0.99999.....=1

@markr,

Quote:

The problem is that you're running into a contradiction and assuming that lim n->∞ 1/10^n = 0 is what's wrong. I'm not saying that every printed word is correct, but if every modern textbook tells you that the limit is zero, then you might want to start questioning some of your other assumptions/techniques...

The contradiction is that when a string of length 1 is cut into lim n->∞ 10^n pieces of length lim n->∞ 1/10^n, the individual pieces by the limit: lim n->∞ 1/10^n must equal zero length. When these pieces are summed to get back the original length we get zero for the total length. There are two things that contradiction depends on:

1.) That an infinite summation of zeros is zero. If this is not true, then there is no contradiction.

2.) If the limit: lim n->∞ 1/10^n = 0. If this is not true, and the limit represents some infinitely small quantity, then there is no contradiction.

I cannot in my mind find anyway to justify 1.) not being true. At least one of the terms in the infinite sum would have to have a value greater than zero for the sum to be greater than zero.

If the limit in 2.) was an infinitely small number instead of an absolute zero then it would make sense to build up a finite length with infinite many infinitely small pieces.

1 Reply

@tomr,

3) The application of limits in the infinite sum is incorrect.

"If the limit in 2.) was an infinitely small number instead of an absolute zero then it would make sense to build up a finite length with infinite many infinitely small pieces."

That's just it. Each term in the infinite sequence is non-zero. That's why when you multiply the nth term by 10^n, you always get 1. The terms never actually reach zero, but they get arbitrarily close. They get so close that they become indistinguishable from zero - you can't come up with a number that is greater than zero and less than all of the terms in the sequence. Therefore the "limit" is zero.

Your mistake is assigning the value of zero to the terms before multiplying by their count. Each term in the "string length" sequence consists of a count of pieces multiplied by the length of each piece. Yes, at infinity you can't distinguish the length from zero, but you're still multiplying it by a huge number and that product is always 1.

1 Reply

@markr,

Quote:

3) The application of limits in the infinite sum is incorrect.

So the problem you have is with the limit law that says: the limit of sums is equal to the sum of the limits.

Sum Law for limits: lim x->a [f(x) + g(x)] = lim x->a f(x) + lim x->a g(x)

You want me to only use the limit:

lim n->∞ ∑ 1/10^n = lim n->∞ [1/10^n + 1/10^n + 1/10^n + ...] = lim n->∞ 10^n/10^n = lim n->∞ 1 = 1.

and not use:

lim n->∞ ∑ 1/10^n = lim n->∞ 1/10^n + lim n->∞ 1/10^n + lim n->∞ 1/10^n + lim n->∞ 1/10^n +... = 0 + 0 + 0 + 0 +... = 0

Why? Beside the fact that it makes the math work out. There are no exceptions or conditions for the Sum Law of limits. Notice also how the first limit completely avoids any use of the limit and the answer works out the way it should because algebra took care of the work, while the second method uses the sum law which forces us to actually apply a limit in the problem to determine the size of individual pieces and the answer is wrong.

Quote:

That's just it. Each term in the infinite sequence is non-zero. That's why when you multiply the nth term by 10^n, you always get 1. The terms never actually reach zero, but they get arbitrarily close. They get so close that they become indistinguishable from zero - you can't come up with a number that is greater than zero and less than all of the terms in the sequence. Therefore the "limit" is zero.

I am in complete agreement with everything you say here. If that was the same explanation you would give to someone who had questions about whether 1 = .999... then there would be no controversy. For instance, when I gave the expression .999... = 1 - lim n->∞ 1/10 ^n, simply acknowledge that there is something there but it is arbitrarily close to zero and we have no way to represent the difference. So for all practical purposes the difference is zero and we say the numbers are the same.

I am glad we may have come to something we can agree about on this topic.

2 Replies

@tomr,

Your infinite sum is still computing the wrong thing. Let's take a look at some of the early terms in the sequence (i.e. let's let n be small instead of approaching infinity).
n = 1 -> 1/10
n = 2 -> 1/100 + 1/100 = 2/100
n = 3 -> 1/1000 + 1/1000 + 1/1000 = 3/1000
n = m -> m/10^m
As I stated in an earlier post, you're computing lim n->∞ (n / 10^n).
And as I stated then, that is zero.
You've incorrectly manipulated the product into an infinite sum.

