Why Do People Say That 1 and .999 Are Equal? - Comments

[Mark44]

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Yes, what I meant is for a specific Riemann sum, the widths are uniformly bounded; the function Min width is a continuous function on a compact set -- integration is done on a compact interval [a,b] -- so it assumes an actual minimum value on [a,b]. Now, we can consider what happens as we let the widths go to 0, but for a _specific_ Riemann sum, the widths do not go to 0.

Sure they do. Assuming for simplicity's sake that $\Delta x = \frac{b - a} n$, as $n \to \infty, \Delta x \to 0$. There is no positive minimum value that's a lower bound on $\Delta x.$

For the same reason, we can evaluate $\lim_{x \to \infty} x \cdot \frac 1 {2x + 3}$ and get 1/2, even though there is no upper bound on x, and no positive lower bound on $\frac 1 {2x + 3}$.

And then we define convergence using nets.

 

[WWGD]

.Sure they do. Assuming for simplicity's sake that $\Delta x = \frac{b - a} n$, as $n \to \infty, \Delta x \to 0$. There is no positive minimum value that's a lower bound on $\Delta x.$

For the same reason, we can evaluate $\lim_{x \to \infty} x \cdot \frac 1 {2x + 3}$ and get 1/2, even though there is no upper bound on x, and no positive lower bound on $\frac 1 {2x + 3}$.

Yes, but a specific Riemann sum takes on a specific value for $n$. Then you consider different Riemann sums where n becomes larger. So you have a collection of Riemann sums, each with a specific value of n, and we consider what happens as we consider specific sums with increasingly-larger values for $n$. So you end up with a net of Riemann sums , where the order is inclusion of partition set {#x_1,x_2,...,x_k#} , that converges ( under the right conditions) as a net to the value of the integral.

 

[Mark44]

Yes, but a specific Riemann sum takes on a specific value for $n$. Then you consider different Riemann sums where n becomes larger.

But the Riemann integral is the limit as the partition gets finer, of these sums.
https://en.wikipedia.org/wiki/Riemann_integral

Loosely speaking, the Riemann integral is the limit of the Riemann sums of a function as the partitions get finer.

So you have a collection of Riemann sums, each with a specific value of n, and we consider what happens as we consider specific sums with increasingly-larger values for $n$.

In other words, as you take the limit as n goes to infinity, which forces the subinterval width to zero.

 

[WWGD]

But the Riemann integral is the limit as the partition gets finer, of these sums.
https://en.wikipedia.org/wiki/Riemann_integral

In other words, as you take the limit as n goes to infinity, which forces the subinterval width to zero.

This is kind of difficult to explain in words, but my point is that, for any individual Riemann sum, the function (partition width) is a continuous function on a compact set [a,b], so it assumes a minimum value. Of course, we let the minimal values go to 0.

 

[Mark44]

This is kind of difficult to explain in words, but my point is that, for any individual Riemann sum, the function (partition width) is a continuous function on a compact set [a,b], so it assumes a minimum value. Of course, we let the minimal values go to 0.

A particular Riemann sum will involve a finite number of subintervals of nonzero width. The Riemann integral, however, is the limit of these Riemann sums, as the subinterval width goes to zero.

 

[WWGD]

A particular Riemann sum will involve a finite number of subintervals of nonzero width. The Riemann integral, however, is the limit of these Riemann sums, as the subinterval width goes to zero.

So we are saying the same thing. I was addressing a poster question about the fact that an indefinitely-small Standard Real number must equal 0, which led them to believe that it must follow that a Riemann sum must equal 0 , since all the terms in the sum become indefinitely small. But this is not so, because in an individual/specific Riemann sum, the terms do not become indefinitely small; they are uniformly bounded by my previous argument. Of course this bound goes to 0, but the minimal width on a specific Riemann sum does not go to 0 or else the sum itself would be zero. Now, it would be interesting how to do integrals in non-standard analysis. I have seen methods to differentiate but I have not seen integration theory in non-standard analysis.

