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Find out if I/10^i = J/2^j (decadic - dyadic pairing)

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VK

unread,
Mar 15, 2007, 12:36:07 PM3/15/07
to
Given positive decadic rational it is needed to find if there is an
equivalent positive dyadic rational.
Other words it is needed to find for given positive integers I and i
if there are such positive integers J and j that
I/10^i = J/2^j

Positive case:
(0.125) 125/10^3 = 1/8 = 1/2^3

Negative case:
(0.1) 1/10^1

P.S. That would be great to find a formal algorithm - or a hint for a
formal algorithm - without infinite subsets of infinity and stuff.
Cantor is great and I'm saying _nothing_ bad about any of related
topics; but I hope to implement a computer check: if such algorithm
exists and it is up to the capabilities of existing machines.

cbr...@cbrownsystems.com

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Mar 15, 2007, 2:55:21 PM3/15/07
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On Mar 15, 9:36 am, "VK" <schools_r...@yahoo.com> wrote:
> Given positive decadic rational it is needed to find if there is an
> equivalent positive dyadic rational.
> Other words it is needed to find for given positive integers I and i
> if there are such positive integers J and j that
> I/10^i = J/2^j
>

I * 2^j = J * 10^i
= J * (5 * 2)^i
= J * 5^i * 2^i

Therefore, I must be divisible by 5^i for any solutions to exist.
Suppose I = 5^i * x for some positive integer x; then

x * 2^j = J * 2^i

where x and i are given. Then J = 2^k*x and j = i + k are solutions
for all positive integers k; and these are the only solutions.

Cheers - Chas

cbr...@cbrownsystems.com

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Mar 15, 2007, 3:04:05 PM3/15/07
to

Whoops! Not just positive k; in fact, let k be the smallest integer
such that J = 2^k*x is an integer and j = i + k is a positive integer.

Cheers - Chas

VK

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Mar 15, 2007, 3:50:04 PM3/15/07
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VK

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Mar 15, 2007, 5:05:34 PM3/15/07
to
On Mar 15, 10:04 pm, cbr...@cbrownsystems.com wrote:
> > > Given positive decadic rational it is needed to find if there is an
> > > equivalent positive dyadic rational.
> > > Other words it is needed to find for given positive integers I and i
> > > if there are such positive integers J and j that
> > > I/10^i = J/2^j
>
> > I * 2^j = J * 10^i
> > = J * (5 * 2)^i
> > = J * 5^i * 2^i
>
> > Therefore, I must be divisible by 5^i for any solutions to exist.
> > Suppose I = 5^i * x for some positive integer x; then
>
> > x * 2^j = J * 2^i
>
> > where x and i are given. Then J = 2^k*x and j = i + k are solutions
> > for all positive integers k; and these are the only solutions.
>
> Whoops! Not just positive k; in fact, let k be the smallest integer
> such that J = 2^k*x is an integer and j = i + k is a positive integer.

This question may sound as "wise" as say "Who did invent the equation
x^2=32 ?" so sorry in advance if it sounds like that. Any way: was
this proof ever published under some name by some author? Is it your
own solution? Or is it a private case of some proof for N cases? The
"dyadicity of decimal" - as spelled and proved above - is important
for many computer math issues, so it would be convenient for me to
refer to it by name rather than spelling the body over and over again.

cbr...@cbrownsystems.com

unread,
Mar 15, 2007, 10:42:05 PM3/15/07
to

Happy to be of service, but the above proof seemed pretty obvious to
me (and it would certainly be so to many, many others with far more
experience than myself). So, more on the level of "x^2 = 32" than
deserving of being a named theorem.

The only "name-able" theorem being used here (without my stating it
explicitly) is the Fundamental Theorem of Arithemtic:

http://en.wikipedia.org/wiki/Fundamental_theorem_of_arithmetic

which is required, e.g., for my assertion that there are solutions if
and only if I is divisible by 5^i.

If you find the result interesting and useful, then you might want to
study some discrete mathematics or number theory. Questions of
divisibility and integer solutions to equations such as the above are
dealt with extensively in these areas. Then you, too, will be able to
answer such questions easily! Amaze your friends!

Of course, if you like you can call it "Chas's Theorem".
Unfortunately, no one will know what you mean.

Cheers - Chas

VK

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Mar 17, 2007, 1:40:51 PM3/17/07
to
On Mar 16, 5:42 am, cbr...@cbrownsystems.com wrote:
> If you find the result interesting and useful, then you might want to
> study some discrete mathematics or number theory.

I will surely do, actually I'm doing it right now. I passed a starting
"math boot camp" during my bachelory for math linguistics, but is was
more than ten years ago so many things got rusted and dusted :-( :-)
To avoid possible suspicions: I am not searching to (dis)proof - by
using the school math - the continuum hypothesis, last Ferma theorem
or something dishonest like that :-) I an a linguist by education
doing some semantical studies at free of making money time. For one
interesting - for myself - problem I would like to bridge some math
entities with some entities in the discrete computer math.

> Of course, if you like you can call it "Chas's Theorem".
> Unfortunately, no one will know what you mean.

IMO it doesn't have to be a grounds-breaking statement to have a name,
it is only important how often one has to refer to it. This way three
aims at once are achieved:
1) The sequence of one's statements will not look a la "Jack's house"
poetry
2) In case of publishing it helps to preserve the Amazon forest ;-)
3) In order to follow you to the end your readers will have to
memorize the meanings of your terms. This way one implicitly enforces
her own mindset on readers: which is often used as the second hidden
level of proof.
While the 3rd evil purpose :-) is not my aim, especially the 1st one
is in effect. To stay firm yet humble we can call it a "lemma" instead
of a "theorem". IMHO for the time being the meaningfulness has more
priority than the copyrighting so we have

Dyadicity lemma (a.k.a. Chas' lemma)
Given known positive integers I and i and unknown positive integers J
and j the equation I/10^i = J/2^j has the only solution for J and j
when and only when I%i^5 = 0

Did I get anything wrong?

Also: I simplified the situation by removing any computer-related
context where the Peano axioms are not applicable, at least not all of
them. This way I defined natural numbers as "positive integers"
instead of "non-negative integers" as they are in the domain where
I'll be using them. Otherwise that would be -0 and +0 leading to -
infinity or +infinity in operations and similar hardly mathematical
stuff. From the strictly mathematical view: would be the lemma somehow
affected by defining natural numbers as "non-negative integers" so
including 0 ?

