is rational. Otherwise, x is irrational.is
If x is a repeating decimal, proposition "repeating decimal is rational" =
simply false by sematics.By the way, this came to me: =CE=B5-=CE=B4 method was used by people to thi=
=20
=20
On Wed, 2024-03-27 at 12:50 +0100, David Brown wrote:
On 26/03/2024 22:43, wij wrote:
On Tue, 2024-03-26 at 13:13 -0700, Chris M. Thomasson wrote:
On 3/26/2024 7:51 AM, wij wrote:
[...]
Repeating decimals are rational, say
0.142857 142857 142857
That is just 1 / 7 represented in base 10.
Now, think of using a TRNG to create each digit...
That would be, irrational... ;^)
Just repeat the pattern infinitely, then it is irrational.
Nonsense.
As said "∀x,a∈ℚ, x-a∈ℚ", if the subtraction a= 142857/10^(6*i) >>> cannot terminate, 1/7 != 0.(142857)
Nonsense.
I am surprise your math. knowledge is so low worse than teenagers.
Simply stating random things does not make them so.
I recommend you stick to C++ in this C++ newsgroup.
I know. You 'occupied' c/c++ forum and think you are speech police.
For now, this discussion is mainly in comp.theory
But you have shown your knowledge is so so low, don't go there waste our time.
As for your maths, you'd do better learning some basics of the
mathematics of real numbers and rational numbers, and that being able to
find the Unicode characters for some logic symbols does not mean you
understand how to write a proof.
On Wed, 2024-03-27 at 13:57 +0100, Ralf Goertz wrote:
Am Wed, 27 Mar 2024 20:12:38 +0800
schrieb wij <wyniijj5@gmail.com>:
On Wed, 2024-03-27 at 12:50 +0100, David Brown wrote:
On 26/03/2024 22:43, wij wrote:
Just repeat the pattern infinitely, then it is irrational.
Nonsense.
As said "∀x,a∈ℚ, x-a∈ℚ", if the subtraction a= 142857/10^(6*i) >>>>> cannot terminate, 1/7 != 0.(142857)
Nonsense.
I am surprise your math. knowledge is so low worse than teenagers.
Use the standard trick:
x=0.[142857] => 1,000,000*x=142857.[142857]
subtract the first equation from the second:
999,999*x=142857 => x=142857/999,999=1/7
To determine whether a number x is rational or not, we can repeatedly subtract
rational numbers a? from x.
If x-a1-a2-a3-...=0 can be verified in finite
steps, then x is rational.
Otherwise, x is irrational.
If x is a repeating decimal, proposition "repeating decimal is rational" is simply false by sematics.
On Wed, 2024-03-27 at 13:57 +0100, Ralf Goertz wrote:
Am Wed, 27 Mar 2024 20:12:38 +0800
schrieb wij <wyniijj5@gmail.com>:
On Wed, 2024-03-27 at 12:50 +0100, David Brown wrote:
On 26/03/2024 22:43, wij wrote:
Just repeat the pattern infinitely, then it is irrational.
Nonsense.
As said "∀x,a∈ℚ, x-a∈ℚ", if the subtraction a= 142857/10^(6*i) >>>>> cannot terminate, 1/7 != 0.(142857)
Nonsense.
I am surprise your math. knowledge is so low worse than teenagers.
Use the standard trick:
x=0.[142857] => 1,000,000*x=142857.[142857]
subtract the first equation from the second:
999,999*x=142857 => x=142857/999,999=1/7
To determine whether a number x is rational or not, we can repeatedly subtract
rational numbers a? from x. If x-a1-a2-a3-...=0 can be verified in finite steps, then x is rational. Otherwise, x is irrational.
If x is a repeating decimal, proposition "repeating decimal is rational" is simply false by sematics.
On 27/03/2024 14:32, wij wrote:, if the subtraction a=3D 142857/10^(6*i)
On Wed, 2024-03-27 at 13:57 +0100, Ralf Goertz wrote:
Am Wed, 27 Mar 2024 20:12:38 +0800
schrieb wij <wyniijj5@gmail.com>:
=20
On Wed, 2024-03-27 at 12:50 +0100, David Brown wrote:
On 26/03/2024 22:43, wij wrote:
=20=20
Just repeat the pattern infinitely, then it is irrational.
Nonsense.
=C2=A0=C2=A0=20
As said "=E2=88=80x,a=E2=88=88=E2=84=9A, x-a=E2=88=88=E2=84=9A"=
ubtract=20=20=20cannot terminate, 1/7 !=3D 0.(142857)=20
=C2=A0=C2=A0=20
Nonsense.
=C2=A0=C2=A0=20
I am surprise your math. knowledge is so low worse than teenagers.
Use the standard trick:
=20
x=3D0.[142857] =3D> 1,000,000*x=3D142857.[142857]
=20
subtract the first equation from the second:
=20
999,999*x=3D142857 =3D> x=3D142857/999,999=3D1/7
=20
To determine whether a number x is rational or not, we can repeatedly s=
niterational numbers a? from x. If x-a1-a2-a3-...=3D0 can be verified in fi=
" issteps, then x is rational. Otherwise, x is irrational.
If x is a repeating decimal, proposition "repeating decimal is rational=
rationalsimply false by sematics.=20
=20
Let me just ask you two simple questions:
=20
Do you think 1/7 is a rational number or an irrational number?
=20
What do you think the decimal expansion of 1/7 is?
=20
On 26/03/2024 21:13, Chris M. Thomasson wrote:
On 3/26/2024 7:51 AM, wij wrote:
[...]
Repeating decimals are rational, say
0.142857 142857 142857
That is just 1 / 7 represented in base 10.
Obviously (to everyone except perhaps wij).
Now, think of using a TRNG to create each digit...
That would be, irrational... ;^)
That would not be a defined number. I am not convinced it is meaningful
to talk about its properties at all.
On Wed, 2024-03-27 at 13:57 +0100, Ralf Goertz wrote:
Am Wed, 27 Mar 2024 20:12:38 +0800
schrieb wij <wyniijj5@gmail.com>:
On Wed, 2024-03-27 at 12:50 +0100, David Brown wrote:
On 26/03/2024 22:43, wij wrote:
Just repeat the pattern infinitely, then it is irrational.
Nonsense.
As said "∀x,a∈ℚ, x-a∈ℚ", if the subtraction a= 142857/10^(6*i) >>>>> cannot terminate, 1/7 != 0.(142857)
Nonsense.
I am surprise your math. knowledge is so low worse than teenagers.
Use the standard trick:
x=0.[142857] => 1,000,000*x=142857.[142857]
subtract the first equation from the second:
999,999*x=142857 => x=142857/999,999=1/7
To determine whether a number x is rational or not,[...]
On 27/03/2024 14:32, wij wrote:
On Wed, 2024-03-27 at 13:57 +0100, Ralf Goertz wrote:
Am Wed, 27 Mar 2024 20:12:38 +0800
schrieb wij <wyniijj5@gmail.com>:
On Wed, 2024-03-27 at 12:50 +0100, David Brown wrote:
On 26/03/2024 22:43, wij wrote:
Just repeat the pattern infinitely, then it is irrational.
Nonsense.
As said "∀x,a∈ℚ, x-a∈ℚ", if the subtraction a= 142857/10^(6*i) >>>>>> cannot terminate, 1/7 != 0.(142857)
Nonsense.
I am surprise your math. knowledge is so low worse than teenagers.
Use the standard trick:
x=0.[142857] => 1,000,000*x=142857.[142857]
subtract the first equation from the second:
999,999*x=142857 => x=142857/999,999=1/7
To determine whether a number x is rational or not, we can repeatedly
subtract
rational numbers a? from x.
I assume that when you say "rational numbers a?", you mean numbers with finite decimal expansions?
Your method could, I suppose, be used to prove that x is rational - but
not to prove that it is irrational. It is not particularly helpful,
unless you are using it as some way to build up the rationals
inductively from a starting point of "assumed" rationals.
If x-a1-a2-a3-...=0 can be verified in finite
steps, then x is rational.
Correct.
Otherwise, x is irrational.
Incorrect.
