Simple mathematical problem (?)

[Suggestologist]

If you are suggesting that 0.999... is not equal to one because of some "representational limitation", then let's hear what you think that limitation is. Otherwise, this is just another one of your irrelevant points.

One of MY irrelevant points? What about all of YOUR irrelevant, failing to even try to consider anything outside of your cramped viewpoint points?

Pi Pi.

 

[Suggestologist]

It's not an assumption. It's a fact. That is precisely the point of this whole thread.

0.24999... = 0.25000... for exactly the same reason that 0.999... = 1.000...

Where do you get it from the numeric representational system? Without going to a different logical level?

 

[xouper]

Suggestologist: One of MY irrelevant points? What about all of YOUR irrelevant, failing to even try to consider anything outside of your cramped viewpoint points?

The conversation in this thread is about mathematics and the real numbers. If you choose to characterize that as a "cramped viewpoint", then you have just removed any remaining doubt that you are a crackpot.

 

[Suggestologist]

The conversation in this thread is about mathematics and the real numbers. If you choose to characterize that as a "cramped viewpoint", then you have just removed any remaining doubt that you are a crackpot.

It was YOU who engaged in the ad hominem nonsense. While failing to consider anything outside of your CRAMPED viewpoint.

 

[xouper]

xouper: It's not an assumption. It's a fact. That is precisely the point of this whole thread.

0.24999... = 0.25000... for exactly the same reason that 0.999... = 1.000...

Suggestologist: Where do you get it from the numeric representational system? Without going to a different logical level?

That's what we've been trying to beat into your thick skull, you moron. It's a direct consequence of the definitions of decimal representation. Sheesh, we've been over this point a bazllion times in this thread already. Are you really that dense?

And not just decimal notation, it works the same in any other number base. In base 8, for example -

0.777... = 1
and
0.2777... = 0.3000...

In base two,

0.111... = 1

 

[xouper]

Suggestologist: While failing to consider anything outside of your CRAMPED viewpoint.

It is not MY cramped viewpoint. It is the viewpoint of mathematicians in general. YOU are the one who doesn't get it. You have completely failed to make any case at all that mathematicians are wrong about this question. Your criticisms of the mathematical proofs in this thread are completely devoid of any valid point.

I've said this before and I'll say it again. If you were to post your nonsense on the sci.math newsgroup, you would have long ago been dismissed as a crackpot. Here on JREF we have been quite a bit more patient, but you are rapidly wearing out your welcome here with your crap. Either make some legitimate mathematical points or STFU.

 

[Suggestologist]

That's what we've been trying to beat into your thick skull, you moron. It's a direct consequence of the definitions of decimal representation. Sheesh, we've been over this point a bazllion times in this thread already. Are you really that dense?

And not just decimal notation, it works the same in any other number base. In base 8, for example -

0.777... = 1
and
0.2777... = 0.3000...

In base two,

0.111... = 1

You're engaging in the same circular logic you have from the beginning. Obviously, you're a p-zombie who can't think in more than one way. Your obnoxious style is not acceptable. You will be ignored from now on. Bye Bye.

 

[Cabbage]

I haven't posted to this thread in a while, honestly, because I think to do so would be a complete waste of my time. However, what the hell, I'll give it one last try.

Suggestologist, I find it interesting that you keep asking people for the definition of decimal notation and so forth--everyone here but you is using the standard definitions, and these can be looked up elsewhere on the internet and in various books. A far more interesting question is what is your definition of decimal representations is--why not give us your definitions, when that is so obviously the crux of this whole disagreement. I will say (and I have never denied it, in fact) that by your definition, it's entirely possible that .999... is not equal to 1. Understand, however, that you are alone in this definition, and the whole point of the thread is that .999... is exactly equal to 1 under the standard definitions, not yours.

One other thing, I asked this question quite a while back, and never received an answer to it (though I'm dying to see what your answer may be):

You had said:

I was asked to show how "Infinity + 1" was relevant to the discussion earlier in this thread and this is where it is relevant. .99999..... has omega nines following the decimal point. If multiplying by 10 produces 9.99999...... such that 9.99999..... - .9..... = 9; then you have to assume that omega+1 9's were actually following the decimal point in .99999.....; and that omega+1 = omega which is not correct in the transfinite system. Now someone will say, who cares about the transfinites, we're talking about the reals. And my reply is that we are talking about the properties of a real number, and so the transfinites apply.

I replied with the comment/question:

There is no (omega+1)th digit in a real number.

