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Why Do People Say That 1 And .999 Are Equal?

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Why Do People Say That 1 And .999 Are Equal?

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While there are several **arguments** here that people might accept as proofs, they all, as stated, are non-rigorous, involving some basic, unstated assumptions. For example, in argument two, it is assumed that multiplying x= 0.999... by 10 gives 10x= 9.999... and then that subtracting x give 9x= 0.999... again. Both of those assume the usual arithmetic properties are true for 0.999... That is true but exactly the sort of thing people who object to "0.999...= 1" would object to anyway. In argument four, it is accepted that 0.333...= 1/3. Why would a person who objects to "0.999...= 1" accept that?

It is not too difficult to give a**rigorous** proof using "geometric series". It is easily shown that the sum [itex]\sum_{i= 0}^n ar^i[/itex] is [tex]\frac{a(1- r^n)}{1- r}[/tex]. To do that, let [itex]S_n= \sum_{i= 0}^n ar^i= a+ ar+ ar^2+ \cdot\cdot\cdot+ ar^n)[/itex]. Since that is a **finite** sum, the usual properties of arithmetic hold and we can write [itex]S_n- a= ar+ ar^2+ \cdot\cdot\cdot+ ar^n= r(a+ ar+ \cdot\cdot\cdot+ ar^{n-1})[/itex]. The quantity in the last parentheses is **almost** [itex]S_n[/itex] itself. It is only missing the last term, [itex]ar^n[/itex]. Restore that by adding [itex]ar^{n+1}[/itex] to both sides: [itex]S_n- a+ ar^n= r(a+ ar+ \cdot\cdot\cdot+ ar^{n-1})+ ar^{n+ 1}[/itex]. Taking that last term inside the parentheses we have [itex]S_n- a+ ar^n= r(a+ ar+ \cdot\cdot\cdot+ ar^{n-1}+ ar^n)= rS_n[/itex]. We can write that as [itex]rS_n- S_n= (r- 1)S)_n= ar^n- a= a(r^n- 1)[/itex] and dividing both sides by r- 1, [itex]S_n= \frac{a(r^n- 1)}{r- 1}[/itex] which, for r< 1 can be written more as [itex]S_n= \frac{a(1- r^n)}{1- r}[/itex]

The geometric series, [itex]\sum_{i= 0}^\infty ar^i[/itex], is, by definition, the limit of the "partial sums", [itex]S_n= \sum_{i= 0}^n ar^i[/itex], as n goes to infinity. As long as [itex]|r|< 1[/itex], [itex]\lim_{n\to\infty} r^n= 0[/itex] so that limit is easily calculated as [itex]\sum_{i= 0}^\infty ar^i= \frac{a}{1- r}[/itex].

Now, turning to the problem at hand. 0.999... is, by definition of the decimal numeration system, [itex]0.9+ 0.09+ 0.009+ \cdot\cdot\cdot= 0.9+ 0.9(0.1)+ 0.9(0.01)+ \cdot\cdot\cdot= 0.9+ 0.9(.1)+ 0.9(.1^2)+ \cdot\cdot\cdot[/itex], precisely a geometric series with a= 0.9 and r= 0.1. So the sum, and the value of 0.9999..., is [itex]\frac{0.9}{1- 0.1}= \frac{0.9}{0.9}= 1[/itex].

It is not too difficult to give a

The geometric series, [itex]\sum_{i= 0}^\infty ar^i[/itex], is, by definition, the limit of the "partial sums", [itex]S_n= \sum_{i= 0}^n ar^i[/itex], as n goes to infinity. As long as [itex]|r|< 1[/itex], [itex]\lim_{n\to\infty} r^n= 0[/itex] so that limit is easily calculated as [itex]\sum_{i= 0}^\infty ar^i= \frac{a}{1- r}[/itex].