So now with the dyadicity lemma it is possible to determine the
dyadicity factor (d-factor) of each given decadic rational. For say
125/10^3 d-factor is 1 and for say 1/10^1 d-factor is 0

Now I'm getting on a shaky ground because I'm not sure what the
intersection would be of decadic rationals with reals?
125/1000 = 0.125 is fine, but 0.125 is also 0.124(9)

cbr...@cbrownsystems.com

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Mar 17, 2007, 6:14:56 PM3/17/07
to

Slight typo: I % 5^i, not I % i^5.

But since "%" as a symbol for "modulo" is rather computer-language
specific, it would be clearer to say:

Lemma: Given known positive integers I and i and unknown positive


integers J
and j the equation I/10^i = J/2^j has the only solution for J and j

when and only when 5^i divides I.

In your subsequent arguments, it would then be normal to say something
like "..., since in this case 5^i does not divide I, then by the
lemma, there are no solutions". Or if there are many lemmas at work in
the argument, one might number them; so that the above lemma would be
"lemma 2.1", and one would say something like "..., since by lemma
1.6, 5^i does not divide I, then by lemma 2.1, there are no
solutions".

One is welcome to name the lemma; but it's usually unnecessary.

>
> Also: I simplified the situation by removing any computer-related
> context where the Peano axioms are not applicable, at least not all of
> them. This way I defined natural numbers as "positive integers"
> instead of "non-negative integers" as they are in the domain where
> I'll be using them. Otherwise that would be -0 and +0 leading to -
> infinity or +infinity in operations and similar hardly mathematical
> stuff.

Erm? There is no -0 and +0; there is only 0.

> From the strictly mathematical view: would be the lemma somehow
> affected by defining natural numbers as "non-negative integers" so
> including 0 ?
>

Well, go back and consider the proof I gave. If I*2^j = J*10^0, is I
divisible by 5^0? Does J = I*2^j, j = any non-negative integer provide
a solution? If 0*2^j = J*10^i, is 0 divisible by 5^i? Does J = 0, j =
any non-negative integer provide a solution?

> So now with the dyadicity lemma it is possible to determine the
> dyadicity factor (d-factor) of each given decadic rational. For say
> 125/10^3 d-factor is 1 and for say 1/10^1 d-factor is 0
>

Instead of giving examples of "determining", it would be more
mathematical if you /define/ exactly what you mean by "x is the d-
factor of I/10^i". Can you do this?

> Now I'm getting on a shaky ground because I'm not sure what the
> intersection would be of decadic rationals with reals?
> 125/1000 = 0.125 is fine, but 0.125 is also 0.124(9)

This is a common confusion for people as they begin exploring
mathematics beyond "calculation", which you can resolve by thinking
linguistically.

Despite our informal usage, "0.125" and "0.124(9)" are /not/, strictly
speaking, real numbers at all in a mathematical sense.

They are instead different /names/ for the /same/ real number, just as
sqrt(4) and 6/3 are different /names/ for the /same/ rational number
(which is also named "2", "two", and "II").

It is true that if r is a real number, it /can/ be named using an
(infinite) sequence of digits such as "0.1234567891011...", but that
is /not/ the definition of "r is a real number".

It is instead a definition of how to use the /name/
"0.1234567891011..." to /identify/ a particular real number; whose
existence and properties are independent of how we propose to name it.

This is really no different than the situation with the natural number
named "10". To say that by "10", we mean (using the usual base 10
convention) the number "1*10 + 0" is not particularly helpful as an
answer to the question, "which natural number is 10"? The name is not
the number itself.

As such, all numbers of the form I/10^i, i >= 0 are also real numbers
by the /definition/ of "real number"; so the intersection of the
decadic rationals and the reals is the decadic rationals, however we
may choose to /write/ those numbers (i.e., "give them names").

Mathematics specializes in abstracting away the "naming" of various
objects, in favor of defining mathematical objects (such as numbers,
triangles, sequences, sets, functions, etc.) solely by the properties
they posses. Two mathematical objects are the same if and only if they
have the same /properties/; regardless of whether they have been
identified using the same /names/.

And that is what is generally meant mathematically by "x = y"; not
that the symbol "x" is the same symbol as "y", but that the object
named by the symbol "x" has exactly the same /properties/ as the
object named by the symbol "y".

Cheers - Chas

VK

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Mar 17, 2007, 6:42:50 PM3/17/07
to
http://groups.google.com/group/sci.math/msg/d11e214a2525c119

I'll be reading what you just posted and I'll be making necessary
corrections in my statements - and in my mind. I really appreciate
your help.

VK

unread,
Mar 18, 2007, 1:42:44 PM3/18/07
to
> > Given known positive integers I and i and unknown positive integers J
> > and j the equation I/10^i = J/2^j has the only solution for J and j
> > when and only when I%i^5 = 0
>
> > Did I get anything wrong?
>
> Slight typo: I % 5^i, not I % i^5.

Oops... Dyslexia - a regular default of a linguist. :-)

> But since "%" as a symbol for "modulo" is rather computer-language
> specific, it would be clearer to say:
>
> Lemma: Given known positive integers I and i and unknown positive
> integers J
> and j the equation I/10^i = J/2^j has the only solution for J and j
> when and only when 5^i divides I.

Got it.

> Erm? There is no -0 and +0; there is only 0.

<off topic>That depends on the used abstraction system ;-) Where I'm
going to work there are -0/+0 respectively and a possible set such as
{-0, 0, 1} and 1/-0 = -infinity, 1/0 = +infinity; so effectively two
0s with different valency towards infinity; thus bye-bye Peano 7th and
8th and Peano 9th gets a rather unusual behavior. Side effect of that
zero as an "absolute nothing" is unknown to computer. It is simply a
denormalized number smaller than ulp - so still something but beyond
the given granularity. For the standard math model we are discussing
in this thread it is really irrelevant, just an OT blah-blah :-)</off
topic>

> > From the strictly mathematical view: would be the lemma somehow
> > affected by defining natural numbers as "non-negative integers" so
> > including 0 ?
>
> Well, go back and consider the proof I gave. If I*2^j = J*10^0, is I
> divisible by 5^0? Does J = I*2^j, j = any non-negative integer provide
> a solution? If 0*2^j = J*10^i, is 0 divisible by 5^i? Does J = 0, j =
> any non-negative integer provide a solution?

I guess it was an over-check question from my side, because I already
was answering to myself to this question - but in another newsgroup
and in another thread. By redefining naturals as "non-negative
integers" we are including into the set of solvable cases all natural
numbers as well as a particular case of decadic rationals: any natural
N is also N/10^0 is also N/2^0 is representable as dyadic rational.

