Is 1 = 0.999... ? Really?

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Is it true that 1 = 0.999...? And Exactly Why or Why Not?

Yes, 1 = 0.999...
12
40%
No, 1 ≠ 0.999...
15
50%
Other
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10%
 
Total votes : 30

Re: Is 1 = 0.999... ? Really?

Postby Magnus Anderson » Sun Feb 02, 2020 6:10 pm

gib wrote:If L is the largest number, then it is a number. If it's a number, it represents a certain quantity of things. So if you had a set consisting of that many things, there's no reason to say you cannot add one more thing to the set. If you add one more thing to it, you will have L + 1 things (this was the argument you so adamantly insisted was true about adding to infinity). But if L is the largest number, you cannot have a larger number L + 1. Therefore, adding one more item to a set of items whose quantity is L will not give you L + 1 items. <-- A Contradiction.


This is based on the erroneous assumption that every number has a number greater than it. This holds true for naturals, integers, rationals and standard reals but it does not hold true for all numbers. Indeed, it holds true for all numbers EXCEPT for one: the largest number.

Of course I have. There's no other possibility. Unfortunately, you refuse to explain to me what your concept of number is. The best I can do, therefore, is to respond within the context of the standard definition of number.


Well, you could try, oh I don't know, explaining what you mean by "number"?


Anything that is more or less than something else is a number. The largest number is a number because it's something that is greater than every other number. The same applies to infinity: it's a number because it's something that is greater than every integer (or standard real, if you will.)

Yeah, well, when the only example you can dig up for the definition of infinity has to be described as "closer" to the definition you have in mind, you know you're reaching. On the other hand, if you type "definition of infinity" into google, you get a whole ream of exact definitions, and none of them mention integers.


Why do I have to use the same exact words as other people do? Back when I was in school, it was highly desirable for students to use their own words for the simple reason that by doing so they prove they aren't parrots.

"Greater than every integer" and "Greater than every standard real" are two different (and only slightly so) expressions of one and the same thing. There's no difference. Nothing "new and exotic" about the former.

Wikipedia wrote:Infinity (often denoted by the symbol \(\infty\) or Unicode ∞) represents something that is boundless or endless or else something that is larger than any real or natural number.


The Wiki says "Greater than every natural number". Still not "Greater than every integer", the "new and exotic" definition that I'm putting forward.

That you're better off defining infinity in terms of reals rather than integers. You stated earlier: "Infinity is a number greater than every integer, so it's greater than 1,000 as well as 1,000,000. And yes, infinity (if it refers to a specific quantity greater than every integer, and not merely to any such quantity) has a place on the number line." <-- If you mean that infinity falls on the hyperreal section of the number line, you could state this more clearly by saying infinity is greater than all reals. Otherwise, it leads one to wonder whether you think there are real numbers not only greater than infinity but all integers (and given some of the ideas you've defended in this thread, I wouldn't put it passed you).


If something is greater than all integers, it's also greater than all standard reals.

The problem is that you refuse to acknowledge you're talking about hyperreals.


That's not true.
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Re: Is 1 = 0.999... ? Really?

Postby iambiguous » Sun Feb 02, 2020 7:14 pm

This is based on the erroneous assumption that every number has a number greater than it. This holds true for naturals, integers, rationals and standard reals but it does not hold true for all numbers. Indeed, it holds true for all numbers EXCEPT for one: the largest number.


Or, only somewhat tongue in cheek:

This is based on the erroneous assumption that every God has a God greater than it. This holds true for Christians, Muslims, Hindus, Shintos and standard Gods but it does not hold true for all Gods. Indeed, it holds true for all Gods EXCEPT for one: the largest God.

The one true God that you believe in. If, for example, you're a Zoroastrianist, that would be Ahura Mazda.
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Re: Is 1 = 0.999... ? Really?

Postby Magnus Anderson » Sun Feb 02, 2020 8:51 pm

gib wrote:Woaw, now that's an odd concept. This scares me because I'm going to have to ask, once again, how you're thinking about this, and we've seen how that goes down. But I'm going to venture a guess: you're thinking of the elipses as saying something like "goes off to infinity", as in the way we might write: 1, 2, 3,... Which is fine, but then you still want to add another number "after" infinity.


In which case, you must have a serious problem with the expression \(0.9 < 0.99 < 0.999 < \cdots < 1\).

it become especially deceptive when we think of the reals as belonging to a set


I have no idea how to respond to the statement that there is no such thing as the set of real numbers.

When we talk about sets, we are talking about a collection of members, and members are conceptualized as discrete units. It's all fine and dandy to talk about the set of integers because these are already conceptualized as discrete units. But when it comes to reals, the idea of the set of real numbers presupposes that you can add them to the set as discrete units.


Maybe you should start with the definition of the term "discrete unit"?

Why are you saying that an integer such as \(9\) is a "discrete unit" whereas a real number such as \(9.249238491\) is not?