Now, turning to the problem at hand. 0.999... is, by definition of the decimal numeration system, [itex]0.9+ 0.09+ 0.009+ \cdot\cdot\cdot= 0.9+ 0.9(0.1)+ 0.9(0.01)+ \cdot\cdot\cdot= 0.9+ 0.9(.1)+ 0.9(.1^2)+ \cdot\cdot\cdot[/itex], precisely a geometric series with a= 0.9 and r= 0.1. So the sum, and the value of 0.9999..., is [itex]\frac{0.9}{1- 0.1}= \frac{0.9}{0.9}= 1[/itex].

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Here is the previous FAQWhile there are severalargumentshere that people might accept as proofs, they all, as stated, are non-rigorous, involving some basic, unstated assumptions.

https://www.physicsforums.com/insights/is-there-a-rigorous-proof-of-1-0-999/

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I'm not an expert on these things, but you can easily tell that a proof is wrong when it proves 0.9999...=1 on surreal numbers. Which most of the above do.

I'm not entirely sure about HallsofIvy's proof, it might be OK, even if I think it's swapping ##\infty## with ##\omega## somewhere along the way.

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You can easily prove it by contradiction, as I indicated. Once that is done, you know that they are equal. Then I would to be careful about saying that other proofs are wrong unless I completely understood the other proofs and can pinpoint the error.I'm not an expert on these things, but you can easily tell that a proof is wrong when it proves 0.9999...=1 on surreal numbers. Which most of the above do.

I'm not entirely sure about HallsofIvy's proof, it might be OK, even if I think it's swapping ##\infty## with ##\omega## somewhere along the way.

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How is this assumption insulting. Just by ordinary division you can show that the 3 digit keeps repeating.Proofs #2 & #3 are what convinced me that 1 and .999... are NOT equal. It's just insulting to assume that .333... = 1/3 in this context.

To the right of the decimal point, both have the same number of 9 digits. Also I don't understand what you are saying about "finite numbers". Both .999... and 9.999... are finite numbers; i.e., strictly less than infinity.Algr said:And why assume that .999... and 9.999... have different numbers of nines when this would never happen for any finite number?

I did a quick scan of all of the proofs. None of them mentioned surreal numbers. If you can point out where I might have missed this, I would appreciate it.I'm not an expert on these things, but you can easily tell that a proof is wrong when it proves 0.9999...=1 on surreal numbers. Which most of the above do.

Please show me where.SlowThinker said:I'm not entirely sure about HallsofIvy's proof, it might be OK, even if I think it's swapping ##\infty## with ##\omega## somewhere along the way.

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My idea was: if the proof, applied over surreal numbers, proves that 0.9999...=1, then the proof is not correct. There must be some implicit assumption.I did a quick scan of all of the proofs. None of them mentioned surreal numbers. If you can point out where I might have missed this, I would appreciate it.

Please show me where.

I would say that, by definition of the decimal numeration system, 0.9999...=##\sum_{n=1}^\omega{}9\ 0.1^n## rather than ##\sum_{n=1}^\infty{}9\ 0.1^n##. The difference is, obviously, 0 in real numbers.Now, turning to the problem at hand. 0.999... is, by definition of the decimal numeration system, ##0.9+ 0.09+ 0.009+ \cdot\cdot\cdot## = ##0.9+ 0.9(0.1)+ 0.9(0.01)+ \cdot\cdot\cdot## = ##0.9+ 0.9(.1)+ 0.9(.1^2)+ \cdot\cdot\cdot##,precisely a geometric series with a= 0.9 and r= 0.1. So the sum, and the value of 0.9999..., is [itex]\frac{0.9}{1- 0.1}= \frac{0.9}{0.9}= 1[/itex].

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I did a quick scan of all of the proofs. None of them mentioned surreal numbers. If you can point out where I might have missed this, I would appreciate it.

Why would the proof need to invoke surreal numbers? The proofs are all concerned with two representations of the same real number.SlowThinker said:My idea was: if the proof, applied over surreal numbers, proves that 0.9999...=1, then the proof is not correct. There must be some implicit assumption.

Please show me where.

