Let (X) be a class of decimal / base-10 numerals that have no fractional part and that stand for a positive number.
Any two decimal expressions that are instances of (X) represent one and the same number if and only if every explicit and implicit digit in one expression is the same as the explicit or implicit digit located at the same place in the other expression.
(999\dotso) is an instance of (X).
(999\dotso) has no explicit digit associated with (10^0) but it has an implicit one and that digit is (0).
The decimal expression of (\sum_{i=0}^{i->\infty} 9\times10^i) is an instance of (X).
The decimal expression of (\sum_{i=0}^{i->\infty} 9\times10^i) has an explicit digit associated with (10^0) and that digit is (9).
Therefore, (999\dotso) and (\sum_{i=0}^{i->\infty} 9\times10^i) represent two different numbers.
How else can we argue that something isn’t present other than to simply state that it isn’t - implying that if you still contend that it is there - show it.
I didn’t say that it is impossible. I said that when someone simply says that something is there and we do not see it there - there is nothing to say but - “It is not there - so show us why you think it is there”.
If you insist that it is there without being able to provide proving evidence - we jump down on the chart to a categorical disagreement assignment (“Responder Expires” on the next to last flowchart line) - meaning a new class of people who can choose whichever branch of the debate they prefer to believe. Future debates can take it up from there.
Alternatively the responder can offer a new argument, expiring only that one argument first offered but can offer a different one in a similar manner as before (looping back up toward the top of the chart).
How about presenting an argument that concludes with “Therefore, it is not there”?
So you think the burden of proof is entirely on responders? Responders have to prove the proposer wrong whereas the proposer merely has to propose ideas and accept/reject counter-arguments he is offered?
One way to argue against (4) is by arguing in favor of your belief that the digit associated with (10^0) in (999\dotso) is (9).
If there was an implicit 0 digit there, I would see it
I don’t see an implicit 0 digit indicated.
Therefore there is no implicit 0 digit present.
That is my evidentiary argument.
It is not an issue of finding logic flaws when dealing with a premise. Premises must be agreed to by both parties. If the responder cannot change the proposer’s mind or vise versa, the argument ends there. A new argument must be presented by the responder.
You haven’t proposed that is what you are talking about. If it is, then that should be your response. Otherwise your statement that a 0 digit is implied is simply not agreeable.
I’ve realized something. (4) is wrong though not entirely wrong. It is possible for the digit associated with (10^0) in (999\dotso) to be (9), it is merely not necessarily so. The key insight is that the position of the first digit in (999\dotso) is not specified. It can be literally any. (I think it was phyllo who hinted at something similar in the past.)
For example, if the index of the first digit is (0), which means that its weight is (10^0), then the digit associated with (10^0) is (9). The only problem is that it is not necessarily so. For example, the index of the first digit can also be (infA^2), in which case, if the number of (9)s is (infA), the digit associated with (10^0) would be (0) since the endless chain of (9)s would never reach that position.
(There’s another problem too. If the weight of the first digit is (10^0) then what is the weight of the next (9)? There must be “the next (9)” since we’re dealing with an endless series of (9)s in the direction of the least significant digit. It should be (10^{-1}) but that would mean that (999\dotso) has a fractional part which we have previously agreed that it does not have. Thus, in such a case, the expression would represent a conceptually impossible number. And even if it were not a conceptually impossible number, it still wouldn’t be (\dotso999) simply because the digit associated with (10^{-1}) in (\dots999) is (0).)
With all that said, I have to revise my argument. Like so:
Probably no longer taught. When I’m made emperor, first thing I’m going to do is send all the math education bureaucrats to a reeducation camp. Math education is a disaster. Not the students’ fault.
Of course pi is finite. It’s a real number between 3 and 4 on the number line.
0 1 2 3 4 5 ...
^
|
pi
You are right that the decimal representation of pi has infinitely many digits. But that doesn’t make pi infinite. In fact pi only contains a finite amount of information. For example, the famous Leibniz series for pi is
Note that I completely expressed pi using only 15 characters. In other words if I wanted to completely transmit all the digits of pi to a friend on Mars, I would not have to send infinitely many digits; I’d only have to send the 15 character formula and they could use that to calculate as many digits as they want.
Google says nothing of the kind. Pi is a finite number, it’s a real number between 3 and 4.
Well 1/3 is exactly expressed with only three characters. Again you are confusing a number with its representation.
If you prefer, and I explained this several times in my previous post and urge you to slow down and comprehend this point: I do not need to talk about infinitely many 3’s at all, and from now on I won’t do that anymore.
What I am saying is that I can get AS CLOSE AS I WANT to 1/3 by taking a large enough FINITE number of 3’s, such as 0.333333333 or 0.3333333333333333. I never have to talk about infinitely many 3’s and from now on I won’t do so.
It’s not one that I’m making. I am not talking about infinitely many 3’s at all. I already explained that several times. I am asking you to slow down and focus on what I am saying to you. I am NOT talking about infinitely many 3’s.
I am no longer talking about that because it’s more precise to talk about arbitrary closeness of finitely many 3’s.
1/3 is never 'hit." Rather, we can get as close as we want to 1/3 by taking a sufficiently large but still FINITE number of 3’s.
I am asking you to slow down and engage with what I’m saying.
Well, 0.1(_3) is as finite a representation as can be. Why don’t you take that to heart?
Not so. Any instance of X contains only finitely many digits. That’s because the natural numbers are generated by starting with 0 and repeatedly adding 1. Every number so generated has only finitely many digits.
Once you stipulate that your expressions “stand for a positive number,” you are limited to finite-length strings of decimal digits. Those are the rules for positive numbers.
Of course you can make up your own notation if you can make sense of it (you haven’t yet done so) but you can’t call them positive numbers as if they’re natural numbers. They’re not.
I think the left most digit is necessarily indexed at (infA) because of your “…”. But that is another issue.
Your index (i) cannot exceed (infA) because it has to remain countable - the natural numbers. The “…” indicates a countable natural number series only.
Again the “…” is the natural numbers - not negative. So your index could not become negative even if it began at (infA) counting down.
So I have to still disagree with (4).
I still cannot identify an implicit 0 digit possible anywhere (other than your index (i)).
We haven’t said anything about counting to it.
Every index less than (infA) is “countable”.
So “(infA-1)” is technically countable - just not identifiable.
There are implicit (0)s everywhere. For example, there’s an implicit digit right before the first digit in (99\dot9). That digit is (0). There is also an implicit digit at (-1) (which is the index of the first digit after the decimal point.) That digit is also (0).
I assume what you’re saying here is that you do not think that the digit associated with (10^0) is (0) but rather (9).
Can I assume that the following argument of yours is applicable to Version 4 of my argument?
We cannot identify (infA-1) because we cannot know what number or position it would be other than not quite being at infinity (the numerically non existent end of the endless). It is something less - but what we can’t tell. And it is technically “countable” merely because it is less than the total number of natural numbers.