You would be right to say there is no second last real number between 0 and 1, but I question whether it makes sense to talk about them as members of a set.


The fact is, absolutely anything can be a member of a set.

I was ok talking about the sequence of all integers, or even a sequence of real numbers so long as we picked out specific real numbers to belong to the sequence, but when it comes to all the real numbers between two points on the number line (or the whole number line), I'm not even sure we can talk about them as belonging to a set. They aren't discrete enough; they form a smooth continuum. The number line is composed of arbitrary segments, each one merging into its neighbors, no definite point where it begins and the next ends--all except if we divide it up a specific way--but we can't do that infinitely.


In other words, there is no set of all reals between \(0\) and \(1\) but there is a set of finite number of reals between \(0\) and \(1\)? And that's because reals are not "discrete enough"?
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Re: Is 1 = 0.999... ? Really?

Postby gib » Mon Feb 03, 2020 2:59 am

Magnus Anderson wrote:
gib wrote:I can't address this without understand what the ... means to you. We've already ruled out a continuum, at least on the high end, but then I'm at a loss to understand what this could possibly mean.


The ellipsis indicates that the pattern continues indefinitely. In the case of \(a_1 < a_2 < a_3 < \cdots < b\), the pattern is \(a_n < a_{n+1}\).

The first part of the expression, namely \(a_1 < a_2 < a_3 < \cdots\), is equivalent to the following list of statements:

\(a_1 < a_2\)
\(a_2 < a_3\)
\(a_3 < a_4\)
\(\dotso\)

The list is made out of an infinite number of statements. It contains every \(a_n, n \in N\) and it basically says that every \(a_n, n \in N\) is less than \(a_{n+1}\).

The second part of the expression, namely \(\cdots < b\), is equivalent to the following list of statements:

\(a_1 < b\)
\(a_2 < b\)
\(a_3 < b\)
\(\dotso\)

Like the previous one, the list is made out of an infinite number of statements. It basically says that every \(a_n, n \in N\) is less than \(b\).

You can take every variable in that expression and replace it with a standard real while preserving the truth value.

For example:

\(0.9 < 0.99 < 0.999 < \cdots < 1\)

In this case, \(a_n = \sum_{i=1}^{n} \frac{9}{10^i}, n \in \{1, 2, 3, \dotso\}\) and \(b = 1\).


Ah, there we go. Now was that so hard? You see what a little explanation can do (and an example to boot!)? I understand what you mean!

My confusion lay in the assumption that we were talking about the largest number (because that's how this started), or the last number in the sequence of all integers--you know, at the infinitely big end--but you're right, \(\sum_{i=1}^{n} \frac{9}{10^i}\) does represent an endless sequence of sums, and 1 does come after it. I stand corrected.

In this case, the "end" of the sequence has a double meaning and can lead to equivocation. In the one sense, the sequence \(\sum_{i=1}^{n} \frac{9}{10^i}\) has no end. You just keep adding \(\frac{9}{10^i}\) forever. On the other hand, it does have an end on the number line--that is to say, there is a definite limit which we can call the "end" (whether that's 1 or < 1 is the central dispute of this thread, but let's not worry about that now). In other words, there is no end term of the infinite sum but there is an end location on the number line--two different things.

It's still problematic when we think of it as a sequence, however (at least for me), since the word "end" in that case has only one meaning: the last item in the sequence. And I am still troubled by the idea of the last item coming "after" a sub-sequence with no end. It might be argued that this isn't a proper sequence. Maybe there are two sequences: a parent sequence and a child sequence. Maybe the parent sequence consists of only two items: P = (\(\sum_{i=1}^{n} \frac{9}{10^i}\), 1), where the child sequence is \(\sum_{i=1}^{n} \frac{9}{10^i}\): C = (0.9, 0.09, 0.009, ...). In other words, the child sequence is treated as a single discrete item and placed in the first position of the parent sequence. This would work for me. This would permit that 1 still comes after all terms in \(\sum_{i=1}^{n} \frac{9}{10^i}\) without requiring that it be expressed as a "sketchy" sequence.

Magnus Anderson wrote:
gib wrote:It doesn't mention \(a_4\) either yet you agreed \(a_4\) was implicit.


It does mention \(a_4\) but not explicitly i.e. it mentions it implicitly. \(a_{L-1}\), on the other hand, is not mentioned at all -- explicitly or implicitly.


And how does one tell? Don't you think there ought to be something in the notation that indicates the existence of implicit terms? Take the notation representing the continuity between the reals and the infinitely large hyperreals for example:

1, 2, 3, ... , R-2, R-1, R, R+1, R+2, ...

Here, the ... represents implicit terms on both sides: 4, 5, 6, ... on the left, ... R-5, R-4, R-3, on the right.

Is it the fact that it's obviously a sequence on both sides (i.e. the hyperreals are listed out as a sequence rather than a single number)?

Magnus Anderson wrote:
I didn't say there couldn't be order without "first", "last", and "second last". I said "It's impossible to have an order with positions in the order missing."