So, if we're in agreement on the limit, you've got to find another explanation for the contradiction. I've just provided it.

1 Reply

@tomr,

By the way, I think Kolyo did a fine job of explaining it multiple ways. Unfortunately, he was trying to explain it to someone (not you) who kept making it a finite problem and felt compelled to link it to political leaning.

I think you and I went down this rat hole because of the contradiction you encountered with your string example and your insistence that your technique is valid, and therefore the claimed value of the limit must be wrong. It remains to be seen if you agree with that.

1 Reply

@markr,

You are having trouble adding lim n->∞ 10^n pieces. So let me try to show that it is in fact the limit that is causing the contradiction and not some error in the way I am computing the sum.

Let us instead of cutting our string of length 1 into 10 ^n pieces, cut the string into n pieces as n->∞. So now each individual piece is lim n->∞ 1/n in length.

Now take the infinite sum by the two methods above:

lim n->∞ ∑ 1/n = lim n->∞ (1/n + 1/n + 1/n + 1/n + ...) = lim n->∞ (n/n) = lim n->∞ 1 = 1. (by Algebra)

lim n->∞ ∑ 1/n = lim n->∞ 1/n + lim n->∞ 1/n + lim n->∞ 1/n + lim n->∞ 1/n + ... = 0 + 0 + 0 + 0 +... = 0. (by the Sum Law of Limits)

1≠0.

The exact same thing that I showed you when we index to 10^n happens when we index to n as n->∞. You have to admit that I am doing the infinite sum correctly now. There are no exceptions to the Sum Law. So what is going on here?

Quote:

As I stated in an earlier post, you're computing lim n->∞ (n / 10^n).
And as I stated then, that is zero.
You've incorrectly manipulated the product into an infinite sum.

Do you still feel this way now?

Quote:

So, if we're in agreement on the limit, you've got to find another explanation for the contradiction. I've just provided it.

I am unsure what we are in agreement of. You will admit that the limit comes arbitrarily close but not equal to zero, but at the same time you say there is no contradiction. You have not proved anything because I was not indexing to n in the previous example. I was indexing to 10^n as n->∞ in the infinite sum before because thats how many pieces we had. It does not matter how many pieces of string there are though because if every term is the same and all terms evaluate to zero then the result of the sum must be zero. Surely you see that.

2 Replies

@markr,

Quote:

By the way, I think Kolyo did a fine job of explaining it multiple ways. Unfortunately, he was trying to explain it to someone (not you) who kept making it a finite problem and felt compelled to link it to political leaning.

I think you and I went down this rat hole because of the contradiction you encountered with your string example and your insistence that your technique is valid, and therefore the claimed value of the limit must be wrong. It remains to be seen if you agree with that.

I went back and read Kolyo's posts. I think Kolyo should come back and talk more about the hyperreals. I have seen that notation before but never completely understood how it works. I think it would be interesting as Kolyo suggested to see if there was a way using hyperreal numbers to show there is a difference between 1 and .999...

1 Reply

@tomr,

Sum Law: Just because a law is true for a finite number of terms doesn't *necessarily* mean that it is true for an infinite number of terms.

Again, I would argue that you're mishandling limits by separating n from 1/n in the computation of the string length. Take a look at the early terms of your new sequence:

1: 1/1 = 1
2: 1/2 + 1/2 = 1
3: 1/3 + 1/3 + 1/3 = 1
m: m*(1/m) = 1

Now, at what point does this sequence of ones suddenly decide to jump to zero (or any number other than 1)? This isn't an asymptotic curve that is approaching zero or anything other than 1. It's a flat line. You continue to make assumptions, arrive at a contradiction, and blame the one fact that every sane mathematician will agree on. Open your mind to the fact that something else is wrong with what you're assuming/doing.

If this doesn't demonstrate that your method is incorrect, then I don't know what will:

Surely you agree that there are functions whose limit as x approaches infinity is something other than zero - either a finite value or infinity (e.g. f(x) = 1 or f(x) = x^2).

But wait, that's not true. For if we take the limit of x*f(x)/x (which is really still f(x) if x != 0) and apply tomr's Infinite Sum Law of Limits (for which there are no exceptions), we get an infinite sum of zeros when we separate x*f(x) from 1/x.