 

[Mark44]

I'm not sure that we were. Isaac0427 said

wouldn't a Riemann sum not mater at all, and just be equal to zero?

I believe he really meant Riemann integral here.

Your response to him was

I think the point is that the rectangles, while going to 0, are uniformly bounded , by compactness of the region of integration, so these widths do not become indefinitely-close to 0; they become close to 0 in a controlled way.

Your explanation used a lot of fairly high-powered analysis, but was flawed; i.e., that the (areas of) the rectangles are going to zero while their widths miraculously remain positive, or that they "become close to zero in a controlled way."

 

[WWGD]

I'm not sure that we were. Isaac0427 said I believe he really meant Riemann integral here.

Your response to him was
Your explanation used a lot of fairly high-powered analysis, but was flawed; i.e., that the (areas of) the rectangles are going to zero while their widths miraculously remain positive, or that they "become close to zero in a controlled way."

Do you see something specifically wrong with the claim that rectangle width is a continuous function on a compact set so it achieves a minimum, which is a uniform lower bound? Is this _not_ a controlled way? What flaw do you see here? Or do you claim that in a specific Riemann sum the widths, and specifically the minimal width, becomes indefinitely close to 0. Convince me then that the integral is not zero.

 

[Mark44]

Do you see something specifically wrong with the claim that rectangle width is continuous on a compact set so it achieves a minimum, which is a uniform lower bound? Is this _not_ a controlled way? What flaw do you see here?

This part, which I just quoted:

so these widths do not become indefinitely-close to 0;

 

[WWGD]

This part, which I just quoted:

In a _specific sum_ , the function interval width is continuous and defined on a compact set. What is wrong with this argument? If you let your minimal width go to 0 in a _specific sum_ without bound, your integral will equal 0. What you end up getting is a net of Riemann sums, which will converge netwise to the value of the integral under the right conditions.

 

[WWGD]

I also see it like this: The collection {$x_1=a, x_2,..., x_n=b$} covers [a,b]. By compactness of [a,b], the collection has a finite subcover. Now, consider min $(x_{i+1}-x_i )$:=m on the finite subcover. A finite set will have a minimum value , say m. The value $m=0$ will never be realized. Maybe I am not explaining myself well here.

 

[WWGD]

Sorry, could not edit. Just to say I need to get to sleep, I will address any other disagreement tomorrow.

 

[Mark44]

In a _specific sum_ , the function interval width is continuous and defined on a compact set. What is wrong with this argument? If you let your minimal width go to 0 in a _specific sum_ without bound, your integral will equal 0. What you end up getting is a net of Riemann sums, which will converge netwise to the value of the integral under the right conditions.

We are not talking about a specific Riemann sum. My complaint is that you are talking about a limit, but don't seem to realize it. Emphasis added.

I think the point is that the rectangles, while going to 0, are uniformly bounded , by compactness of the region of integration, so these widths do not become indefinitely-close to 0

In other words, as $\Delta x \to 0$ -- that's a limit. All the other stuff about compact sets, continuity, and subcovers etc. is just word salad that I'm 99% sure would be over the head of the person you were responding to.

 

[madness]

There's a difference between what you say above and what I think you're trying to say.
First: limit of .999 ... is meaningless, as you aren't saying what is changing.
Second, $\lim_{n \to \infty}\sum_{j = 1}^n \frac 9 {10^j} = 1$. The sum can be made arbitrarily close to 1 just by taking more terms in the sum.

But what I actually said was "limit of 0.999... as you keep on adding more decimal places", which is not the same as what you quoted from me. I'm not sure what point you are making in the second part, as it seems to agree with what I wrote. The mathematical statement simply says that as you make n bigger and bigger you get closer and closer to 1, which I don't think anybody would disagree with.

 

[PeroK]

But what I actually said was "limit of 0.999... as you keep on adding more decimal places", which is not the same as what you quoted from me. I'm not sure what point you are making in the second part, as it seems to agree with what I wrote. The mathematical statement simply says that as you make n bigger and bigger you get closer and closer to 1, which I don't think anybody would disagree with.

