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

Discussion in 'Alternative Theories' started by chinglu, Oct 27, 2013.

1. ### TrippyALEA IACTA ESTStaff Member

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Promises promises.

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.

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3. ### chingluValued Senior Member

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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.

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5. ### chingluValued Senior Member

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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.

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7. ### Billy TUse Sugar Cane Alcohol car FuelValued Senior Member

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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."

8. ### Billy TUse Sugar Cane Alcohol car FuelValued Senior Member

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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.

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.

9. ### chingluValued Senior Member

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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.

10. ### chingluValued Senior Member

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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.

11. ### Quantum QuackLife's a tease...Valued Senior Member

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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?

12. ### arfa branecall me arfValued Senior Member

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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.

13. ### TrippyALEA IACTA ESTStaff Member

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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?

14. ### TrippyALEA IACTA ESTStaff Member

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As Arfa says...

15. ### chingluValued Senior Member

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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.

16. ### arfa branecall me arfValued Senior Member

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So, there is no induction in mathematics. Right.

17. ### chingluValued Senior Member

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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.

18. ### chingluValued Senior Member

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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.

19. ### TrippyALEA IACTA ESTStaff Member

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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?

20. ### chingluValued Senior Member

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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?

21. ### chingluValued Senior Member

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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.

22. ### arfa branecall me arfValued Senior Member

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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?

23. ### chingluValued Senior Member

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Sure, I know what it is.

#1577 proves the Taylor series is not an infinite addition.