Maybe you should try explaining what it means "to have an order with positions in the order missing".

It's ludicrous to say there is a queue of 10 people but the last one in the queue is the 11th one because there is no 7th person. If there were 11 people initially, and person #7 left, then the 8th person simply becomes the 7th, the 9th becomes the 8th, and so on.


And how exactly does that relate to what I'm saying?


Hard to tell when I don't even know what you're talking about. But now that you've elucidated what you're talking about, we can put this line of argument to rest.

Magnus Anderson wrote:
Max wrote:"The largest number" does not mean "A natural, integer, rational or a standard real greater than every other number". It means "A number greater than every other number".


gib wrote:Ok, well, I don't believe in it.


Saying that you do not understand a position is not an argument against that position. I know, I'm just stating where I stand. It's merely an expression of ignorance. Or enlightenment. You have to understand an argument before you can say it's wrong. Indeed, most of what you're doing in this thread amounts to "I don't understand it, therefore it's wrong".


Fine, if you mean something incredibly insightful and profound by "the largest number" and my brain is just too stupid to grasp it, I can't possibly know whether you're right or wrong. But given your lack of explanation for what kind of number the "largest number" is, I fall back on the ordinary concept of "number" and say that I know no such number in that sense exists. And chances are pretty good you don't have a special meaning of "number" in mind and just misunderstand how numbers work (even if we include hyperreals in the mix, you can't have a largest number).
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Re: Is 1 = 0.999... ? Really?

Postby Magnus Anderson » Mon Feb 03, 2020 10:57 am

gib wrote:My confusion lay in the assumption that we were talking about the largest number (because that's how this started), or the last number in the sequence of all integers--you know, at the infinitely big end--but you're right, \(\sum_{i=1}^{n} \frac{9}{10^i}\) does represent an endless sequence of sums, and 1 does come after it. I stand corrected.


That's an example of a sequence that has a beginning (first element, which is \(a_1 = 0.9\)) and an end (last element, which is \(b = a_L = 1\)) but no second last element, no third last element and so on.

On the other hand, it does have an end on the number line--that is to say, there is a definite limit which we can call the "end" (whether that's 1 or < 1 is the central dispute of this thread, but let's not worry about that now). In other words, there is no end term of the infinite sum but there is an end location on the number line--two different things.


Note that there is no \(0.\dot9\) in \(0.9 < 0.99 < 0.999 < \cdots\). The expression only contains finite decimals. So we both agree that \(0.9 < 0.99 < 0.999 < \cdots < 1\) is true.

It's still problematic when we think of it as a sequence, however (at least for me), since the word "end" in that case has only one meaning: the last item in the sequence.


What does "the last item in the sequence" mean?

I've explained this myself on multiple occasions. Basically, what it means is "the item occupying the position that has the highest index in the sequence" or "the item that comes after all other items in the sequence, which means, there is no item in the sequence that comes after it". Thus, the last item in the sequence \((0.9, 0.99, 0.999, \dotso, 1)\) is \(1\) since it comes after all other items in the sequence. The fact that it's located an infinite number of positions away from all other items does not change that fact.

It might be argued that this isn't a proper sequence.


In which case it is necessary to explain what the word "sequence" means to you, what makes a sequence proper and what makes it improper, as well as the relevance of all of that.

Maybe there are two sequences: a parent sequence and a child sequence. Maybe the parent sequence consists of only two items: P = (\(\sum_{i=1}^{n} \frac{9}{10^i}\), 1), where the child sequence is \(\sum_{i=1}^{n} \frac{9}{10^i}\): C = (0.9, 0.09, 0.009, ...). In other words, the child sequence is treated as a single discrete item and placed in the first position of the parent sequence. This would work for me. This would permit that 1 still comes after all terms in \(\sum_{i=1}^{n} \frac{9}{10^i}\) without requiring that it be expressed as a "sketchy" sequence.


It's fine either way.

And how does one tell? Don't you think there ought to be something in the notation that indicates the existence of implicit terms?


The first three elements in \((a_1, a_2, a_3, \dotso, a_L)\) reveal the pattern. If I wanted to say the sequence is bi-infinite, I would have written something like \((a_1, a_2, a_3, \dotso, a_{L-2}, a_{L-1}, a_L)\) instead.

And chances are pretty good you don't have a special meaning of "number" in mind and just misunderstand how numbers work (even if we include hyperreals in the mix, you can't have a largest number).


There is nothing about naturals, integers, rationals, standard and non-standard reals that says "Look, there can't be a number greater every other number."
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Re: Is 1 = 0.999... ? Really?

Postby Carleas » Mon Feb 03, 2020 6:09 pm

Magnus Anderson wrote:This is based on the erroneous assumption that every number has a number greater than it. This holds true for naturals, integers, rationals and standard reals but it does not hold true for all numbers. Indeed, it holds true for all numbers EXCEPT for one: the largest number.

