MagnusMagnus 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?
CarleasCarleas 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.