All you have proven is that you have not picked appropriate rationals in
the sequence, or that x is a number with a non-finite decimal expansion.
You haven't demonstrated that it is irrational.
Your method here doesn't give you anything new. It boils down to saying that if we assume that all rationals have finite decimal expansions, we
can prove that numbers without finite decimal expansions are not
rational - and that's a simple tautology. The assumption is, of course, wrong.
If x is a repeating decimal, proposition "repeating decimal is
rational" is
simply false by sematics.
Incorrect.
I don't know what you meant to say, but repeating (a.k.a. periodic)
decimals are most certainly rational numbers.
I think that proving it
would be rather easy although I didn't try to do it in rigorous
manner.
The idea of proof is multiplying repeating decimal with period P
by (10**P-1) will produce finite decimal. Which is obviously rational.
On Wed, 2024-03-27 at 16:02 +0100, David Brown wrote:
On 27/03/2024 14:32, wij wrote:rational
On Wed, 2024-03-27 at 13:57 +0100, Ralf Goertz wrote:
Am Wed, 27 Mar 2024 20:12:38 +0800
schrieb wij <wyniijj5@gmail.com>:
On Wed, 2024-03-27 at 12:50 +0100, David Brown wrote:
On 26/03/2024 22:43, wij wrote:
Just repeat the pattern infinitely, then it is irrational.
Nonsense.
As said "∀x,a∈ℚ, x-a∈ℚ", if the subtraction a= 142857/10^(6*i)
cannot terminate, 1/7 != 0.(142857)
Nonsense.
I am surprise your math. knowledge is so low worse than teenagers.
Use the standard trick:
x=0.[142857] => 1,000,000*x=142857.[142857]
subtract the first equation from the second:
999,999*x=142857 => x=142857/999,999=1/7
To determine whether a number x is rational or not, we can repeatedly subtract
rational numbers a? from x. If x-a1-a2-a3-...=0 can be verified in finite >>> steps, then x is rational. Otherwise, x is irrational.
If x is a repeating decimal, proposition "repeating decimal is rational" is >>> simply false by sematics.
Let me just ask you two simple questions:
Do you think 1/7 is a rational number or an irrational number?
What do you think the decimal expansion of 1/7 is?
When converting 1/7 to decimal, the result ≒ 0.(142857), the procedure never terminates which means the conversion is never complete.
On 27/03/2024 10:47, Paavo Helde wrote:
In mathematics, a notion like pi actually defines an irrational number
by fixing its properties. For some such irrational numbers it is
possible to give an infinite algorithm which produces the sequence of
its digits. The funny thing is that after fixing the number there is
no randomness any more, so e.g. a machine computing subsequent digits
of pi would make a pretty poor RNG ;-)
The digits would actually make quite a good RNG - if you had a practical
way to keep getting the digits.
It is not proven, but it is strongly
suspected that pi is normal in all bases - you will get an even
distribution of digits over time, and any finite sequence of digits will occur equally often as other sequences of the same length. This is the
key characteristic of an unbiased random source. No test of the
sequence of digits could determine that it is not from a random source.
There is an algorithm that lets you find digits of the hexadecimal
expansion of pi without finding all the previous digits, which could be useful here.
But it won't be very good for security purposes or other such uses of
random numbers!
On 3/27/2024 9:05 AM, wij wrote:=9A", if the subtraction a=3D 142857/10^(6*i)
On Wed, 2024-03-27 at 16:02 +0100, David Brown wrote:
On 27/03/2024 14:32, wij wrote:
On Wed, 2024-03-27 at 13:57 +0100, Ralf Goertz wrote:
Am Wed, 27 Mar 2024 20:12:38 +0800
schrieb wij <wyniijj5@gmail.com>:
=20
On Wed, 2024-03-27 at 12:50 +0100, David Brown wrote:
On 26/03/2024 22:43, wij wrote:
=20=20
Just repeat the pattern infinitely, then it is irrational.
Nonsense.
=C2=A0=C2=A0=C2=A0=20
As said "=E2=88=80x,a=E2=88=88=E2=84=9A, x-a=E2=88=88=E2=84=
rs.=20cannot terminate, 1/7 !=3D 0.(142857)=20
=C2=A0=C2=A0=C2=A0=20
Nonsense.
=C2=A0=C2=A0=C2=A0=20
I am surprise your math. knowledge is so low worse than teenage=
ly subtract=20=20
Use the standard trick:
=20
x=3D0.[142857] =3D> 1,000,000*x=3D142857.[142857]
=20
subtract the first equation from the second:
=20
999,999*x=3D142857 =3D> x=3D142857/999,999=3D1/7
=20
To determine whether a number x is rational or not, we can repeated=
n finiterational numbers a? from x. If x-a1-a2-a3-...=3D0 can be verified i=
onal" issteps, then x is rational. Otherwise, x is irrational.
If x is a repeating decimal, proposition "repeating decimal is rati=
ocedurerationalsimply false by sematics.=20
=20
Let me just ask you two simple questions:
=20
Do you think 1/7 is a rational number or an irrational number?
=20
=20
What do you think the decimal expansion of 1/7 is?=20
=20
When converting 1/7 to decimal, the result =E2=89=92 0.(142857), the pr=
=20never terminates which means the conversion is never complete.=20
=20
=20
=20
You can stop iteration as soon as you detect a cycle, or period if you=
will. In 1/7, say it took 6 iterations to hit the period... Sound okay?
On Wed, 2024-03-27 at 13:40 -0700, Chris M. Thomasson wrote:
On 3/27/2024 9:05 AM, wij wrote:
On Wed, 2024-03-27 at 16:02 +0100, David Brown wrote:
On 27/03/2024 14:32, wij wrote:rational
On Wed, 2024-03-27 at 13:57 +0100, Ralf Goertz wrote:
Am Wed, 27 Mar 2024 20:12:38 +0800
schrieb wij <wyniijj5@gmail.com>:
On Wed, 2024-03-27 at 12:50 +0100, David Brown wrote:Use the standard trick:
On 26/03/2024 22:43, wij wrote:
Just repeat the pattern infinitely, then it is irrational.
Nonsense.
As said "∀x,a∈ℚ, x-a∈ℚ", if the subtraction a= 142857/10^(6*i)
cannot terminate, 1/7 != 0.(142857)
Nonsense.
I am surprise your math. knowledge is so low worse than teenagers. >>>>>>
x=0.[142857] => 1,000,000*x=142857.[142857]
subtract the first equation from the second:
999,999*x=142857 => x=142857/999,999=1/7
To determine whether a number x is rational or not, we can repeatedly subtract
rational numbers a? from x. If x-a1-a2-a3-...=0 can be verified in finite >>>>> steps, then x is rational. Otherwise, x is irrational.
If x is a repeating decimal, proposition "repeating decimal is rational" is
simply false by sematics.
Let me just ask you two simple questions:
Do you think 1/7 is a rational number or an irrational number?
What do you think the decimal expansion of 1/7 is?
When converting 1/7 to decimal, the result ≒ 0.(142857), the procedure >>> never terminates which means the conversion is never complete.
You can stop iteration as soon as you detect a cycle, or period if you
will. In 1/7, say it took 6 iterations to hit the period... Sound okay?
Stupid! It is an infinite string.
Cycle or period can only be determined for
finite string.
On Wed, 2024-03-27 at 13:40 -0700, Chris M. Thomasson wrote:
On 3/27/2024 9:05 AM, wij wrote:
On Wed, 2024-03-27 at 16:02 +0100, David Brown wrote:
On 27/03/2024 14:32, wij wrote:rational
On Wed, 2024-03-27 at 13:57 +0100, Ralf Goertz wrote:
Am Wed, 27 Mar 2024 20:12:38 +0800
schrieb wij <wyniijj5@gmail.com>:
On Wed, 2024-03-27 at 12:50 +0100, David Brown wrote:Use the standard trick:
On 26/03/2024 22:43, wij wrote:
Just repeat the pattern infinitely, then it is irrational.
Nonsense.
As said "∀x,a∈ℚ, x-a∈ℚ", if the subtraction a= 142857/10^(6*i)
cannot terminate, 1/7 != 0.(142857)
Nonsense.