There is precedent for this, in that division works for all numbers except \(0\). We could similarly say that addition is undefined with respect to \(L\). I suspect that changing the definition of addition in that way would have other repercussions, but I'm not sure. I think all operations would have to be undefined with respect to \(L\).


If I might reply to one of Magnus' objections that I don't think was fully addressed.

If I understand your objection to the mainline proof, you think that this is false:
\(10 * .999... = 9 + .999...\)

This might be a good place to address this, because the question of what \(.999...\) means seems to be live right now.

How do you feel about a recursive definition: \(.999...\) is a number that satisfies two conditions: the first decimal place is a \(9\), and for the \(n\)th decimal place, if the \(n-1\)th decimal place is a \(9\), then the \(n\)th decimal place is also a \(9\).

Are those conditions sufficient to uniquely define \(.999...\)? I'd guess Magnus and Gib will disagree.
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Re: Is 1 = 0.999... ? Really?

Postby Ecmandu » Mon Feb 03, 2020 6:37 pm

I think “convergence theory” is a load of crap.

“Convergence theory” is the theory that infinite sums equal (not approximately equal) to a whole number.!!
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Re: Is 1 = 0.999... ? Really?

Postby gib » Mon Feb 03, 2020 11:50 pm

Magnus

Magnus Anderson wrote:Note that there is no \(0.\dot9\) in \(0.9 < 0.99 < 0.999 < \cdots\). The expression only contains finite decimals. So we both agree that \(0.9 < 0.99 < 0.999 < \cdots < 1\) is true.


Yes.

Magnus Anderson wrote:
It's still problematic when we think of it as a sequence, however (at least for me), since the word "end" in that case has only one meaning: the last item in the sequence.


What does "the last item in the sequence" mean?


Well, I think it's best to define "sequence" in order to answer that question. To me, a sequence is a series of objects with a defininte order. These objects can obviously be absract but they must be discrete (i.e. the objects have clearly defined identities, no ambiguity about where one object ends and another begins, no blurring of identities, etc.). For example, sections of a river would be difficult to describe as a sequence as there is no clear border between one section and another, and if you wanted to define each section as a collection of \(H_2O\) molecules, the molecules would blend in with and over take the molecules of the other sections, thereby making it unclear where one section ends and another begins (to the point where it can be unclear whether one section is always "before" another). On the other hand, carts in a train can much more easily be described as a sequence as they are definitely discrete and always maintain a specific order.

Each member of the sequence can be represented by a unique and unambiguous symbol, and the sequence itself can be represented by a linear arrangement of symbols (ex. \(a_1\), \(a_2\), \(a_3\), ...) demonstrating the order each one comes in. This linear arrangement necessitates that there are no gaps in the sequence. So \(a_1\), \(a_2\), , \(a_3\), ... is just bad notation, and more or less means the same as \(a_1\), \(a_2\), \(a_3\), ... (or if you want to say the gap really is a member of the sequence, it should be represented with a symbol--like 0 or 'null'--thereby making it at least an abstract object).

The last item in the sequence would just mean the symbol in the sequence taking the last position. So if you wrote out the sequence from left or right, it would be the rightmost item. In the sequence of the first 10 integers after 0:

1, 2, 3, 4, 5, 6, 7, 8, 9, 10

... 10 is the last item in the sequence.

I don't dismiss the possibility of infinite sequences, so I'm not saying sequences must have an end. If we were to extend the above sequence to include all integers after 0, we would express it as:

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, ...

...and it wouldn't have an end.

Infinite sequences can be infinite in both directions, as in the case of the sequence of all integers (not just the ones coming after 0), and in that case it would have neither a beginning nor an end.

Magnus Anderson wrote:I've explained this myself on multiple occasions. Basically, what it means is "the item occupying the position that has the highest index in the sequence" or "the item that comes after all other items in the sequence, which means, there is no item in the sequence that comes after it". Thus, the last item in the sequence \((0.9, 0.99, 0.999, \dotso, 1)\) is \(1\) since it comes after all other items in the sequence. The fact that it's located an infinite number of positions away from all other items does not change that fact.


Any time you talk about an item that comes after the end of an endless sequence, I have a problem with that. I hope you see the problem. If the sequence is endless, it doesn't have an end. But if the last item comes after the end of the endless sequence, we are essentially saying it comes after something that doesn't exist. Please tell me you see the problem here.

To talk about 1 coming after the end of the sequence (0.9, 0.99, 0.999, ...) isn't a problem because we're not quite saying it's the last item in the sequence, but that it comes after the end of the region on the number line covered by 0.9, 0.99, 0.999... which isn't infinite. It actually has an end. But when you say 1 is the last item in the sequence, you're no longer talking about a point on the number line but a position in the sequence. You're saying the sequence is infinite (and therefore doesn't have an end), yet 1 shows up at the end. <-- There is a clear contradiction there which is why I have a problem with it (and why I brought up the two layered sequence model of parent/child).

