…at which point we have hit an infinitely repeating pattern, and it becomes clear that this will never end. No matter how long we divide, we will find the repeating pattern of 0.33. That’s how we reach the conclusion that the fraction 1/3 has a decimal representation: we divide it. In the same way 1/4 = 0.25: we divided 1 by 4.
It’s exactly the same with 1/7. It’s a more obnoxious and long process to show how 1/7 is a repeating pattern, so I’ll save myself the trouble, but feel free to do it yourself if you feel like.
If it’s any help, we do have a widely accepted axiom of closure over multiplication … I’m sure this is new to netzweltler so I’ll repeat it here … “Multiplying any two real number results in yet another real number” … this gives us a corollary over division if we include a small little exception … “Dividing any two real number results in yet another real number, except when dividing by zero” … and of course we’ve not been dividing by zero anywhere in this discussion.
Therefore … 1 ÷ 3, 1 ÷ 7 and so on … all must result in a real number … and all are on the real number line … so when we see Mr. netzweltler claim these are not numbers, then we must assume Mr. netzweltler rejects these closure laws … and has spiraled off into some la-la land where he creates a unique set of rules governing these not-quite-as-real-as-the-real-numbers-are and … apparently … closure need not be required … very strange indeed … however, we can clearly state axiomatically that “multiplying any two surreal numbers results in anything that makes netzweltler’s claims true” … and the corollary “dividing any two surreal numbers can result in exactly what netzweltler says it does” … thus non-repeating decimals are elements of the surreals, but repeating decimals are not elements of the surreals … cute …
Correct me if I’m wrong, and please feel free to provide a citation, but:
[0, 0.9)∪[0.9, 0.99)∪[0.99, 0.999)∪… = [0, 1)
I’m sorry, but if our functional notation clearly gives the answer (0.999…) = 1 … and our Cauchy sequence notation clearly gives the answer (0.999…) = 1 … but this “union” notation somehow gives the answer (0.999…) ≠ 1 … then I think we should suspect the “union” notation as being inadequate for the task we are expecting it to perform.
In short, does [0, 0.9]∪[0.9, 0.99]∪[0.99, 0.999]∪… include the infinitely short line segment we’ll eventually have? I’m perfectly aware that [0[sub]s[/sub], 0.9[sub]s[/sub]]∪[0.9[sub]s[/sub], 0.99[sub]s[/sub]]∪[0.99[sub]s[/sub], 0.999[sub]s[/sub]]∪… doesn’t (where subscript “s” means an element of the surreal numbers) … and I’m reserving comment on this until Mr. netzweltler provides the definitions of the four basic arithmetic operators and any associated axioms for his own personal set of surreal numbers.
I’ve been taking “…” to mean and “this is continued to the limit”.
In which case it seems to me that [0, 0.9]∪[0.9, 0.99]∪[0.99, 0.999]∪… = [0, 1], because that formulation is not essentially different to 0.9 + 0.99 + 0.999 + … = 0.999… = 1 (and the essential similarity is what netzweltler seems to be driving at, so I’m just running with that).
It’s possible that I am missing something, that’s happened before.
But everything we know about infinities show that they must be treated specially, and defining “infinitely close” as a form of “the same” gives the best math. It’s not that an infinite convergent series “is” its limit, it’s that it’s less useful to define a rather abstract “difference” than to define it as being equal.
0.333… is a notation. We can describe it as an infinite series or as the result of completing the infinite task of doing 1/3 by long division. It is useful to operate with rules that say 0.333… = 1/3, and, to my knowledge, completely useless to operate with rules that say 0.333… exists, but not as a number.
If our beloved correspondent had said that, I’d have said he’s full of beans… But maybe you can explain where the difference is? If you get an infinite number of numbers, then you have an infinite number of nines, and thus 0.999~
(If septimus agrees with you, then I’ll concede without demur.)
ETA: Oh! You said “If you include finite n.” I don’t! The proof by induction does not require a finite number. (Does it?) Like the proof of the infinity of primes, it uses induction to say something about infinity.
The set ℕ, the natural numbers, DOES contain an infinite number of numbers, but it does NOT contain ∞ .
In writing a series with … (or using Σ notation with ∞ as a bound), there is an implicit limit involved. This (tiny) abuse of notation has become key to the confusions in this thread, and is why I rewrote the key claim without any … or limit: 1 ∉ Union {{ r | 0 ≤ r ≤ 1 - 10-j } | j ∊ ℕ}
(ETA: This is not to say that Mr. Netz’s position is sensical. He “moves his pen” an infinite number of times (whatever that means), to 0.99 at t=.99, to 0.9999 at t=.9999 and continues until t=1 … but doesn’t get to 1.)
The equal sign “=” has a unique meaning in math. For two numbers to be equal their position on the number line must be identical. There is no additional meaning of “=” like “they are equal because it’s useful” or “they are equal because we axiomatically define it” or “they are equal because there is no way to deal with infinitesimals otherwise”.
While The Great Unwashed and watchwolf49 are wrong if they claim
[0, 0.9]∪[0.9, 0.99]∪[0.99, 0.999]∪… = [0, 1]
they are right that it should be [0, 1] to make 0.999… = 1 true.
One is an infinite list of numbers that each have finitely many 9s, and the other has all those, but also 0.9… .
