1 is 0.9999999999999............

Undefined said:
Hi Trippy. I am running out of time again, so I will be as brief as I must.
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Promises promises.

Undefined said:
It's not your treatment's USAGE benefits I am pointing to; it is the logical NEED for it above and beyond the fundamental operations involved which give the initial long-division result/string.

See? I am only pointing out that all your 'overlay' for 'generic' case is merely re-formatting to more generic cases which lose sight of the fundamentals involved. Yes, they are convenient usages, and I already stated that your use of the 'powers' representation/treatment merely replicates the 'decimal location' involved in each part of the string. No biggie. No great 'revelation' or 'insight' gained from that 'overlay' of new formatting/manipulation. It does not actually do anything different except introduce more assumptive/formatting overlays which hide the simple operations which they are even MORE ABSTRACTLY referring to.

See? It always gets back to whether or not the UNITARY 'starting point' is used as part of the 'proof/treatment'....or if the convenient treatment you use gets us back to such a unity when YOU introduce the '9' to effectively RECONSTRUCT the facile/trivial unitary 'like/like 9/9 factor which, no surprise, in your treatment as designed WILL lead back to the 'result' YOU designed the abstraction/formatting system/treatment for. Circuitous. Since it does not 'prove' anything outside its own pre-determined result designed into your trivial treatments/overlays.

Look, Trippy, I have to log out soon. If we cannot get beyond the obvious misunderstandings between us in this aspect, then I humbly suggest we leave it between us as "agree to disagree"....for th reasons we each have respectively posted. Ok?

PS: Really, I have to log out very soon. :)
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You've failed to retain the context of the post I was replying to haven't you.

Also, inspite of all of your bluster and blither, you haven't answered the two very direct questions I asked you.
 
Trippy said:
What's to prove?

Please excuse the formatting of this first step if the columns don't quite line up.

$$
\hspace{37 pt}0.111\overline{1} \\
\frac{1}{9}= 9 \overline{\big) 1.0000...}\\
\hspace{37 pt}-9 \\
\hspace{42pt}\overline{\hspace{7 pt} 1}0 \\
\hspace{43 pt} -9 \\
\hspace{47 pt} \overline{\hspace{7}1}0 \\
\hspace{49 pt} -9 \\
\hspace{56 pt}\overline{\hspace{7}1}0 \\
\hspace{56 pt} -9 \\
\hspace{65 pt} \overline{\hspace{7}1...}
$$

$$ \frac{1}{9}= 0.111\overline{1}$$

$$ \frac{1}{9}=1 \times 10^{-1} + 1 \times 10^{-2} + 1 \times 10^{-3} + 1 \times 10^{-4} + ... + 1 \times 10^{-(n-1)} + 1 \times 10^{-n} + 1 \times 10^{-(n+1)}+ ...$$

$$ 9 \times \frac{1}{9}= 9 \times (1 \times 10^{-1} + 1 \times 10^{-2} + 1 \times 10^{-3} + 1 \times 10^{-4} + ... + 1 \times 10^{-(n-1)} + 1 \times 10^{-n} + 1 \times 10^{-(n+1)}+ ...)$$

$$ 9 \times \frac{1}{9}= 9 \times 10^{-1} + 9 \times 10^{-2} + 9 \times 10^{-3} + 9 \times 10^{-4} + ... + 9 \times 10^{-(n-1)} + 9 \times 10^{-n} + 9 \times 10^{-(n+1)}+ ...$$

$$ \frac{9}{9}= 0.999\overline{9}$$
Click to expand...

No problem with the lining up.

Trippy, long division is recursive. All recursive procedures are finite. You cannot generate a completed infinite sequence of digits using the division algorithm.

All you can make is a finite sequence of digits or steps for all n of a recursive algorithm. Therefore, you do not have an infinite sequence of digits as a consequence of the algorithm.

This is basic recursion theory.
 
Billy T said:
Math is not done by "common sense" nor by popular vote, nor by opinion, BUT BY PROOFS.

Several different proofs that 1 =0.999... have been presented. Some are even simpler than mine, but not as rigorous as require multiplying an infinitely long decimal expression to conclude that 1 and 0.999... are identical values, both rational and finite as they are the same , identical, number, only two different names for that same number. My proof, without need of multiplying an infinitely long decimal string, nor any limiting procedure that some valid proofs do use, and every step is logically derived from stated definitions, so it is quite rigorous.

In contrast you give ONLY your opinion and no supporting proof; further more you can not find any error in the proof I gave and I even number the steps for you to tell what step you did not think was valid.
SUMMARY: YOU HAVE ZERO UNDERSTANTING OF MATH (and very poor comprehension of English)

I will admit you now show a slight improvement in you written text: I. e. no longer write "infinite value 0.9999..." but still think that in same old false claim (only you ignorantly assert)* "that finite 1 can not equal infinite 0.999..." That is of course false as 0.999... is not infinite in value, only has an infinitely long decimal expression that has the value of very finite 1. (Much like 1/3 =0.3333.... has an infinitely long decimal expression for finite, rational fraction 1/3.)

