On 6/5/2022 10:16 AM, Alan Mackenzie wrote:
> Mike Terry <news.dead.person.stones@darjeeling.plus.com> wrote:
>> On 05/06/2022 13:14, Alan Mackenzie wrote:
>>> olcott <NoOne@nowhere.com> wrote:
>>>> On 6/5/2022 6:12 AM, Alan Mackenzie wrote:
>>>>> olcott <NoOne@nowhere.com> wrote:
>>>>>> On 6/5/2022 5:14 AM, Mikko wrote:
>>>>>>> On 2022-06-04 19:28:19 +0000, olcott said:
>
>>>>>>>> A Turing machine is said to halt whenever it reaches a
>>>>>>>> configuration for which δ is not defined; this is possible because
>>>>>>>> δ is a partial function. In fact, we will assume that no
>>>>>>>> transitions are defined for any final state so the Turing machine
>>>>>>>> will halt whenever it enters a final state.  (Linz:1990:234)
>
>>>>>>>> Linz, Peter 1990. An Introduction to Formal Languages and Automata.
>>>>>>>> Lexington/Toronto: D. C. Heath and Company.
>
>>>>>>>> When translated into ordinary software engineering terms this means
>>>>>>>> terminated normally. In a C function this means reaching the "ret"
>>>>>>>> instruction.
>
>>>>>>> The best equivalent to "not defined" is not "ret". Instead, "not
>>>>>>> defined" should include at least:
>>>>>>> - HLT or any other instruction that means 'halt'
>>>>>>> - any undefined op code
>>>>>>> - any return or pop instruction if the stack is empty
>>>>>>> - an instruction fetch from a location that is not specifiec by the
>>>>>>> program
>>>>>>> That way the analogy to Linz' definition is much better.
>
>>>>>>> Mikko
>
>>>>>> Reaching a final state is merely the Turing machine way of saying
>>>>>> terminated normally. "ret" is the C way of saying the same thing.
>
>>>>> Sophistry. What would be the turing machine equivalent of an
>>>>> "abnormal termination" in C?
>
>>>> An aborted simulation.
>
>>> There's no such thing on a turing machine. It either runs and halts,
>>> or it runs forever.
>
>>> Your aborted simulation is just one final state of a turing machine,
>>> which has thus halted.
>
>> A TM "aborting" a simulation is just the TM ceasing to calculate
>> computation steps for some computation, and going on to calculate
>> something else instead. It does not mean:
>> a) that the TM (doing the simulation) has halted
>> b) that the simulated computation halts
>> c) that the simulated computation never halts
>
> OK. I've a feeling we're talking more about nice shades of words than
> computer science here, but ....
>
> If the simulation is the entire turing machine, aborting it will bring
> the TM to a halt state. If that simulation is merely part of the TM,
> then the word "halt" has a different meaning when applied to that
> simulation part from when applied to the entire TM. I'm not even sure
> what you mean when you say a part of a TM has halted or not halted.
>
>> Regards,
>> Mike.
>

This is a complete example of what it meant.

void Infinite_Loop()
{ HERE: goto HERE;
}

int main()
{ Output("Input_Halts = ", H0(Infinite_Loop));
}

_Infinite_Loop()
[00001342](01) 55 push ebp
[00001343](02) 8bec mov ebp,esp
[00001345](02) ebfe jmp 00001345
[00001347](01) 5d pop ebp
[00001348](01) c3 ret
Size in bytes:(0007) [00001348]

_main()
[00001372](01) 55 push ebp
[00001373](02) 8bec mov ebp,esp
[00001375](05) 6842130000 push 00001342
[0000137a](05) e833fdffff call 000010b2
[0000137f](03) 83c404 add esp,+04
[00001382](01) 50 push eax
[00001383](05) 6823040000 push 00000423
[00001388](05) e8e5f0ffff call 00000472
[0000138d](03) 83c408 add esp,+08
[00001390](02) 33c0 xor eax,eax
[00001392](01) 5d pop ebp
[00001393](01) c3 ret
Size in bytes:(0034) [00001393]

machine stack stack machine assembly
address address data code language
======== ======== ======== ========= =============
[00001372][0010228f][00000000] 55 push ebp
[00001373][0010228f][00000000] 8bec mov ebp,esp
[00001375][0010228b][00001342] 6842130000 push 00001342 // push
_Infinite_Loop
[0000137a][00102287][0000137f] e833fdffff call 000010b2 // call H0

Begin Local Halt Decider Simulation Execution Trace Stored at:212343
[00001342][00212333][00212337] 55 push ebp
[00001343][00212333][00212337] 8bec mov ebp,esp
[00001345][00212333][00212337] ebfe jmp 00001345
[00001345][00212333][00212337] ebfe jmp 00001345
Local Halt Decider: Infinite Loop Detected Simulation Stopped

[0000137f][0010228f][00000000] 83c404 add esp,+04
[00001382][0010228b][00000000] 50 push eax
[00001383][00102287][00000423] 6823040000 push 00000423
[00001388][00102287][00000423] e8e5f0ffff call 00000472
Input_Halts = 0
[0000138d][0010228f][00000000] 83c408 add esp,+08
[00001390][0010228f][00000000] 33c0 xor eax,eax
[00001392][00102293][00100000] 5d pop ebp
[00001393][00102297][00000004] c3 ret
Number of Instructions Executed(680)

--
Copyright 2022 Pete Olcott

"Talent hits a target no one else can hit;
Genius hits a target no one else can see."
Arthur Schopenhauer

On Sun, 5 Jun 2022 11:07:38 -0500
olcott <NoOne@NoWhere.com> wrote:

> On 6/5/2022 9:46 AM, Mike Terry wrote:
> > On 05/06/2022 13:14, Alan Mackenzie wrote:
> >> olcott <NoOne@nowhere.com> wrote:
> >>> On 6/5/2022 6:12 AM, Alan Mackenzie wrote:
> >>>> olcott <NoOne@nowhere.com> wrote:
> >>>>> On 6/5/2022 5:14 AM, Mikko wrote:
> >>>>>> On 2022-06-04 19:28:19 +0000, olcott said:
> >>
> >>>>>>> A Turing machine is said to halt whenever it reaches a
> >>>>>>> configuration for which δ is not defined; this is possible
> >>>>>>> because δ is a partial function. In fact, we will assume that
> >>>>>>> no transitions are defined for any final state so the Turing
> >>>>>>> machine will halt whenever it enters a final state.
> >>>>>>> (Linz:1990:234)
> >>
> >>>>>>> Linz, Peter 1990. An Introduction to Formal Languages and
> >>>>>>> Automata. Lexington/Toronto: D. C. Heath and Company.
> >>
> >>>>>>> When translated into ordinary software engineering terms this
> >>>>>>> means terminated normally. In a C function this means
> >>>>>>> reaching the "ret" instruction.
> >>
> >>>>>> The best equivalent to "not defined" is not "ret". Instead,
> >>>>>> "not defined" should include at least:
> >>>>>> - HLT or any other instruction that means 'halt'
> >>>>>> - any undefined op code
> >>>>>> - any return or pop instruction if the stack is empty
> >>>>>> - an instruction fetch from a location that is not specifiec
> >>>>>> by the program
> >>>>>> That way the analogy to Linz' definition is much better.
> >>
> >>>>>> Mikko
> >>
> >>>>> Reaching a final state is merely the Turing machine way of
> >>>>> saying terminated normally. "ret" is the C way of saying the
> >>>>> same thing.
> >>
> >>>> Sophistry.  What would be the turing machine equivalent of an
> >>>> "abnormal termination" in C?
> >>
> >>> An aborted simulation.
> >>
> >> There's no such thing on a turing machine.  It either runs and
> >> halts, or it runs forever.
> >>
> >> Your aborted simulation is just one final state of a turing
> >> machine, which has thus halted.
> >
> > A TM "aborting" a simulation is just the TM ceasing to calculate
> > computation steps for some computation, and going on to calculate
> > something else instead.  It does not mean:
> > a)  that the TM (doing the simulation) has halted
> > b)  that the simulated computation halts
> > c)  that the simulated computation never halts
> >
> > Regards,
> > Mike.
> >
>
> That an aborted simulated has not reached a final state has not
> halted is proven by the fact that your screwy reasoning would have to
> conclude that an infinite loop halts.
>
> *This is a Stipulative definition*
> Computation that halts ... the Turing machine will halt whenever it
> enters a final state. (Linz:1990:234)
>
> A stipulative definition is a type of definition in which a new or
> currently existing term is given a new specific meaning for the
> purposes of argument or discussion in a given context.
> https://en.wikipedia.org/wiki/Stipulative_definition
>
> void Infinite_Loop()
> {
> HERE: goto HERE;
> }
>
> int main()
> {
> Output("Input_Halts = ", H0(Infinite_Loop));
> }
>
> _Infinite_Loop()
> [00001342](01) 55 push ebp
> [00001343](02) 8bec mov ebp,esp
> [00001345](02) ebfe jmp 00001345
> [00001347](01) 5d pop ebp
> [00001348](01) c3 ret
> Size in bytes:(0007) [00001348]

You only call Infinite_Loop() if you detect a recursion
into your decider something that the proofs you are attempting to refute
do not do. What you have has nothing to do with the Halting Problem.

/Flibble

On 6/5/2022 11:12 AM, Mr Flibble wrote:
> On Sun, 5 Jun 2022 11:07:38 -0500
> olcott <NoOne@NoWhere.com> wrote:
>
>> On 6/5/2022 9:46 AM, Mike Terry wrote:
>>> On 05/06/2022 13:14, Alan Mackenzie wrote:
>>>> olcott <NoOne@nowhere.com> wrote:
>>>>> On 6/5/2022 6:12 AM, Alan Mackenzie wrote:
>>>>>> olcott <NoOne@nowhere.com> wrote:
>>>>>>> On 6/5/2022 5:14 AM, Mikko wrote:
>>>>>>>> On 2022-06-04 19:28:19 +0000, olcott said:
>>>>
>>>>>>>>> A Turing machine is said to halt whenever it reaches a
>>>>>>>>> configuration for which δ is not defined; this is possible
>>>>>>>>> because δ is a partial function. In fact, we will assume that
>>>>>>>>> no transitions are defined for any final state so the Turing
>>>>>>>>> machine will halt whenever it enters a final state.
>>>>>>>>> (Linz:1990:234)
>>>>
>>>>>>>>> Linz, Peter 1990. An Introduction to Formal Languages and
>>>>>>>>> Automata. Lexington/Toronto: D. C. Heath and Company.
>>>>
>>>>>>>>> When translated into ordinary software engineering terms this
>>>>>>>>> means terminated normally. In a C function this means
>>>>>>>>> reaching the "ret" instruction.
>>>>
>>>>>>>> The best equivalent to "not defined" is not "ret". Instead,
>>>>>>>> "not defined" should include at least:
>>>>>>>> - HLT or any other instruction that means 'halt'
>>>>>>>> - any undefined op code
>>>>>>>> - any return or pop instruction if the stack is empty
>>>>>>>> - an instruction fetch from a location that is not specifiec
>>>>>>>> by the program
>>>>>>>> That way the analogy to Linz' definition is much better.
>>>>
>>>>>>>> Mikko
>>>>
>>>>>>> Reaching a final state is merely the Turing machine way of
>>>>>>> saying terminated normally. "ret" is the C way of saying the
>>>>>>> same thing.
>>>>
>>>>>> Sophistry.  What would be the turing machine equivalent of an
>>>>>> "abnormal termination" in C?
>>>>
>>>>> An aborted simulation.
>>>>
>>>> There's no such thing on a turing machine.  It either runs and
>>>> halts, or it runs forever.
>>>>
>>>> Your aborted simulation is just one final state of a turing
>>>> machine, which has thus halted.
>>>
>>> A TM "aborting" a simulation is just the TM ceasing to calculate
>>> computation steps for some computation, and going on to calculate
>>> something else instead.  It does not mean:
>>> a)  that the TM (doing the simulation) has halted
>>> b)  that the simulated computation halts
>>> c)  that the simulated computation never halts
>>>
>>> Regards,
>>> Mike.
>>>
>>
>> That an aborted simulated has not reached a final state has not
>> halted is proven by the fact that your screwy reasoning would have to
>> conclude that an infinite loop halts.
>>
>> *This is a Stipulative definition*
>> Computation that halts ... the Turing machine will halt whenever it
>> enters a final state. (Linz:1990:234)
>>
>> A stipulative definition is a type of definition in which a new or
>> currently existing term is given a new specific meaning for the
>> purposes of argument or discussion in a given context.
>> https://en.wikipedia.org/wiki/Stipulative_definition
>>
>> void Infinite_Loop()
>> {
>> HERE: goto HERE;
>> }
>>
>> int main()
>> {
>> Output("Input_Halts = ", H0(Infinite_Loop));
>> }
>>
>> _Infinite_Loop()
>> [00001342](01) 55 push ebp
>> [00001343](02) 8bec mov ebp,esp
>> [00001345](02) ebfe jmp 00001345
>> [00001347](01) 5d pop ebp
>> [00001348](01) c3 ret
>> Size in bytes:(0007) [00001348]
>
> You only call Infinite_Loop() if you detect a recursion
> into your decider something that the proofs you are attempting to refute
> do not do. What you have has nothing to do with the Halting Problem.
>
> /Flibble
>

That you are simply not bright enough to see that the Linz proof does
specify infinitely nested simulation is really no actual rebuttal at all.