(So one might construct the following list, for each n of the natural numbers, 0.9 to n places is on the nth place of the list. So you get the following list:
1 – 0.9
2 – 0.99
3 – 0.999
…
But 0.9… isn’t on that list.)
Compare the following argument:
"If 999[to n terms] is a number, then 999[to n1 terms] is a number. (Closure property of addition.)
So there is a number 999…"
But there isn’t a natural number 999…, so it looks like something must have gone wrong with your argument.
The proof about the infinity of primes says that there is infinitely many primes, and your proof shows that there are infinitely many numbers that start 0.999[to n terms]. But the proof about the infinity of primes doesn’t show that there is some infinite prime (for lack of a better term).
It is well-defined what “an infinite number of times” means in math. There is one move for each n ∈ ℕ.
n = 1, t = 0: I move my pen from point 0 to point 0.9 of the number line
n = 2, t = 0.9: I move my pen from point 0.9 to point 0.99 of the number line
n = 3, t = 0.99: I move my pen from point 0.99 to point 0.999 of the number line
…
Please please do as I asked and rewrite this as
[0, 0.3]∪[0.3, 0.33]∪[0.33, 0.333]∪… = [0, 1/3]
they are right that it should be [0, 1/3] to make 0.333… = 1/3 true.
Otherwise, contestants will still be confused whether you don’t understand that 1 = 0.999, or just object to all non-finite decimal representations.
You are correct. But this isn’t about defining =, it’s about defining the result of infinite processes.
If I divide 1/3 by long division I get however many 3s behind the decimal point I have the patience to make. A reasonable interpretation of that is that if I “complete” that infinite process I will have 1/3 = 0.333…
One can then quibble, as you have done, with various summations and unions, that it isn’t, but you are then still using summation rules that do not have inherent meaning for infinite sums. This is evidenced by the “result” not being numbers according to you. Now I have shown, and you have acknowledged, that these “non-numbers” behave just like they would if they were numbers, in fact they behave like the fractions we used to generate them. Therefore, when deciding what it means to sum something to infinity, mathematicians equate the sum to its limit, which gives us a definition of summing infinities that correspond to reality. You would rather define it as giving “NaN”, which makes absolutely no sense.
You keep asking for which step brings us to 1/3 for 0.333, or 1 for 0.999…, despite knowing full well that this makes no sense in term of infinities. But I am now going to turn that around, despite having said that it was a nonsense question before, and ask you: If stating what step brings us to completion is necessary for you to accept 1/3 = 0.333…, why is it not necessary to specify such a step when you state that the end of a process where every step is a number is something that is not?
If we put aside what we think what is “reasonable”, what we think what “makes sense”, what we think what is “useful”, if we are dealing with the hard facts, you don’t get around acknowledging that 0.999… can be represented by a step-by-step process which - as a result after infinitely many steps - shows that we don’t reach a defined position (as shown in my “copy and paste” examples). If you think these examples are wrong you need to show where they go wrong and what is undefined or ill-defined with these examples.
Where you go wrong is here “as a result after infinitely many steps - shows that we don’t reach a defined position”. You insist that we do reach something definite, such as [0, 1). That has a defined end point, so it’s just as correct to say if “we haven’t reached it, we haven’t completed the infinity” as “when we complete the infinity we are somewhere undefined”. In fact it is more correct. At no point do we move off the number line, and at every point we get closer to one. Saying “but then if we complete the infinity we’re not on the number line” is not logical.
I’m curious as to what principle you base this claim on, other than “I say so”? So far you’ve only stated that they are not equal because you axiomatically define it that way.
Infinity is not a number, we can only approach infinity with numbers n ∈ ℕ.
You’re putting aside what we have proven to be reasonable, what we have proven makes sense, what we have proven is useful (cf. post #700). You have no made comment on this proof and so we can assume that you’re “putting it aside” and ignoring it. Although it is very convenient for you to not acknowledge this, that doesn’t make it go away. If you believe there is an error in that proof, please state your objections. Right now it looks like you don’t understand it, and that’s an illogical objection.
Also, please clarify “as a result after infinitely many steps” … specifically what do you mean by “after” something that never stops … to have a condition of “after” requires that you do stop at some point, “after” a very very large finite number of steps …
It might not be “logical”, it might not be “reasonable”, it might not be what “makes sense”, it might not be “useful”, but it is what I’d call the hard facts.
We have completed infinity, because we took a step for each n ∈ ℕ. To disprove that you need to show which step/steps is/are missing to complete infinity. Nevertheless we didn’t reach 1, because none of the steps brought us there.
Your claim is that we can do all the math even with the non-numbers. Let’s take 0.333… (as a favour to septimus). We can represent 0.333… by the union [0, 0.3]∪[0.3, 0.33]∪[0.33, 0.333]∪…
Do you agree that for a real number to be greater than a set of smaller numbers we need to show that the real number contains a segment which cannot be found in one of the smaller numbers?
This is your notation, doesn’t it read “from the endpoint at, and including zero; to the endpoint at, but not including one”. The length of this line segment is _____ ?
If 1 > .999̇ then 1 - .999̇ = x, where*** x > 0***
But we also know that for any positive number y, x < y, because we can always add as many more 9s as we need to make x smaller until x is less than y
Therefore, since x is a positive number, x < x
(Proving that x < x implies that 1 = 2 is left as an exercise for the student)
∴ the original assumption that 1 > .999̇ is false