Therefore, the limit of all functions is zero as x approaches infinity. QED

--------
"You will admit that the limit comes arbitrarily close but not equal to zero..."

No, the limit is zero - period. Individual terms (finite) are arbitrarily close to zero, but the limit IS zero. Look up the definition of limit.

1 Reply

@markr,

Quote:

Sum Law: Just because a law is true for a finite number of terms doesn't *necessarily* mean that it is true for an infinite number of terms.

You are absolutely right. It doesn't necessarily mean that. But you should be able to give me a reason (or a proof) why I can't. And there is none except that you do not get the answer you expect. It is the fault of the limit process equating things to zero that are infinitely small. No problem will ever occur in finite sums because it takes an infinite number of terms to make that infinitesimal difference between the limit value and the actual value apparent.

Quote:

If this doesn't demonstrate that your method is incorrect, then I don't know what will:

Surely you agree that there are functions whose limit as x approaches infinity is something other than zero - either a finite value or infinity (e.g. f(x) = 1 or f(x) = x^2).

But wait, that's not true. For if we take the limit of x*f(x)/x (which is really still f(x) if x != 0) and apply tomr's Infinite Sum Law of Limits (for which there are no exceptions), we get an infinite sum of zeros when we separate x*f(x) from 1/x.

Therefore, the limit of all functions is zero as x approaches infinity. QED

I definitely see the absurdity of the conclusion you are coming to. Yes you could manipulate any function in an infinite series by multiplying x/x and then pulling f(x)*x out of the sum. But you would need to use the product rule for limits as well. I am not saying that makes it any different than what I described before but it is an extra step.

In math, you have to be able to justify why something works the way it does. If you are given rules, and then you make exceptions to those rules you need to be able to say why. Beside the fact that you do not get the expected answer when you apply those rules to a new class of problems. There just is not any satisfactory explanation why the Sum Law must stop with finite numbers, except that it only works for finite numbers and we end up getting into a bunch of trouble when you apply it to infinite sums (for the reason that the limit is actually a value larger than zero, though we cannot represent its value).

I would never actually use the Sum Law in the way I did if I was not trying to show a contradiction. I know it gives the wrong answer.

Quote:

"You will admit that the limit comes arbitrarily close but not equal to zero..."

No, the limit is zero - period. Individual terms (finite) are arbitrarily close to zero, but the limit IS zero. Look up the definition of limit.

I know the definition of a limit:

lim x->a f(x) = L
The limit of f(x), as x approaches a, equals L. We can make the values of f(x) arbitrarily close to L by taking x to be sufficiently close to a but not equal to a.

We can get into epsilon/delta definitions if needed. But I am unsure now what you mean when you say the individual terms (finite) are arbitrarily close to zero. Do you mean for "term" something like lim n->∞ 1/n is arbitrarily close to zero, but the value after you evaluate the limit is zero?

3 Replies

@tomr,

Quote:

In math, you have to be able to justify why something works the way it does. If you are given rules, and then you make exceptions to those rules you need to be able to say why. Beside the fact that you do not get the expected answer when you apply those rules to a new class of problems. There just is not any satisfactory explanation why the Sum Law must stop with finite number [of terms], except that it only works for finite number [of terms] and we end up getting into a bunch of trouble when we apply it to infinite sums (for the reason that the limit is actually a value larger than zero, though we cannot represent its value).

I edit my previous post late.

0 Replies

@tomr,

tomr wrote:

Now take the infinite sum by the two methods above:

lim n->∞ ∑ 1/n = lim n->∞ (1/n + 1/n + 1/n + 1/n + ...) = lim n->∞ (n/n) = lim n->∞ 1 = 1. (by Algebra)
correct

lim n->∞ ∑ 1/n = lim n->∞ 1/n + lim n->∞ 1/n + lim n->∞ 1/n + lim n->∞ 1/n + ... = 0 + 0 + 0 + 0 +... = 0. (by the Sum Law of Limits)
wrong, see explanation below

The "Sum Law" (Addition Rule) works for finite sums. It certainly does not allow you to switch the order of

lim n->∞ ...and... ∑(i = 1 to n)

because the number of summands is not fixed and finite; it is approaching ∞ as n approaches infinity.

***

If the move you've just described were legal, Calculus would not work, because you could "prove" every integral was equal to zero.