I am surprise your math. knowledge is so low worse than teenagers. >>>>>>
x=0.[142857] => 1,000,000*x=142857.[142857]
subtract the first equation from the second:
999,999*x=142857 => x=142857/999,999=1/7
To determine whether a number x is rational or not, we can repeatedly subtract
rational numbers a? from x. If x-a1-a2-a3-...=0 can be verified in finite >>>>> steps, then x is rational. Otherwise, x is irrational.
If x is a repeating decimal, proposition "repeating decimal is rational" is
simply false by sematics.
Let me just ask you two simple questions:
Do you think 1/7 is a rational number or an irrational number?
What do you think the decimal expansion of 1/7 is?
When converting 1/7 to decimal, the result ≒ 0.(142857), the procedure >>> never terminates which means the conversion is never complete.
You can stop iteration as soon as you detect a cycle, or period if you
will. In 1/7, say it took 6 iterations to hit the period... Sound okay?
Stupid! It is an infinite string. Cycle or period can only be determined for finite string.
On 26/03/2024 21:13, Chris M. Thomasson wrote:
On 3/26/2024 7:51 AM, wij wrote:
[...]
Repeating decimals are rational, say
0.142857 142857 142857
That is just 1 / 7 represented in base 10.
Obviously (to everyone except perhaps wij).
Now, think of using a TRNG to create each digit...
That would be, irrational... ;^)
That would not be a defined number. I am not convinced it is meaningful
to talk about its properties at all.
On Wed, 2024-03-27 at 12:50 +0100, David Brown wrote:
On 26/03/2024 22:43, wij wrote:
On Tue, 2024-03-26 at 13:13 -0700, Chris M. Thomasson wrote:
On 3/26/2024 7:51 AM, wij wrote:
[...]
Repeating decimals are rational, say
0.142857 142857 142857
That is just 1 / 7 represented in base 10.
Now, think of using a TRNG to create each digit...
That would be, irrational... ;^)
Just repeat the pattern infinitely, then it is irrational.
Nonsense.
As said "∀x,a∈ℚ, x-a∈ℚ", if the subtraction a= 142857/10^(6*i) >>> cannot terminate, 1/7 != 0.(142857)
Nonsense.
I am surprise your math. knowledge is so low worse than teenagers.
Simply stating random things does not make them so.
I recommend you stick to C++ in this C++ newsgroup.
I know. You 'occupied' c/c++ forum and think you are speech police.
For now, this discussion is mainly in comp.theory
But you have shown your knowledge is so so low, don't go there waste our time.
As for your maths, you'd do better learning some basics of the
mathematics of real numbers and rational numbers, and that being able to
find the Unicode characters for some logic symbols does not mean you
understand how to write a proof.
On Wed, 2024-03-27 at 16:02 +0100, David Brown wrote:
On 27/03/2024 14:32, wij wrote:rational
On Wed, 2024-03-27 at 13:57 +0100, Ralf Goertz wrote:
Am Wed, 27 Mar 2024 20:12:38 +0800
schrieb wij <wyniijj5@gmail.com>:
On Wed, 2024-03-27 at 12:50 +0100, David Brown wrote:
On 26/03/2024 22:43, wij wrote:
Just repeat the pattern infinitely, then it is irrational.
Nonsense.
As said "∀x,a∈ℚ, x-a∈ℚ", if the subtraction a= 142857/10^(6*i)
cannot terminate, 1/7 != 0.(142857)
Nonsense.
I am surprise your math. knowledge is so low worse than teenagers.
Use the standard trick:
x=0.[142857] => 1,000,000*x=142857.[142857]
subtract the first equation from the second:
999,999*x=142857 => x=142857/999,999=1/7
To determine whether a number x is rational or not, we can repeatedly subtract
rational numbers a? from x. If x-a1-a2-a3-...=0 can be verified in finite >>> steps, then x is rational. Otherwise, x is irrational.
If x is a repeating decimal, proposition "repeating decimal is rational" is >>> simply false by sematics.
Let me just ask you two simple questions:
Do you think 1/7 is a rational number or an irrational number?
What do you think the decimal expansion of 1/7 is?
When converting 1/7 to decimal, the result ≒ 0.(142857), the procedure never terminates which means the conversion is never complete.
On Wed, 2024-03-27 at 13:40 -0700, Chris M. Thomasson wrote:
On 3/27/2024 9:05 AM, wij wrote:
On Wed, 2024-03-27 at 16:02 +0100, David Brown wrote:
On 27/03/2024 14:32, wij wrote:rational
On Wed, 2024-03-27 at 13:57 +0100, Ralf Goertz wrote:
Am Wed, 27 Mar 2024 20:12:38 +0800
schrieb wij <wyniijj5@gmail.com>:
On Wed, 2024-03-27 at 12:50 +0100, David Brown wrote:Use the standard trick:
On 26/03/2024 22:43, wij wrote:
Just repeat the pattern infinitely, then it is irrational.
Nonsense.
As said "∀x,a∈ℚ, x-a∈ℚ", if the subtraction a= 142857/10^(6*i)
cannot terminate, 1/7 != 0.(142857)
Nonsense.
I am surprise your math. knowledge is so low worse than teenagers. >>>>>>
x=0.[142857] => 1,000,000*x=142857.[142857]
subtract the first equation from the second:
999,999*x=142857 => x=142857/999,999=1/7
To determine whether a number x is rational or not, we can repeatedly subtract
rational numbers a? from x. If x-a1-a2-a3-...=0 can be verified in finite >>>>> steps, then x is rational. Otherwise, x is irrational.
If x is a repeating decimal, proposition "repeating decimal is rational" is
simply false by sematics.
Let me just ask you two simple questions:
Do you think 1/7 is a rational number or an irrational number?
What do you think the decimal expansion of 1/7 is?
When converting 1/7 to decimal, the result ≒ 0.(142857), the procedure >>> never terminates which means the conversion is never complete.
You can stop iteration as soon as you detect a cycle, or period if you
will. In 1/7, say it took 6 iterations to hit the period... Sound okay?
Stupid! It is an infinite string. Cycle or period can only be determined for finite string.
On 3/27/2024 3:31 AM, David Brown wrote:
On 26/03/2024 21:13, Chris M. Thomasson wrote:
On 3/26/2024 7:51 AM, wij wrote:
[...]
Repeating decimals are rational, say
0.142857 142857 142857
That is just 1 / 7 represented in base 10.
Obviously (to everyone except perhaps wij).
;^)
Now, think of using a TRNG to create each digit...
That would be, irrational... ;^)
That would not be a defined number. I am not convinced it is
meaningful to talk about its properties at all.
Well, it would be a "number" at any finite view of it? Or, is that just moronic thinking?
For some reason it makes me think of infinite
convergents of continued fractions.
Can we gain a rational that can
approximate any irrational up to a certain precision, so to speak?
On 27/03/2024 17:05, wij wrote:=9A", if the subtraction a=3D 142857/10^(6*i)
On Wed, 2024-03-27 at 16:02 +0100, David Brown wrote:
On 27/03/2024 14:32, wij wrote:
On Wed, 2024-03-27 at 13:57 +0100, Ralf Goertz wrote:
Am Wed, 27 Mar 2024 20:12:38 +0800
schrieb wij <wyniijj5@gmail.com>:
=20
On Wed, 2024-03-27 at 12:50 +0100, David Brown wrote:
On 26/03/2024 22:43, wij wrote:
=20=20
Just repeat the pattern infinitely, then it is irrational.
Nonsense.
=C2=A0=C2=A0=C2=A0=20
As said "=E2=88=80x,a=E2=88=88=E2=84=9A, x-a=E2=88=88=E2=84=
rs.=20cannot terminate, 1/7 !=3D 0.(142857)=20
=C2=A0=C2=A0=C2=A0=20
Nonsense.
