Originally posted by: ZeroNine8
It seems like what was proven is that the limit of 0.999... = 1 as the number of repeated digits goes to infinity. I have no problem believing that 1 is the limit of the convergent series 0.999..., however just because it is the limit does not mean it is equal. Because infinity isn't a number, merely a concept, you can't claim that 0.999... with infinite digits actually equals 1, only that it approaches it as the limit with more and more 9's.
In plenty of cases, the limit of a convergent series is never actually reached by that series.
The notation 0.999... has nothing to do with a finite series. The fact that some finite subseries of that series is not equal to 0.999... is irrelevant. Your defintion of "actuality" as synonymous with "finiteness" makes the notation 0.999... invalid, as it is defined as an infinite geometric sum. You have abdicated the right to use that notation, so objecting to the value of it is equivalent to arguing about how many angels fit on the head of a pin.Originally posted by: ZeroNine8
It seems like what was proven is that the limit of 0.999... = 1 as the number of repeated digits goes to infinity. I have no problem believing that 1 is the limit of the convergent series 0.999..., however just because it is the limit does not mean it is equal. Because infinity isn't a number, merely a concept, you can't claim that 0.999... with infinite digits actually equals 1, only that it approaches it as the limit with more and more 9's.
In plenty of cases, the limit of a convergent series is never actually reached by that series.
Originally posted by: Wiktor
Ok, I clicked your link, now maybe I'll even read it, but still IMO (lol)
the 999.../1000... number (decimal: 0.999...) is just as good as 1/1000... (decimal: 0.000...1).
So if you say "There is NO NUMBER with an infinite number of zeros followed by a 1" (not that I disagree), then I ask what is this number: 1/1000... in decimal.
See what my problem is? Whatever you say about 0.000.....1 - same goes for 0.999... just because the notation makes more sense for 999.../1000... doesn't mean it's any better than 1/1000...
I claim that 0.999... and 0.0....1 have a lot more in common than 0.999... and 0.333... No?
(And the whole point of discussing this is that if 0.0....1 exists then 0.9... does not equal 1)
Originally posted by: ZeroNine8
In your 'proof' you claim that for ANY N>0, .99...+.1^N>1 and claim that this is a strict inequality. However, since you are allowed to pick .99... then I'm allowed to pick N=infinity. In this case, you claim that 1>1 therefore, your assertion that the strict inequality holds for any N>0 fails, does it not? After this point, being able to nest .99... between 1+.1^N and 1-.1^N also fails, doesn't it? By my reasoning this allows 1-.1^N<=.99..., thus you have not constrained it to 1, only .99... or 1.
If this is not the case, what is the explanation for when N=infinity?
Consequently, by restricting N to real numbers (not infinity), you also kind of have to restrict .99... to some finite amount of digits at which point it's easy to prove .99... != 1
Originally posted by: RossGr
N is restricted to the set of counting numbers, it must be specfied, specify a vaild N, infinity is not a counting number. The heart of the matter is that it is true for ANY specified N. This is what it makes it work and is one of the key concepts of the proof.
It ties into the fact that you must be able to specify the postion of every digit in a real number. This is another reason that the concept of an infinte number of zeros followed by a 1 is not a vaild number. You must be able to specify the value of the digit in the Nth position. For .999.... it is is easy if d(n) is the digit in the nth postion of the number I have d(n)=9 for all n. This is a basic part of the definintion of real numbers as an infinite geometric series. For the NON number .000...(infinity)1, you cannot find a d(n)=1 because you have already said d(n)=0 for all n. That is what is meant by an infinite number of something.
Originally posted by: ZeroNine8
edit:Originally posted by: RossGr
N is restricted to the set of counting numbers, it must be specfied, specify a vaild N, infinity is not a counting number. The heart of the matter is that it is true for ANY specified N. This is what it makes it work and is one of the key concepts of the proof.