* Still persisting in the ignorant claim that the value must be infinite if the decimal expression of it is infinitely long.
You don't even realize / understand that 1/4 = 0.25 is also infinitely long decimal expression given more correctly (no assumption about less significant decimal place / locations being zero needed as that is explicitly so stated.) as 1/4 = 0.25000000000000000000 ....
Click to expand...

See my response to RPenner.

There is no such thing in mathematics that permits infinite addition.

Addition is defined recursively, whether it is natural numbers or real numbers.

All recursive definitions only produce finite elements. They may continue on and on, but they always remain finite and never form an infinite sequence.

See, that is the task at hand for those that think they may use infinite addition. There is no such thing in mathematics at the foundations yet some use this claiming they have a valid proof.
 
chinglu said:
No problem with the lining up.

Trippy, long division is recursive. All recursive procedures are finite. You cannot generate a completed infinite sequence of digits using the division algorithm.

All you can make is a finite sequence of digits or steps for all n of a recursive algorithm. Therefore, you do not have an infinite sequence of digits as a consequence of the algorithm.

This is basic recursion theory.
Click to expand...
I think you are basically correct - it assumed, not proven, than this, what I would call an induction process, can be validly assumed to give the infinitely long string: 0.999... but in fact we only have proven it for finite lengths.

I.e. It is really the "limiting process" in disguise. We can do it for a very large number of places and then note that the part going on for an infinite number of places more has very small value so "in the limit" 0.999... is one."
 
chinglu said:
See my response to RPenner.
Click to expand...
which thread and number?
chinglu said:
There is no such thing in mathematics that permits infinite addition.
Click to expand...
Again I agree, if wanting to PROVE something, but usually not in practice a problem. I getting tired of "beating my own drum," but again my proof makes not use of infinitely long addition strings.

I'm going to bed now, but so I don't need to add a comment about division later, I'll note now, my use of it is only to claim a/a =1 and to indicate BUT NOT DO, division by the notation a/b, and even then only when a< b as I recall, but would need to check the proof to be sure.

My subtraction Always results in either zero when equals are subtracted or some positive power of the base less unity. I.e. in base 10 thing like 99999, or 99, etc.

SUMMARY: I don't do multiplications, don't add infinitely long strings (or even finite ones), take no limits, do very limited divisions and subtractions. Oh yes, I forgot to tell: I don't do windows either.
 
Billy T said:
which thread and number?Again I agree, if wanting to PROVE something, but usually not in practice a problem. I getting tired of "beating my own drum," but again my proof makes not use of infinitely long addition strings.
Click to expand...

Sure, #1532 for RPenner. He has not responded.

If you are interested, I can prove why any recursive definition/procedure/algorithm produces only finite length elements. This includes but is not limited to addition, division and multiplication.
 
Billy T said:
I think you are basically correct - it assumed, not proven, than this, what I would call an induction process, can be validly assumed to give the infinitely long string: 0.999... but in fact we only have proven it for finite lengths.

I.e. It is really the "limiting process" in disguise. We can do it for a very large number of places and then note that the part going on for an infinite number of places more has very small value so "in the limit" 0.999... is one."
Click to expand...

A limit is not actually the number the sequence becomes. So, this is not the answer. The sequence goes on and on but will not ever cross the limit. They are two different ideas.
 
Can I suggest that the number one is only a balloon full of decimal places?
You know .....the number one is a boundary for 0.999...

imagines a balloon full of o.999...
"the surface of the balloon is the number one."
so therefore 0.999... = 1 balloon &
1.999... = 2 balloons etc etc...

Who says that the number one is finite any way?
 
The only problem with long division then, is that we do it stepwise. But with 1 divided by 3, say, it's inductively true that the remainder given by each step is the same, and the partial result is the same. So we must conclude that, if the "process" continues forever the result is an infinite string, otherwise known as a repeating or recurring decimal.

Chinglu's claim: "all recursive procedures are finite", seems to be predicated on nothing more than some "ability" to perform only a finite number of steps. Numbers however, are not constrained by "abilities" of humans or machines they can build. At least, I don't think they are.
 
chinglu said:
No problem with the lining up.

Trippy, long division is recursive. All recursive procedures are finite. You cannot generate a completed infinite sequence of digits using the division algorithm.

All you can make is a finite sequence of digits or steps for all n of a recursive algorithm. Therefore, you do not have an infinite sequence of digits as a consequence of the algorithm.

This is basic recursion theory.
Click to expand...