> > So now with the dyadicity lemma it is possible to determine the
> > dyadicity factor (d-factor) of each given decadic rational. For say
> > 125/10^3 d-factor is 1 and for say 1/10^1 d-factor is 0
>
> Instead of giving examples of "determining", it would be more
> mathematical if you /define/ exactly what you mean by "x is the d-
> factor of I/10^i". Can you do this?

I'll try right now.

For each given decadic rational I/10^i its d-factor is 1 when and only
when the said rational has dyadic solution by the lemma 1.0 If such
solution doesn't exist then d-factor of the said rational is 0.

But there is no direct transformation between all of "aliases", or am
I still missing something? Given { 0.125 125/10^3 0.124(9)
0.1249999999(9) }
It is a trivia to arrive to 0.125 0.124(9) 0.1249999999(9) from
125/10^3 or to 125/10^3 from 0.125
I don't see how though to arrive to 125/10^3 from 0.124(9) or
0.1249999999(9)
Or am still I putting too much of arithmetics into math?

cbr...@cbrownsystems.com

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Mar 18, 2007, 8:19:56 PM3/18/07
to
On Mar 18, 10:42 am, "VK" <schools_r...@yahoo.com> wrote:
> > > Given known positive integers I and i and unknown positive integers J
> > > and j the equation I/10^i = J/2^j has the only solution for J and j
> > > when and only when I%i^5 = 0
>
> > > Did I get anything wrong?
>
> > Slight typo: I % 5^i, not I % i^5.
>
> Oops... Dyslexia - a regular default of a linguist. :-)
>

Dyslexic Atheist: "I do not believe in Dog!"

> > But since "%" as a symbol for "modulo" is rather computer-language
> > specific, it would be clearer to say:
>
> > Lemma: Given known positive integers I and i and unknown positive
> > integers J
> > and j the equation I/10^i = J/2^j has the only solution for J and j
> > when and only when 5^i divides I.
>
> Got it.
>
> > Erm? There is no -0 and +0; there is only 0.
>
> <off topic>That depends on the used abstraction system ;-) Where I'm
> going to work there are -0/+0 respectively and a possible set such as
> {-0, 0, 1} and 1/-0 = -infinity, 1/0 = +infinity; so effectively two
> 0s with different valency towards infinity; thus bye-bye Peano 7th and
> 8th and Peano 9th gets a rather unusual behavior.

Yes, it seems pretty odd. I assume we are not talking about integer
types here, but instead some sort of floating point number. The
floating point numbers don't even form a field (technical jargon
apology).

But on topic, I would say that then the names "-0" and "+0" are /not/
names of real numbers; because there are no real numbers with the
properties you describe. Thus, the floating point "reals" are not real
numbers themselves - instead they are elements of some other number
system (and there are many, many such different systems).

That's perfect. This is sometimes called "the indicator function on a
set": f(s) = 1 if s is in the set, and f(s) = 0 if s is not in the
set.

> But there is no direct transformation between all of "aliases"...

Perhaps not in practice (because "direct" is a bit non-mathematical,
and "all" is a lot of aliases!), but in principle, yes, we could
transform between these aliases.

>, or am
> I still missing something?

To see that you are not alone in your uncertainty, search for
"0.999... = 1?" in sci.math. It is a perennial subject of confusion
and general usenet flaming.

> Given { 0.125 125/10^3 0.124(9)
> 0.1249999999(9) }
> It is a trivia to arrive to 0.125 0.124(9) 0.1249999999(9) from
> 125/10^3 or to 125/10^3 from 0.125
> I don't see how though to arrive to 125/10^3 from 0.124(9) or
> 0.1249999999(9)
> Or am still I putting too much of arithmetics into math?

I think the latter (although to cheat, you might say that one could
more obviously "arrive" at 125 from 124.9999....)

What is the /meaning/ of saying "the real number 0.124(9)"? It is
"that real number which is the limit of the sequence of rational
numbers (0, 0.1, 0.12, 0.124, 0.1249, 0.12499, ...)".

What is the /meaning/ of saying "the real number 0.125(0)"? It is
"that real number which is the limit of the sequence of rational
numbers (0, 0.1, 0.12, 0.124, 0.125, 0.1250, 0.12500, ...)".

We can /prove/ the following, from the definition of the real numbers:

(1) We are formally correct above when we say /the/ limit of the
sequence of rational numbers (as opposed to "a limit, if it even
exists"); because there exists exactly one real number with this
property.

(2) If two different sequences of digits have the same limit in the
above definitions, then for some natural number m, all digits at
positions greater than m are 0 in one of the sequences of digits, and
all digits at positions greater than m are 9's in the other sequence.

So the latter observation makes it easy to "see" that 0.999... =
1.000... is the only sort of special case we need to look out for,
easily recognized and dealt with.

Of course there are many other sequences of rational numbers which
also have 125/10^3 as their limit. For example, the sequence of
rational numbers (1 + 125/10^3, 1/2 + 125/10^3, 1/4 + 125/10^3, ...,
1/2^n + 125/10^3, ...) has as its limit our same pal 125/10^3.

In fact, one common definition of the real numbers basically defines a
real number as the /set/ of /all sequences of rationals/ having the
same limit. (real numbers as equivalence classes of Cauchy sequences
of rationals being the buzzwords).

On a more philosophical level:

Suppose x is a real number satisfying x^2 = 2 and x > 0; and y is also
a real number satisfying y^2 = 2 and y > 0. Do you also need to know /
in principle/ what the 100,000th digit in the decimal expansions of
both x and y are in order to say "x = y"? In other words, is sqrt(2) a
sufficient description of a /unique/ real number?

Suppose z is the length of the hypotenuse of a right triangle having
sides length 1. Is "z = sqrt(2)" somehow lacking as a precise
description of this length?

Cheers - Chas

VK

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Mar 19, 2007, 4:04:06 PM3/19/07
to
On Mar 19, 3:19 am, cbr...@cbrownsystems.com wrote:
> > <off topic>That depends on the used abstraction system ;-) Where I'm
> > going to work there are -0/+0 respectively and a possible set such as
> > {-0, 0, 1} and 1/-0 = -infinity, 1/0 = +infinity; so effectively two
> > 0s with different valency towards infinity; thus bye-bye Peano 7th and
> > 8th and Peano 9th gets a rather unusual behavior.
>
> Yes, it seems pretty odd. I assume we are not talking about integer
> types here, but instead some sort of floating point number. The
> floating point numbers don't even form a field (technical jargon
> apology).