Ĥ.q0 ⟨Ĥ⟩ ⊢* H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qy
If the correctly simulated input ⟨Ĥ⟩ ⟨Ĥ⟩ to H would reach its own final
state of ⟨Ĥ.qy⟩ or ⟨Ĥ.qn⟩.

Ĥ.q0 ⟨Ĥ⟩ ⊢* H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qn
If the correctly simulated input ⟨Ĥ⟩ ⟨Ĥ⟩ to H would never reach its own
final state of ⟨Ĥ.qy⟩ or ⟨Ĥ.qn⟩.
When Ĥ is applied to ⟨Ĥ⟩ // subscripts indicate unique finite strings
Ĥ copies its input ⟨Ĥ0⟩ to ⟨Ĥ1⟩ then H simulates ⟨Ĥ0⟩ ⟨Ĥ1⟩

Then these steps would keep repeating: (unless their simulation is aborted)
Ĥ0 copies its input ⟨Ĥ1⟩ to ⟨Ĥ2⟩ then H0 simulates ⟨Ĥ1⟩ ⟨Ĥ2⟩
Ĥ1 copies its input ⟨Ĥ2⟩ to ⟨Ĥ3⟩ then H1 simulates ⟨Ĥ2⟩ ⟨Ĥ3⟩
Ĥ2 copies its input ⟨Ĥ3⟩ to ⟨Ĥ4⟩ then H2 simulates ⟨Ĥ3⟩ ⟨Ĥ4⟩...

--
Copyright 2022 Pete Olcott

"Talent hits a target no one else can hit;
Genius hits a target no one else can see."
Arthur Schopenhauer

On 6/5/22 11:34 AM, Mr Flibble wrote:
> On Sun, 5 Jun 2022 16:28:05 +0100
> Andy Walker <anw@cuboid.co.uk> wrote:
>
>> On 05/06/2022 14:47, Mr Flibble wrote:
>>> On Sun, 5 Jun 2022 07:58:42 -0400
>>> Richard Damon <Richard@Damon-Family.org> wrote:
>>>> [...] Sort of like how the number Pi has an
>>>> exact value, but you can never actually express it (because it
>>>> takes an infinite number of digits).
>>> PI does not have an exact value; no irrational number has an exact
>>> value.
>>
>> Of course "pi" has an exact value; as do [eg] "sqrt(2)",
>> "e", and all the other computable real [and complex] numbers.
>> Whether that value can be expressed in finite terms in some
>> particular representation is quite another matter. That in turn
>> depends on the representation; standard decimals is merely one
>> [common] choice. Note that in symbolic computer systems, those
>> computable reals are typically written "pi" [or whatever], and the
>> computer works with that exactly, so that [eg] "sin^2 (pi/3) == 3/4",
>> not 0.7499...; and also that in decimal-type notations most rationals
>> equally have no terminating expansion. Numbers such as "pi" and
>> "sqrt(2)" are not defined as decimal expansions but via their
>> properties [eg that "sqrt(2)" is the unique positive real whose
>> square is 2, or equivalently that it is the ratio of the diagonal of
>> a square to its side, and "pi" is the least positive real whose sine
>> is zero]. Those properties are exact, and tell you all you ever need
>> to know about those numbers.
>>
>> [I have removed my name from the "Subject:"; I don't know why
>> anyone saw fit to attach it to this debate, such as it is, on the HP.]
>
> What has decimal (base 10) expansion got to do with anything? An
> irrational number has a non-terminating sequence in ANY base. I am
> sorry but you are simply mistaken: irrational numbers do NOT have an
> exact value; this is obvious to anyone who understands logic and uses a
> sane definition for infinity.
>
> /Flibble
>

How about in base pi? then it is the number 10

Base pi is an interesting base for some problems.

What is your definition of "an exact value"?

Maybe the problem is you don't quite understand the meaning of that term.

On 6/5/22 11:49 AM, Mr Flibble wrote:
> On Sun, 5 Jun 2022 15:44:32 -0000 (UTC)
> Alan Mackenzie <acm@muc.de> wrote:
>
>> Mr Flibble <flibble@reddwarf.jmc> wrote:
>>> On Sun, 5 Jun 2022 16:28:05 +0100
>>> Andy Walker <anw@cuboid.co.uk> wrote:
>>
>>>> On 05/06/2022 14:47, Mr Flibble wrote:
>>>>> On Sun, 5 Jun 2022 07:58:42 -0400
>>>>> Richard Damon <Richard@Damon-Family.org> wrote:
>>>>>> [...] Sort of like how the number Pi has an
>>>>>> exact value, but you can never actually express it (because it
>>>>>> takes an infinite number of digits).
>>>>> PI does not have an exact value; no irrational number has an
>>>>> exact value.
>>
>>>> Of course "pi" has an exact value; as do [eg] "sqrt(2)",
>>>> "e", and all the other computable real [and complex] numbers.
>>>> Whether that value can be expressed in finite terms in some
>>>> particular representation is quite another matter. That in turn
>>>> depends on the representation; standard decimals is merely one
>>>> [common] choice. Note that in symbolic computer systems, those
>>>> computable reals are typically written "pi" [or whatever], and the
>>>> computer works with that exactly, so that [eg] "sin^2 (pi/3) ==
>>>> 3/4", not 0.7499...; and also that in decimal-type notations most
>>>> rationals equally have no terminating expansion. Numbers such as
>>>> "pi" and "sqrt(2)" are not defined as decimal expansions but via
>>>> their properties [eg that "sqrt(2)" is the unique positive real
>>>> whose square is 2, or equivalently that it is the ratio of the
>>>> diagonal of a square to its side, and "pi" is the least positive
>>>> real whose sine is zero]. Those properties are exact, and tell
>>>> you all you ever need to know about those numbers.
>>
>> [ .... ]
>>
>>> What has decimal (base 10) expansion got to do with anything? An
>>> irrational number has a non-terminating sequence in ANY base. I am
>>> sorry but you are simply mistaken: irrational numbers do NOT have an
>>> exact value; this is obvious to anyone who understands logic and
>>> uses a sane definition for infinity.
>>
>> That irrational numbers are exact values is clear to anybody with a
>> degree in maths. Definitions of "infinity" (of which there are many)
>> have nothing to do with this.
>
> You are wrong and fractally so so your degree in maths appears to be
> worthless. An irrational number's sequence is statistically random,
> has no fixed point on the number line ergo has no exact representation.
> Any number with no exact representation has, by definition, no exact
> value, only an approximation. Infinity has everything to do with this
> as an irrational's sequence ("digits") never terminates (i.e. it is an
> INFINITELY long sequence).
>
> /Flibble
>

Nope. Irrational numbers DO have exact points on the number line.

And what does representation have to do with exact value?

Also, irrational numbers sequence of digits are not necessarily
statistically random, in some representations, they can be VERY
predictible for some numbers.

One simple construction to show exact position, draw a box with sides
exactly 1.

Draw a line though opposite corners and make one point the value 0.

The other corner will be EXACTLY at the point sqrt(2), so that
irrational number has an exact point on the number line.

You just don't understand what an exact value means, likely because you
can't understand things that are somewhat abstract.

On 6/5/2022 7:38 AM, Ben wrote:
> Alan Mackenzie <acm@muc.de> writes:
>
>> olcott <NoOne@nowhere.com> wrote:
>>> On 6/5/2022 6:12 AM, Alan Mackenzie wrote:
>
>>>> ... What would be the turing machine equivalent of an
>>>> "abnormal termination" in C?
>>
>>> An aborted simulation.
>>
>> There's no such thing on a turing machine. It either runs and halts, or
>> it runs forever.
>>
>> Your aborted simulation is just one final state of a turing machine,
>> which has thus halted.
>
> A year ago I tried to get PO to accept a few basic facts about the
> topic. One of these was
>
> (B) Every computation that halts, for whatever reason, is a halting
> computation.
>
> After much ducking a diving, PO replied "OK".
>

Within the context of this quote that I have repeated hundreds of times:

Computation that halts ... the Turing machine will halt whenever it
enters a final state. (Linz:1990:234)

I would count your reply as flat out dishonest.

--
Copyright 2022 Pete Olcott

"Talent hits a target no one else can hit;
Genius hits a target no one else can see."
Arthur Schopenhauer

On Sun, 5 Jun 2022 12:22:45 -0400
Richard Damon <Richard@Damon-Family.org> wrote:

> On 6/5/22 11:49 AM, Mr Flibble wrote:
> > On Sun, 5 Jun 2022 15:44:32 -0000 (UTC)
> > Alan Mackenzie <acm@muc.de> wrote:
> >
> >> Mr Flibble <flibble@reddwarf.jmc> wrote:
> >>> On Sun, 5 Jun 2022 16:28:05 +0100
> >>> Andy Walker <anw@cuboid.co.uk> wrote:
> >>
> >>>> On 05/06/2022 14:47, Mr Flibble wrote:
> >>>>> On Sun, 5 Jun 2022 07:58:42 -0400
> >>>>> Richard Damon <Richard@Damon-Family.org> wrote:
> >>>>>> [...] Sort of like how the number Pi has an
> >>>>>> exact value, but you can never actually express it (because it
> >>>>>> takes an infinite number of digits).
> >>>>> PI does not have an exact value; no irrational number has an
> >>>>> exact value.
> >>
> >>>> Of course "pi" has an exact value; as do [eg] "sqrt(2)",
> >>>> "e", and all the other computable real [and complex] numbers.
> >>>> Whether that value can be expressed in finite terms in some
> >>>> particular representation is quite another matter. That in turn
> >>>> depends on the representation; standard decimals is merely one
> >>>> [common] choice. Note that in symbolic computer systems, those
> >>>> computable reals are typically written "pi" [or whatever], and
> >>>> the computer works with that exactly, so that [eg] "sin^2 (pi/3)
> >>>> == 3/4", not 0.7499...; and also that in decimal-type notations
> >>>> most rationals equally have no terminating expansion. Numbers
> >>>> such as "pi" and "sqrt(2)" are not defined as decimal expansions
> >>>> but via their properties [eg that "sqrt(2)" is the unique
> >>>> positive real whose square is 2, or equivalently that it is the
> >>>> ratio of the diagonal of a square to its side, and "pi" is the
> >>>> least positive real whose sine is zero]. Those properties are
> >>>> exact, and tell you all you ever need to know about those
> >>>> numbers.
> >>
> >> [ .... ]
> >>
> >>> What has decimal (base 10) expansion got to do with anything? An
> >>> irrational number has a non-terminating sequence in ANY base. I
> >>> am sorry but you are simply mistaken: irrational numbers do NOT
> >>> have an exact value; this is obvious to anyone who understands
> >>> logic and uses a sane definition for infinity.
> >>
> >> That irrational numbers are exact values is clear to anybody with a
> >> degree in maths. Definitions of "infinity" (of which there are
> >> many) have nothing to do with this.
> >
> > You are wrong and fractally so so your degree in maths appears to be
> > worthless. An irrational number's sequence is statistically random,
> > has no fixed point on the number line ergo has no exact
> > representation. Any number with no exact representation has, by
> > definition, no exact value, only an approximation. Infinity has
> > everything to do with this as an irrational's sequence ("digits")
> > never terminates (i.e. it is an INFINITELY long sequence).
> >
> > /Flibble
> >
>
> Nope. Irrational numbers DO have exact points on the number line.
>
> And what does representation have to do with exact value?
>
> Also, irrational numbers sequence of digits are not necessarily
> statistically random, in some representations, they can be VERY
> predictible for some numbers.
>
> One simple construction to show exact position, draw a box with sides
> exactly 1.
>
> Draw a line though opposite corners and make one point the value 0.
>
> The other corner will be EXACTLY at the point sqrt(2), so that
> irrational number has an exact point on the number line.
>
> You just don't understand what an exact value means, likely because
> you can't understand things that are somewhat abstract.

An irrational number does not have an exact point on the number line as
it will move about as you "zoom in", you can keep "zooming in" forever
(i.e. infinitely) and it will keep moving about because the number
never terminates.

If I couldn't understand things that are somewhat abstract then I
wouldn't have a computer science degree (BSc Hons) and 30 years of
industry experience.

/Flibble

On 6/5/2022 11:22 AM, Richard Damon wrote:
> On 6/5/22 11:49 AM, Mr Flibble wrote:
>> On Sun, 5 Jun 2022 15:44:32 -0000 (UTC)
>> Alan Mackenzie <acm@muc.de> wrote:
>>
>>> Mr Flibble <flibble@reddwarf.jmc> wrote:
>>>> On Sun, 5 Jun 2022 16:28:05 +0100
>>>> Andy Walker <anw@cuboid.co.uk> wrote:
>>>
>>>>> On 05/06/2022 14:47, Mr Flibble wrote:
>>>>>> On Sun, 5 Jun 2022 07:58:42 -0400
>>>>>> Richard Damon <Richard@Damon-Family.org> wrote:
>>>>>>> [...] Sort of like how the number Pi has an
>>>>>>> exact value, but you can never actually express it (because it
>>>>>>> takes an infinite number of digits).
>>>>>> PI does not have an exact value; no irrational number has an
>>>>>> exact value.
>>>
>>>>>        Of course "pi" has an exact value;  as do [eg] "sqrt(2)",
>>>>> "e", and all the other computable real [and complex] numbers.
>>>>> Whether that value can be expressed in finite terms in some
>>>>> particular representation is quite another matter.  That in turn
>>>>> depends on the representation;  standard decimals is merely one
>>>>> [common] choice.  Note that in symbolic computer systems, those
>>>>> computable reals are typically written "pi" [or whatever], and the
>>>>> computer works with that exactly, so that [eg] "sin^2 (pi/3) ==
>>>>> 3/4", not 0.7499...; and also that in decimal-type notations most
>>>>> rationals equally have no terminating expansion.  Numbers such as
>>>>> "pi" and "sqrt(2)" are not defined as decimal expansions but via
>>>>> their properties [eg that "sqrt(2)" is the unique positive real
>>>>> whose square is 2, or equivalently that it is the ratio of the
>>>>> diagonal of a square to its side, and "pi" is the least positive
>>>>> real whose sine is zero].  Those properties are exact, and tell
>>>>> you all you ever need to know about those numbers.
>>>
>>> [ .... ]
>>>
>>>> What has decimal (base 10) expansion got to do with anything? An
>>>> irrational number has a non-terminating sequence in ANY base.  I am
>>>> sorry but you are simply mistaken: irrational numbers do NOT have an
>>>> exact value; this is obvious to anyone who understands logic and
>>>> uses a sane definition for infinity.
>>>
>>> That irrational numbers are exact values is clear to anybody with a
>>> degree in maths.  Definitions of "infinity" (of which there are many)
>>> have nothing to do with this.
>>
>> You are wrong and fractally so so your degree in maths appears to be
>> worthless.  An irrational number's sequence is statistically random,
>> has no fixed point on the number line ergo has no exact representation.
>> Any number with no exact representation has, by definition, no exact
>> value, only an approximation.  Infinity has everything to do with this
>> as an irrational's sequence ("digits") never terminates (i.e. it is an
>> INFINITELY long sequence).
>>
>> /Flibble
>>
>
> Nope. Irrational numbers DO have exact points on the number line.

In this we agree. Also every identifiable point on a number line has
another identifiable point that is immediately adjacent to it with no
other points inbetween.

[0,1] is exactly one geometric point longer than [0,1)

--
Copyright 2022 Pete Olcott

"Talent hits a target no one else can hit;
Genius hits a target no one else can see."
Arthur Schopenhauer

On 6/5/22 11:59 AM, olcott wrote:
> On 6/5/2022 10:17 AM, Ben wrote:
>> Ben <ben.usenet@bsb.me.uk> writes:
>>
>>> Alan Mackenzie <acm@muc.de> writes:
>>>
>>>> olcott <NoOne@nowhere.com> wrote:
>>>>> On 6/5/2022 6:12 AM, Alan Mackenzie wrote:
>>>
>>>>>> ...  What would be the turing machine equivalent of an
>>>>>> "abnormal termination" in C?
>>>>
>>>>> An aborted simulation.
>>>>
>>>> There's no such thing on a turing machine.  It either runs and
>>>> halts, or
>>>> it runs forever.
>>>>
>>>> Your aborted simulation is just one final state of a turing machine,
>>>> which has thus halted.
>>>
>>> A year ago I tried to get PO to accept a few basic facts about the
>>> topic.  One of these was
>>>
>>> (B) Every computation that halts, for whatever reason, is a halting
>>>      computation.
>>>
>>> After much ducking [and] diving, PO replied "OK".
>>
>> I should explain that the purpose of this question was because, at the
>> time, PO was claiming that the reason H_Hat(H_Hat) halts was "special":
>> the consequence of a simulation being stopped.  The fact that
>> H_Hat(H_Hat) halts for some special reason used to be feature of PO's
>> posts.  The phrasing "H(H_Hat, H_Hat) == 0 is correct because
>> H_Hat(H_Hat) only halts because..." was the mantra if the day.
>> Obviously some new, less clear, wording was called for.
>>
>
> Computation that halts ... the Turing machine will halt whenever it
> enters a final state. (Linz:1990:234)
>
> It is true that the correctly simulated input to H(P,P) never reaches
> its own final state thus never halts.
>

Except that it does (if H(P,P) returns 0), and you have even posted a
trace of it.

What can't reach that point is a trace of the simulation by H.

Either the design of the one H we are currently looking at never aborts,
and thus, yes, the input to H(P,P) does never halt, but H can't
correctly return that answer, because it also never halts, or

The design of the one H we are currently looking at does abort its
simulation, and thus H's simulation isn't actually "correct" because it
is incomplete, but if we give that same input to an actual correct
simulator, we see that the correct simulation of the input has P call
H(P,P), that H simulating its input, and after some time deciding to
abort its simulation (incorrectly) and returning 0 to P, and then P halting.

Thus, the MACHINE P(P) does reach its final state and thus is halting.

The fact that in INCOMPLETE trace doesn't (yet) reach a final state does
NOT prove non-halting.

You are just showing that you don't understand what the Turing Machine
that is being talked about actually is.

On 6/5/22 11:57 AM, olcott wrote:
> On 6/5/2022 7:38 AM, Ben wrote:
>> Alan Mackenzie <acm@muc.de> writes:
>>
>>> olcott <NoOne@nowhere.com> wrote:
>>>> On 6/5/2022 6:12 AM, Alan Mackenzie wrote:
>>
>>>>> ...  What would be the turing machine equivalent of an
>>>>> "abnormal termination" in C?
>>>
>>>> An aborted simulation.
>>>
>>> There's no such thing on a turing machine.  It either runs and halts, or
>>> it runs forever.
>>>
>>> Your aborted simulation is just one final state of a turing machine,
>>> which has thus halted.
>>
>> A year ago I tried to get PO to accept a few basic facts about the
>> topic.  One of these was
>>
>> (B) Every computation that halts, for whatever reason, is a halting
>>      computation.
>>
>> After much ducking a diving, PO replied "OK".
>>
>
> Whatever I said years ago has been superseded by my current understanding:
>
> Computation that halts ... the Turing machine will halt whenever it
> enters a final state. (Linz:1990:234)
>
>

And H^ applied to <H^> will Halt if H applied to <H^> <H^> rejects its
input as non-halting, thus showing that the H was wrong.

The fact that H aborted its simulation of that input doesn't mean a
thing, as the definition of Halting only refers to the ACTUAL MACHINE,
not some partial simulation of it.

FAIL.

Mr Flibble <flibble@reddwarf.jmc> wrote:
> On Sun, 5 Jun 2022 15:44:32 -0000 (UTC)
> Alan Mackenzie <acm@muc.de> wrote:

>> Mr Flibble <flibble@reddwarf.jmc> wrote:
>> > On Sun, 5 Jun 2022 16:28:05 +0100
>> > Andy Walker <anw@cuboid.co.uk> wrote:

>> >> On 05/06/2022 14:47, Mr Flibble wrote:
>> >> > On Sun, 5 Jun 2022 07:58:42 -0400
>> >> > Richard Damon <Richard@Damon-Family.org> wrote:
>> >> >> [...] Sort of like how the number Pi has an
>> >> >> exact value, but you can never actually express it (because it
>> >> >> takes an infinite number of digits).
>> >> > PI does not have an exact value; no irrational number has an
>> >> > exact value.

>> >> Of course "pi" has an exact value; as do [eg] "sqrt(2)",
>> >> "e", and all the other computable real [and complex] numbers.
>> >> Whether that value can be expressed in finite terms in some
>> >> particular representation is quite another matter. That in turn
>> >> depends on the representation; standard decimals is merely one
>> >> [common] choice. Note that in symbolic computer systems, those
>> >> computable reals are typically written "pi" [or whatever], and the
>> >> computer works with that exactly, so that [eg] "sin^2 (pi/3) ==
>> >> 3/4", not 0.7499...; and also that in decimal-type notations most
>> >> rationals equally have no terminating expansion. Numbers such as
>> >> "pi" and "sqrt(2)" are not defined as decimal expansions but via
>> >> their properties [eg that "sqrt(2)" is the unique positive real
>> >> whose square is 2, or equivalently that it is the ratio of the
>> >> diagonal of a square to its side, and "pi" is the least positive
>> >> real whose sine is zero]. Those properties are exact, and tell
>> >> you all you ever need to know about those numbers.

>> [ .... ]

>> > What has decimal (base 10) expansion got to do with anything? An
>> > irrational number has a non-terminating sequence in ANY base. I am
>> > sorry but you are simply mistaken: irrational numbers do NOT have
>> > an exact value; this is obvious to anyone who understands logic and
>> > uses a sane definition for infinity.