It is important, though, to realize that a limit is a number, and not a process. It's a small point, perhaps, but it sharpens your understanding of these things.

 

[madness]

It is important, though, to realize that a limit is a number, and not a process. It's a small point, perhaps, but it sharpens your understanding of these things.

Yes, but it's also important to remember that this number (the limit) is the number that you get closer and closer to as you apply the process of increasing n. Or, if you don't like this process, it's the number that you can get as close to as you like by choosing a high enough value of n.

 

[PeroK]

Yes, but it's also important to remember that this number (the limit) is the number that you get closer and closer to as you apply the process of increasing n. Or, if you don't like this process, it's the number that you can get as close to as you like by choosing a high enough value of n.

Neither of those is the definition of a limit.

 

[HallsofIvy]

Neither of those is the definition of a limit.

Neither is a rigorous definition of a limit but they are a statement of what is implied by such a definition. In particular, the limit of the sequence 0.9, 0.99, 0.999, ... is the number 1, not the sequence itself.

 

[PeroK]

Neither is a rigorous definition of a limit but they are a statement of what is implied by such a definition. In particular, the limit of the sequence 0.9, 0.99, 0.999, ... is the number 1, not the sequence itself.

Perhaps, but my original point was that there are those who see the limit as a number (full stop) and those who see the limit as the hypothetical result of a never-ending process. Breaking away from this latter view is perhaps an important step in grasping rigorous maths.

 

[jbriggs444]

Perhaps, but my original point was that there are those who see the limit as a number (full stop) and those who see the limit as the hypothetical result of a never-ending process. Breaking away from this latter view is perhaps an important step in grasping rigorous maths.

The phrasing that madness used was careful and avoided speaking of the result of a never-ending process.

Yes, but it's also important to remember that this number (the limit) is the number that you get closer and closer to as you apply the process of increasing n. Or, if you don't like this process, it's the number that you can get as close to as you like by choosing a high enough value of n.

 

[PeroK]

The phrasing that madness used was careful and avoided speaking of the result of a never-ending process.

Fair enough. It's not the way I read it, but my apologies to madness.

 

[WWGD]

We are not talking about a specific Riemann sum. My complaint is that you are talking about a limit, but don't seem to realize it. Emphasis added.

In other words, as $\Delta x \to 0$ -- that's a limit. All the other stuff about compact sets, continuity, and subcovers etc. is just word salad that I'm 99% sure would be over the head of the person you were responding to.

No, I think you are not realizing my point. I am talking about an individua Riemann sum not being 0, addressing the question. I do know that the Riemann sum is the limit ( if it exists) of the Riemann sums. I never said nor implied otherwise. I did reply to the OP that for individual sums, the width is controlled, so it does not go to zero. The rest of the exchanges were with and for you.

 

[Mark44]

No, I think you are not realizing my point. I am talking about an individua Riemann sum not being 0, addressing the question. I do know that the Riemann sum is the limit ( if it exists) of the Riemann sums.

I think you mean Riemann integral being the limit of the Riemann sums.

I never said nor implied otherwise. I did reply to the OP that for individual sums, the width is controlled, so it does not go to zero. The rest of the exchanges were with and for you.

My complaint is in the following. I interpret the phrase "while going to 0" as equivalent to saying "the limit of the Riemann sums as n goes to infinity." You can't say that the widths of the rectangles are going to zero, but at the same time do not become close to zero. Here again is what you said, with emphasis added.

I think the point is that the rectangles, while going to 0, are uniformly bounded , by compactness of the region of integration, so these widths do not become indefinitely-close to 0

The rectangle widths are either going to zero or they're not -- you can't have it both ways.

 

[WWGD]

I think you mean Riemann integral being the limit of the Riemann sums.My complaint is in the following. I interpret the phrase "while going to 0" as equivalent to saying "the limit of the Riemann sums as n goes to infinity." You can't say that the widths of the rectangles are going to zero, but at the same time do not become close to zero. Here again is what you said, with emphasis added.

The rectangle widths are either going to zero or they're not -- you can't have it both ways.