Magnus Anderson wrote:In which case it is necessary to explain what the word "sequence" means to you, what makes a sequence proper and what makes it improper, as well as the relevance of all of that.


Picking up from my definition above, what makes this kind of sequence improper (maybe) is just the problem noted above. It suggests something contradictory. It says the last item of the sequence comes after the end of an endless sub-sequence. The relevance is that this decides whether we can call (0.9, 0.99, 0.999, ... , 1) a sequence or not.

Magnus Anderson wrote:
Maybe there are two sequences: a parent sequence and a child sequence. Maybe the parent sequence consists of only two items: P = (\(\sum_{i=1}^{n} \frac{9}{10^i}\), 1), where the child sequence is \(\sum_{i=1}^{n} \frac{9}{10^i}\): C = (0.9, 0.09, 0.009, ...). In other words, the child sequence is treated as a single discrete item and placed in the first position of the parent sequence. This would work for me. This would permit that 1 still comes after all terms in \(\sum_{i=1}^{n} \frac{9}{10^i}\) without requiring that it be expressed as a "sketchy" sequence.


It's fine either way.


Good. Then we have at least one way to think about it that we can agree upon.

Magnus Anderson wrote:The first three elements in \((a_1, a_2, a_3, \dotso, a_L)\) reveal the pattern. If I wanted to say the sequence is bi-infinite, I would have written something like \((a_1, a_2, a_3, \dotso, a_{L-2}, a_{L-1}, a_L)\) instead.


That's what I thought. But now here's a challenge: what if we wanted to add 2 and 3 to the sequence above:

(0.9, 0.99, 0.999, ... , 1, 2, 3)

Then it looks as though we are saying there is a number that comes just before 1.

If you were to ask me, I'd say we need a new notation. Maybe a bar:

(0.9, 0.99, 0.999, ...| , 1, 2, 3)

The bar after the ... indicates that there is no continuim between the endless sub-sequence represented by ... and the sub-sequence immediately following it (no item immediately preceding 1).

Magnus Anderson wrote:There is nothing about naturals, integers, rationals, standard and non-standard reals that says "Look, there can't be a number greater every other number."


This is the first time I've seen you use the term "non-standard reals". Are these the kinds of strange numbers you have in mind? Can there be a largest non-standard real?

I'm not sure if you meant to lump them together with all other number types in this statement, but it sounds like you all of a sudden are saying there can be a largest number greater than all naturals, integers, rationals, and standard reals. So let me get this straight: you're saying there is no largest natural, no largest integer, no largest rational, and no largest standard real, but there is a largest non-standard real which is greater than all natural, integers, rationals, and standard reals. Is that correct? Are you willing to explain to me what a non-standard real is?

Carleas

Carleas wrote:If I understand your objection to the mainline proof, you think that this is false:
\(10 * .999... = 9 + .999...\)

This might be a good place to address this, because the question of what \(.999...\) means seems to be live right now.

How do you feel about a recursive definition: \(.999...\) is a number that satisfies two conditions: the first decimal place is a \(9\), and for the \(n\)th decimal place, if the \(n-1\)th decimal place is a \(9\), then the \(n\)th decimal place is also a \(9\).

Are those conditions sufficient to uniquely define \(.999...\)? I'd guess Magnus and Gib will disagree.


I'm not sure what Magnus's objection to this step was, but I seem to recall something about a 0 appearing at the end of the string of 9s. Magnus is of the camp that believes you can have a last position of an endless sequence, which I would guess implies he thinks there's a last 9. If you multiply \(0.\dot9\) by 10, therefore, there ends up being a 0 at the end of the sequence. This means the number you get is not \(9 + 0.\dot9\), and therefore the step that follows (where we replace \(0.\dot9\) with X) is unjustified.

I'm putting words in his mouth, obviously, but if this is correct, he *might* not agree with your definition since it insists that for every 9, there is a 9 that follows, and he would say this is true except for the infinitieth 9.
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Re: Is 1 = 0.999... ? Really?

Postby Mowk » Tue Feb 04, 2020 1:05 am

Is .111... = 0 ?
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Re: Is 1 = 0.999... ? Really?

Postby iambiguous » Tue Feb 04, 2020 1:28 am

Mowk wrote:Is .111... = 0 ?



More to the point [mine], is 1 apple = 0.999...apple?
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Re: Is 1 = 0.999... ? Really?

Postby Mowk » Tue Feb 04, 2020 1:40 am

More to the point [mine]


Yes, for all intents and purposes. But why do you consider it more to the point? Oh. That's right... cause it is [yours].
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Re: Is 1 = 0.999... ? Really?

Postby promethean75 » Tue Feb 04, 2020 1:52 am

ecmandu, silhouette, andy and gib; you're fired.

Biggs and mowk will take it from here.
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Re: Is 1 = 0.999... ? Really?

Postby iambiguous » Tue Feb 04, 2020 2:33 am

Mowk wrote:
More to the point [mine]


Yes, for all intents and purposes. But why do you consider it more to the point? Oh. That's right... cause it is [yours].