Now, turning to the problem at hand. 0.999... is, by definition of the decimal numeration system, ##0.9+ 0.09+ 0.009+ \cdot\cdot\cdot## = ##0.9+ 0.9(0.1)+ 0.9(0.01)+ \cdot\cdot\cdot## = ##0.9+ 0.9(.1)+ 0.9(.1^2)+ \cdot\cdot\cdot##,precisely a geometric series with a= 0.9 and r= 0.1. So the sum, and the value of 0.9999..., is [itex]\frac{0.9}{1- 0.1}= \frac{0.9}{0.9}= 1[/itex].

And that's the set we're concerned with here.SlowThinker said:I would say that, by definition of the decimal numeration system, 0.9999...=##\sum_{n=1}^\omega{}9\ 0.1^n## rather than ##\sum_{n=1}^\infty{}9\ 0.1^n##. The difference is, obviously, 0 in real numbers.

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If the proof can be used to prove a false statement, then the proof must be wrong.Why would the proof need to invoke surreal numbers? The proofs are all concerned with two representations of the same real number.

0.9999...##\neq##1 in surreal numbers, so if the proof works with them too, then it's wrong.

I dare say that all of the "informal proofs" from the article immediately work with surreal numbers, so they must be wrong.

This quote from the original article,

seems to be the answer that people are looking for, together with "Can we define number systems such that 1=0.999… does not hold?Of course!

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So does this mean that .99999... = .99999...**8** ?

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No, because the right hand side is nonsense. You can't have a non-terminating run of nines and then terminate it.So does this mean that .99999... = .99999...8?

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Okay... but it seems like to make .999 = 1 you also have to terminate the run of nines. Anyway I know this has been discussed several times so I'll read up some more.No, because the right hand side is nonsense. You can't have a non-terminating run of nines and then terminate it.

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As already mentioned, the right side is meaningless for two reasons. On the left side theSo does this mean that .99999... = .99999...8?

Don't confuse .999 with .999... They mean two very different things. The first is identical to 999/1000 which is smaller than 1, and the second is equal to 1.Okay... but it seems like to make .999 = 1 you also have to terminate the run of nines.

Ernest S Walton said:Anyway I know this has been discussed several times so I'll read up some more.

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*Both .999… and 9.999… are finite numbers; i.e., strictly less than infinity.*How is this assumption insulting. Just by ordinary division you can show that the 3 digit keeps repeating.Proofs #2 & #3 are what convinced me that 1 and .999... are NOT equal. It's just insulting to assume that .333... = 1/3 in this context.

To the right of the decimal point, both have the same number of 9 digits. Also I don't understand what you are saying about "finite numbers". Both .999... and 9.999... are finite numbers; i.e., strictly less than infinity.Algr said:And why assume that .999... and 9.999... have different numbers of nines when this would never happen for any finite number?

I did a quick scan of all of the proofs. None of them mentioned surreal numbers. If you can point out where I might have missed this, I would appreciate it.I'm not an expert on these things, but you can easily tell that a proof is wrong when it proves 0.9999...=1 on surreal numbers. Which most of the above do.

Please show me where.SlowThinker said:I'm not entirely sure about HallsofIvy's proof, it might be OK, even if I think it's swapping ##infty## with ##omega## somewhere along the way.

How can you call 0.999... or 9.999... as finite?

If these are finite then infinity can also be considered finite!

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*Both .999… and 9.999… are finite numbers; i.e., strictly less than infinity.*

How can you call 0.999... or 9.999... as finite?

If these are finite then infinity can also be considered finite!

Please look up the definition of a finite number. There is a difference between an infinite _decimal expansion_ and an infinite number.

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*Both .999… and 9.999… are finite numbers; i.e., strictly less than infinity.*

How can you call 0.999... or 9.999... as finite?

If these are finite then infinity can also be considered finite!

WWGD is correct. Finite numbers are bounded; infinite numbers are unbounded.Please look up the definition of a finite number. There is a difference between an infinite _decimal expansion_ and an infinite number.

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They are clearly different but if you're approximating, they're almost equal.

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Who are "they"? Are you referring to .999 and .999... ?They are clearly different but if you're approximating, they're almost equal.

Clearly these two numbers are close, but that's not the point. What we're saying is that .999... is identically equal to 1, and .999 is smaller than 1 by .001.