>> That irrational numbers are exact values is clear to anybody with a
>> degree in maths. Definitions of "infinity" (of which there are many)
>> have nothing to do with this.

> You are wrong and fractally so so your degree in maths appears to be
> worthless.

No, I am right, along with the world's other mathematics graduates. You
are stuck in the distant (100s of years) past, when mathematicians were
still puzzling over what you're puzzling over. The fundamentals of
maths have been worked out, and you are in the position of an alchemist
faced with modern chemistry.

> An irrational number's sequence is statistically random, has no fixed
> point on the number line ergo has no exact representation. Any number
> with no exact representation has, by definition, no exact value, only
> an approximation. Infinity has everything to do with this as an
> irrational's sequence ("digits") never terminates (i.e. it is an
> INFINITELY long sequence).

> /Flibble

--
Alan Mackenzie (Nuremberg, Germany).

On 6/5/2022 11:28 AM, Mr Flibble wrote:
> On Sun, 5 Jun 2022 12:22:45 -0400
> Richard Damon <Richard@Damon-Family.org> wrote:
>
>> On 6/5/22 11:49 AM, Mr Flibble wrote:
>>> On Sun, 5 Jun 2022 15:44:32 -0000 (UTC)
>>> Alan Mackenzie <acm@muc.de> wrote:
>>>
>>>> Mr Flibble <flibble@reddwarf.jmc> wrote:
>>>>> On Sun, 5 Jun 2022 16:28:05 +0100
>>>>> Andy Walker <anw@cuboid.co.uk> wrote:
>>>>
>>>>>> On 05/06/2022 14:47, Mr Flibble wrote:
>>>>>>> On Sun, 5 Jun 2022 07:58:42 -0400
>>>>>>> Richard Damon <Richard@Damon-Family.org> wrote:
>>>>>>>> [...] Sort of like how the number Pi has an
>>>>>>>> exact value, but you can never actually express it (because it
>>>>>>>> takes an infinite number of digits).
>>>>>>> PI does not have an exact value; no irrational number has an
>>>>>>> exact value.
>>>>
>>>>>> Of course "pi" has an exact value; as do [eg] "sqrt(2)",
>>>>>> "e", and all the other computable real [and complex] numbers.
>>>>>> Whether that value can be expressed in finite terms in some
>>>>>> particular representation is quite another matter. That in turn
>>>>>> depends on the representation; standard decimals is merely one
>>>>>> [common] choice. Note that in symbolic computer systems, those
>>>>>> computable reals are typically written "pi" [or whatever], and
>>>>>> the computer works with that exactly, so that [eg] "sin^2 (pi/3)
>>>>>> == 3/4", not 0.7499...; and also that in decimal-type notations
>>>>>> most rationals equally have no terminating expansion. Numbers
>>>>>> such as "pi" and "sqrt(2)" are not defined as decimal expansions
>>>>>> but via their properties [eg that "sqrt(2)" is the unique
>>>>>> positive real whose square is 2, or equivalently that it is the
>>>>>> ratio of the diagonal of a square to its side, and "pi" is the
>>>>>> least positive real whose sine is zero]. Those properties are
>>>>>> exact, and tell you all you ever need to know about those
>>>>>> numbers.
>>>>
>>>> [ .... ]
>>>>
>>>>> What has decimal (base 10) expansion got to do with anything? An
>>>>> irrational number has a non-terminating sequence in ANY base. I
>>>>> am sorry but you are simply mistaken: irrational numbers do NOT
>>>>> have an exact value; this is obvious to anyone who understands
>>>>> logic and uses a sane definition for infinity.
>>>>
>>>> That irrational numbers are exact values is clear to anybody with a
>>>> degree in maths. Definitions of "infinity" (of which there are
>>>> many) have nothing to do with this.
>>>
>>> You are wrong and fractally so so your degree in maths appears to be
>>> worthless. An irrational number's sequence is statistically random,
>>> has no fixed point on the number line ergo has no exact
>>> representation. Any number with no exact representation has, by
>>> definition, no exact value, only an approximation. Infinity has
>>> everything to do with this as an irrational's sequence ("digits")
>>> never terminates (i.e. it is an INFINITELY long sequence).
>>>
>>> /Flibble
>>>
>>
>> Nope. Irrational numbers DO have exact points on the number line.
>>
>> And what does representation have to do with exact value?
>>
>> Also, irrational numbers sequence of digits are not necessarily
>> statistically random, in some representations, they can be VERY
>> predictible for some numbers.
>>
>> One simple construction to show exact position, draw a box with sides
>> exactly 1.
>>
>> Draw a line though opposite corners and make one point the value 0.
>>
>> The other corner will be EXACTLY at the point sqrt(2), so that
>> irrational number has an exact point on the number line.
>>
>> You just don't understand what an exact value means, likely because
>> you can't understand things that are somewhat abstract.
>
> An irrational number does not have an exact point on the number line as
> it will move about as you "zoom in", you can keep "zooming in" forever
> (i.e. infinitely) and it will keep moving about because the number
> never terminates.
>
> If I couldn't understand things that are somewhat abstract then I
> wouldn't have a computer science degree (BSc Hons) and 30 years of
> industry experience.
>
> /Flibble
>

You haven't studied the specific matter at hand thoroughly enough.
With the same credentials as me you could learn these things.
It has taken me at least 15,000 hours since 2004.

You can find the 86612 postings of my prior work right here in this
forum. My work prior to 2004 has not yet been fully restored to the
USENET archives.

--
Copyright 2022 Pete Olcott

"Talent hits a target no one else can hit;
Genius hits a target no one else can see."
Arthur Schopenhauer

On Sun, 5 Jun 2022 12:17:48 -0400
Richard Damon <Richard@Damon-Family.org> wrote:

> On 6/5/22 11:34 AM, Mr Flibble wrote:
> > On Sun, 5 Jun 2022 16:28:05 +0100
> > Andy Walker <anw@cuboid.co.uk> wrote:
> >
> >> On 05/06/2022 14:47, Mr Flibble wrote:
> >>> On Sun, 5 Jun 2022 07:58:42 -0400
> >>> Richard Damon <Richard@Damon-Family.org> wrote:
> >>>> [...] Sort of like how the number Pi has an
> >>>> exact value, but you can never actually express it (because it
> >>>> takes an infinite number of digits).
> >>> PI does not have an exact value; no irrational number has an exact
> >>> value.
> >>
> >> Of course "pi" has an exact value; as do [eg] "sqrt(2)",
> >> "e", and all the other computable real [and complex] numbers.
> >> Whether that value can be expressed in finite terms in some
> >> particular representation is quite another matter. That in turn
> >> depends on the representation; standard decimals is merely one
> >> [common] choice. Note that in symbolic computer systems, those
> >> computable reals are typically written "pi" [or whatever], and the
> >> computer works with that exactly, so that [eg] "sin^2 (pi/3) ==
> >> 3/4", not 0.7499...; and also that in decimal-type notations most
> >> rationals equally have no terminating expansion. Numbers such as
> >> "pi" and "sqrt(2)" are not defined as decimal expansions but via
> >> their properties [eg that "sqrt(2)" is the unique positive real
> >> whose square is 2, or equivalently that it is the ratio of the
> >> diagonal of a square to its side, and "pi" is the least positive
> >> real whose sine is zero]. Those properties are exact, and tell
> >> you all you ever need to know about those numbers.
> >>
> >> [I have removed my name from the "Subject:"; I don't know
> >> why anyone saw fit to attach it to this debate, such as it is, on
> >> the HP.]
> >
> > What has decimal (base 10) expansion got to do with anything? An
> > irrational number has a non-terminating sequence in ANY base. I am
> > sorry but you are simply mistaken: irrational numbers do NOT have an
> > exact value; this is obvious to anyone who understands logic and
> > uses a sane definition for infinity.
> >
> > /Flibble
> >
>
> How about in base pi? then it is the number 10

how about base banana? then it is the number 10.

PI, like banana, is just a symbol representing an irrational number
that has no exact value. To use it here is circular and therefor
erroneous.

>
> Base pi is an interesting base for some problems.
>
> What is your definition of "an exact value"?
>
> Maybe the problem is you don't quite understand the meaning of that
> term.

Of course I understand the fucking term. For the purposes of this
discussion an exact value is a real number (non-integer) that
terminates in a base that is not a multiple of itself.

/Flibble

On Sun, 5 Jun 2022 16:34:18 -0000 (UTC)
Alan Mackenzie <acm@muc.de> wrote:

> Mr Flibble <flibble@reddwarf.jmc> wrote:
> > On Sun, 5 Jun 2022 15:44:32 -0000 (UTC)
> > Alan Mackenzie <acm@muc.de> wrote:
>
> >> Mr Flibble <flibble@reddwarf.jmc> wrote:
> >> > On Sun, 5 Jun 2022 16:28:05 +0100
> >> > Andy Walker <anw@cuboid.co.uk> wrote:
>
> >> >> On 05/06/2022 14:47, Mr Flibble wrote:
> >> >> > On Sun, 5 Jun 2022 07:58:42 -0400
> >> >> > Richard Damon <Richard@Damon-Family.org> wrote:
> >> >> >> [...] Sort of like how the number Pi has an
> >> >> >> exact value, but you can never actually express it (because
> >> >> >> it takes an infinite number of digits).
> >> >> > PI does not have an exact value; no irrational number has an
> >> >> > exact value.
>
> >> >> Of course "pi" has an exact value; as do [eg] "sqrt(2)",
> >> >> "e", and all the other computable real [and complex] numbers.
> >> >> Whether that value can be expressed in finite terms in some
> >> >> particular representation is quite another matter. That in turn
> >> >> depends on the representation; standard decimals is merely one
> >> >> [common] choice. Note that in symbolic computer systems, those
> >> >> computable reals are typically written "pi" [or whatever], and
> >> >> the computer works with that exactly, so that [eg] "sin^2
> >> >> (pi/3) == 3/4", not 0.7499...; and also that in decimal-type
> >> >> notations most rationals equally have no terminating expansion.
> >> >> Numbers such as "pi" and "sqrt(2)" are not defined as decimal
> >> >> expansions but via their properties [eg that "sqrt(2)" is the
> >> >> unique positive real whose square is 2, or equivalently that it
> >> >> is the ratio of the diagonal of a square to its side, and "pi"
> >> >> is the least positive real whose sine is zero]. Those
> >> >> properties are exact, and tell you all you ever need to know
> >> >> about those numbers.
>
> >> [ .... ]
>
> >> > What has decimal (base 10) expansion got to do with anything? An
> >> > irrational number has a non-terminating sequence in ANY base. I
> >> > am sorry but you are simply mistaken: irrational numbers do NOT
> >> > have an exact value; this is obvious to anyone who understands
> >> > logic and uses a sane definition for infinity.
>
> >> That irrational numbers are exact values is clear to anybody with a
> >> degree in maths. Definitions of "infinity" (of which there are
> >> many) have nothing to do with this.
>
> > You are wrong and fractally so so your degree in maths appears to be
> > worthless.
>
> No, I am right, along with the world's other mathematics graduates.
> You are stuck in the distant (100s of years) past, when
> mathematicians were still puzzling over what you're puzzling over.
> The fundamentals of maths have been worked out, and you are in the
> position of an alchemist faced with modern chemistry.

Pure assertion with NOTHING to back it up.

>
> > An irrational number's sequence is statistically random, has no
> > fixed point on the number line ergo has no exact representation.
> > Any number with no exact representation has, by definition, no
> > exact value, only an approximation. Infinity has everything to do
> > with this as an irrational's sequence ("digits") never terminates
> > (i.e. it is an INFINITELY long sequence).

Ignored this part I see.