2 Replies

@tomr,

tomr wrote:

markr wrote:
Sum Law: Just because a law is true for a finite number of terms doesn't *necessarily* mean that it is true for an infinite number of terms.

You are absolutely right. It doesn't necessarily mean that. But you should be able to give me a reason (or a proof) why I can't.

Tom, you yourself have provided the proof.

You've disproven the infinite case of the "sum rule" with a counter-example!

0 Replies

@tomr,

"Do you mean for "term" something like lim n->∞ 1/n is arbitrarily close to zero, but the value after you evaluate the limit is zero?"

By individual terms, I mean pick a finite (as large as you want) value for n. The resulting "term", 1/n, will never be zero, but you can get as close to zero as you want. Although no result is ever zero, the limit is zero - per the definition of limit.

0 Replies

@Kolyo,

Ah, some relief - finally. I'm done with this one...

1 Reply

@OmSigDAVID,

OmSigDAVID wrote:

.OOOOOOOOOOOOOOO1 lies between them.

That is not enough zeros. The discussion was about an infinite number of 9s, not 16 9s.

0 Replies

@Kolyo,

Quote:

The "Sum Law" (Addition Rule) works for finite sums. It certainly does not allow you to switch the order of

lim n->∞ ...and... ∑(i = 1 to n)

because the number of summands is not fixed and finite; it is approaching ∞ as n approaches infinity.

***

If the move you've just described were legal, Calculus would not work, because you could "prove" every integral was equal to zero.

lim n->∞ ∑(i = 1 to 3) 1/n can equal = lim 1/n + lim 1/n + lim 1/n.
But lim n->∞ ∑(i = 1 to ∞) 1/n cannot equal ≠ lim n->∞ 1/n + lim n->∞ 1/n + lim n->∞ 1/n + ...

Now ask why this is so.

Are you resting the blame of getting the wrong answer with my application of the Sum Law of limits to the infinite sum? If so, you could have just as easily put the blame with the limiting process itself or even to the concept of an infinite sum. Without a reason to explain why I am not allowed to apply the Sum Law to an infinite sum, beside the fact that the math doesn't work, you do not know where to put the blame here. You automatically assume it is with the application of the limit law to the infinite sum but it could just as easily be the limit itself that is producing the incorrect result. So we all agree that the result of using an infinite sum and the Sum Law gives us wrong answers, it is now a matter of pinpointing the problem. Saying that I am not allowed to do as I did merely by stating that Sum Law can only be used for finite sums is a statement that is good if all you are interested in is producing correct results. However if you are investigating what actually is going wrong it is not satifactory. Do you see that distinction?

About the calculus not working, you are right and it is an absurd idea that every integral equals zero. I think I had Riemann Sums in mind when I started thinking along these lines.

By the way is there a hyperreal notation for the lim->∞ 1/n. I'd be curious to see if converting the problem to hyperreal notation changes the result.

0 Replies

@markr,

Hey you're alright. Thanks for hanging in there..

0 Replies

@tomr,

tomr wrote:

I think Kolyo should come back and talk more about the hyperreals. I have seen that notation before but never completely understood how it works. I think it would be interesting as Kolyo suggested to see if there was a way using hyperreal numbers to show there is a difference between 1 and .999...

Maybe later when I've read more about them myself. The only mention I've seen of hyperreals was that online paper I linked to from this thread. And I only read through about 70% of the paper myself.

1 Reply

@Kolyo,

I did not realize you had left a link. I will check it out also.

0 Replies

@Kolyo,

Well, now I see why you got so mad at my comment about mathematics. You are good at math. I haven't read through your explanation of why .99999999999 repeated = 1, but from what I remember, oh gosh, 15 years ago, about limits, it approaches 1 but never actually reaches 1. For practical purposes, it's so close to 1 that you could use it as 1 in most mathematical problems, but that still doesn't actually makes it one.
I'm just basing this off of what I remember about limits. The limit of 1/x, as x approaches infinity is zero, and as x approaches zero the limit is infinity, but x can never actually reach infinity, and x can never actually be zero, it's undefined...anyways.
It seems to me if we took the limit of a function that, when you took x to infinity, was 1, that would mean the limit is 1, but in reality the function never actually is exactly 1, (because you can never actually count to infinity). It just approaches 1.
I will read what you wrote though and research limits. I could be remembering limits incorrectly.

1 Reply

0.99999.....=1

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