They do - if I got the "field" term in math properly (60/40 chance :-)
Actually my note about computer +/-0 was incomplete up to being wrong.
0 can be as nothing as nothing can be on computer as well. The catch
is that it is a finite system here, so 0 is "undecidable": it may be
the real nothing or it may be such a small result that it is beyond
the given granularity. So the involved operations have to accommodate
both options: if from the "practical point of view" -1 *
10^-99999999999999999999999999 is mostly the same as nothing, from the
math view it is still a negative value.

> > For each given decadic rational I/10^i its d-factor is 1 when and only
> > when the said rational has dyadic solution by the lemma 1.0 If such
> > solution doesn't exist then d-factor of the said rational is 0.
>
> That's perfect. This is sometimes called "the indicator function on a
> set": f(s) = 1 if s is in the set, and f(s) = 0 if s is not in the
> set.

Great, so another step is finished.

> To see that you are not alone in your uncertainty, search for
> "0.999... = 1?" in sci.math. It is a perennial subject of confusion
> and general usenet flaming.

I'm staying - or at least always trying to stay - on the principle
"Don't come in the stranger monastery with your own Charter". That
doesn't prohibit to add new entities or to re-interpret some existing
ones if the necessity is obviously here - but for internal use only.
Like if one pipe is oval and the other pipe is circle then one either
makes a coupling by her own or she stays without the water.

So AFAICT the current math system doesn't make a difference between
"going to" and "being at", "close to the point" and "be at the point".
I see the possibility of such abstraction, it doesn't make me upset or
dizzy or something :-) It also provides an elegant workaround for many
ancient paradox. But see further.

Here is the dead lock. No, no problem with the math, it's great! But
the system as spelled has zero valency with the computer math - or
discrete math if one wants. You see, 1 + 2 has a solution both on
paper and on any computer. At the same time 0.9(9) + 1.9(9) has no
algorithmic solution on any of existing systems: unless we cheat and
introduce some (period) numbers so
0.9(9) + 1.9(9) = 0.9 + 1.9 + 2 * (period9)
Respectively we'll have to add some SqrtMantissa2 such as
1 + SqrtMantissa2 = sqrt(2)
and a whole set of "infinite mantissa holders" of the latter kind.
This way the finite operations will be made by the computer, and
infinite parts simply accumulated/negated and returned to human
receiver for further considerations. That is the only way otherwise 1
+ 2 in some conditions may take the Infinity to calculate - even if
someone manages to create an ideal Turing machine.

btw a skeleton in the programming closet: is a problem solvable if it
is algorithmic but it requires infinity to complete the algorithm?

Sorry for all this long wording but I had to show what the particular
problem is. And yes, of cause I'm aware of rounding, error correction
and restoring value methods. But I'm not interested in them at this
point - otherwise I wouldn't by "harassing" people on sci.math.*

So to get anywhere further, I need to find an extra criterion to
separate
1) all numbers having (0) in period
2) all numbers having any other period
3) all numbers with unknown/no period

(3) is easy - these are irrationals.
(1) and (2) ? 0.125 and 0.124(9), 1.3 and 1.(3)? Does the part of
science we are talking about - its official part I mean - has any
criteria of distinction?

Michael Press

unread,
Mar 19, 2007, 8:15:41 PM3/19/07
to
In article
<1174263596.0...@d57g2000hsg.googlegroups.com>,
cbr...@cbrownsystems.com wrote:

> Dyslexic Atheist:

Agnostic dyslexic insomniac lay
awake all night wondering if there is a dog.
--
Michael Press

cbr...@cbrownsystems.com

unread,
Mar 20, 2007, 12:17:58 AM3/20/07
to
On Mar 19, 1:04 pm, "VK" <schools_r...@yahoo.com> wrote:
> On Mar 19, 3:19 am, cbr...@cbrownsystems.com wrote:
>
> > > <off topic>That depends on the used abstraction system ;-) Where I'm
> > > going to work there are -0/+0 respectively and a possible set such as
> > > {-0, 0, 1} and 1/-0 = -infinity, 1/0 = +infinity; so effectively two
> > > 0s with different valency towards infinity; thus bye-bye Peano 7th and
> > > 8th and Peano 9th gets a rather unusual behavior.
>
> > Yes, it seems pretty odd. I assume we are not talking about integer
> > types here, but instead some sort of floating point number. The
> > floating point numbers don't even form a field (technical jargon
> > apology).
>
> They do - if I got the "field" term in math properly (60/40 chance :-)

Hmmm... I'll take that bet on the opposing side :).

A field needs to be associative over both + and *. But for finite
precision floating point numbers, it is not always the case that

(a + b) + c = a + (b + c)
(a*b)*c = a*(b*c)

There's a long and interesting thread on this with examples at "Are
floating-point numbers rational numbers?":

http://groups.google.com/group/sci.math/browse_frm/thread/eca5a6e57432c648/e2057165e9ffe331

that touches on many of the topics we are discussing regarding the
difference between the "name" of a number and that number.

The latter is better stated as "getting forever closer to the point"
and "is the point"; but I see your meaning. The reals are sometimes
defined as the closure of this "idea" applied to the rationals.

> I see the possibility of such abstraction, it doesn't make me upset or
> dizzy or something :-) It also provides an elegant workaround for many
> ancient paradox.

It also makes basic arithmetic manipulations "error-free" in a way
that the floating-point numbers can't do. I know that sqrt(2)^2 = 2
and I can make that substitution in an equation without adding any
error whatsoever to my (symbolic) calculation.

> But see further.
>

<snip>

> Here is the dead lock. No, no problem with the math, it's great! But
> the system as spelled has zero valency with the computer math - or
> discrete math if one wants. You see, 1 + 2 has a solution both on
> paper and on any computer.

Yes, the integer type on a computer is "very much the same" as the
integers modulo some really large power of 2. So short of overflow,
they form a ring; and thus the usual manipulations of +, -, * obey the
usual rules of a ring (associative, distributive, unique additive
inverse).

Ubfortunately, this is /not/ the case with finite precision floats
(although one /can/ make a field of rationals using a different
notation).

> At the same time 0.9(9) + 1.9(9) has no
> algorithmic solution on any of existing systems: unless we cheat and
> introduce some (period) numbers so
> 0.9(9) + 1.9(9) = 0.9 + 1.9 + 2 * (period9)

But this is simply a matter of one's definition of "floating point
number in computers" versus "real number in computers".

For /finite precision/ floating point numbers (fixed precision or
not), there is no such thing as "0.999(9)". Simply becuase "0.999(9)"
is not a /finite/ precision floating point; it is an /infinite
precision/ floating point, and as such not an IEEE standard :).

> Respectively we'll have to add some SqrtMantissa2 such as
> 1 + SqrtMantissa2 = sqrt(2)
> and a whole set of "infinite mantissa holders" of the latter kind.