My point on this thread has revolved around those like me who admittedly don't have the capacity to really grasp the math here.

Yet it is things like math and science and actual empirical/material facts which, to me, make-up the bulk of the either/or world. So, it would seem that eventually in a discussion of this sort one or the other assessment and conclusion would finally prevail.

And yet the exchanges themselves revolve almost entirely around words discussing numbers. So, it seems reasonable to me to wonder how the points being made would translate in regard to an actual physical thing like an apple.

And, in emphasizing that this point is mine, I am only trying to reinforce the fact that I am not trying to suggest that it should be yours or anyone else's.

In part because I don't have a solid background in math and in part because it seems strange that, since mathematics is not something relevant to the manner in which I construe the meaning of dasein, there are still such fierce disagreements about this issue.

It makes me more uncertain in regard to the either/or world as well. In other words, how much of that might be dependent on a subjective frame of mind? I mean look at interactions in the quantum world.

Hope that helped.
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Re: Is 1 = 0.999... ? Really?

Postby Ecmandu » Tue Feb 04, 2020 2:45 am

promethean75 wrote:ecmandu, silhouette, andy and gib; you're fired.

Biggs and mowk will take it from here.


I’d be happy to see that.

You know, the point I’ve raised about “convergent series” really hasn’t been addressed here.

0.999...
Divided by 1/3 equals

0.333... * 3

Which divided by 1/3 again equals

0.111 *9

Eventually this sequence will equal:

0.000...

Which means that 0.999... equals zero at convergence


Now take a number like 1:

It’s 1/2 + 1/2 if divided by 2

1/2 + 1/2 divided by 2

Is 1/4+1/4+1/4+1/4....

This means that at convergence:

0.000... = 1 !!!

How’s the math of convergence theory look to you?

To me it looks like bullshit!

In order for 0.999... to equal 1, EVERY number must equal zero!!!

Make sense to you?

Me neither!
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Re: Is 1 = 0.999... ? Really?

Postby phyllo » Tue Feb 04, 2020 3:29 am

Where does the boundary of an apple begin or end?

It's as real or imagined as 1 or 0.999...
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Re: Is 1 = 0.999... ? Really?

Postby gib » Tue Feb 04, 2020 3:38 am

promethean75 wrote:ecmandu, silhouette, andy and gib; you're fired.

Biggs and mowk will take it from here.


I expect a hefty severance and at least two good reference.
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Re: Is 1 = 0.999... ? Really?

Postby iambiguous » Tue Feb 04, 2020 3:40 am

phyllo wrote:Where does the boundary of an apple begin or end?

It's as real or imagined as 1 or 0.999...


True. On the subatomic level in particular.
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Re: Is 1 = 0.999... ? Really?

Postby gib » Tue Feb 04, 2020 3:41 am

phyllo wrote:Where does the boundary of an apple begin or end?

It's as real or imagined as 1 or 0.999...


How does one imagine the missing piece of 0.999... of an apple. Anything you imagine, it must be even less than that. Hyperreal enthusiasts would tell you there is an infinitesimal piece of apple missing--if that's not enough to say the whole apple, for all practical purposes, is there, then going passed that infinitesimal (because infinitesimal can be divided) should put the debate to rest.
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Re: Is 1 = 0.999... ? Really?

Postby Carleas » Tue Feb 04, 2020 4:54 am

gib wrote:I'm not sure what Magnus's objection to this step was, but I seem to recall something about a 0 appearing at the end of the string of 9s. Magnus is of the camp that believes you can have a last position of an endless sequence, which I would guess implies he thinks there's a last 9. If you multiply \(0.\dot9\) by 10, therefore, there ends up being a 0 at the end of the sequence. This means the number you get is not \(9 + 0.\dot9\), and therefore the step that follows (where we replace \(0.\dot9\) with X) is unjustified.


I don't know that Magnus has taken that position, but I think that is something like the usual idea when people object to that line in the proof. But that doesn't really make sense. Why would we add a zero, as opposed to another 9?

I also think a more rigorous construction of \(0.\dot9\) would resolve the issue. Suppose we create \(0.\dot9\) by taking \(3 * 0.\dot3\), and we construct \(0.\dot3\) by dividing 3 into 1. Assuming long division is a reliable algorithm, we get an infinite string of threes (similar to my recursive definition of \(0.\dot9\)).

If we use a construction like that, we have an alternative way of asking whether \(10 * 0.\dot9 = 9 + 0.\dot9\), because we can multiply 10 elsewhere in the construction (by the associative property) and get to the same repeating long division. I don't think that construction would actually work, but maybe we could find one (if not, we're just not talking about the same thing, so there's no actual disagreement about what equals or doesn't equal 1).

I see this question of whether one can create a coherent mathematics that has a largest number as a red herring. Even if there is a largest number, \(0.\dot9\) still equals 1.
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Re: Is 1 = 0.999... ? Really?