=C2=A0=C2=A0=C2=A0=20
I am surprise your math. knowledge is so low worse than teenage=
ly subtract=20=20
Use the standard trick:
=20
x=3D0.[142857] =3D> 1,000,000*x=3D142857.[142857]
=20
subtract the first equation from the second:
=20
999,999*x=3D142857 =3D> x=3D142857/999,999=3D1/7
=20
To determine whether a number x is rational or not, we can repeated=
n finiterational numbers a? from x. If x-a1-a2-a3-...=3D0 can be verified i=
onal" issteps, then x is rational. Otherwise, x is irrational.
If x is a repeating decimal, proposition "repeating decimal is rati=
ocedurerationalsimply false by sematics.=20
=20
Let me just ask you two simple questions:
=20
Do you think 1/7 is a rational number or an irrational number?
=20
=20
What do you think the decimal expansion of 1/7 is?=20
=20
When converting 1/7 to decimal, the result =E2=89=92 0.(142857), the pr=
gree=20never terminates which means the conversion is never complete.=20
=20
It is a repeating decimal.=C2=A0 If you try to write it all out, then I a=
you will not finish.=C2=A0 That does not mean it is not the decimal expan=sion=20
of 1/7 - the list of multiples of (negative) powers of 10 which sum up==20
to 1/7.=C2=A0 You just need a better notation so that you can finish the =task=20
- and 0.(142857), as you wrote, is one such notation.'s=20
=20
(I have no idea what you think the symbol "=E2=89=92" might mean.)
=20
But you agree that 0.(142857) is the decimal expansion of 1/7, even=20
though you could not write it out long-hand, and you agree that 1/7 i=20 rational.=C2=A0 And clearly 0.(142857) is a repeating decimal, since that=
what the notation means.
=20
I can't see how you can still misunderstand this.
=20
On 27/03/2024 22:39, wij wrote:al.
On Wed, 2024-03-27 at 13:40 -0700, Chris M. Thomasson wrote:
On 3/27/2024 9:05 AM, wij wrote:
On Wed, 2024-03-27 at 16:02 +0100, David Brown wrote:
On 27/03/2024 14:32, wij wrote:
On Wed, 2024-03-27 at 13:57 +0100, Ralf Goertz wrote:
Am Wed, 27 Mar 2024 20:12:38 +0800
schrieb wij <wyniijj5@gmail.com>:
=20
On Wed, 2024-03-27 at 12:50 +0100, David Brown wrote:
On 26/03/2024 22:43, wij wrote:
=20
Just repeat the pattern infinitely, then it is irration=
=E2=84=9A", if the subtraction a=3D 142857/10^(6*i)=20
Nonsense.
=C2=A0=C2=A0=C2=A0=C2=A0=20
As said "=E2=88=80x,a=E2=88=88=E2=84=9A, x-a=E2=88=88=
nagers.=20cannot terminate, 1/7 !=3D 0.(142857) =C2=A0=C2=A0=C2=A0=C2=A0=20=20
Nonsense.
=C2=A0=C2=A0=C2=A0=C2=A0=20
I am surprise your math. knowledge is so low worse than tee=
atedly subtract=20=20
Use the standard trick:
=20
x=3D0.[142857] =3D> 1,000,000*x=3D142857.[142857]
=20
subtract the first equation from the second:
=20
999,999*x=3D142857 =3D> x=3D142857/999,999=3D1/7
=20
To determine whether a number x is rational or not, we can repe=
ed in finiterational numbers a? from x. If x-a1-a2-a3-...=3D0 can be verifi=
rational" issteps, then x is rational. Otherwise, x is irrational.
If x is a repeating decimal, proposition "repeating decimal is =
e procedurerationalsimply false by sematics.=20
=20
Let me just ask you two simple questions:
=20
Do you think 1/7 is a rational number or an irrational number?
=20
=20
What do you think the decimal expansion of 1/7 is?=20
=20
When converting 1/7 to decimal, the result =E2=89=92 0.(142857), th=
unever terminates which means the conversion is never complete.=20
=20
=20
=20
You can stop iteration as soon as you detect a cycle, or period if yo=
y?will. In 1/7, say it took 6 iterations to hit the period... Sound oka=
d for=20
Stupid! It is an infinite string. Cycle or period can only be determine=
=20finite string.=20
=20
Nonsense.
=20
You /know/ the cycle for the infinite decimal expansion for 1/7 - it is=
the digits "142857", repeated every 6 digits in the decimal expansion.==20
Again, that's what the notation 0.(142857) - /your/ choice of notation,==20
so presumably familiar to you - means."0.(142857)" is pre-determined and specified not detected. (unless I misund= erstood
=20
On 27/03/2024 21:17, Chris M. Thomasson wrote:
On 3/27/2024 3:31 AM, David Brown wrote:
On 26/03/2024 21:13, Chris M. Thomasson wrote:
On 3/26/2024 7:51 AM, wij wrote:
[...]
Repeating decimals are rational, say
0.142857 142857 142857
That is just 1 / 7 represented in base 10.
Obviously (to everyone except perhaps wij).
;^)
Now, think of using a TRNG to create each digit...
That would be, irrational... ;^)
That would not be a defined number. I am not convinced it is
meaningful to talk about its properties at all.
Well, it would be a "number" at any finite view of it? Or, is that
just moronic thinking?
It is not a defined number in any way - you are explicitly using a
procedure which does not give any defined or deterministic results. How could it be a number when no two people could agree to its value?
Now, if you were to take a TRNG and generate a series of digits, writing each one down as you go along, then at any given time you would have a finite decimal expansion which would be a number (a rational, if that matters to you). But that is a particular instance of creating such a number - the procedure you describe does not define a number nor is it a number in itself.
For some reason it makes me think of infinite convergents of continued
fractions.
I am not sure you are helping yourself or anyone else there.
A sequence
of digits obtained randomly does not converge.
Can we gain a rational that can approximate any irrational up to a
certain precision, so to speak?
Of course we can. That is a completely separate issue.
If you have a real number "x" (positive for simplicity), and a precision "e", then we can easily find a rational number q such that |x - q| < e.
We do so by letting "i" be an integer greater than 1/e. Let "j" be the first integer less than or equal to i * x.
So
j <= (i * x) < (j + 1)
and thus
j / i <= x < (j + 1) / i
q0 = j / i and q1 = (j + 1) / i are both rational numbers, and "x" lies between them.
q1 - q0 = 1 / i, which is less than "e" since "i" is greater than 1/e.
So since (q1 - q0) < e, and q0 <= x < q1, then we can see that "x" is "trapped" between two rationals that are both less than "e" from "x".
You can approximate any real number as closely as you like with a series
of rational numbers. That is basically what it means to say that the
real numbers "complete" the rationals - the fill in the gaps that can be expressed as the limits of bounded sequences of rationals where the
limit is not itself a rational.
Snipet from https://sourceforge.net/projects/cscall/files/MisFiles/RealNumber-en.txt/download
...
Real Nunmber(ℝ)::= {x| x is represented by n-ary <fixed_point_number>, the
digits may be infinitely long }
Note: This definition implies that repeating decimals are irrational number.
Let's list a common magic proof in the way as a brief explanation:
(1) x= 0.999...
(2) 10x= 9+x // 10x= 9.999...
(3) 9x=9
(4) x=1
Ans: There is no axiom or theorem to prove (1) => (2).
Note: If the steps of converting a number x to <fixed_point_number> is not
finite, x is not a ratio of two integers, because the following
statement is always true: ∀x,a∈ℚ, x-a∈ℚ
---End of quote
wij <wyniijj5@gmail.com> writes:Number-en.txt/download
Snipet from https://sourceforge.net/projects/cscall/files/MisFiles/Real=
_number>, the=20
...
Real Nunmber(=E2=84=9D)::=3D {x| x is represented by n-ary <fixed_point=
irrational number.=C2=A0=C2=A0 digits may be infinitely long }
=20
=C2=A0=C2=A0 Note: This definition implies that repeating decimals are =
gic proof in the way as a brief explanation:=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0 Let's list a common ma=
..999...=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0 (1) x=3D 0=
9+x=C2=A0 // 10x=3D 9.999...=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0 (2) 10x=3D=
=C2=A0=C2=A0=C2=A0=20=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0 (3) 9x=3D9=
or theorem to prove (1) =3D> (2).=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0 (4) x=3D1 =C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0 Ans: There is no axiom=
t_number> is not=20
=C2=A0=C2=A0 Note: If the steps of converting a number x to <fixed_poin=
io of two integers, because the following=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0 finite, x is not a rat=
ue: =E2=88=80x,a=E2=88=88=E2=84=9A, x-a=E2=88=88=E2=84=9A=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0 statement is always tr=
=20=20
---End of quote
What does the arity of the "n-ary <fixed_point_number>" refer to? The
base or the number of digits?