/Flibble

On 6/5/2022 11:31 AM, Richard Damon wrote:
> On 6/5/22 11:57 AM, olcott wrote:
>> On 6/5/2022 7:38 AM, Ben wrote:
>>> Alan Mackenzie <acm@muc.de> writes:
>>>
>>>> olcott <NoOne@nowhere.com> wrote:
>>>>> On 6/5/2022 6:12 AM, Alan Mackenzie wrote:
>>>
>>>>>> ...  What would be the turing machine equivalent of an
>>>>>> "abnormal termination" in C?
>>>>
>>>>> An aborted simulation.
>>>>
>>>> There's no such thing on a turing machine.  It either runs and
>>>> halts, or
>>>> it runs forever.
>>>>
>>>> Your aborted simulation is just one final state of a turing machine,
>>>> which has thus halted.
>>>
>>> A year ago I tried to get PO to accept a few basic facts about the
>>> topic.  One of these was
>>>
>>> (B) Every computation that halts, for whatever reason, is a halting
>>>      computation.
>>>
>>> After much ducking a diving, PO replied "OK".
>>>
>>
>> Whatever I said years ago has been superseded by my current
>> understanding:
>>
>> Computation that halts ... the Turing machine will halt whenever it
>> enters a final state. (Linz:1990:234)
>>
>>
>
> And H^ applied to <H^> will Halt if H applied to <H^> <H^> rejects its
> input as non-halting, thus showing that the H was wrong.

You are a brain dead moron on this point.

--
Copyright 2022 Pete Olcott

"Talent hits a target no one else can hit;
Genius hits a target no one else can see."
Arthur Schopenhauer

On 6/5/2022 11:38 AM, Mr Flibble wrote:
> On Sun, 5 Jun 2022 16:34:18 -0000 (UTC)
> Alan Mackenzie <acm@muc.de> wrote:
>
>> Mr Flibble <flibble@reddwarf.jmc> wrote:
>>> On Sun, 5 Jun 2022 15:44:32 -0000 (UTC)
>>> Alan Mackenzie <acm@muc.de> wrote:
>>
>>>> Mr Flibble <flibble@reddwarf.jmc> wrote:
>>>>> On Sun, 5 Jun 2022 16:28:05 +0100
>>>>> Andy Walker <anw@cuboid.co.uk> wrote:
>>
>>>>>> On 05/06/2022 14:47, Mr Flibble wrote:
>>>>>>> On Sun, 5 Jun 2022 07:58:42 -0400
>>>>>>> Richard Damon <Richard@Damon-Family.org> wrote:
>>>>>>>> [...] Sort of like how the number Pi has an
>>>>>>>> exact value, but you can never actually express it (because
>>>>>>>> it takes an infinite number of digits).
>>>>>>> PI does not have an exact value; no irrational number has an
>>>>>>> exact value.
>>
>>>>>> Of course "pi" has an exact value; as do [eg] "sqrt(2)",
>>>>>> "e", and all the other computable real [and complex] numbers.
>>>>>> Whether that value can be expressed in finite terms in some
>>>>>> particular representation is quite another matter. That in turn
>>>>>> depends on the representation; standard decimals is merely one
>>>>>> [common] choice. Note that in symbolic computer systems, those
>>>>>> computable reals are typically written "pi" [or whatever], and
>>>>>> the computer works with that exactly, so that [eg] "sin^2
>>>>>> (pi/3) == 3/4", not 0.7499...; and also that in decimal-type
>>>>>> notations most rationals equally have no terminating expansion.
>>>>>> Numbers such as "pi" and "sqrt(2)" are not defined as decimal
>>>>>> expansions but via their properties [eg that "sqrt(2)" is the
>>>>>> unique positive real whose square is 2, or equivalently that it
>>>>>> is the ratio of the diagonal of a square to its side, and "pi"
>>>>>> is the least positive real whose sine is zero]. Those
>>>>>> properties are exact, and tell you all you ever need to know
>>>>>> about those numbers.
>>
>>>> [ .... ]
>>
>>>>> What has decimal (base 10) expansion got to do with anything? An
>>>>> irrational number has a non-terminating sequence in ANY base. I
>>>>> am sorry but you are simply mistaken: irrational numbers do NOT
>>>>> have an exact value; this is obvious to anyone who understands
>>>>> logic and uses a sane definition for infinity.
>>
>>>> That irrational numbers are exact values is clear to anybody with a
>>>> degree in maths. Definitions of "infinity" (of which there are
>>>> many) have nothing to do with this.
>>
>>> You are wrong and fractally so so your degree in maths appears to be
>>> worthless.
>>
>> No, I am right, along with the world's other mathematics graduates.
>> You are stuck in the distant (100s of years) past, when
>> mathematicians were still puzzling over what you're puzzling over.
>> The fundamentals of maths have been worked out, and you are in the
>> position of an alchemist faced with modern chemistry.
>
> Pure assertion with NOTHING to back it up.
>
>>
>>> An irrational number's sequence is statistically random, has no
>>> fixed point on the number line ergo has no exact representation.
>>> Any number with no exact representation has, by definition, no
>>> exact value, only an approximation. Infinity has everything to do
>>> with this as an irrational's sequence ("digits") never terminates
>>> (i.e. it is an INFINITELY long sequence).
>
> Ignored this part I see.
>
> /Flibble
>

You are utterly clueless on these things.
The square-root of 2 does not jump around on the number line.

--
Copyright 2022 Pete Olcott

"Talent hits a target no one else can hit;
Genius hits a target no one else can see."
Arthur Schopenhauer

On 6/5/22 12:07 PM, olcott wrote:
> On 6/5/2022 9:46 AM, Mike Terry wrote:
>> On 05/06/2022 13:14, Alan Mackenzie wrote:
>>> olcott <NoOne@nowhere.com> wrote:
>>>> On 6/5/2022 6:12 AM, Alan Mackenzie wrote:
>>>>> olcott <NoOne@nowhere.com> wrote:
>>>>>> On 6/5/2022 5:14 AM, Mikko wrote:
>>>>>>> On 2022-06-04 19:28:19 +0000, olcott said:
>>>
>>>>>>>> A Turing machine is said to halt whenever it reaches a
>>>>>>>> configuration for which δ is not defined; this is possible because
>>>>>>>> δ is a partial function. In fact, we will assume that no
>>>>>>>> transitions are defined for any final state so the Turing machine
>>>>>>>> will halt whenever it enters a final state.  (Linz:1990:234)
>>>
>>>>>>>> Linz, Peter 1990. An Introduction to Formal Languages and Automata.
>>>>>>>> Lexington/Toronto: D. C. Heath and Company.
>>>
>>>>>>>> When translated into ordinary software engineering terms this means
>>>>>>>> terminated normally. In a C function this means reaching the "ret"
>>>>>>>> instruction.
>>>
>>>>>>> The best equivalent to "not defined" is not "ret". Instead, "not
>>>>>>> defined" should include at least:
>>>>>>> - HLT or any other instruction that means 'halt'
>>>>>>> - any undefined op code
>>>>>>> - any return or pop instruction if the stack is empty
>>>>>>> - an instruction fetch from a location that is not specifiec by the
>>>>>>>     program
>>>>>>> That way the analogy to Linz' definition is much better.
>>>
>>>>>>> Mikko
>>>
>>>>>> Reaching a final state is merely the Turing machine way of saying
>>>>>> terminated normally. "ret" is the C way of saying the same thing.
>>>
>>>>> Sophistry.  What would be the turing machine equivalent of an
>>>>> "abnormal termination" in C?
>>>
>>>> An aborted simulation.
>>>
>>> There's no such thing on a turing machine.  It either runs and halts, or
>>> it runs forever.
>>>
>>> Your aborted simulation is just one final state of a turing machine,
>>> which has thus halted.
>>
>> A TM "aborting" a simulation is just the TM ceasing to calculate
>> computation steps for some computation, and going on to calculate
>> something else instead.  It does not mean:
>> a)  that the TM (doing the simulation) has halted
>> b)  that the simulated computation halts
>> c)  that the simulated computation never halts
>>
>> Regards,
>> Mike.
>>
>
> That an aborted simulated has not reached a final state has not halted
> is proven by the fact that your screwy reasoning would have to conclude
> that an infinite loop halts.
>
> *This is a Stipulative definition*
> Computation that halts ... the Turing machine will halt whenever it
> enters a final state. (Linz:1990:234)
>
> A stipulative definition is a type of definition in which a new or
> currently existing term is given a new specific meaning for the purposes
> of argument or discussion in a given context.
> https://en.wikipedia.org/wiki/Stipulative_definition

So, since you stipulated definition refers to the behavior of the actual
Turing Machine, and not the simulation, the simualation, aborted or
otherwise doesn't apply to showing if the machine is halting or not.

Thus H's aborting of its simulation of Infinte_Loop doesn't actually
provide proof or refutation of Infinite_Loop halting or not. Yes, if H
simulated long enough before aborting, there IS a logically sound and
valid arguement from the data to prove that Infinite_Loop is actually
non-halting, but ultimately, we need to go back to your definition in
that proof.

Also, we can say thus H's simulation of H^/P says NOTHING about the
actual halting status of that input, only the actual running of it, or
an actually sound and valid proof from the information obtained.

The problem is, there is no such proof in the case of its simulation of
H^/P, and the proof that you have shown has had its errors pointed out
and never corrected (one of the errors is not showing an actual correct
trace of what H has simulated, at least not correct if H is actually a
computation)

Since, it is shown that H^ applied to <H^> halts if H applied to <H^>
<H^> rejects its input by claiming it is non-halting, that PROVES, BY
THE DEFINITION, that H was wrong in its decision.

The TURING MACHINE Halted, thus BY THE DEFINITION of a HALT decider, H
MUST accept the input of the representation of that machine/input
combination.

Thus, you FAIL in your proof BECAUSE of your stipulated definition.

>
> void Infinite_Loop()
> {
>   HERE: goto HERE;
> }
>
> int main()
> {
>   Output("Input_Halts = ", H0(Infinite_Loop));
> }
>
> _Infinite_Loop()
> [00001342](01)  55              push ebp
> [00001343](02)  8bec            mov ebp,esp
> [00001345](02)  ebfe            jmp 00001345
> [00001347](01)  5d              pop ebp
> [00001348](01)  c3              ret
> Size in bytes:(0007) [00001348]
>
>
>

On Sun, 5 Jun 2022 11:41:05 -0500
olcott <NoOne@NoWhere.com> wrote:

> On 6/5/2022 11:38 AM, Mr Flibble wrote:
> > On Sun, 5 Jun 2022 16:34:18 -0000 (UTC)
> > Alan Mackenzie <acm@muc.de> wrote:
> >
> >> Mr Flibble <flibble@reddwarf.jmc> wrote:
> >>> On Sun, 5 Jun 2022 15:44:32 -0000 (UTC)
> >>> Alan Mackenzie <acm@muc.de> wrote:
> >>
> >>>> Mr Flibble <flibble@reddwarf.jmc> wrote:
> >>>>> On Sun, 5 Jun 2022 16:28:05 +0100
> >>>>> Andy Walker <anw@cuboid.co.uk> wrote:
> >>
> >>>>>> On 05/06/2022 14:47, Mr Flibble wrote:
> >>>>>>> On Sun, 5 Jun 2022 07:58:42 -0400
> >>>>>>> Richard Damon <Richard@Damon-Family.org> wrote:
> >>>>>>>> [...] Sort of like how the number Pi has an
> >>>>>>>> exact value, but you can never actually express it (because
> >>>>>>>> it takes an infinite number of digits).
> >>>>>>> PI does not have an exact value; no irrational number has an
> >>>>>>> exact value.
> >>
> >>>>>> Of course "pi" has an exact value; as do [eg]
> >>>>>> "sqrt(2)", "e", and all the other computable real [and
> >>>>>> complex] numbers. Whether that value can be expressed in
> >>>>>> finite terms in some particular representation is quite
> >>>>>> another matter. That in turn depends on the representation;
> >>>>>> standard decimals is merely one [common] choice. Note that in
> >>>>>> symbolic computer systems, those computable reals are
> >>>>>> typically written "pi" [or whatever], and the computer works
> >>>>>> with that exactly, so that [eg] "sin^2 (pi/3) == 3/4", not
> >>>>>> 0.7499...; and also that in decimal-type notations most
> >>>>>> rationals equally have no terminating expansion. Numbers such
> >>>>>> as "pi" and "sqrt(2)" are not defined as decimal expansions
> >>>>>> but via their properties [eg that "sqrt(2)" is the unique
> >>>>>> positive real whose square is 2, or equivalently that it is
> >>>>>> the ratio of the diagonal of a square to its side, and "pi" is
> >>>>>> the least positive real whose sine is zero]. Those properties
> >>>>>> are exact, and tell you all you ever need to know about those
> >>>>>> numbers.
> >>
> >>>> [ .... ]
> >>
> >>>>> What has decimal (base 10) expansion got to do with anything? An
> >>>>> irrational number has a non-terminating sequence in ANY base. I
> >>>>> am sorry but you are simply mistaken: irrational numbers do NOT
> >>>>> have an exact value; this is obvious to anyone who understands
> >>>>> logic and uses a sane definition for infinity.
> >>
> >>>> That irrational numbers are exact values is clear to anybody
> >>>> with a degree in maths. Definitions of "infinity" (of which
> >>>> there are many) have nothing to do with this.
> >>
> >>> You are wrong and fractally so so your degree in maths appears to
> >>> be worthless.
> >>
> >> No, I am right, along with the world's other mathematics graduates.
> >> You are stuck in the distant (100s of years) past, when
> >> mathematicians were still puzzling over what you're puzzling over.
> >> The fundamentals of maths have been worked out, and you are in the
> >> position of an alchemist faced with modern chemistry.
> >
> > Pure assertion with NOTHING to back it up.
> >
> >>
> >>> An irrational number's sequence is statistically random, has no
> >>> fixed point on the number line ergo has no exact representation.
> >>> Any number with no exact representation has, by definition, no
> >>> exact value, only an approximation. Infinity has everything to do
> >>> with this as an irrational's sequence ("digits") never terminates
> >>> (i.e. it is an INFINITELY long sequence).
> >
> > Ignored this part I see.
> >
> > /Flibble
> >
>
> You are utterly clueless on these things.

Projection.

> The square-root of 2 does not jump around on the number line.
Yes it does, all irrational numbers do.

/Flibble

On 6/5/22 12:15 PM, olcott wrote:
> On 6/5/2022 11:12 AM, Mr Flibble wrote:
>> On Sun, 5 Jun 2022 11:07:38 -0500
>> olcott <NoOne@NoWhere.com> wrote:
>>
>>> On 6/5/2022 9:46 AM, Mike Terry wrote:
>>>> On 05/06/2022 13:14, Alan Mackenzie wrote:
>>>>> olcott <NoOne@nowhere.com> wrote:
>>>>>> On 6/5/2022 6:12 AM, Alan Mackenzie wrote:
>>>>>>> olcott <NoOne@nowhere.com> wrote:
>>>>>>>> On 6/5/2022 5:14 AM, Mikko wrote:
>>>>>>>>> On 2022-06-04 19:28:19 +0000, olcott said:
>>>>>>>>>> A Turing machine is said to halt whenever it reaches a
>>>>>>>>>> configuration for which δ is not defined; this is possible
>>>>>>>>>> because δ is a partial function. In fact, we will assume that
>>>>>>>>>> no transitions are defined for any final state so the Turing
>>>>>>>>>> machine will halt whenever it enters a final state.
>>>>>>>>>> (Linz:1990:234)
>>>>>>>>>> Linz, Peter 1990. An Introduction to Formal Languages and
>>>>>>>>>> Automata. Lexington/Toronto: D. C. Heath and Company.
>>>>>>>>>> When translated into ordinary software engineering terms this
>>>>>>>>>> means terminated normally. In a C function this means
>>>>>>>>>> reaching the "ret" instruction.
>>>>>>>>> The best equivalent to "not defined" is not "ret". Instead,
>>>>>>>>> "not defined" should include at least:
>>>>>>>>> - HLT or any other instruction that means 'halt'
>>>>>>>>> - any undefined op code
>>>>>>>>> - any return or pop instruction if the stack is empty
>>>>>>>>> - an instruction fetch from a location that is not specifiec
>>>>>>>>> by the program
>>>>>>>>> That way the analogy to Linz' definition is much better.
>>>>>>>>> Mikko
>>>>>>>> Reaching a final state is merely the Turing machine way of
>>>>>>>> saying terminated normally. "ret" is the C way of saying the
>>>>>>>> same thing.
>>>>>>> Sophistry.  What would be the turing machine equivalent of an
>>>>>>> "abnormal termination" in C?
>>>>>> An aborted simulation.
>>>>>
>>>>> There's no such thing on a turing machine.  It either runs and
>>>>> halts, or it runs forever.
>>>>>
>>>>> Your aborted simulation is just one final state of a turing
>>>>> machine, which has thus halted.
>>>>
>>>> A TM "aborting" a simulation is just the TM ceasing to calculate
>>>> computation steps for some computation, and going on to calculate
>>>> something else instead.  It does not mean:
>>>> a)  that the TM (doing the simulation) has halted
>>>> b)  that the simulated computation halts
>>>> c)  that the simulated computation never halts
>>>>
>>>> Regards,
>>>> Mike.
>>>
>>> That an aborted simulated has not reached a final state has not
>>> halted is proven by the fact that your screwy reasoning would have to
>>> conclude that an infinite loop halts.
>>>
>>> *This is a Stipulative definition*
>>> Computation that halts ... the Turing machine will halt whenever it
>>> enters a final state. (Linz:1990:234)
>>>
>>> A stipulative definition is a type of definition in which a new or
>>> currently existing term is given a new specific meaning for the
>>> purposes of argument or discussion in a given context.
>>> https://en.wikipedia.org/wiki/Stipulative_definition
>>>
>>> void Infinite_Loop()
>>> {
>>>     HERE: goto HERE;
>>> }
>>>
>>> int main()
>>> {
>>>     Output("Input_Halts = ", H0(Infinite_Loop));
>>> }
>>>
>>> _Infinite_Loop()
>>> [00001342](01)  55              push ebp
>>> [00001343](02)  8bec            mov ebp,esp
>>> [00001345](02)  ebfe            jmp 00001345
>>> [00001347](01)  5d              pop ebp
>>> [00001348](01)  c3              ret
>>> Size in bytes:(0007) [00001348]
>>
>> You only call Infinite_Loop() if you detect a recursion
>> into your decider something that the proofs you are attempting to refute
>> do not do.  What you have has nothing to do with the Halting Problem.
>>
>> /Flibble
>>
>
> That you are simply not bright enough to see that the Linz proof does
> specify infinitely nested simulation is really no actual rebuttal at all.
>
> Ĥ.q0 ⟨Ĥ⟩ ⊢* H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qy
> If the correctly simulated input ⟨Ĥ⟩ ⟨Ĥ⟩ to H would reach its own final
> state of ⟨Ĥ.qy⟩ or ⟨Ĥ.qn⟩.
>
> Ĥ.q0 ⟨Ĥ⟩ ⊢* H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qn
> If the correctly simulated input ⟨Ĥ⟩ ⟨Ĥ⟩ to H would never reach its own
> final state of ⟨Ĥ.qy⟩ or ⟨Ĥ.qn⟩.
> When Ĥ is applied to ⟨Ĥ⟩      // subscripts indicate unique finite strings
> Ĥ copies its input ⟨Ĥ0⟩ to ⟨Ĥ1⟩ then H simulates ⟨Ĥ0⟩ ⟨Ĥ1⟩
>
> Then these steps would keep repeating: (unless their simulation is aborted)
> Ĥ0 copies its input ⟨Ĥ1⟩ to ⟨Ĥ2⟩ then H0 simulates ⟨Ĥ1⟩ ⟨Ĥ2⟩
> Ĥ1 copies its input ⟨Ĥ2⟩ to ⟨Ĥ3⟩ then H1 simulates ⟨Ĥ2⟩ ⟨Ĥ3⟩
> Ĥ2 copies its input ⟨Ĥ3⟩ to ⟨Ĥ4⟩ then H2 simulates ⟨Ĥ3⟩ ⟨Ĥ4⟩...
>
>

And, if it DOES keep repeating as you claim, then H NEVER aborts its
simulation, and thus fails to answer.

If H does abort its simulation, then the pattern does NOT keep
repeating, because the H used by H^ will also abort its own simulation
and break the infinte chain.

The H simulating that input just doesn't get to that point before it
aborts it simulation.

You proof somewhat boils down to claiming there exists a N > N+5, which
just isn't true.

On 6/5/22 12:28 PM, Mr Flibble wrote:
> On Sun, 5 Jun 2022 12:22:45 -0400
> Richard Damon <Richard@Damon-Family.org> wrote:
>
>> On 6/5/22 11:49 AM, Mr Flibble wrote:
>>> On Sun, 5 Jun 2022 15:44:32 -0000 (UTC)
>>> Alan Mackenzie <acm@muc.de> wrote:
>>>
>>>> Mr Flibble <flibble@reddwarf.jmc> wrote:
>>>>> On Sun, 5 Jun 2022 16:28:05 +0100
>>>>> Andy Walker <anw@cuboid.co.uk> wrote:
>>>>
>>>>>> On 05/06/2022 14:47, Mr Flibble wrote:
>>>>>>> On Sun, 5 Jun 2022 07:58:42 -0400
>>>>>>> Richard Damon <Richard@Damon-Family.org> wrote:
>>>>>>>> [...] Sort of like how the number Pi has an
>>>>>>>> exact value, but you can never actually express it (because it
>>>>>>>> takes an infinite number of digits).
>>>>>>> PI does not have an exact value; no irrational number has an
>>>>>>> exact value.
>>>>
>>>>>> Of course "pi" has an exact value; as do [eg] "sqrt(2)",
>>>>>> "e", and all the other computable real [and complex] numbers.
>>>>>> Whether that value can be expressed in finite terms in some
>>>>>> particular representation is quite another matter. That in turn
>>>>>> depends on the representation; standard decimals is merely one
>>>>>> [common] choice. Note that in symbolic computer systems, those
>>>>>> computable reals are typically written "pi" [or whatever], and
>>>>>> the computer works with that exactly, so that [eg] "sin^2 (pi/3)
>>>>>> == 3/4", not 0.7499...; and also that in decimal-type notations
>>>>>> most rationals equally have no terminating expansion. Numbers
>>>>>> such as "pi" and "sqrt(2)" are not defined as decimal expansions
>>>>>> but via their properties [eg that "sqrt(2)" is the unique
>>>>>> positive real whose square is 2, or equivalently that it is the
>>>>>> ratio of the diagonal of a square to its side, and "pi" is the
>>>>>> least positive real whose sine is zero]. Those properties are
>>>>>> exact, and tell you all you ever need to know about those
>>>>>> numbers.
>>>>
>>>> [ .... ]
>>>>
>>>>> What has decimal (base 10) expansion got to do with anything? An
>>>>> irrational number has a non-terminating sequence in ANY base. I
>>>>> am sorry but you are simply mistaken: irrational numbers do NOT
>>>>> have an exact value; this is obvious to anyone who understands
>>>>> logic and uses a sane definition for infinity.
>>>>
>>>> That irrational numbers are exact values is clear to anybody with a
>>>> degree in maths. Definitions of "infinity" (of which there are
>>>> many) have nothing to do with this.
>>>
>>> You are wrong and fractally so so your degree in maths appears to be
>>> worthless. An irrational number's sequence is statistically random,
>>> has no fixed point on the number line ergo has no exact
>>> representation. Any number with no exact representation has, by
>>> definition, no exact value, only an approximation. Infinity has
>>> everything to do with this as an irrational's sequence ("digits")
>>> never terminates (i.e. it is an INFINITELY long sequence).
>>>
>>> /Flibble
>>>
>>
>> Nope. Irrational numbers DO have exact points on the number line.
>>
>> And what does representation have to do with exact value?
>>
>> Also, irrational numbers sequence of digits are not necessarily
>> statistically random, in some representations, they can be VERY
>> predictible for some numbers.
>>
>> One simple construction to show exact position, draw a box with sides
>> exactly 1.
>>
>> Draw a line though opposite corners and make one point the value 0.
>>
>> The other corner will be EXACTLY at the point sqrt(2), so that
>> irrational number has an exact point on the number line.
>>
>> You just don't understand what an exact value means, likely because
>> you can't understand things that are somewhat abstract.
>
> An irrational number does not have an exact point on the number line as
> it will move about as you "zoom in", you can keep "zooming in" forever
> (i.e. infinitely) and it will keep moving about because the number
> never terminates.
>
> If I couldn't understand things that are somewhat abstract then I
> wouldn't have a computer science degree (BSc Hons) and 30 years of
> industry experience.
>
> /Flibble
>

Then why do you think irrational numbers don't have an exact location?