Well, this is exactly what symbolic packages such as Mathematica do.
They define a (smallish) set of operators such a "sqrt" symbolically;
and perform (essentially) string substitutions such as "(sqrt(x))^2 ->
x".

They can also evaluate expressions like "lim n->oo (1/x^n)", again /
symbolically/ not computationally; exactly as you yourself might
justify that the limit of the previous expression for any x^2 < 1 is,
indeed, the real number named "0", and no other real number.

So for these kind of packages, there is "cheating" way of representing
a number by 0.9(9) which is distinct from representing the /same/
number as 1.0(0). These two symbols (or "symbolic computations") yield
the same result: they are demonstrably "equal".

> This way the finite operations will be made by the computer, and
> infinite parts simply accumulated/negated and returned to human
> receiver for further considerations. That is the only way otherwise 1
> + 2 in some conditions may take the Infinity to calculate - even if
> someone manages to create an ideal Turing machine.
>

"In some conditions" seems to mean "using certain algorithms". But
even if those algorithms are so badly designed as to /not/ halt on
such a calculation, it doesn't stop us from noting that there /are/
algorithms that /do/ halt with the output of a symbol (marks on tape,
ascii character, whatever) which corresponds to the real number 3, if
and only if the real numbers 1 + 2 = 3.

> btw a skeleton in the programming closet: is a problem solvable if it
> is algorithmic but it requires infinity to complete the algorithm?
>

Don't get me started! It's an interesting set of problems: Turing's
Halting problem, the uncomputability of the list of all computable
numbers, Godel's incompleteness theorem, recursively enumerable sets,
etc.

> Sorry for all this long wording but I had to show what the particular
> problem is. And yes, of cause I'm aware of rounding, error correction
> and restoring value methods. But I'm not interested in them at this
> point - otherwise I wouldn't by "harassing" people on sci.math.*
>

Heaven's! I feel more like you are permitting /me/ to harass /you/ :).
Plus, there is a whole area of "abstract" mathematics related to these
questions - numerical analysis. It even has its own newsgroup in the
sci.math.* heirarchy.

> So to get anywhere further,

... ah, yes; back to work! This was a paying gig if I recall :) ...

> I need to find an extra criterion to
> separate
> 1) all numbers having (0) in period
> 2) all numbers having any other period
> 3) all numbers with unknown/no period
>
> (3) is easy - these are irrationals.
> (1) and (2) ? 0.125 and 0.124(9), 1.3 and 1.(3)? Does the part of
> science we are talking about - its official part I mean - has any
> criteria of distinction?

These less philosophical questions are much easier to answer
succinctly and unequivocally.

You are correct regarding (3)(but can you prove it?).

As regards (1) and (2): informally stated, it's pretty easy (although
a little long-winded) to prove that x.(y) = z.(0) if and only if y =
b-1 and z = x+1; where x and z are integers, y is a natural with 0 <=
y < b, and b is the base we are working in for our sequence of digits
(e.g., 2 or 10).

(Proof available on request; just don't name it after me!)

Cheers - Chas

cbr...@cbrownsystems.com

unread,
Mar 20, 2007, 12:59:49 AM3/20/07
to
On Mar 19, 9:17 pm, cbr...@cbrownsystems.com wrote:

Self nit-pick:

> As regards (1) and (2): informally stated, it's pretty easy (although

> a little long-winded) to prove that x.(y) = z.(0) ...

... iff y = b^n - 1 for n in N ; and z = x + 1 where x and z are
integers, and b is the base we are working in for our sequence of


digits (e.g., 2 or 10).

Cheers - Chas

cbr...@cbrownsystems.com

unread,
Mar 20, 2007, 1:16:03 AM3/20/07
to

Arrrgh! One more time, as is required for precision:

Base b, "x.(y)" = "z.(0)" iff (y = 0 and z = x) or (y = b-1 and z = x
+ 1)

Cheers - Chbs, I mean, Chas

cbr...@cbrownsystems.com

unread,
Mar 20, 2007, 1:28:42 AM3/20/07
to
On Mar 19, 5:15 pm, Michael Press <rub...@pacbell.net> wrote:
> In article
> <1174263596.070777.320...@d57g2000hsg.googlegroups.com>,

>
> cbr...@cbrownsystems.com wrote:
> > Dyslexic Atheist:
>
> Agnostic dyslexic insomniac lay
> awake all night wondering if there is a dog.
> --
> Michael Press

A man, a dog, a CAG, Odanama!

Cheers - Chas

cbr...@cbrownsystems.com

unread,
Mar 20, 2007, 1:33:22 AM3/20/07
to
On Mar 19, 10:28 pm, cbr...@cbrownsystems.com wrote:
> On Mar 19, 5:15 pm, Michael Press <rub...@pacbell.net> wrote:

> > Agnostic dyslexic insomniac lay
> > awake all night wondering if there is a dog.

>


> A man, a dog, a CAG, Odanama!
>

Oooh! Wait! Even better!

A dog, a plan, a canal: Pagoda!

Cheers - Chas


VK

unread,
Mar 20, 2007, 2:57:03 PM3/20/07
to
On Mar 20, 7:17 am, cbr...@cbrownsystems.com wrote:

<snip - and yes, I didn't get fields in math right, thank you>

> There's a long and interesting thread on this with examples at "Are
> floating-point numbers rational numbers?":
>

> http://groups.google.com/group/sci.math/browse_frm/thread/eca5a6e5743...


>
> that touches on many of the topics we are discussing regarding the
> difference between the "name" of a number and that number.

Thanks for source, I'll be reading it.

<snip>

> > So AFAICT the current math system doesn't make a difference between
> > "going to" and "being at", "close to the point" and "be at the point".
>
> The latter is better stated as "getting forever closer to the point"
> and "is the point"; but I see your meaning. The reals are sometimes
> defined as the closure of this "idea" applied to the rationals.

ACK (Got it)

<snip>

> "In some conditions" seems to mean "using certain algorithms". But
> even if those algorithms are so badly designed as to /not/ halt on
> such a calculation, it doesn't stop us from noting that there /are/
> algorithms that /do/ halt with the output of a symbol (marks on tape,
> ascii character, whatever) which corresponds to the real number 3, if
> and only if the real numbers 1 + 2 = 3.

Please define "badly designed algorithm" :-)
Lambda calculus negative proof to Entscheidungsproblem /Decision
problem/ means that the infinite nature of an algorithm cannot be
concluded from the algorithm itself. This way by giving to a computer
a problem such as 0.(9) + 1.(9) = ? we have only two options:
1) deus ex machina - so we inform the program in advance that this
algorithm leads to infinity, so use a workaround.
2) we are imposing Tmax period and after that the algorithm is either
completed or aborted.