What's the (omega+1)th digit of pi? What if you multiply it (pi) by 10? Divide it (pi) by 10?

Well?

 

[xouper]

Suggestologist: You're engaging in the same circular logic you have from the beginning.

So you say. Where's your proof, big-mouth. Oh that's right, you have none.

Obviously, you're a p-zombie who can't think in more than one way. Your obnoxious style is not acceptable. You will be ignored from now on. Bye Bye.

No loss to me to be ignored by a crackpot.

 

[T'ai Chi]

Still having problems with math, T'ai? (yes, I saw the smilie, too. I think, though, we need a new, lame smilie for attempts at answers so lame they need to fly under the cover of smilies so nobody can challenge them. Eh?)

*shrug* Whatever.

 

[Suggestologist]

I haven't posted to this thread in a while, honestly, because I think to do so would be a complete waste of my time. However, what the hell, I'll give it one last try.

Suggestologist, I find it interesting that you keep asking people for the definition of decimal notation and so forth--everyone here but you is using the standard definitions, and these can be looked up elsewhere on the internet and in various books. A far more interesting question is what is your definition of decimal representations is--why not give us your definitions, when that is so obviously the crux of this whole disagreement. I will say (and I have never denied it, in fact) that by your definition, it's entirely possible that .999... is not equal to 1.

Well, I wanted to fit my idea into their framework, because it was obvious that most other posters either could not understand or were not willing to consider the description I have already described. That's why I asked for theirs.

Ok. If you look at the expansion of the fractional part of a number as: "0."d1d2d3...d(omega-1)d(omega); Where each d is a digit between 0 and 9; then there is a granularity limitation of the decimal real numbers at the omega digit. In the same sense as the granularity limitation of integers is 1; the limitation for real numbers is: for i = 1 to (omega-1), d(i)=0; d(omega)=1 -- which I'll call iota here and which I earlier represented as 0.00000...00001, where "..." stands for (omega minus 10) "0" digits.

This is consistent when looking at the .999... representation of 1/1 that someone presented earlier in the thread. As I commented then, there is a remainder of 1 iota after the omega digit is calculated from that division. And, as I have written before, this means that the .999.... representation is ambiguous; it may mean 1 and it may mean 1 MINUS iota, depending on where it came from.

There are no real decimal number representations between .999... and 1; but this does mean that the distance between them is 0 and that they are the same number; the distance between them is iota and they are different numbers.

Otherwise .99999...99998 would also equal one; and you could do an (omega) infinite regress to show that all fractional decimal numbers of the form I described above equal 1. This would be like turning the decimal numbers into the integers.

Now, think about what this conceptualization means for calculating (1+.999...)/2; it means that we need to know where .999... came from; and if it came from (1 minus iota), then the answer will result in a loss of information unless we make a side-note about the manipulations we have put the number through.

Understand, however, that you are alone in this definition, and the whole point of the thread is that .999... is exactly equal to 1 under the standard definitions, not yours.

If you mean the mathematical propositions one solves to earn a grade on a mathematics test, then I agree that those are the standard definitions one learns to regurgitate. But perhaps this is because people don't use the information at the omega digit... of course if they learned to use infinitessimals to do their calculus instead of the epsilon-delta limit crap that is taught....

If people couldn't play with numbers and actually use their brains to think about the consequences for efficient computation, we'd still be using the natural number system for everything.

One other thing, I asked this question quite a while back, and never received an answer to it (though I'm dying to see what your answer may be):

You had said:

I replied with the comment/question:
Well?

What would happen for some number is that the information at the omega digit would be lost. Just like taking 22 DIV 10 *10 = 20 in the integer system. That information may or may not be needed.

 

[Earthborn]

That's why I asked for theirs.

That makes no sense. You already know their definition:

If you mean the mathematical propositions one solves to earn a grade on a mathematics test

That's it! It doesn't make much sense to use definitions that would cause you to fail, does it?

Well, I wanted to fit my idea into their framework

I guess as a child you also stubbornly tried to fit the square block in the round hole!

because it was obvious that most other posters either could not understand or were not willing to consider the description I have already described.

I think it is more obvious that if you continue to argue something without explaining that you use completely different definitions than anyone else, you look silly.

Could you explain what in under your definitions is 3 times a third and what the decimal representation of a third is?

it means that we need to know where .999...

Doesn't this just mean that your definition is not self consistent in a mathematical sense?

 

[LW]

I'll call iota here and which I earlier represented as 0.00000...00001, where "..." stands for (omega minus 10) "0" digits.

Are you using some arithmetic where 'omega - 10' is not equal to 'omega'?