I know people with degrees (even with honors) and industry experiance
that still show that they don't really understand what they are talking
about.

The "width" of the point representing the location of an irrational
number is just as much "0" as that of a rational number, so specifies
just as exact of a location.

The fact that we can't write it in a rational base with a finite number
of digits doesn't actally mean anything.

On 6/5/22 12:29 PM, olcott wrote:
> On 6/5/2022 11:22 AM, Richard Damon wrote:
>> On 6/5/22 11:49 AM, Mr Flibble wrote:
>>> On Sun, 5 Jun 2022 15:44:32 -0000 (UTC)
>>> Alan Mackenzie <acm@muc.de> wrote:
>>>
>>>> Mr Flibble <flibble@reddwarf.jmc> wrote:
>>>>> On Sun, 5 Jun 2022 16:28:05 +0100
>>>>> Andy Walker <anw@cuboid.co.uk> wrote:
>>>>
>>>>>> On 05/06/2022 14:47, Mr Flibble wrote:
>>>>>>> On Sun, 5 Jun 2022 07:58:42 -0400
>>>>>>> Richard Damon <Richard@Damon-Family.org> wrote:
>>>>>>>> [...] Sort of like how the number Pi has an
>>>>>>>> exact value, but you can never actually express it (because it
>>>>>>>> takes an infinite number of digits).
>>>>>>> PI does not have an exact value; no irrational number has an
>>>>>>> exact value.
>>>>
>>>>>>        Of course "pi" has an exact value;  as do [eg] "sqrt(2)",
>>>>>> "e", and all the other computable real [and complex] numbers.
>>>>>> Whether that value can be expressed in finite terms in some
>>>>>> particular representation is quite another matter.  That in turn
>>>>>> depends on the representation;  standard decimals is merely one
>>>>>> [common] choice.  Note that in symbolic computer systems, those
>>>>>> computable reals are typically written "pi" [or whatever], and the
>>>>>> computer works with that exactly, so that [eg] "sin^2 (pi/3) ==
>>>>>> 3/4", not 0.7499...; and also that in decimal-type notations most
>>>>>> rationals equally have no terminating expansion.  Numbers such as
>>>>>> "pi" and "sqrt(2)" are not defined as decimal expansions but via
>>>>>> their properties [eg that "sqrt(2)" is the unique positive real
>>>>>> whose square is 2, or equivalently that it is the ratio of the
>>>>>> diagonal of a square to its side, and "pi" is the least positive
>>>>>> real whose sine is zero].  Those properties are exact, and tell
>>>>>> you all you ever need to know about those numbers.
>>>>
>>>> [ .... ]
>>>>
>>>>> What has decimal (base 10) expansion got to do with anything? An
>>>>> irrational number has a non-terminating sequence in ANY base.  I am
>>>>> sorry but you are simply mistaken: irrational numbers do NOT have an
>>>>> exact value; this is obvious to anyone who understands logic and
>>>>> uses a sane definition for infinity.
>>>>
>>>> That irrational numbers are exact values is clear to anybody with a
>>>> degree in maths.  Definitions of "infinity" (of which there are many)
>>>> have nothing to do with this.
>>>
>>> You are wrong and fractally so so your degree in maths appears to be
>>> worthless.  An irrational number's sequence is statistically random,
>>> has no fixed point on the number line ergo has no exact representation.
>>> Any number with no exact representation has, by definition, no exact
>>> value, only an approximation.  Infinity has everything to do with this
>>> as an irrational's sequence ("digits") never terminates (i.e. it is an
>>> INFINITELY long sequence).
>>>
>>> /Flibble
>>>
>>
>> Nope. Irrational numbers DO have exact points on the number line.
>
> In this we agree. Also every identifiable point on a number line has
> another identifiable point that is immediately adjacent to it with no
> other points inbetween.
>
> [0,1] is exactly one geometric point longer than [0,1)
>
>

Well yes, there is one more point (the point 1.0000) is [0, 1] then in
[0, 1), but that doesn't make it "longer" as both lines have a
measurement lenght of "1.00" (the missing point is immeasurable in the
Reals).

On 6/5/22 12:42 PM, Mr Flibble wrote:
> On Sun, 5 Jun 2022 11:41:05 -0500
> olcott <NoOne@NoWhere.com> wrote:
>
>> On 6/5/2022 11:38 AM, Mr Flibble wrote:
>>> On Sun, 5 Jun 2022 16:34:18 -0000 (UTC)
>>> Alan Mackenzie <acm@muc.de> wrote:
>>>
>>>> Mr Flibble <flibble@reddwarf.jmc> wrote:
>>>>> On Sun, 5 Jun 2022 15:44:32 -0000 (UTC)
>>>>> Alan Mackenzie <acm@muc.de> wrote:
>>>>
>>>>>> Mr Flibble <flibble@reddwarf.jmc> wrote:
>>>>>>> On Sun, 5 Jun 2022 16:28:05 +0100
>>>>>>> Andy Walker <anw@cuboid.co.uk> wrote:
>>>>
>>>>>>>> On 05/06/2022 14:47, Mr Flibble wrote:
>>>>>>>>> On Sun, 5 Jun 2022 07:58:42 -0400
>>>>>>>>> Richard Damon <Richard@Damon-Family.org> wrote:
>>>>>>>>>> [...] Sort of like how the number Pi has an
>>>>>>>>>> exact value, but you can never actually express it (because
>>>>>>>>>> it takes an infinite number of digits).
>>>>>>>>> PI does not have an exact value; no irrational number has an
>>>>>>>>> exact value.
>>>>
>>>>>>>> Of course "pi" has an exact value; as do [eg]
>>>>>>>> "sqrt(2)", "e", and all the other computable real [and
>>>>>>>> complex] numbers. Whether that value can be expressed in
>>>>>>>> finite terms in some particular representation is quite
>>>>>>>> another matter. That in turn depends on the representation;
>>>>>>>> standard decimals is merely one [common] choice. Note that in
>>>>>>>> symbolic computer systems, those computable reals are
>>>>>>>> typically written "pi" [or whatever], and the computer works
>>>>>>>> with that exactly, so that [eg] "sin^2 (pi/3) == 3/4", not
>>>>>>>> 0.7499...; and also that in decimal-type notations most
>>>>>>>> rationals equally have no terminating expansion. Numbers such
>>>>>>>> as "pi" and "sqrt(2)" are not defined as decimal expansions
>>>>>>>> but via their properties [eg that "sqrt(2)" is the unique
>>>>>>>> positive real whose square is 2, or equivalently that it is
>>>>>>>> the ratio of the diagonal of a square to its side, and "pi" is
>>>>>>>> the least positive real whose sine is zero]. Those properties
>>>>>>>> are exact, and tell you all you ever need to know about those
>>>>>>>> numbers.
>>>>
>>>>>> [ .... ]
>>>>
>>>>>>> What has decimal (base 10) expansion got to do with anything? An
>>>>>>> irrational number has a non-terminating sequence in ANY base. I
>>>>>>> am sorry but you are simply mistaken: irrational numbers do NOT
>>>>>>> have an exact value; this is obvious to anyone who understands
>>>>>>> logic and uses a sane definition for infinity.
>>>>
>>>>>> That irrational numbers are exact values is clear to anybody
>>>>>> with a degree in maths. Definitions of "infinity" (of which
>>>>>> there are many) have nothing to do with this.
>>>>
>>>>> You are wrong and fractally so so your degree in maths appears to
>>>>> be worthless.
>>>>
>>>> No, I am right, along with the world's other mathematics graduates.
>>>> You are stuck in the distant (100s of years) past, when
>>>> mathematicians were still puzzling over what you're puzzling over.
>>>> The fundamentals of maths have been worked out, and you are in the
>>>> position of an alchemist faced with modern chemistry.
>>>
>>> Pure assertion with NOTHING to back it up.
>>>
>>>>
>>>>> An irrational number's sequence is statistically random, has no
>>>>> fixed point on the number line ergo has no exact representation.
>>>>> Any number with no exact representation has, by definition, no
>>>>> exact value, only an approximation. Infinity has everything to do
>>>>> with this as an irrational's sequence ("digits") never terminates
>>>>> (i.e. it is an INFINITELY long sequence).
>>>
>>> Ignored this part I see.
>>>
>>> /Flibble
>>>
>>
>> You are utterly clueless on these things.
>
> Projection.
>
>> The square-root of 2 does not jump around on the number line.
>
> Yes it does, all irrational numbers do.
>
> /Flibble
>

So, what other values does it jump to besides the actual value of sqrt(2)?