This way Church - Cole negative lambda calculus solution eliminate any
"philosophical dependencies" from the physical computer limitations.
The problem is not in memory storage space or such but in the
undecidable nature of the algorithm.
In case (1) the algorithm is infinite because we said it is infinite.
In case (2) the algorithm is not finished because it was not finished
in Tmax period. But is it really infinite? Maybe it is or maybe it
would be finished at Tmax + 1 moment.

In either case there is not any machine fault or a precision problem.

Note: "Entscheidungsproblem" instead of "Decision problem" is not to
show up. The original German word is used in the areas I'm working
with. So kind of "slang exchange" with you. :-)

> > btw a skeleton in the programming closet: is a problem solvable if it
> > is algorithmic but it requires infinity to complete the algorithm?
>
> Don't get me started! It's an interesting set of problems: Turing's
> Halting problem, the uncomputability of the list of all computable
> numbers, Godel's incompleteness theorem, recursively enumerable sets,
> etc.

{ Entscheidungsproblem, lambda calculus, Goedel's incompleteness
theorem }

Hah! Goedel in both sets! :-)

"The thing that hath been, it is that which shall be;
and that which is done is that which shall be done:
and there is no new thing under the sun.

Is there any thing whereof it may be said, See, this is new?
it hath been already of old time, which was before us."
Ecclesiastes 1-1
:-) :-|
btw Lester Zick in another thread is puzzled with the "true nature" of
the point just like Euclid was about 2,250 years ago - and with the
same doubts in mind: "can nothing be something and if it can then how
can it be?"

> > I need to find an extra criterion to
> > separate
> > 1) all numbers having (0) in period
> > 2) all numbers having any other period
> > 3) all numbers with unknown/no period
>
> > (3) is easy - these are irrationals.
> > (1) and (2) ? 0.125 and 0.124(9), 1.3 and 1.(3)? Does the part of
> > science we are talking about - its official part I mean - has any
> > criteria of distinction?
>

> You are correct regarding (3)(but can you prove it?).

Oh gosh, I remember I have to be careful here :-)
sqrt(2) is an irrational or a rational with unknown - yet - period?
Other words is sqrt(2) irrational or is it a number with undecidable
by current tools status: maybe rational with a very long period, maybe
indeed an irrational.
What do you personally think: is everything clear about sqrt(2) or say
PI - or there is still a question sign here?

> As regards (1) and (2): informally stated, it's pretty easy (although
> a little long-winded) to prove that x.(y) = z.(0) if and only if y =
> b-1 and z = x+1; where x and z are integers, y is a natural with 0 <=
> y < b, and b is the base we are working in for our sequence of digits
> (e.g., 2 or 10).
>
> (Proof available on request; just don't name it after me!)

I don't think "request" is a suitable word here, but if you have time
for that second lemma proof then it would be the most great.
btw a paper sheet scan copy or MathML or LaTeX is fine as well _if_ it
saves your time. Expressing higher math in pure ASCII should be a
torture.

cbr...@cbrownsystems.com

unread,
Mar 21, 2007, 7:25:40 PM3/21/07
to
On Mar 20, 11:57 am, "VK" <schools_r...@yahoo.com> wrote:
> On Mar 20, 7:17 am, cbr...@cbrownsystems.com wrote:
>
<snip>

> > "In some conditions" seems to mean "using certain algorithms". But


> > even if those algorithms are so badly designed as to /not/ halt on
> > such a calculation, it doesn't stop us from noting that there /are/
> > algorithms that /do/ halt with the output of a symbol (marks on tape,
> > ascii character, whatever) which corresponds to the real number 3, if
> > and only if the real numbers 1 + 2 = 3.
>
> Please define "badly designed algorithm" :-)
> Lambda calculus negative proof to Entscheidungsproblem /Decision
> problem/ means that the infinite nature of an algorithm cannot be
> concluded from the algorithm itself. This way by giving to a computer
> a problem such as 0.(9) + 1.(9) = ? we have only two options:
> 1) deus ex machina - so we inform the program in advance that this
> algorithm leads to infinity, so use a workaround.
> 2) we are imposing Tmax period and after that the algorithm is either
> completed or aborted.
>

But this seems to confuse the algorithm with the underlying
mathematical question to which we hope the algorithm will provide
answer.

As algorithms, we can perform multiplication of x * y by sequential
addition of x, y times; or we can use "long multiplication" to speed
things up considerably. Both algorithms halt providing the same
result; although a skeptic might say we require a proof that both
algorithms do indeed calculate x*y.

Similarly, one /can/ add 0.(9) + 1.(9) by adding digits one at a time;
but one can also write an algorithm that does not proceed this way;
and which will halt for the sum of any pair of repeating decimals of
the form x.(y).

Of course, a skeptic would require proof - in both cases - that the
algorithm is actually carrying out the mathematical sum in question;
but having been provided that proof, both algorithms do indeed sum
these numbers.

The fact that the "digit by digit" algorithm fails to halt doesn't
mean the problem is somehow "infinitely complex"; because there /does/
exist an algoirthm which /always/ halts and is only "finitely"
complex.

In other words: the sum of x.(y) + z.(w) is always decidable; despite
the fact that there are "bad" algorithms that fail to halt in some (or
all) situations.

What Church et al are concerned with is the fact that there exist
certain types of problems which are /not/ decidable; i.e., for which /
no/ algorithm exists which is guaranteed to halt.

> This way Church - Cole negative lambda calculus solution eliminate any
> "philosophical dependencies" from the physical computer limitations.
> The problem is not in memory storage space or such but in the
> undecidable nature of the algorithm.
> In case (1) the algorithm is infinite because we said it is infinite.
> In case (2) the algorithm is not finished because it was not finished
> in Tmax period. But is it really infinite? Maybe it is or maybe it
> would be finished at Tmax + 1 moment.
>
> In either case there is not any machine fault or a precision problem.
>
> Note: "Entscheidungsproblem" instead of "Decision problem" is not to
> show up. The original German word is used in the areas I'm working
> with. So kind of "slang exchange" with you. :-)
>
> > > btw a skeleton in the programming closet: is a problem solvable if it
> > > is algorithmic but it requires infinity to complete the algorithm?
>
> > Don't get me started! It's an interesting set of problems: Turing's
> > Halting problem, the uncomputability of the list of all computable
> > numbers, Godel's incompleteness theorem, recursively enumerable sets,
> > etc.
>
> { Entscheidungsproblem, lambda calculus, Goedel's incompleteness
> theorem }
>
> Hah! Goedel in both sets! :-)
>

Yeah, that guy gets around.