It ties into the fact that you must be able to specify the postion of every digit in a real number. This is another reason that the concept of an infinte number of zeros followed by a 1 is not a vaild number. You must be able to specify the value of the digit in the Nth position. For .999.... it is is easy if d(n) is the digit in the nth postion of the number I have d(n)=9 for all n. This is a basic part of the definintion of real numbers as an infinite geometric series. For the NON number .000...(infinity)1, you cannot find a d(n)=1 because you have already said d(n)=0 for all n. That is what is meant by an infinite number of something.
How do you justifying not allowing N=infinity because infinity isn't a real counting number (I don't disagree) yet allowing your .99... to have infinite 9's? How do you have a Non-number'th digit? Because we're playing with the concept of infinity, it seems to me that you have just as much right to say d(n<infinity)=9 and d(infinity)=9 as I do to say d(n<infinity)=0 and d(infinity)=1. If you say that d(infinity) doesn't exist, then .99... must be truncated at some finite value and can easily be proven !=1, yet if you allow d(infinity) to exist then I have just specified .000...1, haven't I?
yet if you allow d(infinity) to exist then I have just specified .000...1, haven't I?
Originally posted by: ZeroNine8
Actually, the way I see it, you get
0 < 1-.99...<= 0.00...1
thus have not constrained .99...=1 yet.
.99... + .00...1 = 1 (infinite'th 9 and infinite'th place 1 added together)
my argument is that if I can specify some .00...1 such that .00...1 +.99... =1, then .99...!=1. I believe I have specified such a .00...1 using your assumptions.
Originally posted by: rjain
0.00...1 is the limit as n -> infinity of 10^-n. This is 0.
ZeroNine, you have just proven that .99... = 1. Congrats.
No, because the "..." notation is shorthand for a limit.Originally posted by: ZeroNine8
uh, right, because limit is the same as equals...yeah. I thought we covered that.Originally posted by: rjain
0.00...1 is the limit as n -> infinity of 10^-n. This is 0.
ZeroNine, you have just proven that .99... = 1. Congrats.
Originally posted by: ZeroNine8
Well, after looking over it again and having a chance to brush up on my proof logic, indeed you can prove that .99... =1.
To everyone, like me, who think this just can't possibly be true, just take comfort in the fact that it only happens because 0.99... isn't a real number. Because .99... is not a real number, the way it relates to real numbers doesn't exactly make sense with what is intuitive. This proof works because of the nature of the concept of infinity and the fact that .99... does not exist as a real number (.99... is neither rational nor irrational, thus is not a real number, the set defined by the union of all rational and irrational numbers). Ahh, the quirky things you can do with infinity.
.99... is a real number because it is the sum of rational numbers (therefore it is a rational number, moreover). A real number cannot equal an imaginary number and a rational number cannot equal an irrational number. None of this is quirky, you just don't understand the notation being used. Please get over your ego and accept that you might have to learn something some time in your life.Originally posted by: ZeroNine8
Well, after looking over it again and having a chance to brush up on my proof logic, indeed you can prove that .99... =1.
To everyone, like me, who think this just can't possibly be true, just take comfort in the fact that it only happens because 0.99... isn't a real number. Because .99... is not a real number, the way it relates to real numbers doesn't exactly make sense with what is intuitive. This proof works because of the nature of the concept of infinity and the fact that .99... does not exist as a real number (.99... is neither rational nor irrational, thus is not a real number, the set defined by the union of all rational and irrational numbers). Ahh, the quirky things you can do with infinity.
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Originally posted by: ZeroNine8
In your 'proof' you claim that for ANY N>0, .99...+.1^N>1 and claim that this is a strict inequality. However, since you are allowed to pick .99... then I'm allowed to pick N=infinity. In this case, you claim that 1>1 therefore, your assertion that the strict inequality holds for any N>0 fails, does it not? After this point, being able to nest .99... between 1+.1^N and 1-.1^N also fails, doesn't it? By my reasoning this allows 1-.1^N<=.99..., thus you have not constrained it to 1, only .99... or 1.