Yes, I meant Riemann integral, my bad. I was referring to individual Riemann sums, not to the Riemann integral itself. I was addressing the question of whether a Riemann sum is zero or not, because the general term goes to zero in an uncontrolled way. I made no reference to the Riemann integral itself; that is a whole other topic. But yes, the Riemann integral is the limit as the partition width goes to zero. I don't see why the two contradict or are incompatible with each other.

 

[Mark44]

I was referring to individual Riemann sums, not to the Riemann integral itself. I was addressing the question of whether a Riemann sum is zero or not, because the general term goes to zero in an uncontrolled way.

When you say, "the rectangles, while going to 0, ..." you're no longer talking about a specific Riemann sum -- you're talking about the limit of the Riemann sums (as the number of terms goes to infinity), which means you're talking about the Riemann integral.

These were your exact words:

I think the point is that the rectangles, while going to 0, are uniformly bounded , by compactness of the region of integration, so these widths do not become indefinitely-close to 0

In short, you're saying that the rectangles are going to zero and that their widths are not going to zero. That's impossible.

Also, I don't know what you mean "goes to zero in an uncontrolled way" ...

I made no reference to the Riemann integral itself; that is a whole other topic. But yes, the Riemann integral is the limit as the partition width goes to zero. I don't see why the two contradict or are incompatible with each other.

 

[WWGD]

When you say, "the rectangles, while going to 0, ..." you're no longer talking about a specific Riemann sum -- you're talking about the limit of the Riemann sums (as the number of terms goes to infinity), which means you're talking about the Riemann integral.

These were your exact words:
In short, you're saying that the rectangles are going to zero and that their widths are not going to zero. That's impossible.

Also, I don't know what you mean "goes to zero in an uncontrolled way" ...

No, I am definitely talking about an individual Riemann sum. By a controlled way, I mean there is a lower bound for the widths, which there is by the usual arguments. I think the confusion here is you are referring to the Riemann integral, while I am referring to specific Riemann sums.

 

[Mark44]

No, I am definitely talking about an individual Riemann sum. By a controlled way, I mean there is a lower bound for the widths, which there is by the usual arguments. I think the confusion here is you are referring to the Riemann integral, while I am referring to specific Riemann sums.

Please explain to me how the widths of the rectangles can be "going to 0" and at the same time "these widths do not become indefinitely-close to 0" -- your words.

I maintain that you can't possibly be talking about a specific Riemann sum with a specific number of terms (hence a specific number of rectangles, all of whose areas are positive numbers).

 

[lavinia]

We are not talking about a specific Riemann sum. My complaint is that you are talking about a limit, but don't seem to realize it. Emphasis added.

In other words, as $\Delta x \to 0$ -- that's a limit. All the other stuff about compact sets, continuity, and subcovers etc. is just word salad that I'm 99% sure would be over the head of the person you were responding to.

I think that WWGD is trying to point out that on an open interval one would need to divide it into an infinite sequence of closed intervals whose lengths would become arbitrarily small near the edge of the interval. On a closed interval one can choose a subdivision that is finite.

 

[WWGD]

Please explain to me how the widths of the rectangles can be "going to 0" and at the same time "these widths do not become indefinitely-close to 0" -- your words.

I maintain that you can't possibly be talking about a specific Riemann sum with a specific number of terms (hence a specific number of rectangles, all of whose areas are positive numbers).

The maximal width goes to 0 if we consider a net of Riemann sums. In order to compute the Riemann integral, we consider what happens as the width of partitions go to zero.
But any sum has a minimal width.

I don't mean to be disrespectful, but this conversation does not seem to be going anywhere productive and frankly, I would like to leave it at this. I am not blaming it on you, but we either drop it or take a different tack to it, because otherwise we will be going ( at least seemingly) in circles for a long time here. We are just not communicating effectively here; I suggest we leave it at this, at least for now. No disrespect intended.

 

[Mark44]

The maximal width goes to 0 if we consider a net of Riemann sums. In order to compute the Riemann integral, we consider what happens as the width of partitions go to zero.
But any sum has a minimal width.