On Sun, 5 Jun 2022 12:50:13 -0400
Richard Damon <Richard@Damon-Family.org> wrote:

> On 6/5/22 12:28 PM, Mr Flibble wrote:
> > On Sun, 5 Jun 2022 12:22:45 -0400
> > Richard Damon <Richard@Damon-Family.org> wrote:
> >
> >> On 6/5/22 11:49 AM, Mr Flibble wrote:
> >>> On Sun, 5 Jun 2022 15:44:32 -0000 (UTC)
> >>> Alan Mackenzie <acm@muc.de> wrote:
> >>>
> >>>> Mr Flibble <flibble@reddwarf.jmc> wrote:
> >>>>> On Sun, 5 Jun 2022 16:28:05 +0100
> >>>>> Andy Walker <anw@cuboid.co.uk> wrote:
> >>>>
> >>>>>> On 05/06/2022 14:47, Mr Flibble wrote:
> >>>>>>> On Sun, 5 Jun 2022 07:58:42 -0400
> >>>>>>> Richard Damon <Richard@Damon-Family.org> wrote:
> >>>>>>>> [...] Sort of like how the number Pi has an
> >>>>>>>> exact value, but you can never actually express it (because
> >>>>>>>> it takes an infinite number of digits).
> >>>>>>> PI does not have an exact value; no irrational number has an
> >>>>>>> exact value.
> >>>>
> >>>>>> Of course "pi" has an exact value; as do [eg]
> >>>>>> "sqrt(2)", "e", and all the other computable real [and
> >>>>>> complex] numbers. Whether that value can be expressed in
> >>>>>> finite terms in some particular representation is quite
> >>>>>> another matter. That in turn depends on the representation;
> >>>>>> standard decimals is merely one [common] choice. Note that in
> >>>>>> symbolic computer systems, those computable reals are
> >>>>>> typically written "pi" [or whatever], and the computer works
> >>>>>> with that exactly, so that [eg] "sin^2 (pi/3) == 3/4", not
> >>>>>> 0.7499...; and also that in decimal-type notations most
> >>>>>> rationals equally have no terminating expansion. Numbers such
> >>>>>> as "pi" and "sqrt(2)" are not defined as decimal expansions
> >>>>>> but via their properties [eg that "sqrt(2)" is the unique
> >>>>>> positive real whose square is 2, or equivalently that it is
> >>>>>> the ratio of the diagonal of a square to its side, and "pi" is
> >>>>>> the least positive real whose sine is zero]. Those properties
> >>>>>> are exact, and tell you all you ever need to know about those
> >>>>>> numbers.
> >>>>
> >>>> [ .... ]
> >>>>
> >>>>> What has decimal (base 10) expansion got to do with anything? An
> >>>>> irrational number has a non-terminating sequence in ANY base. I
> >>>>> am sorry but you are simply mistaken: irrational numbers do NOT
> >>>>> have an exact value; this is obvious to anyone who understands
> >>>>> logic and uses a sane definition for infinity.
> >>>>
> >>>> That irrational numbers are exact values is clear to anybody
> >>>> with a degree in maths. Definitions of "infinity" (of which
> >>>> there are many) have nothing to do with this.
> >>>
> >>> You are wrong and fractally so so your degree in maths appears to
> >>> be worthless. An irrational number's sequence is statistically
> >>> random, has no fixed point on the number line ergo has no exact
> >>> representation. Any number with no exact representation has, by
> >>> definition, no exact value, only an approximation. Infinity has
> >>> everything to do with this as an irrational's sequence ("digits")
> >>> never terminates (i.e. it is an INFINITELY long sequence).
> >>>
> >>> /Flibble
> >>>
> >>
> >> Nope. Irrational numbers DO have exact points on the number line.
> >>
> >> And what does representation have to do with exact value?
> >>
> >> Also, irrational numbers sequence of digits are not necessarily
> >> statistically random, in some representations, they can be VERY
> >> predictible for some numbers.
> >>
> >> One simple construction to show exact position, draw a box with
> >> sides exactly 1.
> >>
> >> Draw a line though opposite corners and make one point the value 0.
> >>
> >> The other corner will be EXACTLY at the point sqrt(2), so that
> >> irrational number has an exact point on the number line.
> >>
> >> You just don't understand what an exact value means, likely because
> >> you can't understand things that are somewhat abstract.
> >
> > An irrational number does not have an exact point on the number
> > line as it will move about as you "zoom in", you can keep "zooming
> > in" forever (i.e. infinitely) and it will keep moving about because
> > the number never terminates.
> >
> > If I couldn't understand things that are somewhat abstract then I
> > wouldn't have a computer science degree (BSc Hons) and 30 years of
> > industry experience.
> >
> > /Flibble
> >
>
> Then why do you think irrational numbers don't have an exact location?
>
> I know people with degrees (even with honors) and industry experiance
> that still show that they don't really understand what they are
> talking about.
>
> The "width" of the point representing the location of an irrational
> number is just as much "0" as that of a rational number, so specifies
> just as exact of a location.
>
> The fact that we can't write it in a rational base with a finite
> number of digits doesn't actally mean anything.

3.1415xxxxxxxxx (a)
3.14159xxxxxxxx (b)

(b) is nearer to 3.14160 than 3.14150 that (a) indicates hence its value
"moves about" as accuracy increases.

/Flibble

On 6/5/22 12:37 PM, Mr Flibble wrote:
> On Sun, 5 Jun 2022 12:17:48 -0400
> Richard Damon <Richard@Damon-Family.org> wrote:
>
>> On 6/5/22 11:34 AM, Mr Flibble wrote:
>>> On Sun, 5 Jun 2022 16:28:05 +0100
>>> Andy Walker <anw@cuboid.co.uk> wrote:
>>>
>>>> On 05/06/2022 14:47, Mr Flibble wrote:
>>>>> On Sun, 5 Jun 2022 07:58:42 -0400
>>>>> Richard Damon <Richard@Damon-Family.org> wrote:
>>>>>> [...] Sort of like how the number Pi has an
>>>>>> exact value, but you can never actually express it (because it
>>>>>> takes an infinite number of digits).
>>>>> PI does not have an exact value; no irrational number has an exact
>>>>> value.
>>>>
>>>> Of course "pi" has an exact value; as do [eg] "sqrt(2)",
>>>> "e", and all the other computable real [and complex] numbers.
>>>> Whether that value can be expressed in finite terms in some
>>>> particular representation is quite another matter. That in turn
>>>> depends on the representation; standard decimals is merely one
>>>> [common] choice. Note that in symbolic computer systems, those
>>>> computable reals are typically written "pi" [or whatever], and the
>>>> computer works with that exactly, so that [eg] "sin^2 (pi/3) ==
>>>> 3/4", not 0.7499...; and also that in decimal-type notations most
>>>> rationals equally have no terminating expansion. Numbers such as
>>>> "pi" and "sqrt(2)" are not defined as decimal expansions but via
>>>> their properties [eg that "sqrt(2)" is the unique positive real
>>>> whose square is 2, or equivalently that it is the ratio of the
>>>> diagonal of a square to its side, and "pi" is the least positive
>>>> real whose sine is zero]. Those properties are exact, and tell
>>>> you all you ever need to know about those numbers.
>>>>
>>>> [I have removed my name from the "Subject:"; I don't know
>>>> why anyone saw fit to attach it to this debate, such as it is, on
>>>> the HP.]
>>>
>>> What has decimal (base 10) expansion got to do with anything? An
>>> irrational number has a non-terminating sequence in ANY base. I am
>>> sorry but you are simply mistaken: irrational numbers do NOT have an
>>> exact value; this is obvious to anyone who understands logic and
>>> uses a sane definition for infinity.
>>>
>>> /Flibble
>>>
>>
>> How about in base pi? then it is the number 10
>
> how about base banana? then it is the number 10.
>
> PI, like banana, is just a symbol representing an irrational number
> that has no exact value. To use it here is circular and therefor
> erroneous.
>
>>
>> Base pi is an interesting base for some problems.
>>
>> What is your definition of "an exact value"?
>>
>> Maybe the problem is you don't quite understand the meaning of that
>> term.
>
> Of course I understand the fucking term. For the purposes of this
> discussion an exact value is a real number (non-integer) that
> terminates in a base that is not a multiple of itself.
>
> /Flibble
>

Where do you get that definition from?

So 1/3 isn't an exact value?

The only bases you can express its value in as a finite number are
multiples of 3, which is a multiple of 1/3.

I suppose even 1/2 becomes non-exact by your definition, as what base
would you use that isn't a multiple of it?

On Sun, 5 Jun 2022 12:54:22 -0400
Richard Damon <Richard@Damon-Family.org> wrote:

> On 6/5/22 12:42 PM, Mr Flibble wrote:
> > On Sun, 5 Jun 2022 11:41:05 -0500
> > olcott <NoOne@NoWhere.com> wrote:
> >
> >> On 6/5/2022 11:38 AM, Mr Flibble wrote:
> >>> On Sun, 5 Jun 2022 16:34:18 -0000 (UTC)
> >>> Alan Mackenzie <acm@muc.de> wrote:
> >>>
> >>>> Mr Flibble <flibble@reddwarf.jmc> wrote:
> >>>>> On Sun, 5 Jun 2022 15:44:32 -0000 (UTC)
> >>>>> Alan Mackenzie <acm@muc.de> wrote:
> >>>>
> >>>>>> Mr Flibble <flibble@reddwarf.jmc> wrote:
> >>>>>>> On Sun, 5 Jun 2022 16:28:05 +0100
> >>>>>>> Andy Walker <anw@cuboid.co.uk> wrote:
> >>>>
> >>>>>>>> On 05/06/2022 14:47, Mr Flibble wrote:
> >>>>>>>>> On Sun, 5 Jun 2022 07:58:42 -0400
> >>>>>>>>> Richard Damon <Richard@Damon-Family.org> wrote:
> >>>>>>>>>> [...] Sort of like how the number Pi has an
> >>>>>>>>>> exact value, but you can never actually express it (because
> >>>>>>>>>> it takes an infinite number of digits).
> >>>>>>>>> PI does not have an exact value; no irrational number has an
> >>>>>>>>> exact value.
> >>>>
> >>>>>>>> Of course "pi" has an exact value; as do [eg]
> >>>>>>>> "sqrt(2)", "e", and all the other computable real [and
> >>>>>>>> complex] numbers. Whether that value can be expressed in
> >>>>>>>> finite terms in some particular representation is quite
> >>>>>>>> another matter. That in turn depends on the representation;
> >>>>>>>> standard decimals is merely one [common] choice. Note that
> >>>>>>>> in symbolic computer systems, those computable reals are
> >>>>>>>> typically written "pi" [or whatever], and the computer works
> >>>>>>>> with that exactly, so that [eg] "sin^2 (pi/3) == 3/4", not
> >>>>>>>> 0.7499...; and also that in decimal-type notations most
> >>>>>>>> rationals equally have no terminating expansion. Numbers such
> >>>>>>>> as "pi" and "sqrt(2)" are not defined as decimal expansions
> >>>>>>>> but via their properties [eg that "sqrt(2)" is the unique
> >>>>>>>> positive real whose square is 2, or equivalently that it is
> >>>>>>>> the ratio of the diagonal of a square to its side, and "pi"
> >>>>>>>> is the least positive real whose sine is zero]. Those
> >>>>>>>> properties are exact, and tell you all you ever need to know
> >>>>>>>> about those numbers.
> >>>>
> >>>>>> [ .... ]
> >>>>
> >>>>>>> What has decimal (base 10) expansion got to do with anything?
> >>>>>>> An irrational number has a non-terminating sequence in ANY
> >>>>>>> base. I am sorry but you are simply mistaken: irrational
> >>>>>>> numbers do NOT have an exact value; this is obvious to anyone
> >>>>>>> who understands logic and uses a sane definition for
> >>>>>>> infinity.
> >>>>
> >>>>>> That irrational numbers are exact values is clear to anybody
> >>>>>> with a degree in maths. Definitions of "infinity" (of which
> >>>>>> there are many) have nothing to do with this.
> >>>>
> >>>>> You are wrong and fractally so so your degree in maths appears
> >>>>> to be worthless.
> >>>>
> >>>> No, I am right, along with the world's other mathematics
> >>>> graduates. You are stuck in the distant (100s of years) past,
> >>>> when mathematicians were still puzzling over what you're
> >>>> puzzling over. The fundamentals of maths have been worked out,
> >>>> and you are in the position of an alchemist faced with modern
> >>>> chemistry.
> >>>
> >>> Pure assertion with NOTHING to back it up.
> >>>
> >>>>
> >>>>> An irrational number's sequence is statistically random, has no
> >>>>> fixed point on the number line ergo has no exact representation.
> >>>>> Any number with no exact representation has, by definition, no
> >>>>> exact value, only an approximation. Infinity has everything to
> >>>>> do with this as an irrational's sequence ("digits") never
> >>>>> terminates (i.e. it is an INFINITELY long sequence).
> >>>
> >>> Ignored this part I see.
> >>>
> >>> /Flibble
> >>>
> >>
> >> You are utterly clueless on these things.
> >
> > Projection.
> >
> >> The square-root of 2 does not jump around on the number line.
> >
> > Yes it does, all irrational numbers do.
> >
> > /Flibble
> >
>
> So, what other values does it jump to besides the actual value of
> sqrt(2)?

It jumps to the next slightly more accurate value as the approximation
increases in accuracy.

/Flibble