For the 10^n th time. .999... does NOT represent a limit, it represents an infinte number of 9's the limit has been taken. If you would please take the time to click on the link in my sig, should be obvious which one! You will see how mathematicians deal with this number.
For the 10^n th time. .999... does NOT represent a limit, it represents an infinte number of 9's the limit has been taken. If you would please take the time to click on the link in my sig, should be obvious which one! You will see how mathematicians deal with this number.
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.
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)
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)
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
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: 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...
(And the whole point of discussing this is that if 0.0....1 exists then 0.9... does not equal 1)
How many times must I respond to this same idea? Please read my previous post.
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?
So if you take N= infinty, rearrange the inequality and you get that 0<1-.999.... < 0 this is a statement of exact equality , where is the problem? It is not necessary to take N to infinity, and not recomended, since infinity is not a valid integer, but it does not, as you seem to think, create a contradiction.
No, as above. the one MUST occupy a specified position there must be some n such that d(n)=1 that n is finite. Once again the whole point of this proof is to avoid the use of questionable "infinitly" large values, As soon as you bring this to the table you are in fishy territory. In my proof I lump the infinite tail of 9's into a single value and use only my knowledge that it is greater then zero. Why do you wish to introduce dicy quanitys when I have gone to great effort to avoid them.
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.
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.
0.00...1 is the limit as n -> infinity of 10^-n. This is 0.
ZeroNine, you have just proven that .99... = 1. Congrats.
ZeroNine, you have just proven that .99... = 1. Congrats.
So until you get past this bad concept that infinity is simply a very large finite number you will not be able to understand your error.
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.
uh, right, because limit is the same as equals...yeah. I thought we covered that.
In this case, the limit at infinity *does* equal zero. I think. I'm a CS major because I like to have the computer do the math for me.
Anyhow, here's a nice pattern for those of you who dislike algebra:
1/9 = 0.11111...
2/9 = 0.22222...
3/9 = 1/3 = 0.33333...
4/9 = 0.44444...
5/9 = 0.55555...
6/9 = 2/3 = 0.66666...
7/9 = 0.77777...
8/9 = 0.88888...
9/9 = 1/1 = 1 = 0.99999...
It's not really a good proof, but it came up in math class once when my high school calc teacher was proving this.
And I paraphrase badly my teacher's explanation, but the problem with "An infinite number of zeroes followed by a one" is that, by definition, that number is closer to zero than all the real and rational numbers, and so it must be equal to zero.
Anyhow, here's a nice pattern for those of you who dislike algebra:
1/9 = 0.11111...
2/9 = 0.22222...
3/9 = 1/3 = 0.33333...
4/9 = 0.44444...
5/9 = 0.55555...
6/9 = 2/3 = 0.66666...
7/9 = 0.77777...
8/9 = 0.88888...
9/9 = 1/1 = 1 = 0.99999...
It's not really a good proof, but it came up in math class once when my high school calc teacher was proving this.
And I paraphrase badly my teacher's explanation, but the problem with "An infinite number of zeroes followed by a one" is that, by definition, that number is closer to zero than all the real and rational numbers, and so it must be equal to zero.
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.
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.
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.
Udder nonsense.
If .99... is not a real number then what is? Do you have your own definiton? It is not magic, it is simply a facet of the construction of the real number line.
.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.
sao123
Lifer
- May 27, 2002
- 12,653
- 205
- 106
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...
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.
The reason I was thinking it wasn't a rational number was obviously wrong, as you so kindly pointed out, so it is just an interesting infinite decimal construct in math. If it were easy and intuitive it wouldn't make a good sig I guess
edit: (Though I am still not completely 100% convinced for the same reasons mentioned in the above post)
edit: (Though I am still not completely 100% convinced for the same reasons mentioned in the above post)
Though, on the topic of .99... being rational, humor me if you would.
There are no finite integers that can, themselves, be divided to give .99... so the next argument is that .99... is the sum of rational numbers, but you still cannot express it as a finite sum of rational numbers. What is the justification for the argument holding true for an infinite sum vs. a finite sum? I can't get past the fact that you can claim .99... is rational yet are unable to define two integers that the rational is constructed from, which seems to contradict the fact that it is rational.
I have read, in several places, that an irrational number can be expressed as the infinite sum of rational numbers, so the fact that it is an infinite sum of rationals does not justify that it is actually a rational.
There are no finite integers that can, themselves, be divided to give .99... so the next argument is that .99... is the sum of rational numbers, but you still cannot express it as a finite sum of rational numbers. What is the justification for the argument holding true for an infinite sum vs. a finite sum? I can't get past the fact that you can claim .99... is rational yet are unable to define two integers that the rational is constructed from, which seems to contradict the fact that it is rational.
I have read, in several places, that an irrational number can be expressed as the infinite sum of rational numbers, so the fact that it is an infinite sum of rationals does not justify that it is actually a rational.
sao123
Lifer
- May 27, 2002
- 12,653
- 205
- 106
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.
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.
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.
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.
That is your equation not mine, you need to reexamin what I did, seems like you are neglecting a lot of terms in the sum. When I let N go to infinity I get 0 + .999.... + 0 =1.
Perhaps you need to brush up on sigma notation.
See my previous post which answers this question.
Wrong, My proof is valid, your objections don' t hold any water.
Edit:
I guess 1/3 !=.333... cause there is always a remainder.
Try again.
DrPizza
Administrator Elite Member Goat Whisperer
For the love of God...
.999 repeating *IS* a rational number...
a rational number means: "can be written as a ratio"
a repeating decimal is a rational number.
Are some of you arguing that .333repeating does not equal 1/3 ?!?
.999... is a repeating decimal. ALL repeating decimals are rational numbers.
If your intuition tells you that .999 repeating doesn't = 1, well, your intuition is wrong. Get over it.
And, for the next person who posts that .999... <> 1 for the reason of "IMO" or "IMHO," I hope you get struck by lightening.
I wish I could say this for the final time... .999 repeating EQUALS EXACTLY 1.
RossGr (and others who know what they're talking about) are correct.
For those of you without the math background, that would be like me arguing with someone fluent in Japanese about the definition of some word in Japanese. Only an idiot continues to argue that he's right when countless people have shown valid proof that he's wrong. Some of you are as bad as those people who still believe the earth is flat and moon landings are a big conspiracy.
/end rant
.999 repeating *IS* a rational number...
a rational number means: "can be written as a ratio"
a repeating decimal is a rational number.
Are some of you arguing that .333repeating does not equal 1/3 ?!?
.999... is a repeating decimal. ALL repeating decimals are rational numbers.
If your intuition tells you that .999 repeating doesn't = 1, well, your intuition is wrong. Get over it.
And, for the next person who posts that .999... <> 1 for the reason of "IMO" or "IMHO," I hope you get struck by lightening.
I wish I could say this for the final time... .999 repeating EQUALS EXACTLY 1.
RossGr (and others who know what they're talking about) are correct.
For those of you without the math background, that would be like me arguing with someone fluent in Japanese about the definition of some word in Japanese. Only an idiot continues to argue that he's right when countless people have shown valid proof that he's wrong. Some of you are as bad as those people who still believe the earth is flat and moon landings are a big conspiracy.
/end rant
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