=20
Any rational number can be represented with a single fractional digit
when represented in the base of its denominator.
=20
I.e., 3/7 =3D 0.3 (base 7)
=20
And as far as I am aware, the rationality of a number does not depend on
its representation...
=20
-Stefan
=20
On Thu, 2024-03-28 at 18:17 +0100, David Brown wrote:
On 27/03/2024 22:39, wij wrote:"0.(142857)" is pre-determined and specified not detected. (unless I misunderstood
On Wed, 2024-03-27 at 13:40 -0700, Chris M. Thomasson wrote:
On 3/27/2024 9:05 AM, wij wrote:Stupid! It is an infinite string. Cycle or period can only be determined for
On Wed, 2024-03-27 at 16:02 +0100, David Brown wrote:
On 27/03/2024 14:32, wij wrote:rational
On Wed, 2024-03-27 at 13:57 +0100, Ralf Goertz wrote:
Am Wed, 27 Mar 2024 20:12:38 +0800
schrieb wij <wyniijj5@gmail.com>:
On Wed, 2024-03-27 at 12:50 +0100, David Brown wrote:Use the standard trick:
On 26/03/2024 22:43, wij wrote:
Nonsense.
Just repeat the pattern infinitely, then it is irrational. >>>>>>>>>>
As said "∀x,a∈ℚ, x-a∈ℚ", if the subtraction a= 142857/10^(6*i)
cannot terminate, 1/7 != 0.(142857)
Nonsense.
I am surprise your math. knowledge is so low worse than teenagers. >>>>>>>>
x=0.[142857] => 1,000,000*x=142857.[142857]
subtract the first equation from the second:
999,999*x=142857 => x=142857/999,999=1/7
To determine whether a number x is rational or not, we can repeatedly subtract
rational numbers a? from x. If x-a1-a2-a3-...=0 can be verified in finite
steps, then x is rational. Otherwise, x is irrational.
If x is a repeating decimal, proposition "repeating decimal is rational" is
simply false by sematics.
Let me just ask you two simple questions:
Do you think 1/7 is a rational number or an irrational number?
What do you think the decimal expansion of 1/7 is?
When converting 1/7 to decimal, the result ≒ 0.(142857), the procedure >>>>> never terminates which means the conversion is never complete.
You can stop iteration as soon as you detect a cycle, or period if you >>>> will. In 1/7, say it took 6 iterations to hit the period... Sound okay? >>>
finite string.
Nonsense.
You /know/ the cycle for the infinite decimal expansion for 1/7 - it is
the digits "142857", repeated every 6 digits in the decimal expansion.
Again, that's what the notation 0.(142857) - /your/ choice of notation,
so presumably familiar to you - means.
what Chris said)
On Thu, 2024-03-28 at 18:16 +0100, David Brown wrote:
On 27/03/2024 17:05, wij wrote:
On Wed, 2024-03-27 at 16:02 +0100, David Brown wrote:
On 27/03/2024 14:32, wij wrote:rational
On Wed, 2024-03-27 at 13:57 +0100, Ralf Goertz wrote:
Am Wed, 27 Mar 2024 20:12:38 +0800
schrieb wij <wyniijj5@gmail.com>:
On Wed, 2024-03-27 at 12:50 +0100, David Brown wrote:Use the standard trick:
On 26/03/2024 22:43, wij wrote:
Just repeat the pattern infinitely, then it is irrational.
Nonsense.
As said "∀x,a∈ℚ, x-a∈ℚ", if the subtraction a= 142857/10^(6*i)
cannot terminate, 1/7 != 0.(142857)
Nonsense.
I am surprise your math. knowledge is so low worse than teenagers. >>>>>>
x=0.[142857] => 1,000,000*x=142857.[142857]
subtract the first equation from the second:
999,999*x=142857 => x=142857/999,999=1/7
To determine whether a number x is rational or not, we can repeatedly subtract
rational numbers a? from x. If x-a1-a2-a3-...=0 can be verified in finite >>>>> steps, then x is rational. Otherwise, x is irrational.
If x is a repeating decimal, proposition "repeating decimal is rational" is
simply false by sematics.
Let me just ask you two simple questions:
Do you think 1/7 is a rational number or an irrational number?
What do you think the decimal expansion of 1/7 is?
When converting 1/7 to decimal, the result ≒ 0.(142857), the procedure >>> never terminates which means the conversion is never complete.
It is a repeating decimal. If you try to write it all out, then I agree
you will not finish. That does not mean it is not the decimal expansion
of 1/7 - the list of multiples of (negative) powers of 10 which sum up
to 1/7. You just need a better notation so that you can finish the task
- and 0.(142857), as you wrote, is one such notation.
(I have no idea what you think the symbol "≒" might mean.)
But you agree that 0.(142857) is the decimal expansion of 1/7, even
though you could not write it out long-hand, and you agree that 1/7 i
rational. And clearly 0.(142857) is a repeating decimal, since that's
what the notation means.
I can't see how you can still misunderstand this.
You are restating your assertion without proof, again. I have provided mine. (If you say that is you proof, I will say it is invalid).
On 3/28/2024 10:47 AM, David Brown wrote:
On 27/03/2024 21:17, Chris M. Thomasson wrote:
On 3/27/2024 3:31 AM, David Brown wrote:
On 26/03/2024 21:13, Chris M. Thomasson wrote:
On 3/26/2024 7:51 AM, wij wrote:
[...]
Repeating decimals are rational, say
0.142857 142857 142857
That is just 1 / 7 represented in base 10.
Obviously (to everyone except perhaps wij).
;^)
Now, think of using a TRNG to create each digit...
That would be, irrational... ;^)
That would not be a defined number. I am not convinced it is
meaningful to talk about its properties at all.
Well, it would be a "number" at any finite view of it? Or, is that
just moronic thinking?
It is not a defined number in any way - you are explicitly using a
procedure which does not give any defined or deterministic results.
How could it be a number when no two people could agree to its value?
Now, if you were to take a TRNG and generate a series of digits,
writing each one down as you go along, then at any given time you
would have a finite decimal expansion which would be a number (a
rational, if that matters to you). But that is a particular instance
of creating such a number - the procedure you describe does not define
a number nor is it a number in itself.
Would it be fair to say it defines a random number that is an actual
number at every finite step... Is that a stupid thought?
For some reason it makes me think of infinite convergents of
continued fractions.
I am not sure you are helping yourself or anyone else there.
Well, shit happens. Just some fun with numbers? ;^)
A sequence of digits obtained randomly does not converge.
True, but at every "step", it can be a rational? I just think it can be
a fun study, so to speak...
Can we gain a rational that can approximate any irrational up to a
certain precision, so to speak?
Of course we can. That is a completely separate issue.
Yeah. Well, okay. I agree.
If you have a real number "x" (positive for simplicity), and a
precision "e", then we can easily find a rational number q such that
|x - q| < e. We do so by letting "i" be an integer greater than 1/e.
Let "j" be the first integer less than or equal to i * x.
Using a TRNG to "generate" x, and a precision, say when we stop drawing digits from the TRNG, would create a rational. Right?
So
j <= (i * x) < (j + 1)
and thus
j / i <= x < (j + 1) / i
q0 = j / i and q1 = (j + 1) / i are both rational numbers, and "x"
lies between them.
q1 - q0 = 1 / i, which is less than "e" since "i" is greater than 1/e.
So since (q1 - q0) < e, and q0 <= x < q1, then we can see that "x" is
"trapped" between two rationals that are both less than "e" from "x".
You can approximate any real number as closely as you like with a
series of rational numbers. That is basically what it means to say
that the real numbers "complete" the rationals - the fill in the gaps
that can be expressed as the limits of bounded sequences of rationals
where the limit is not itself a rational.