If this is not the case, what is the explanation for when N=infinity?
Consequently, by restricting N to real numbers (not infinity), you also kind of have to restrict .99... to some finite amount of digits at which point it's easy to prove .99... != 1
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N is restricted to the set of counting numbers, it must be specfied, specify a vaild N, infinity is not a counting number. The heart of the matter is that it is true for ANY specified N. This is what it makes it work and is one of the key concepts of the proof.
It ties into the fact that you must be able to specify the postion of every digit in a real number. This is another reason that the concept of an infinte number of zeros followed by a 1 is not a vaild number. You must be able to specify the value of the digit in the Nth position. For .999.... it is is easy if d(n) is the digit in the nth postion of the number I have d(n)=9 for all n. This is a basic part of the definintion of real numbers as an infinite geometric series. For the NON number .000...(infinity)1, you cannot find a d(n)=1 because you have already said d(n)=0 for all n. That is what is meant by an infinite number of something.
Sorry no point for stating the obvious. That is pretty much the definition of .999... that I have been pushing. It is a simply fact that every postion of a real number is finitly indexable.Originally posted by: sao123
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Originally posted by: ZeroNine8
In your 'proof' you claim that for ANY N>0, .99...+.1^N>1 and claim that this is a strict inequality. However, since you are allowed to pick .99... then I'm allowed to pick N=infinity. In this case, you claim that 1>1 therefore, your assertion that the strict inequality holds for any N>0 fails, does it not? After this point, being able to nest .99... between 1+.1^N and 1-.1^N also fails, doesn't it? By my reasoning this allows 1-.1^N<=.99..., thus you have not constrained it to 1, only .99... or 1.
If this is not the case, what is the explanation for when N=infinity?
Consequently, by restricting N to real numbers (not infinity), you also kind of have to restrict .99... to some finite amount of digits at which point it's easy to prove .99... != 1
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N is restricted to the set of counting numbers, it must be specfied, specify a vaild N, infinity is not a counting number. The heart of the matter is that it is true for ANY specified N. This is what it makes it work and is one of the key concepts of the proof.
It ties into the fact that you must be able to specify the postion of every digit in a real number. This is another reason that the concept of an infinte number of zeros followed by a 1 is not a vaild number. You must be able to specify the value of the digit in the Nth position. For .999.... it is is easy if d(n) is the digit in the nth postion of the number I have d(n)=9 for all n. This is a basic part of the definintion of real numbers as an infinite geometric series. For the NON number .000...(infinity)1, you cannot find a d(n)=1 because you have already said d(n)=0 for all n. That is what is meant by an infinite number of something.
Based on this fact alone...your theory has a hole.
If in .1^N, and N must be a finite number and must be restricted to the set of counting numbers, then the same must apply to your .9999....
I can define .999.... as 9 * E( .1^N ) {N > 0} N here must also be a finite number.
9 * E( .1 ^ N) [from 1 to N] {N > 0} + (.1^N) = 1
3 points please...
This is where you go wrong, it is not necessary to prove .1^N = 0, all I need to do is show that the relationship holds for all N.A) Using the identity laws of addition... in order for your claim to be true then (.1^N) must be zero. However because N is a finite number, the boundries of this value are [1,0) (notice that zero is noninclusive.) Therefore your claim is false.
B) If you indeed chose to use N = infinity or any other nonfinite value... then your equation reduces to
9 * 0 + 0 = 1
0 = 1
which is also false.
C) In order for your number to be a rational number it must fit the following criteria...
Any rational number must be able to be written in the form of one integer divided by another integer. Knowing that all integers are countable and finite,
In your case of .99999999..... repeating forever. What are your unique finite integers producing this combination?
Im not totally convinced its rational yet either.
Originally posted by: sao123
yes...with the little cheat.
what you then forget to mention it that there is exactly a remainder of
(.1 ^ N) with N being the number of repetitions carried out in your division.
Further demonstrating my hole in your theory.
Anything else?
I think your claim is over.