Why can no one in sci.math understand my simple point?
in [Logic]
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From: Tim Little on 26 Jun 2010 23:33 On 2010-06-27, Jesse F. Hughes <jesse(a)phiwumbda.org> wrote: > The usual definitions of computable real require that the machine > accept input n, halts and outputs the first n digits, or something > like that. There are a number of definitions, all of which end up being provably equivalent. The definition CB has so far provided is a particularly bizarre one, but does still appear to be equivalent. - Tim
From: Virgil on 27 Jun 2010 00:08 In article <slrni2dgpo.jrj.tim(a)soprano.little-possums.net>, Tim Little <tim(a)little-possums.net> wrote: > On 2010-06-26, Charlie-Boo <shymathguy(a)gmail.com> wrote: > > It is trivial to calculate in binary (using only 0 and 1) and output > > any desired string surrounded by 2 whenever we want > > You use the singular here, by which your statement is correct. Any > single finite string can be so produced. What's more, any finite > sequence of finite strings can be produced. > > However, you need an infinite sequence of finite strings to define > most real numbers, and there are not enough finite algorithms to > produce them all. It is well known that for 'most' reals no such algorithm can exist, the non-computable transcendentals.
From: Newberry on 27 Jun 2010 01:57 On Jun 26, 8:19 pm, Tim Little <t...(a)little-possums.net> wrote: > On 2010-06-26, Charlie-Boo <shymath...(a)gmail.com> wrote: > > > It is trivial to calculate in binary (using only 0 and 1) and output > > any desired string surrounded by 2 whenever we want > > You use the singular here, by which your statement is correct. Any > single finite string can be so produced. What's more, any finite > sequence of finite strings can be produced. > > However, you need an infinite sequence of finite strings to define > most real numbers, and there are not enough finite algorithms to > produce them all. > > - Tim If I understand Charlie's definition correctly some of the machines will keep outputing digits forever. That is what I meant when when i said that pi and e are generated by an algorithm. There are enough algorithms to produce all the compressible infinite strings.
From: Newberry on 27 Jun 2010 02:00 On Jun 26, 8:12 pm, Tim Little <t...(a)little-possums.net> wrote: > On 2010-06-26, Newberry <newberr...(a)gmail.com> wrote: > > > If the Turing machine is hacked such that it outputs a digit on any > > state transition does it not represent a real? > > Suppose I have a Turing machine that repeatedly marks and erases a > symbol at the start position. What real are you saying that this > machine represents? > > - Tim Well, the Turing machine includes a state machine. On each state transition it outputs a digit. Let us say that the machine outputs 1 on every state transition. The it represents the number 0.111111111111111111111111....1111111...........
From: Newberry on 27 Jun 2010 02:14
On Jun 26, 8:13 pm, Virgil <Vir...(a)home.esc> wrote: > In article > <382a8f19-157c-4c0a-9ed2-b9c295ce0...(a)5g2000yqz.googlegroups.com>, > > > > > > WM <mueck...(a)rz.fh-augsburg.de> wrote: > > On 26 Jun., 23:14, "Mike Terry" > > <news.dead.person.sto...(a)darjeeling.plus.com> wrote: > > > "WM" <mueck...(a)rz.fh-augsburg.de> wrote in message > > > >news:511dd6ef-6fe7-4664-bda0-082971c2f8db(a)z10g2000yqb.googlegroups.com.... > > > > > On 26 Jun., 00:32, "Mike Terry" > > > > > > > Please let me know: Does the list consisting of A0, A1, A2, A3, ... > > > > > > contain its antidiagonal or not? > > > > > > No, of course not. > > > > > The set of all these antidiagonals A0, A1, A2, A3, ... constructed > > > > according to my construction process and including the absent one is a > > > > countable set. > > > > Yes, I've agreed that several times. > > > > > > > That is not a problem. It shows however, that there are countable sets > > > > > > that cannot be listed. > > > > > > How exactly does it show that there are countable sets that cannot be > > > > > listed? > > > > > See the one above: The results of my construction process. > > > > No. The above shows that you've constructed a countable sequence of numbers > > > (A0, A1, A2, A3,...) which CAN be listed, namely consider the list (A0, A1, > > > A2, A3,...). > > > > Read carefully! > > > In order to do so, I posed the question: Does the list consisting of > > A0, A1, A2, A3, ... contain its antidiagonal or not? > > No list contains its antidiagonal, nevertheless, any list together with > its antidiagonal can be listed, and in a large variety of ways. > > > > > You said no. Therefore your assertion "which CAN be listed" is plainly > > wrong. > > Nonsense. > > > > > Of course it is not possible to list a list and its antidiagonal. > > Whyever not? > > Given a list {b0, b1, b2, ...} and its antidiagonal, a, how is the llist > {a, b0,. b12, b2, ...} NOT a list of the original list plus its > antidiagonal? > > > > > > > > OK, but you've still not given any explanation why you think the > > > > > antidiagonals of your process cannot be listed! (I'm genuinely > > > baffled.) > > > > > You said so yourself, few lines above. > > > > Rubbish. I have never said anything other than that (A0, A1, A2, ...) is a > > > countable sequence of reals that CAN be listed > > > but not with all its antidiagonals. > > Since any such list has potentially uncountably many "antidiagonals", of > course not. > > But any list along with up to countably many of its antidiagonals can be > collectively listed. > > > > > > Go on - have one more go, then I'll give up... > > > You'd better give up than make contradictory claims. > > Good advice. Its a shame its author doesn't follow it himself.- Hide quoted text - > > - Show quoted text - If we create a list of ALL the antidiagonals, how come there EXISTS an antidiagonal not on he list. I wonder if you can perform this construction in ZFC. |