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#37
Dec2906, 08:02 AM

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#38
Dec2906, 09:05 AM

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#39
Dec2906, 09:10 AM

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#40
Dec3006, 01:48 AM

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I havent read any of the posts, because we've had this topic far too many times. All i know is that, in the article I quote "These can only approximate 1/3, for example, so we don't have an exact expression for 1/3." and "Clearly 0.9* = 1 + 0¯, so 0¯ is a sort of negative infinitesimal. On the other hand, you can't solve the equation 0.9* + X = 1" God Damn these people, x=0!
Or "Moreover, we have no interpretation for the number 3.14159265." Looks like negative pi to me.... From that, I already know that I know more math than this guy does. This topic has been discussed far too many times, and it is pointless. They are equal, live with it. 


#41
Dec3106, 12:46 AM

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The fact that you can't compute the decimal expansion of a sum from the decimal expansions of its addends is a well known phenomenon that was noticed by Turing. In a fully constructive treatment of the real numbers, this is often stated by saying (informally) that not every positive real number has a decimal expansion. More precisely, there is no constructive proof that every positive real number has a decimal expansion (or at least we don't know of one). 


#42
Dec3106, 02:45 AM

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#43
Dec3106, 05:32 PM

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But saying that Turing proved that not all reals have decimal expansions... that's just crazy talk. 


#44
Dec3106, 05:57 PM

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A real number is a (computable) function f that takes an integer n and returns a fraction r, satisfying the property that: f(n)  f(m) < 2^{n} + 2^{m} (To connect with the "usual" model of the reals, f(n) is a Cauchy sequence) We could define a "decimal number" in a similar fashion  it takes an integer n and returns a decimal digit... and satisfies the property that there exists a bound M such that n > M implies f(n) = 0. (To connect with the "usual" model, f(n) would be the nth place in a decimal number) And I believe that, in fact, there does not exist a (computable) function that takes a real number as input and returns a decimal number as output. 


#45
Dec3106, 11:19 PM

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GibZ do not forget the cardinal rule of the teaching profession, there is a naive student born every minute and we get paid again for teaching the same old claptrap to each new generation.
i.e. just because it bores the teacher, does not mean it seems old to the class. actually this is a license to work forever without learning a new trick. 


#46
Jan107, 12:25 AM

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I don't seem to understand you mathwonk...whose the teacher and whose the student? I'm 14 years old, I'm sure the person who wrote the article is older than me. He should know by now he can't name an alternate system the same as another one, and be surprised that we're confused and insulting him about it.



#47
Jan107, 08:09 PM

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1. There's no computable function that converts arbitrary binary expansions to decimal expansions 2. Turing proved this 3. The article, when writing that Turing proved that no computable function returns the decimal expansion of a real, meant (1) and (2) Alright, perhaps. Let me think about this. Given a binary expansion* adding up to N, where the last term's (binary) exponent is n, the number is in [N, N + 2^n] if repeating 1s are allowed. For no computable function to exist to convert this to decimal form, for almost all natural numbers k there must be some real numbers x_k and y_k such that x_k and y_k differ in the kth decimal place, and such that there exists no computable f(k) such that the first f(k) places suffice to distinguish x_k and y_k. Right? I'll need more time to think about this. * In which every number is either 0 or 1, else the number could change arbitrarily with each additional bit. 


#48
Jan107, 10:26 PM

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I'm trying to be less committal than that! I'm pretty sure I remember a theorem that there are more constructive reals than constructive decimals. I don't know if Turing proved it, but seeing his name in this context is unsurprising.
I have vague recollections that if we compute their values in the "standard" reals, that {values of computabie decimals} is actually proper subset of {values of computable reals}... but I don't remember for sure. But I'm confident about what I did say in my previous post. One important correction to your post: 1. There's no computable function that converts arbitrary binary expansions to decimal expansions That's not what I said  a computable real isn't a "binary expansion": it's a Cauchy sequence whose m and nth terms differ by less than 2^m + 2^n. For example, 1/2, 5/4, 7/8, 17/16, 31/32, ... is a perfectly good computable real number. I stated "fraction", but I think you get something equivalent if you substitute "binary decimal"... but it's important to note that point is that the leading digits of the mth term and the (m+1)th term don't have to agree, such as in the sequence I listed above. If you want to take smaller steps into this stuff, I'm pretty sure that equality is not a computable relation. (For both the computable reals and the computable decimals) 


#49
Jan107, 10:32 PM

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Oh, there's a short proof that equality is not decidable for the computable real numbers. I'll demonstrate with a simpler problem: telling if a real number is equal to 0.
Suppose I had a computable function Z(x) that computes whether the real number x is equal to zero. I can use Z to solve the halting problem as follows: suppose I have some program P, and I want to know if it halts or not. I can construct a computable real number D as follows: D(n) = 0 if P doesn't halt within n steps D(n) = 2^k if P halts after k steps, with k <= n Clearly D is computable  to compute D(n), simply run P for n steps, and see if and when it stops. Clearly, Z(D) = true iff P doesn't halt. So, Z can be used to solve the halting problem. Because the halting problem is incomputable, Z must be an incomputable relation. 


#50
Jan207, 09:01 AM

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Hmm, well, you're quite right. Still, none of this was justified by the link... all of this comes from you, including cleaning up the definition.
Thanks. I concede the point. 


#51
Jan707, 09:42 PM

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While I fully agree with the correct subset of the arguments given above (you know who you are), it seems to me that there is a genuine ambiguity in the notation 0.999.... Caveat: Of course the real number system is defined axiomatically, so questions about the notation resolve ultimately not into questions about about the real number system per se, but about whether the notation is actually referring to the real number system or to some alternate system. This seems to be the difficulty with the original reference that started this thread.
The ambiguity is in the "..."  since the decimal number system was invented long before Cantorian ordinals, the "..." is the "naive infinity" which is usually assumed to by omega, the smallest transfinite ordinal. If you state that assumption explicitly, then there is nothing to discuss, since you have then defined the notation precisely enough that the you can apply Cauchy's theory of limits to it and prove that 0.999... = 1. End of story. However, if you allow the decimal to continue to a higher ordinal number of decimal places, you have a notation that seems not to correspond to the standard reals. Of course, this does not by itself create a number system for the notation to point to. (If that were true, then Anshelm's ontological proof of the existence of God would actually be valid and we should all be convinced by it. I think not.) There are of course various extensions of the reals, particularly the hyperreals and surreals, which seem (at first glance anyway) to be candidates for number fields that would be appropriately expressed by such a largerordinal notation. As far as I am aware, this is an open question (someone correct me if I am wrong). I do know that Conway and Berlekamp showed that one particular notation (the "Hackenbush game") for the surreals had a subset of expressions that had an obvious correspondence with the real numbers expressed in the binary system. I'm not sure that this contribution to the thread advances the discussion in any way; I'm just trying to add something that is perhaps of interest to a discussion on a topic that is extremely tedious to mathematicians who have to deal with cranks on occasion. Stuart Anderson 


#52
Jan807, 12:45 AM

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The hypernatural numbers are not (externally) wellordered, but I don't see why the index set should be expected to be an ordinal. 


#53
Jan807, 01:31 AM

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I wonder how this topic is really number theory...



#54
Jan807, 10:18 AM

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Thanks! Stuart Anderson 


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