So what you're saying is that Long division only works some of the time, becauyse you don't like the implications the rest of the time?

The point is that long division is sufficient to show us that 1/9 repeats without ending, which was the point of the exercise.

Unless you think that 10-9 can give you something other than 1, and 9 can go into 10 as something other than 1r1?
 
arfa brane said:
The only problem with long division then, is that we do it stepwise. But with 1 divided by 3, say, it's inductively true that the remainder given by each step is the same, and the partial result is the same. So we must conclude that, if the "process" continues forever the result is an infinite string, otherwise known as a repeating or recurring decimal.

Chinglu's claim: "all recursive procedures are finite", seems to be predicated on nothing more than some "ability" to perform only a finite number of steps. Numbers however, are not constrained by "abilities" of humans or machines they can build. At least, I don't think they are.
Click to expand...

As Arfa says...
 
arfa brane said:
The only problem with long division then, is that we do it stepwise. But with 1 divided by 3, say, it's inductively true that the remainder given by each step is the same, and the partial result is the same. So we must conclude that, if the "process" continues forever the result is an infinite string, otherwise known as a repeating or recurring decimal.

Chinglu's claim: "all recursive procedures are finite", seems to be predicated on nothing more than some "ability" to perform only a finite number of steps. Numbers however, are not constrained by "abilities" of humans or machines they can build. At least, I don't think they are.
Click to expand...

Your confusion is you do not understand the different between a recursive procedure and a number system. I simply said no recursive procedure produces an infinite set.

Now, if you can prove this false, then do it.

So, there is no infinite addition and there is no infinite division. So, there is no valid proof that .999...=1.
 
Trippy said:
So what you're saying is that Long division only works some of the time, becauyse you don't like the implications the rest of the time?

The point is that long division is sufficient to show us that 1/9 repeats without ending, which was the point of the exercise.

Unless you think that 10-9 can give you something other than 1, and 9 can go into 10 as something other than 1r1?
Click to expand...

I said division only produces a finite number of digits. You cannot take 1/9 using division and end up with an infinite number of digits for an answer.

That does not mean they do not go on and on. That is the old Greek concept of potential infinity. It means, you must prove you have a digit for every possible n, whatever that means. But, that is Cantor's completed infinity.

Recursion cannot get you to Cantor's completed infinity.
 
arfa brane said:
So, there is no induction in mathematics. Right.
Click to expand...

What does this have to do with what I said?

You can prove by induction binary addition is valid.

This does not prove infinite addition exists.

if you think you have a proof validating infinite addition, I would like to see it.
 
chinglu said:
I said division only produces a finite number of digits. You cannot take 1/9 using division and end up with an infinite number of digits for an answer.

That does not mean they go on and on. That is the old Greek concept of potential infinity. It means, you must prove you have a digit for every possible n, whatever that means. But, that is Cantor's completed infinity.

Recursion cannot get you to Cantor's completed infinity.
Click to expand...

I just did. I proved it by induction, although I didn't spell it out explicitly.

Again, we come back to the two questions I asked you.

Can you contradict the assertion?

Can you prove that 10-9 <> 1?

Can you prove that 10/9 <> 1r1?
 
Trippy said:
I just did. I proved it by induction, although I didn't spell it out explicitly.

Again, we come back to the two questions I asked you.

Can you contradict the assertion?

Can you prove that 10-9 <> 1?

Can you prove that 10/9 <> 1r1?
Click to expand...

You did not prove by induction that you created an infinite sequence using long division. Otherwise, state it specifically and prove you created a completed infinite sequence.

I think 10-9 = 1. I only need recursion to prove this.


10/9 <> 1r1. I don' know what you mean since for any n of the division algorithm, 10/9 = 1.1(n-1). That is all I can prove.

Again, can you prove you produced a digit for all possible natural numbers?
 
Why recursion cannot produce infinity

Here is the general recursive procedure that proves any recursive procedure does not produce a completed infinite number of steps.

Define R1 as a natural number register.
Define R2 as an unlimited stack register.

Use the following enhanced Turing algorithm.

R1 = 0;
R2 = empty;

Loop
R2.Push( R1 );
// Do your recursive step here like the nth division step
++R1;
Terminate if R2 contains every natural number.


Now, assume the algorithm terminates. Then R2 contains every natural number. Yet, before termination, we incremented R1, which means it is a finite successor ordinal. Then R2 does not contain R1 based on the algorithm and hence R2 does not contain every natural number, which is a contradiction.

So, no recursive procedure is infinite.
 
chinglu said:
if you think you have a proof validating infinite addition, I would like to see it.
Click to expand...
Do you know what a Taylor series is? My guess is you don't.
Can you prove a Taylor series supports your claim that infinite addition doesn't exist?
 
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