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Both are true. The important part is that real numbers are not precise enough to distinguish an infinitesimal difference - which is, in fact, the reason why we can use infinitesimals to integrate and derive expressions containing real numbers.Wouldn't the difference between .999... and 1 be infinitesimal, not 0.

something infinitesimal is just considered 0

I would think this whole theme is answered by:

The

If you don't like it, use other numbers.

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I think that is exactly my point. If we call limits being equal to, in the equation for derivatives, lim Δx→0 (f(x+Δx)-f(x))/Δx (I know I may not have written it right), the answer would be undefined.limitof 0.999... as we keep adding more decimal places is 1. I think this is also a closer translation of the mathematical statement into plain english.

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They are clearly different but if you're approximating, they're almost equal.

No, they are not different as Real numbers using the definition of Real numbers in terms of Cauchy sequences. This is a technical point that requires the use of precise definitions.

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I think a key point is that .999... is *by definition* a limit, and that limit is equal to 1.

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I think a key point is that .999... isby definitiona limit, and that limit is equal to 1.

True, it is the limit of a Cauchy sequence, which converges by the fact that the Reals are a complete metric space, and the sequence, as you said, converges to 1.

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I have no idea what you're trying to say in the equation above.I would think this whole theme is answered by:

Thedefinitionofrealnumbers says that $$0.999..._{\text{as real number}} \ =_{\text{acting on real numbers}} \ 1_{\text{as real number}}$$

If you don't like it, use other numbers.

There's a difference between what you say above and what I think you're trying to say.limitof 0.999... as we keep adding more decimal places is 1. I think this is also a closer translation of the mathematical statement into plain english.

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.

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There is no need that I can see to bring infinitesimals into the argument. By definition of the notation, .999 ...I have no way to say this is true for sure, but it appears to me as though proof #3 is based off an incorrect assumption. Wouldn't the difference between .999... and 1 be infinitesimal, not 0.

Isaac0427 said:Isn't this one of the reasons we have the concept of infinity, and in this case an infinitesimal value. If something infinitesimal is just considered 0, wouldn't a Riemann sum not mater at all, and just be equal to zero? Wouldn't that make anything using dx, dy, etc. represent a change of zero, as they are defined as an infinitesimaly small change in their respective variable? This is just my thinking, I am not being ignorent, I just want to understand.

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Like someone pointed out, there are no infinitesimals in the Standard Real numbers; these exist only on non-standard models of the Reals. Your other questions are good, let me think about them some more. 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. Good point, though.

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In the Riemann definition of a definite integral, you have the number of rectangles n approaching infinity (i.e., increasing without bound), where the area of each rectangle is approaching zero.If something infinitesimal is just considered 0, wouldn't a Riemann sum not mater at all, and just be equal to zero? Wouldn't that make anything using dx, dy, etc. represent a change of zero, as they are defined as an infinitesimaly small change in their respective variable? This is just my thinking, I am not being ignorent, I just want to understand.

In effect, what you have is the indeterminate form ##[0 \cdot \infty]##. Indeterminate forms are a way to categorize limits that involve a product or quotient or other operation in which one or more of the quantities is unbounded. Indeterminate forms are not number, and different instances of the same indeterminate form can come out to be literally any real number or ##\infty## or ##-\infty##. With Reimann sums, if you have a well-behaved function, the limit comes out to a real number.

See above...Like someone pointed out, there are no infinitesimals in the Standard Real numbers; these exist only on non-standard models of the Reals. Your other questions are good, let me think about them some more. 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. Good point, though.

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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. And then we define convergence using nets.In the Riemann definition of a definite integral, you have the number of rectangles n approaching infinity (i.e., increasing without bound), where the area of each rectangle is approaching zero.

In effect, what you have is the indeterminate form ##[0 \cdot \infty]##. Indeterminate forms are a way to categorize limits that involve a product or quotient or other operation in which one or more of the quantities is unbounded. Indeterminate forms are not number, and different instances of the same indeterminate form can come out to be literally any real number or ##\infty## or ##-\infty##. With Reimann sums, if you have a well-behaved function, the limit comes out to a real number.

See above...

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