Postby gib » Tue Feb 04, 2020 5:56 am

Carleas wrote:I don't know that Magnus has taken that position, but I think that is something like the usual idea whenpeople object to that line in the proof. But that doesn't really make sense. Why would we add a zero, as opposed to another 9?


Because that's what happens when you multiply something by 10: 10 x 123 = 1230.

Obviously, this isn't the rule, it's just a convenient short-cut for deriving the result. The rule isn't: shift all the digits to the left and fill the last place with a 0. The rule is: add the multiplicand to itself a number of times equal to the multiplier. So with 10 x 123, you would add 123 to itself 10 times. You start with 123, then get 246, then 369, etc. until we get 1230. Nowhere in that process are we shifting digits to the left. It's just convenient that, when multiplying by 10, we so happen to get a result that can be derived by a much simpler short-cut: shift the digits to the left and tack on a 0.

This short-cut applies to all numbers with finite decimal expansions, but because it is not the actual rule of multiplication, we cannot necessarily say it carries over to numbers with infinite decimal expansion. We'd have to derive a thorough explanation for why we get this short-cut when multiplying numbers with finite decimal expansions by 10, and then see if that explanation carries over to numbers with infinite decimal expansions.

Carleas wrote:I also think a more rigorous construction of \(0.\dot9\) would resolve the issue. Suppose we create \(0.\dot9\) by taking \(3 * 0.\dot3\), and we construct \(0.\dot3\) by dividing 3 into 1. Assuming long division is a reliable algorithm, we get an infinite string of threes (similar to my recursive definition of \(0.\dot9\)).

If we use a construction like that, we have an alternative way of asking whether \(10 * 0.\dot9 = 9 + 0.\dot9\), because we can multiply 10 elsewhere in the construction (by the associative property) and get to the same repeating long division. I don't think that construction would actually work, but maybe we could find one (if not, we're just not talking about the same thing, so there's no actual disagreement about what equals or doesn't equal 1).


So instead of asking whether \(10 * 0.\dot9 = 9.\dot9\), we ask whether \(10 * (3 * 0.\dot3) = 10 * (3 * \frac{1}{3}) = 9.\dot9\)?

I doubt that would convince the skeptics. Magnus has already ruled out the fact that \(\frac{1}{3} = 0.\dot3\).

Carleas wrote:I see this question of whether one can create a coherent mathematics that has a largest number as a red herring. Even if there is a largest number, \(0.\dot9\) still equals 1.


It's a tangent. This thread is full of tangents. Personally, I'm okay with that as I'm not actually trying to get to the bottom of the main question--does \(0.\dot9\) really equal 1?--I just enjoy a good debate regardless of where it leads.

(But just to address your point, if there is a largest number, it would fundamentally change the way we understand numbers (it would for me at least), and this *could* affect the way we understand the question of does \(0.\dot9\) = 1.)
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Re: Is 1 = 0.999... ? Really?

Postby Mowk » Tue Feb 04, 2020 3:50 pm

Does the "=" sign represents the requirement of a balanced equation. Seems it is tipping ever so .0111.... to one side. Or does .0111...= 0? Seems like a smallest number question as well.
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Re: Is 1 = 0.999... ? Really?

Postby Carleas » Tue Feb 04, 2020 5:30 pm

gib wrote:This short-cut applies to all numbers with finite decimal expansions, but because it is not the actual rule of multiplication, we cannot necessarily say it carries over to numbers with infinite decimal expansion. We'd have to derive a thorough explanation for why we get this short-cut when multiplying numbers with finite decimal expansions by 10, and then see if that explanation carries over to numbers with infinite decimal expansions.

The short-cut isn't necessarily wrong, multiplication by 10 in base 10 shifts the decimal point to the right. But \(123 = 123.0\), so we aren't adding a zero, the zero was already there. That's not the case for \(0.\dot9\) (if there are \(L\) decimal places, there can't be any number in the \(L+1\)th place, because \(L+1\) is undefined).

I also suspect that we can prove the short-cut as a general theorem, since it's the case that for any number in base x, multiplying by x shifts the decimal point one place to the right. (Tangent: is there a base-agnostic word for the 'decimal' point?)

gib wrote:So instead of asking whether \(10 * 0.\dot9 = 9.\dot9\), we ask whether \(10 * (3 * 0.\dot3) = 10 * (3 * \frac{1}{3}) = 9.\dot9\)?

I doubt that would convince the skeptics. Magnus has already ruled out the fact that \(\frac{1}{3} = 0.\dot3\).

I was thinking \(10 * (3 * 0.\dot3) = (3 * \frac{10}{3}) = 9.\dot9\), because \(\frac{10}{3}\) results in the same repeating decimals in the same way.