> "The thing that hath been, it is that which shall be;
> and that which is done is that which shall be done:
> and there is no new thing under the sun.
>
> Is there any thing whereof it may be said, See, this is new?
> it hath been already of old time, which was before us."
> Ecclesiastes 1-1
> :-) :-|
> btw Lester Zick in another thread is puzzled with the "true nature" of
> the point just like Euclid was about 2,250 years ago - and with the
> same doubts in mind: "can nothing be something and if it can then how
> can it be?"
>

Lester claims to be puzzled by many things; somewhat intentionally so
it seems to me.

> > > I need to find an extra criterion to
> > > separate
> > > 1) all numbers having (0) in period
> > > 2) all numbers having any other period
> > > 3) all numbers with unknown/no period
>
> > > (3) is easy - these are irrationals.
> > > (1) and (2) ? 0.125 and 0.124(9), 1.3 and 1.(3)? Does the part of
> > > science we are talking about - its official part I mean - has any
> > > criteria of distinction?
>
> > You are correct regarding (3)(but can you prove it?).
>
> Oh gosh, I remember I have to be careful here :-)
> sqrt(2) is an irrational or a rational with unknown - yet - period?
> Other words is sqrt(2) irrational or is it a number with undecidable
> by current tools status: maybe rational with a very long period, maybe
> indeed an irrational.

Simple proof that sqrt(2) is not rational (known to the Greeks):

http://en.wikipedia.org/wiki/Irrational_number

> What do you personally think: is everything clear about sqrt(2) or say
> PI - or there is still a question sign here?
>

Well, it depends on what you mean by "everything".

Does pi contain a continuous sequence of 10^10 zeroes in its decimal
expansion?

This is not known; and probably not knowable /in practice/ for a
variety of reasons. But, /in principle/, either it is true or it is
false. There exists exactly one sequence of decimal digits which
represents the number pi (see proof below); and either there are 10^10
successive 0's in that sequence or there ain't.

If P(x) is a statement about a real number x such as "x is rational"
or "x satisifies integral(0..1) x*sin(y) dy = 1" which is either true
of a real number or isn't true of that number, then it follows that
P(pi) is either true or false. We may lack a /proof/ of this fact
either in practice (too complex) or in principle (Godel's curse); but
I still consider it a fact.

And that for me is plenty enough to say "pi is unique real number".

> > As regards (1) and (2): informally stated, it's pretty easy (although
> > a little long-winded) to prove that x.(y) = z.(0) if and only if y =
> > b-1 and z = x+1; where x and z are integers, y is a natural with 0 <=
> > y < b, and b is the base we are working in for our sequence of digits
> > (e.g., 2 or 10).
>
> > (Proof available on request; just don't name it after me!)
>
> I don't think "request" is a suitable word here, but if you have time
> for that second lemma proof then it would be the most great.
> btw a paper sheet scan copy or MathML or LaTeX is fine as well _if_ it
> saves your time. Expressing higher math in pure ASCII should be a
> torture.

Well, half the job is setting up exactly what is meant by "x.yzwvu..."
and "x.(y)". But here's an informal proof, assuming we assume the
usual definitions of these representations:

First, note that any digit sequence "0.xyzw..." represents a number x
with 0 <= x <= 1.

By the limit definition of a sequence of digits, the repeating
sequence "0.(b-1)" represents the number 1.0, where b is base.

It should be obvious that no other distinct sequence "0.xyzw..." can
represent the number 1.0; because at least one digit must be less than
(b-1).

Similarly, exactly one sequence represents the number 0: 0.(0).

Supose two digit sequences s = (s_1, s_2, ...) and t = (t_1, t_2, ...)
represent the same number in [0,1). Let n be the first natural where
the nth digit of s differs from the nth digit of t, and without loss
of generality, suppose s_n > t_n.

Letting num(s) and num(t) be the numbers represented by s and t,
resp., we have

num(s) = sum(i = 1 to n-1) (s_i/b^i) + s_n/b^n + sum(i = n+1 to oo)
(s_i/b^i)

num(t) = sum(i = 1 to n-1) (t_i/b^i) + t_n/b^n + sum(i = n+1 to oo)
(t_i/b^i)

By assertion, sum(i = 1 to n-1) (s_i/b^i) = sum(i = 1 to n-1) (t_i/
b^i), so if num(s) = num(t) we must have that

s_n/b^n + sum(i = n+1 to oo) (s_i/b^i) = t_n/b^n + sum(i = n+1 to oo)
(t_i/b^i)

or multiplying through by b^n,

s_n + sum(i = n+1 to oo) (s_i/b^(i-n)) = t_n + sum(i = n+1 to oo) (t_i/
b^(i-n))

Then

s_n - t_n = sum(i = n+1 to oo) (t_i/b^(i-n)) - sum(i = n+1 to oo) (t_i/
b^(i-n))

By our earlier observations, the right hand side is at most 1 (both
sums must be in the closed interval [0,1]). Since s_n <> t_n, and s_n
- t_n must be a natural, therefore s_n - t_n = 1; and we have

sum(i = n+1 to oo) (t_i/b^(i-n)) - sum(i = n+1 to oo) (t_i/b^(i-n)) =
1

Given the restriction of each sum to [0,1], the only way this equality
can be achieved is if

sum (i = n+1 to oo) (t_i/b^(i-n)) = 1

and

sum (i = n+1 to oo) (s_i/b^(i-n)) = 0

which as noted above can only occur if t_i = b-1 for all i > n and s_i
= 0 for all i > n.

Thus, two digit sequences (repeating or not!!!) represent the same
number if, and only if, the first digit they differ at is a difference
of 1, and the sequence with the smaller digit has a repeating tail of
(b-1) and the other sequence a repeating tail of (0).

Cheers - Chas

cbr...@cbrownsystems.com

unread,
Mar 21, 2007, 9:04:12 PM3/21/07
to
On Mar 21, 4:25 pm, cbr...@cbrownsystems.com wrote:

> Well, half the job is setting up exactly what is meant by "x.yzwvu..."
> and "x.(y)". But here's an informal proof, assuming we assume the
> usual definitions of these representations:

Here's a slightly clearer proof which elimates the "it should be
obvious that..." in the previous version:

Suppose s = (s_1, s_2, ..., s_n, ...) is a sequence of digits base b.
Define

num(s) = sum (i = 1 to oo) (s_i/b^i)

where num(s) is "the number represented by the digit sequence s".