Yes, I understand all of this.

What I disagree with, and keep carping on, your assertion that the widths of the rectangles can be "going to 0" and at the same time "these widths do not become indefinitely-close to 0". Obviously, both of these can't be true.

 

[Isaac0427]

I know this discussion came off of my mentioning a Riemann sum. I meant a Riemann integral. My point was that in a Riemann integral, you add up an infinite amount of rectangles with a width of dx. dx, for the integral to have a nonzero value, couldn't be zero, but it is, by definition, less than .00000000001, .00000000000000000000000001, or any number of zeros you put before the one. You would need an infinite amount of zeros. So, I know it may not be the proper notation, but the width of the rectangles would essentially be .000...1. If the fact that it gets infinitely close to zero makes it equal to zero, then a Riemann integral would have a value of zero, which is not the case. If we were to take some fundamental length (in a coordinate system that has one), and have two particles of that size touch with a distance between them of zero, would they be in the same position? I say a fundamental size because the particle's centers would have no distance between them.

 

[lavinia]

I know this discussion came off of my mentioning a Riemann sum. I meant a Riemann integral. My point was that in a Riemann integral, you add up an infinite amount of rectangles with a width of dx. dx, for the integral to have a nonzero value, couldn't be zero, but it is, by definition, less than .00000000001, .00000000000000000000000001, or any number of zeros you put before the one. You would need an infinite amount of zeros. So, I know it may not be the proper notation, but the width of the rectangles would essentially be .000...1. If the fact that it gets infinitely close to zero makes it equal to zero, then a Riemann integral would have a value of zero, which is not the case. If we were to take some fundamental length (in a coordinate system that has one), and have two particles of that size touch with a distance between them of zero, would they be in the same position? I say a fundamental size because the particle's centers would have no distance between them.

This is not what the Riemann integral is. One does not add up an infinite number of rectangles of zero width.

The way I learned it, the Riemann integral is defined in terms of lower and upper sums. For a positive function, one divides a closed interval into finitely many closed segments then takes the length of each segment and multiplies it by the max or the min of the function in the segment. and adds the values up. These give you what are called lower and upper sums. As WWGD said, these segments have finite length and there are always only finitely many terms in the sum The Riemann integral is then the Least Upper Bound of all lower sums obtained by dividing the interval into finitely many segments of finite length. It is also the Greatest Lower Bound of all upper sums. A particular sum can be chosen to have segments of arbitrarily small length, but each sum always has finitely many segments and these segments always have finite length.

This same point is true for the expression $.999...$

The three dots mean that number which is the Least Upper Bound of all finite decimal numbers, .9, .99, 999, .9999 etcetera. It does not stand for some non-standard number in some other number system. It stands for that ordinary real number that is the LUB of the set of decimals ${ .9,.99,.999, .9999}$ etc

The number one is the least Upper Bound of all of these numbers. The proof shows that one can always find a finite decimal sequence of 9's that is arbitrarily close to 1, Another way of saying this is that 1 is the limit of the sequence, $x_1 = .9$, $x_2 = .99$, ... $x_n = .999999$ (n times)

Hallsofivy gives a nice proof of this limit in Post #2.

Every positive number has a decimal expansion. For almost all of them, the decimal expansion is infinitely long. This means that almost every number is the Least Upper Bound of an infinite sequence of increasing decimal numbers. 1 is not special in any way.

 

[WWGD]

Yet another reason, maybe mentioned is that, by the Archimedean property, a Standard Real number cannot be indefinitely small; it must then equal zero. But the difference

1-0.99999... _is_ indefinitely small, so must equal zero. So the two numbers are then equal.

 

[Isaac0427]

1-0.99999... _is_ indefinitely small, so must equal zero.

dx is also infinitesimally small, but if it equaled zero, then any integral or derivative would equal zero or be undefined.

 

[jbriggs444]

dx is also infinitesimally small, but if it equaled zero, then any integral or derivative would equal zero or be undefined.

dx is not a number. It is a notation.