Are you saying that a number constructed digit-by-digit using a TRNG is undefined? Its not a number, however it creates many numbers during the construction process? Fair enough?
Ahhh shit, this is just me having some fun. Sorry.
On 28/03/2024 19:23, wij wrote:al.
On Thu, 2024-03-28 at 18:16 +0100, David Brown wrote:
On 27/03/2024 17:05, wij wrote:
On Wed, 2024-03-27 at 16:02 +0100, David Brown wrote:
On 27/03/2024 14:32, wij wrote:
On Wed, 2024-03-27 at 13:57 +0100, Ralf Goertz wrote:
Am Wed, 27 Mar 2024 20:12:38 +0800
schrieb wij <wyniijj5@gmail.com>:
=20
On Wed, 2024-03-27 at 12:50 +0100, David Brown wrote:
On 26/03/2024 22:43, wij wrote:
=20
Just repeat the pattern infinitely, then it is irration=
=E2=84=9A", if the subtraction a=3D 142857/10^(6*i)=20
Nonsense.
=C2=A0=C2=A0=C2=A0=C2=A0=20
As said "=E2=88=80x,a=E2=88=88=E2=84=9A, x-a=E2=88=88=
nagers.=20cannot terminate, 1/7 !=3D 0.(142857) =C2=A0=C2=A0=C2=A0=C2=A0=20=20
Nonsense.
=C2=A0=C2=A0=C2=A0=C2=A0=20
I am surprise your math. knowledge is so low worse than tee=
atedly subtract=20=20
Use the standard trick:
=20
x=3D0.[142857] =3D> 1,000,000*x=3D142857.[142857]
=20
subtract the first equation from the second:
=20
999,999*x=3D142857 =3D> x=3D142857/999,999=3D1/7
=20
To determine whether a number x is rational or not, we can repe=
ed in finiterational numbers a? from x. If x-a1-a2-a3-...=3D0 can be verifi=
rational" issteps, then x is rational. Otherwise, x is irrational.
If x is a repeating decimal, proposition "repeating decimal is =
e procedurerationalsimply false by sematics.=20
=20
Let me just ask you two simple questions:
=20
Do you think 1/7 is a rational number or an irrational number?
=20
=20
What do you think the decimal expansion of 1/7 is?=20
=20
When converting 1/7 to decimal, the result =E2=89=92 0.(142857), th=
I agreenever terminates which means the conversion is never complete.=20
=20
It is a repeating decimal.=C2=A0 If you try to write it all out, then=
xpansionyou will not finish.=C2=A0 That does not mean it is not the decimal e=
pof 1/7 - the list of multiples of (negative) powers of 10 which sum u=
the taskto 1/7.=C2=A0 You just need a better notation so that you can finish =
that's- and 0.(142857), as you wrote, is one such notation.
=20
(I have no idea what you think the symbol "=E2=89=92" might mean.)
=20
But you agree that 0.(142857) is the decimal expansion of 1/7, even though you could not write it out long-hand, and you agree that 1/7 i rational.=C2=A0 And clearly 0.(142857) is a repeating decimal, since =
mine.what the notation means.=20
=20
I can't see how you can still misunderstand this.
=20
You are restating your assertion without proof, again. I have provided =
=20(If you say that is you proof, I will say it is invalid).=20
=20
=20
There is no point in giving you a rigorous proof that 0.(142857) is the=
decimal expansion of 1/7, if that is what you are contesting.=C2=A0 To be==20
fully rigorous, it requires an understanding of the definition of the=20n't=20
real numbers, sequence limits, and the meaning and validity of=20
operations on infinite sequences.=C2=A0 You have demonstrated that you do=
understand any of that.=C2=A0 You have learned a few of the terms, but fa=iled=20
to understand the concepts.=C2=A0 Oh, and it also requires understanding =what=20
a proof is, which again is clearly outside your expertise.at=20
=20
Ralf gave a proof earlier - it is still in the quoted material above.=20
That is as good as we can get at your level of mathematical=20 understanding.=C2=A0 To be more rigorous, we would need to demonstrate th=
the manipulation (multiplication by a finite integer, and subtraction of==20
sequences) of infinite decimal expansions is valid.=C2=A0 That is all=20 standard stuff, known to mathematics students the world over, but you=20=20
are not nearly ready.
=20
You are going to have to go back-track a long way in what you think you=
know about mathematics.=C2=A0 Somewhere along the line in your education,==20
you've got things badly wrong.=C2=A0 And instead of stopping up and tryin=g to=20
figure out why everyone else is saying something different from you, or==20
asking your teachers for help, you have battered on with your mistakes,==20
leading you to sillier and steadily less logical conclusions.pend=20
=20
I think mathematics is a great hobby.=C2=A0 It's a shame to see someone s=
their time and effort on doing it so badly.
=20
On Fri, 2024-03-29 at 11:53 +0100, David Brown wrote:
On 28/03/2024 19:23, wij wrote:
On Thu, 2024-03-28 at 18:16 +0100, David Brown wrote:
On 27/03/2024 17:05, wij wrote:
On Wed, 2024-03-27 at 16:02 +0100, David Brown wrote:
On 27/03/2024 14:32, wij wrote:rational
On Wed, 2024-03-27 at 13:57 +0100, Ralf Goertz wrote:
Am Wed, 27 Mar 2024 20:12:38 +0800
schrieb wij <wyniijj5@gmail.com>:
On Wed, 2024-03-27 at 12:50 +0100, David Brown wrote:Use the standard trick:
On 26/03/2024 22:43, wij wrote:
Nonsense.
Just repeat the pattern infinitely, then it is irrational. >>>>>>>>>>
As said "∀x,a∈ℚ, x-a∈ℚ", if the subtraction a= 142857/10^(6*i)
cannot terminate, 1/7 != 0.(142857)
Nonsense.
I am surprise your math. knowledge is so low worse than teenagers. >>>>>>>>
x=0.[142857] => 1,000,000*x=142857.[142857]
subtract the first equation from the second:
999,999*x=142857 => x=142857/999,999=1/7
To determine whether a number x is rational or not, we can repeatedly subtract
rational numbers a? from x. If x-a1-a2-a3-...=0 can be verified in finite
steps, then x is rational. Otherwise, x is irrational.
If x is a repeating decimal, proposition "repeating decimal is rational" is
simply false by sematics.
Let me just ask you two simple questions:
Do you think 1/7 is a rational number or an irrational number?
What do you think the decimal expansion of 1/7 is?
When converting 1/7 to decimal, the result ≒ 0.(142857), the procedure >>>>> never terminates which means the conversion is never complete.
It is a repeating decimal. If you try to write it all out, then I agree >>>> you will not finish. That does not mean it is not the decimal expansion >>>> of 1/7 - the list of multiples of (negative) powers of 10 which sum up >>>> to 1/7. You just need a better notation so that you can finish the task >>>> - and 0.(142857), as you wrote, is one such notation.
(I have no idea what you think the symbol "≒" might mean.)
But you agree that 0.(142857) is the decimal expansion of 1/7, even
though you could not write it out long-hand, and you agree that 1/7 i
rational. And clearly 0.(142857) is a repeating decimal, since that's >>>> what the notation means.
I can't see how you can still misunderstand this.
You are restating your assertion without proof, again. I have provided mine.
(If you say that is you proof, I will say it is invalid).
There is no point in giving you a rigorous proof that 0.(142857) is the
decimal expansion of 1/7, if that is what you are contesting. To be
fully rigorous, it requires an understanding of the definition of the
real numbers, sequence limits, and the meaning and validity of
operations on infinite sequences. You have demonstrated that you don't
understand any of that. You have learned a few of the terms, but failed
to understand the concepts. Oh, and it also requires understanding what
a proof is, which again is clearly outside your expertise.
Ralf gave a proof earlier - it is still in the quoted material above.
That is as good as we can get at your level of mathematical
understanding. To be more rigorous, we would need to demonstrate that
the manipulation (multiplication by a finite integer, and subtraction of
sequences) of infinite decimal expansions is valid. That is all
standard stuff, known to mathematics students the world over, but you
are not nearly ready.
