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

[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.

 

[Isaac0427]

dx is not a number. It is a notation.

I thought it represented an infinitely small change of x. I could be wrong, however.

 

[WWGD]

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

Remember I said in previous posts that the length of the rectangles in a fixed Riemann sum is bounded below. You then consider what happens as this approaches 0 in the limit, but this, dealing with limits, is a separate issue.

 

[jbriggs444]

I thought it represented an infinitely small change of x. I could be wrong, however.

In the non-standard analysis, dx can be given meaning as an infinitesimal. In standard analysis, the dx in an integral is pure notation.

 

[Isaac0427]

Ok, I just talked to my math teacher. She explained it to me in person which helped. Thanks guys it really does help. I mostly understand it now, that as something gets so close to 1, for all practical purposes, it is equal to 1.

 

[lavinia]

Ok, I just talked to my math teacher. She explained it to me in person which helped. Thanks guys it really does help. I mostly understand it now, that as something gets so close to 1, for all practical purposes, it is equal to 1.

Ask your math teacher to explain to you what a limit is.

 

[Isaac0427]

Ask your math teacher to explain to you what a limit is.

I know what a limit is. She explained how for practical purposes the limit .999... approaches 1 is considered .999... being equal to one.

 

[jbriggs444]

I know what a limit is. She explained how for practical purposes the limit .999... approaches 1 is considered .999... being equal to one.

It is not "for practical purposes". It is exact. The limit of the sequence .9, .99, .999, ... does not approach 1. Successive terms in the sequence approach 1. The limit is 1. The notation .999... denotes the limit. Hence .999... is 1.

 

[pwsnafu]

I know what a limit is. She explained how for practical purposes the limit .999... approaches 1 is considered .999... being equal to one.

The part where I bolded indicates you don't know what a limit is. Limits don't move and they don't "approach".

 

[gmax137]

Going back to post #2

...
Restore that by adding ar^{n+1} to both sides:
S_n- a+ ar^n= r(a+ ar+ \cdot\cdot\cdot+ ar^{n-1})+ ar^{n+ 1}
...

Maybe (probably) I'm being dense, but didn't you just add ar^n to the left and ar^{n+ 1} to the right?

 

[FactChecker]

Ok, I just talked to my math teacher. She explained it to me in person which helped. Thanks guys it really does help. I mostly understand it now, that as something gets so close to 1, for all practical purposes, it is equal to 1.

You can prove that there is no difference at all because the 9's go forever. This proof may be your first encounter with a mathematical "proof by contradiction". Suppose you assume that there is any difference between .999999... and 1. Say it is over 0.00001 (a formal mathematical proof would use an arbitrarily small ε > 0). Now use enough 9's (0.999999999999) to show that there is less difference than that and that the difference will only decrease as you add more 9's. That contradicts to your original assumption that the difference is greater than 0.00001 (or ε>0). It doesn't matter how small your assumed difference is; you can add enough 9's to get closer to 1 and contradict that assumption. So it proves that there can be no difference between 0.99999... and 1.

 

[HallsofIvy]

Going back to post #2
Maybe (probably) I'm being dense, but didn't you just add ar^n to the left and ar^{n+ 1} to the right?

Yes, that was a typo. It should have been ar^{n+1} on both sides.

 

[Kegan]

i'm 11 i understand that 1 and .999 are two completely different numbers and in math you don't round numbers to get precise answers it just doesn't work like that

 

[WWGD]

i'm 11 i understand that 1 and .999 are two completely different numbers and in math you don't round numbers to get precise answers it just doesn't work like that

@Kegan It is not .999 , it is .9999... with an infinite string of 9s.

 

[Isaac0427]

i'm 11 i understand that 1 and .999 are two completely different numbers and in math you don't round numbers to get precise answers it just doesn't work like that

Instead of thinking of .999... as a number, think of it as the limit of the sequence (.9,.99,.999,.9999,...) as n approaches infinity. That is what the 3 dots mean. The limit of that sequence as n approaches infinity is 1.
Here's another example: (1/0!)+(1/1!)+(1/2!)...=e, however if you stop the sequence at any value of n, the answer will not be e. Does this make sense?

 

[micromass]

Instead of thinking of .999... as a number

No. 0.99999... is a number.

 

[pbuk]

No. 0.99999... is a number.

To be fair, Isaac didn't say it was not a number. He just suggested that it might be easier to see that 0.9999... is the limit of the sequence 0.9, 0.99, 0.999, ..., and that the limit of that sequence is also equal to 1, than it is to make the jump from the recurring decimal to 1 directly.

This illustration is not IMHO any less rigorous than the 9.9999... - 0.9999... illustration of the identical equality to 1. Whether it is easier to comprehend or not is a matter of personal taste.

 

[WWGD]

The confusion seems largely to stem from the implicit assumption that a number can have only one representation. Instead, a number is an equivalence class; one sees this in daily life, e.g., today is Friday, and (ignoring modular issues) any date 7k days from now is also a Friday, so if the difference between the dates (again, re modularity) is a multiple of 7 , then both are the same day of the week. And then there is 2/2, 3/3 , etc.

 

[pbuk]

I don't think days of the week, which are analogous to integers mod 7 is a helpful analogy to the reals. Edit: the rest of this is rubbish And 2/2 is an expression which is equal to 1, it is not a valid representation of a rational number: the numerator and denominator must have no common factor.

 

[micromass]

I don't think days of the week, which are analogous to integers mod 7 is a helpful analogy to the reals. And 2/2 is an expression which is equal to 1, it is not a valid representation of a rational number: the numerator and denominator must have no common factor.

2/2 is a perfect representation of a rational number.

 

[WWGD]

I don't think days of the week, which are analogous to integers mod 7 is a helpful analogy to the reals. And 2/2 is an expression which is equal to 1, it is not a valid representation of a rational number: the numerator and denominator must have no common factor.

Yes, maybe it is not the best example, but the point I wanted to make is that two things don't need to be strictly equal in order to be considered the same. It is more a " to the effects of what we are doing, these two expressions are equal" EDIT: Maybe non-trivial, i.e., non-identity isomorphisms would be a better example.