First, by the limit definition of a sequence of digits, the repeating
sequence (b-1, b-1, b-1, ..., b-1, ...) = "0.(b-1)" represents the
number 1.0.

If s is any sequence of digits distinct from "0.(b-1)", then

num("0.(b-1)") - num(s) = sum(i = 1 to oo) ((b-1) - s_i)/b^i

is a sum of positive numbers and thus > 0; therefore:

(a) num(s) = 1 if and only if s = "0.(b-1)"
(b) every sequence distinct from "0.(b-1)" is less than 1

Similarly, the sequence (0, 0, ..., 0, ...) = "0.(0)" has num("0.(0)")
= 0; and for all sequences s distinct from "0.(0)", num(s) - num("0.
(0)") is the sum of positive numbers, and thus > 0; so

(c) num(s) = 0 if and only if s = "0.(0)"
(d) every sequence distinct from "0.(0)" has num(s) > 0

To summarize:

Every sequence s has num(s) in [0..1].
num(s) = 1 iff s = "0.(b-1)".
num(s) = 0 iff s = "0.(0)".

(where "iff" is shorthand for "if and only if").

Suppose s = (s_1, s_2, ...) and t = (t_1, t_2, ...) are distinct
sequences which represent the same number in [0,1). Let n be the
smallest natural where the nth digit of s differs from the nth digit


of t, and without loss of generality, suppose s_n > t_n.

We then have

num(s) = sum(i = 1 to n-1) (s_i/b^i) + s_n/b^n + sum(i = n+1 to oo)
(s_i/b^i)

num(t) = sum(i = 1 to n-1) (t_i/b^i) + t_n/b^n + sum(i = n+1 to oo)
(t_i/b^i)

By assertion, sum(i = 1 to n-1) (s_i/b^i) = sum(i = 1 to n-1) (t_i/
b^i), so if num(s) = num(t) we must have that

s_n/b^n + sum(i = n+1 to oo) (s_i/b^i) = t_n/b^n + sum(i = n+1 to oo)
(t_i/b^i)

or multiplying through by b^n,

s_n + sum(i = n+1 to oo) (s_i/b^(i-n)) = t_n + sum(i = n+1 to oo)
(t_i/
b^(i-n))

Then

s_n - t_n = sum(i = n+1 to oo) (t_i/b^(i-n)) - sum(i = n+1 to oo)
(t_i/
b^(i-n))

By our earlier observations, the right hand side is at most 1 (both
sums must be in the closed interval [0,1]). Since s_n <> t_n, and s_n
- t_n must be a natural, therefore s_n - t_n = 1; and we have

sum(i = n+1 to oo) (t_i/b^(i-n)) - sum(i = n+1 to oo) (t_i/b^(i-n)) =
1

Given the restriction of each sum to [0,1], this equality
can be achieved if and only if

sum (i = n+1 to oo) (t_i/b^(i-n)) = 1

and

sum (i = n+1 to oo) (s_i/b^(i-n)) = 0

which as noted above can only occur if and only if t_i = b-1 for all i


> n and s_i
= 0 for all i > n.

Thus, two distinct digit sequences (repeating or not!!!) represent the

VK

unread,
Mar 24, 2007, 1:51:47 PM3/24/07
to

We can: but it will not be addition of 0.(9) + 1.(9)
0.999 + 1.999 or even
0.99999999999999999999999999999999999999
+
1.99999999999999999999999999999999999999

is not the same as 0.(9) + 1.(9)

That is why both "binary constraints lemma" you so generously made for
me - they both are taking numbers from an unorthodox point of view:
having (0) in period or having anything else in (period). This way the
actual difference between rationals, irrational, imaginaries or
transcendentals gets irrelevant. It is not in violation of math rules
(?) because we do not redefine any of entities: we just taking into
account contextually important property and disregarding contextually
irrelevant.

> > What do you personally think: is everything clear about sqrt(2) or say
> > PI - or there is still a question sign here?
>
> Well, it depends on what you mean by "everything".

Actually by this question I put both irrationals and transcendentals
in one basket - not out of trickery but by my bad knowledge. So thank
you for taking time to explain instead of simply pointing the error.

Thank you very much once over again. I'm thinking to make a MathML
document out of both lemma. If you have Firefox browser you may be
agree to review it when done to ensure that my "dyslexia" did not pop
up anywhere else :-)

cbr...@cbrownsystems.com

unread,
Mar 25, 2007, 3:36:04 PM3/25/07
to
On Mar 24, 10:51 am, "VK" <schools_r...@yahoo.com> wrote:

> > Similarly, one /can/ add 0.(9) + 1.(9) by adding digits one at a time;
> > but one can also write an algorithm that does not proceed this way;
> > and which will halt for the sum of any pair of repeating decimals of
> > the form x.(y).
>
> We can: but it will not be addition of 0.(9) + 1.(9)
> 0.999 + 1.999 or even
> 0.99999999999999999999999999999999999999
> +
> 1.99999999999999999999999999999999999999
>
> is not the same as 0.(9) + 1.(9)
>

Right. But at the same time, such an algorithm, given a natural n,
does return the nth digit of the sum; and this is (one) defintition of
"computable number".

> That is why both "binary constraints lemma" you so generously made for
> me - they both are taking numbers from an unorthodox point of view:
> having (0) in period or having anything else in (period). This way the
> actual difference between rationals, irrational, imaginaries or
> transcendentals gets irrelevant. It is not in violation of math rules
> (?) because we do not redefine any of entities: we just taking into
> account contextually important property and disregarding contextually
> irrelevant.
>
> > > What do you personally think: is everything clear about sqrt(2) or say
> > > PI - or there is still a question sign here?
>
> > Well, it depends on what you mean by "everything".
>
> Actually by this question I put both irrationals and transcendentals
> in one basket - not out of trickery but by my bad knowledge. So thank
> you for taking time to explain instead of simply pointing the error.
>

All transcendentals are irrationals; but not all irrationals are
transcendental. "x is transcendental" means: there exists no
polynomial p with rational coefficients such that p(x) = 0.

<snip>

> > Thus, two digit sequences (repeating or not!!!) represent the same
> > number if, and only if, the first digit they differ at is a difference
> > of 1, and the sequence with the smaller digit has a repeating tail of
> > (b-1) and the other sequence a repeating tail of (0).
>
> > Cheers - Chas
>
> Thank you very much once over again. I'm thinking to make a MathML
> document out of both lemma. If you have Firefox browser you may be
> agree to review it when done to ensure that my "dyslexia" did not pop
> up anywhere else :-)

Sure. Just post a link here (odds are high that if you e-mail me,
you'll get sucked into a spam trap).

Cheers - Chas

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