You are going to have to go back-track a long way in what you think you
know about mathematics. Somewhere along the line in your education,
you've got things badly wrong. And instead of stopping up and trying to
figure out why everyone else is saying something different from you, or
asking your teachers for help, you have battered on with your mistakes,
leading you to sillier and steadily less logical conclusions.
I think mathematics is a great hobby. It's a shame to see someone spend
their time and effort on doing it so badly.
Have you ever wondered why you cannot prove something you hold true for granted
for so long?
If you cannot provide a proof, what you said above only make you more a sinner.
On 29/03/2024 16:14, wij wrote:tional.
On Fri, 2024-03-29 at 11:53 +0100, David Brown wrote:
On 28/03/2024 19:23, wij wrote:
On Thu, 2024-03-28 at 18:16 +0100, David Brown wrote:
On 27/03/2024 17:05, wij wrote:
On Wed, 2024-03-27 at 16:02 +0100, David Brown wrote:
On 27/03/2024 14:32, wij wrote:
On Wed, 2024-03-27 at 13:57 +0100, Ralf Goertz wrote:
Am Wed, 27 Mar 2024 20:12:38 +0800
schrieb wij <wyniijj5@gmail.com>:
=20
On Wed, 2024-03-27 at 12:50 +0100, David Brown wrote:
On 26/03/2024 22:43, wij wrote:
=20
Just repeat the pattern infinitely, then it is irra=
=88=E2=84=9A", if the subtraction a=3D 142857/10^(6*i)=20
Nonsense.
=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=20
As said "=E2=88=80x,a=E2=88=88=E2=84=9A, x-a=E2=88=
teenagers.=20cannot terminate, 1/7 !=3D 0.(142857) =C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=20=20
Nonsense.
=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=20
I am surprise your math. knowledge is so low worse than=
repeatedly subtract=20=20
Use the standard trick:
=20
x=3D0.[142857] =3D> 1,000,000*x=3D142857.[142857]
=20
subtract the first equation from the second:
=20
999,999*x=3D142857 =3D> x=3D142857/999,999=3D1/7
=20
To determine whether a number x is rational or not, we can =
rified in finiterational numbers a? from x. If x-a1-a2-a3-...=3D0 can be ve=
is rational" issteps, then x is rational. Otherwise, x is irrational.
If x is a repeating decimal, proposition "repeating decimal=
?simply false by sematics.=20
=20
Let me just ask you two simple questions:
=20
Do you think 1/7 is a rational number or an irrational number=
, the procedure=20rational
=20
What do you think the decimal expansion of 1/7 is?=20
=20
When converting 1/7 to decimal, the result =E2=89=92 0.(142857)=
then I agreenever terminates which means the conversion is never complete.=20
=20
It is a repeating decimal.=C2=A0 If you try to write it all out, =
al expansionyou will not finish.=C2=A0 That does not mean it is not the decim=
um upof 1/7 - the list of multiples of (negative) powers of 10 which s=
ish the taskto 1/7.=C2=A0 You just need a better notation so that you can fin=
)- and 0.(142857), as you wrote, is one such notation.
=20
(I have no idea what you think the symbol "=E2=89=92" might mean.=
en=20
But you agree that 0.(142857) is the decimal expansion of 1/7, ev=
/7 ithough you could not write it out long-hand, and you agree that 1=
nce that'srational.=C2=A0 And clearly 0.(142857) is a repeating decimal, si=
ded mine.what the notation means.=20
=20
I can't see how you can still misunderstand this.
=20
You are restating your assertion without proof, again. I have provi=
he(If you say that is you proof, I will say it is invalid).=20
=20
=20
There is no point in giving you a rigorous proof that 0.(142857) is t=
o bedecimal expansion of 1/7, if that is what you are contesting.=C2=A0 T=
u don'tfully rigorous, it requires an understanding of the definition of the real numbers, sequence limits, and the meaning and validity of
operations on infinite sequences.=C2=A0 You have demonstrated that yo=
t failedunderstand any of that.=C2=A0 You have learned a few of the terms, bu=
ing whatto understand the concepts.=C2=A0 Oh, and it also requires understand=
e thata proof is, which again is clearly outside your expertise.
=20
Ralf gave a proof earlier - it is still in the quoted material above. That is as good as we can get at your level of mathematical understanding.=C2=A0 To be more rigorous, we would need to demonstrat=
ofthe manipulation (multiplication by a finite integer, and subtraction=
ousequences) of infinite decimal expansions is valid.=C2=A0 That is all standard stuff, known to mathematics students the world over, but you
are not nearly ready.
=20
You are going to have to go back-track a long way in what you think y=
ion,know about mathematics.=C2=A0 Somewhere along the line in your educat=
rying toyou've got things badly wrong.=C2=A0 And instead of stopping up and t=
orfigure out why everyone else is saying something different from you, =
s,asking your teachers for help, you have battered on with your mistake=
ne spendleading you to sillier and steadily less logical conclusions.
=20
I think mathematics is a great hobby.=C2=A0 It's a shame to see someo=
grantedtheir time and effort on doing it so badly.=20
=20
Have you ever wondered why you cannot prove something you hold true for=
up=20for so long?=20
Yes, regularly.=C2=A0 Sometimes I will then try to find a proof, or look =
and learn about the proofs.=C2=A0 Sometimes I will have to accept that==20
proving the particular thing is beyond my mathematical skills, or my=20sinner.
time and energy, or my interest, and I will defer to accepting that=20
others have proven it.
=20
=20
If you cannot provide a proof, what you said above only make you more a=
t I=20=20=20
In this particular case, I most certainly /can/ provide a proof.=C2=A0 Bu=
can't provide a proof that /you/ would understand.=C2=A0 And since writin=g a=20
proof would be a fair effort, off-topic, and clearly a waste of time=20=20
since you are impervious to mathematical reasoning, I will not bother.=
You can look up such proofs online - I'm sure there are countless=20=20
Youtube videos that will explain it to anyone who is actually interested=
in learning and not merely trying to claim the whole world is wrong=20
except them.
=20
I think mathematics is a great hobby. It's a shame to see someone
spend their time and effort on doing it so badly.
On 28/03/2024 20:41, Chris M. Thomasson wrote:[...]
Are you saying that a number constructed digit-by-digit using a TRNG
is undefined? Its not a number, however it creates many numbers during
the construction process? Fair enough?
Any given finite sequence of digits taken from a TRNG will give you a rational number. But "take 20 digits from a TRNG" does not define a
number - it defines a procedure for generating numbers. Do you see the difference?
Ahhh shit, this is just me having some fun. Sorry.
That's no problem. As long as you are trying to learn, and trying to
have fun, it's fine by me.
Have you ever wondered why you cannot prove something you hold true for granted
for so long?
If you cannot provide a proof, what you said above only make you more a sinner.
Have you wondered if 1/3=0.333..., the conversion algorithm is theoretically flawed?
On Fri, 2024-03-29 at 16:48 +0100, David Brown wrote:
On 29/03/2024 16:14, wij wrote:
On Fri, 2024-03-29 at 11:53 +0100, David Brown wrote:
On 28/03/2024 19:23, wij wrote:
On Thu, 2024-03-28 at 18:16 +0100, David Brown wrote:
On 27/03/2024 17:05, wij wrote:
On Wed, 2024-03-27 at 16:02 +0100, David Brown wrote:
On 27/03/2024 14:32, wij wrote:rational
On Wed, 2024-03-27 at 13:57 +0100, Ralf Goertz wrote:
Am Wed, 27 Mar 2024 20:12:38 +0800
schrieb wij <wyniijj5@gmail.com>:
On Wed, 2024-03-27 at 12:50 +0100, David Brown wrote:Use the standard trick:
On 26/03/2024 22:43, wij wrote:
Nonsense.
Just repeat the pattern infinitely, then it is irrational. >>>>>>>>>>>>
As said "∀x,a∈ℚ, x-a∈ℚ", if the subtraction a= 142857/10^(6*i)
cannot terminate, 1/7 != 0.(142857)
Nonsense.