I agree this isn't the construction we need to convince skeptics, but I'm not sure what construction the skeptics are using to define \(0.\dot9\). Again, without that construction, I'm not sure that we all mean the same thing when we say \(0.\dot9\).

gib wrote:if there is a largest number, it would fundamentally change the way we understand numbers

Absolutely. Numbers aren't closed on addition? A number where addition is defined for only half the number line (all negative numbers, but no positive numbers)? An end to all decimal expansions, such that there is uncertainty about what happens to any infinite expansion when multiplied by 10? Does that make a whole class of rational numbers that aren't closed on multiplication? Is multiplication by \(L\) defined?

My impression is that the existence of \(L\) is ad hoc, unnecessary, and has a lot of unintended consequences. I don't believe it can be proven from other standard axioms, so we'd be adding it as an additional axiom, and it isn't clear why.

gib wrote:It's a tangent. This thread is full of tangents. Personally, I'm okay with that as I'm not actually trying to get to the bottom of the main question--does \(0.9\) really equal 1?--I just enjoy a good debate regardless of where it leads.

Ah, so the red herring is actually sport fishing! Fair enough.

Ecmandu wrote:Eventually this sequence will equal:

0.000...

Which means that 0.999... equals zero at convergence

The limit of the sequence is \(0.0000... * \infty\). My calculus is rusty, but I think that means the sequence doesn't converge.
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Re: Is 1 = 0.999... ? Really?

Postby iambiguous » Tue Feb 04, 2020 5:46 pm

gib wrote:
phyllo wrote:Where does the boundary of an apple begin or end?

It's as real or imagined as 1 or 0.999...


How does one imagine the missing piece of 0.999... of an apple. Anything you imagine, it must be even less than that. Hyperreal enthusiasts would tell you there is an infinitesimal piece of apple missing--if that's not enough to say the whole apple, for all practical purposes, is there, then going passed that infinitesimal (because infinitesimal can be divided) should put the debate to rest.


So, even here someone or something the equivalent of God is necessary. In other words, an omniscient point of view who/that knows everything that can possibly be known about both apples and math.

In the interim, mere mortals such as ourselves carry on as best we can. I'm just curious as to why those here who obviously do have an enormous amount of understanding with respect to math still can't pin this down conclusively.

This tells me something about reality [human or otherwise] that doesn't quite seem to sink in with others. For better or worse.
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Re: Is 1 = 0.999... ? Really?

Postby phyllo » Tue Feb 04, 2020 6:27 pm

So, even here someone or something the equivalent of God is necessary. In other words, an omniscient point of view who/that knows everything that can possibly be known about both apples and math.
No. You can find logical flaws without omniscience or knowing everything.

We can know that 6 times 7 does not equal 43 even when we don't know the true answer. That's because the product must be an even number. We know something about it.
In the interim, mere mortals such as ourselves carry on as best we can. I'm just curious as to why those here who obviously do have an enormous amount of understanding with respect to math still can't pin this down conclusively.
That's because a consistent point of view is not being maintained by some posters. Sometimes they say that infinity is a number and sometimes it's not. Sometimes they say mathematical division works and sometimes it doesn't. Sometimes they talk about numbers and sometimes they talk about sets.

It's like trying to argue against a square-circle. They see a circle when it suits them and a square when it suits them. They don't recognize that you can't have both at the same time.

Demonstrating that inconsistency is difficult.
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Re: Is 1 = 0.999... ? Really?

Postby iambiguous » Tue Feb 04, 2020 6:57 pm

phyllo wrote:
So, even here someone or something the equivalent of God is necessary. In other words, an omniscient point of view who/that knows everything that can possibly be known about both apples and math.
No. You can find logical flaws without omniscience or knowing everything.

We can know that 6 times 7 does not equal 43 even when we don't know the true answer. That's because the product must be an even number. We know something about it.


You know my pitch here. Whether in regard to living matter evolving into apples, the definitive understanding of all things mathematical or human minds capable of discussing either one, we are all embedded in that same profoundly mysterious and problematic gap between what we think we know about anything and all that there is to know about everything.

Besides, the discussion revolves not around whether 6 X 7 = 43 -- who here would get into a fierce debate about that -- but 1 either equaling or not equalling 0.999....


In the interim, mere mortals such as ourselves carry on as best we can. I'm just curious as to why those here who obviously do have an enormous amount of understanding with respect to math still can't pin this down conclusively.


phyllo wrote: That's because a consistent point of view is not being maintained by some posters. Sometimes they say that infinity is a number and sometimes it's not. Sometimes they say mathematical division works and sometimes it doesn't. Sometimes they talk about numbers and sometimes they talk about sets.


Okay, but as long as the exchanges revolve around words talking about numbers pertaining only to more words still, it's hard for folks like me to grasp the relevance of the debate as it might be applicable to, say, technology and engineering. Again, it's actual use value and exchange value in human relationships.

phyllo wrote: It's like trying to argue against a square-circle. They see a circle when it suits them and a square when it suits them. They don't recognize that you can't have both at the same time.

Demonstrating that inconsistency is difficult.


A square circle? Isn't the whole point of the expression "squaring the circle" to suggest something impossible?
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