There are no real decimal number representations between .999... and 1; but this does mean that the distance between them is 0 and that they are the same number; the distance between them is iota and they are different numbers.

So, your point still is that 0.99... is not a real number, even though it looks like a real number, it behaves like a real number, it walks like a real number, and quacks like a real number. [Or the other possibility is that you choose to disregard the basic fact that between any two different real numbers there's an uncountably many real numbers.]

In short, you have taken a real number representation and then redefined it to mean something else than a real number. Similarily, you could define the symbol 2 to mean the rational number 6/4 and then argue that 2 + 1 is not equal to 3.

Otherwise .99999...99998 would also equal one;

That one is not a real number.

 

[LuxFerum]

That one is not a real number.

That one is
sum n=2..inf, 18*10^(- n)

 

[69dodge]

Originally posted by Suggestologist
Ok. If you look at the expansion of the fractional part of a number as: "0."d1d2d3...d(omega-1)d(omega); Where each d is a digit between 0 and 9; then there is a granularity limitation of the decimal real numbers at the omega digit. In the same sense as the granularity limitation of integers is 1; the limitation for real numbers is: for i = 1 to (omega-1), d(i)=0; d(omega)=1 -- which I'll call iota here and which I earlier represented as 0.00000...00001, where "..." stands for (omega minus 10) "0" digits.

I take it this is your impression of the standard definition of decimal expansions of real numbers. Yes?

Your impression is not correct. A decimal expansion does not have a last digit, not even one labelled "omega". Every digit is followed by another. The three dots are at the end, not in the middle. (.999. . . is a decimal expansion; .999. . .999 is not.)

You seem to be treating omega as merely a very large finite number. It is not a finite number. It is a limit ordinal, which means it has no immediate predecessor. You appear to think it has one called "omega - 1".

 

[69dodge]

Originally posted by Suggestologist
Where do you get it from the numeric representational system? Without going to a different logical level?

Do you mean, what is the rule for deciding when two decimal expansions represent the same real number? Or do you mean, why is that the rule?

The rule can be given purely in terms of the numeric representational system (i.e., decimal expansions). The reason for the rule cannot, as it involves the relationship between real numbers and their decimal representations.

I believe I have answered both these questions already. But in response to each answer, you asked for the answer to the other question. When I wrote, in answer to "why":<blockquote>the definition of ".d₁d₂d₃d₄. . .", where the ds are decimal digits, is "the limit of the sequence .d₁, .d₁d₂, .d₁d₂d₃, .d₁d₂d₃d₄, . . .",</blockquote>you asked, "what":<blockquote>Please elaborate on how you turn this into the number line. How do you determine that number A is larger than number B in terms of their digital values?</blockquote>Then when I answered that, you again asked, "why":<blockquote>Where do you get this assumption about dual decimal expansions?</blockquote>You have both answers. Let's stop going around in circles.

Obviously, the digit sequence ".24999. . ." is not the same as the digit sequence ".25000. . .". However, digit sequences like these often appear in contexts in which it is either explicitly stated or implicitly understood that they represent real numbers. In such cases, we use the limit definition above to determine which real number is represented.

Although ".24999. . ." and ".25000. . ." differ as digit sequences, they represent the same real number when interpreted as decimal expansions.

 

[Suggestologist]

I take it this is your impression of the standard definition of decimal expansions of real numbers. Yes?

Your impression is not correct. A decimal expansion does not have a last digit, not even one labelled "omega". Every digit is followed by another. The three dots are at the end, not in the middle. (.999. . . is a decimal expansion; .999. . .999 is not.)

Pretend that they can be in the middle, and look at the consequences.

 

[LuxFerum]

That one is
sum n=2..inf, 18*10^(- n)

0,8+sum n=2..inf, 18*10^(- n)

 

[BillHoyt]

Pretend that they can be in the middle, and look at the consequences.

Hey, why not pretend 4 is really 5 and look at the consequences?


An IQ like a bag of roasted peanuts.

 

[xouper]

Suggestologist: In the same sense as the granularity limitation of integers is 1; the limitation for real numbers is: for i = 1 to (omega-1), d(i)=0; d(omega)=1 -- which I'll call iota here and which I earlier represented as 0.00000...00001, where "..." stands for (omega minus 10) "0" digits.

Why are we discussing this yet again? The notation 0.000...0001 does not represent a real number. Your argument has been soundly refuted earlier in this thread, so why are you bringing it up again? Your behavior in this thread reminds me of Lucianarchy. What a kook.