I am surprise your math. knowledge is so low worse than teenagers. >>>>>>>>>>
x=0.[142857] => 1,000,000*x=142857.[142857]
subtract the first equation from the second:
999,999*x=142857 => x=142857/999,999=1/7
To determine whether a number x is rational or not, we can repeatedly subtract
rational numbers a? from x. If x-a1-a2-a3-...=0 can be verified in finite
steps, then x is rational. Otherwise, x is irrational.
If x is a repeating decimal, proposition "repeating decimal is rational" is
simply false by sematics.
Let me just ask you two simple questions:
Do you think 1/7 is a rational number or an irrational number? >>>>>>>>
What do you think the decimal expansion of 1/7 is?
When converting 1/7 to decimal, the result ≒ 0.(142857), the procedure
never terminates which means the conversion is never complete.
It is a repeating decimal. If you try to write it all out, then I agree >>>>>> you will not finish. That does not mean it is not the decimal expansion >>>>>> of 1/7 - the list of multiples of (negative) powers of 10 which sum up >>>>>> to 1/7. You just need a better notation so that you can finish the task >>>>>> - and 0.(142857), as you wrote, is one such notation.
(I have no idea what you think the symbol "≒" might mean.)
But you agree that 0.(142857) is the decimal expansion of 1/7, even >>>>>> though you could not write it out long-hand, and you agree that 1/7 i >>>>>> rational. And clearly 0.(142857) is a repeating decimal, since that's >>>>>> what the notation means.
I can't see how you can still misunderstand this.
You are restating your assertion without proof, again. I have provided mine.
(If you say that is you proof, I will say it is invalid).
There is no point in giving you a rigorous proof that 0.(142857) is the >>>> decimal expansion of 1/7, if that is what you are contesting. To be
fully rigorous, it requires an understanding of the definition of the
real numbers, sequence limits, and the meaning and validity of
operations on infinite sequences. You have demonstrated that you don't >>>> understand any of that. You have learned a few of the terms, but failed >>>> to understand the concepts. Oh, and it also requires understanding what >>>> a proof is, which again is clearly outside your expertise.
Ralf gave a proof earlier - it is still in the quoted material above.
That is as good as we can get at your level of mathematical
understanding. To be more rigorous, we would need to demonstrate that >>>> the manipulation (multiplication by a finite integer, and subtraction of >>>> sequences) of infinite decimal expansions is valid. That is all
standard stuff, known to mathematics students the world over, but you
are not nearly ready.
You are going to have to go back-track a long way in what you think you >>>> know about mathematics. Somewhere along the line in your education,
you've got things badly wrong. And instead of stopping up and trying to >>>> figure out why everyone else is saying something different from you, or >>>> asking your teachers for help, you have battered on with your mistakes, >>>> leading you to sillier and steadily less logical conclusions.
I think mathematics is a great hobby. It's a shame to see someone spend >>>> their time and effort on doing it so badly.
Have you ever wondered why you cannot prove something you hold true for granted
for so long?
Yes, regularly. Sometimes I will then try to find a proof, or look up
and learn about the proofs. Sometimes I will have to accept that
proving the particular thing is beyond my mathematical skills, or my
time and energy, or my interest, and I will defer to accepting that
others have proven it.
If you cannot provide a proof, what you said above only make you more a sinner.
In this particular case, I most certainly /can/ provide a proof. But I
can't provide a proof that /you/ would understand. And since writing a
proof would be a fair effort, off-topic, and clearly a waste of time
since you are impervious to mathematical reasoning, I will not bother.
You can look up such proofs online - I'm sure there are countless
Youtube videos that will explain it to anyone who is actually interested
in learning and not merely trying to claim the whole world is wrong
except them.
Not the whole world, you can see some on the internet claiming "0.999...!=1", although the proof is also invalid.
And, in every generation, every kid
(developed IQ) in school will keep wondering why 1/3=0.333... 'will stop'
and
why the the number very close to the left of 1 is not 0.999.... !
Have you wondered if 1/3=0.333..., the conversion algorithm is theoretically flawed?
David Brown <david.brown@hesbynett.no> writes:
[...]
I think mathematics is a great hobby. It's a shame to see someone
spend their time and effort on doing it so badly.
It's also a shame to see someone engaging here in a discussion that has nothing to do with C++. David, if you must feed this particular troll,
I suggest doing so in comp.theory.
*You don't have to reply to everything.*
On 29/03/2024 19:35, Keith Thompson wrote:ne
David Brown <david.brown@hesbynett.no> writes:
[...]
I think mathematics is a great hobby.=C2=A0 It's a shame to see someo=
roll,spend their time and effort on doing it so badly.=20
It's also a shame to see someone engaging here in a discussion that has nothing to do with C++.=C2=A0 David, if you must feed this particular t=
to=20I suggest doing so in comp.theory.=20
=20
*You don't have to reply to everything.*
=20
It is Easter, and Usenet traffic is low.=C2=A0 No, I don't have to reply =
everything (and I don't - I have replied to very few of wij's broken=20get=20
maths threads), and this thread will soon die away.=C2=A0 I am trying to =
some idea of why wij thinks the way he does, and perhaps even help him==20
think differently (though that's quite optimistic).
=20
On Sat, 2024-03-30 at 15:49 +0100, David Brown wrote:
On 29/03/2024 19:35, Keith Thompson wrote:
David Brown <david.brown@hesbynett.no> writes:
[...]
I think mathematics is a great hobby. It's a shame to see someone
spend their time and effort on doing it so badly.
It's also a shame to see someone engaging here in a discussion that has
nothing to do with C++. David, if you must feed this particular troll,
I suggest doing so in comp.theory.
*You don't have to reply to everything.*
It is Easter, and Usenet traffic is low. No, I don't have to reply to
everything (and I don't - I have replied to very few of wij's broken
maths threads), and this thread will soon die away. I am trying to get
some idea of why wij thinks the way he does, and perhaps even help him
think differently (though that's quite optimistic).
Persuade me and readers with proof, otherwise you lie or spread lies (from the moment)
On 30/03/2024 16:14, wij wrote:omeone
On Sat, 2024-03-30 at 15:49 +0100, David Brown wrote:
On 29/03/2024 19:35, Keith Thompson wrote:
David Brown <david.brown@hesbynett.no> writes:
[...]
I think mathematics is a great hobby.=C2=A0 It's a shame to see s=
hasspend their time and effort on doing it so badly.=20
It's also a shame to see someone engaging here in a discussion that=
ar troll,nothing to do with C++.=C2=A0 David, if you must feed this particul=
ply toI suggest doing so in comp.theory.=20
=20
*You don't have to reply to everything.*
=20
It is Easter, and Usenet traffic is low.=C2=A0 No, I don't have to re=
to geteverything (and I don't - I have replied to very few of wij's broken maths threads), and this thread will soon die away.=C2=A0 I am trying=
msome idea of why wij thinks the way he does, and perhaps even help hi=
rom the moment)think differently (though that's quite optimistic).=20
=20
Persuade me and readers with proof, otherwise you lie or spread lies (f=
=20=20=20
You were given a proof, but rejected it for no reason other than it=20
showed that your jumble of claims was incorrect.=C2=A0 Thus I don't think=
there is any point in trying to give more detailed proofs.=C2=A0 But if y=ou=20
like, I can give some links to other people's proofs - starting with=20 proving that 0.999... equals 1.=C2=A0 If you agree with these, maybe we c=an=20
move on to proving that 1/3 equals 0.33... repeating, and then further==20
onto showing that repeating decimals are rational.=C2=A0 So let me know w=hich=20
of these links you agree with, or disagree with (preferably with reasons==20
or justification for disagreeing with them).f.)
=20
<https://en.wikipedia.org/wiki/0.999...> <https://www.purplemath.com/modules/howcan1.htm> <https://brilliant.org/wiki/is-0999-equal-1/>
=20
=20
(I don't need to persuade any other readers - they already know this stuf=
=20=20
=20
And if you think I am lying, you can add lying to your list of concepts=
that you don't understand.I cannot read English fast. I will pick the one proof not in my proof. Archimedean property just states that infinitesmal does not exit, IIUC. It = is=20
=20
=20
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