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

Why do people say 1 and 0.999… are equal? Aren’t they two different numbers?

No, they really are the same number, though this is often very counterintuitive to many beginning students. Here are some non-rigourous proofs that 1=0.999…:

Proof #1
For any two unequal numbers, there is always another number in between them. (That is intuitively obvious, and can be pictured on a number line, which will be familiar to many people.) Therefore, if 0.999…. and 1 are different, there must be another number in between them. But there is no way to write a number that is greater that 0.999… and less than 1 in decimal notation.

Proof #2
First, we put
$$\ x = 0.999…$$
Multiplying by 10 gives us
$$\ 10x = 9.999…$$
But then
$$\ 10x-x = 9.999…-0.999…,$$
thus
$$\ 9x = 9$$
Hence we get that
$$\ x = 1$$

Proof #3
If you accept that 0.9999… is a number, then how much less than 1 is it? It’s larger that 0.9999, so it’s less than 0.0001 less than 1. But it’s also larger that 0.9999999, so it’s less than 0.0000001 less than 1. So the difference between 0.9999… and 1 is less than 0.00001, 0.000000001, or 0.any number of 0s followed by 1 … so the difference must be zero. If you accept that two numbers whose difference is 0 must be the same, then that proves that 0.9999… = 1. If you don’t accept that, and you think that two different numbers can have a difference of zero, then you’re in an ‘extended number system’ which has more numbers than we normally use.

Proof #4
First, we have that
$$\ 1/3 = 0.333…$$
If we multiply things by 3, then we get
$$\ 1=3\times (1/3) = 3\times (0.333…) = 0.999…$$
All of these proofs are correct, but they are not rigourous. For example, how do we know that $3 \times (0.333…) = 0.999…$? This is not that obvious if we think about it. A more rigourous proof is given in the post following this one.

Some further questions you might have:

But 1 cannot equal 0.999…, as every number can only have one representation!
Well, the thing is that this is just a misconception that is simply not true. Numbers can have many representations. For example,
$$\frac{1}{3}=\frac{2}{6}=\frac{3}{9}=0.333… ,$$
but somehow, many people don’t have any problems with this thing. The same thing happens to 1=0.999… really, it’s just another way to write the same number. Does this make our number system ugly? I understand that you might think that, but that’s just something we need to accept. Not having that 1=0.999… would make our number system much uglier!

The way I see it, is that 0.999… gets closer and closer to 1, but never quite reaches 1.
This reasoning appears a lot and apparently, many people see 0.999… as some kind of process that gets close to 1. But this is not quite what mathematicians mean with 0.999…
Mathematicians say that 0.999… is a number, just like 2 and 3. So phrases like “it gets close to 1, but never reaches 1” are meaningless. It’s the same as saying 1 gets closer and closer to 2, but never quite reaches 2. This sentence makes no sense, and the same thing happens with 0.999…

Can we define number systems such that 1=0.999… does not hold?
Of course! But these number systems are not as useful, because they don’t conform to our intuition about numbers and limits.

In Proof #2, you say 10x=9.999… But this 9.999… has one fewer nine than 0.999…
Another popular argument. This time, the confusion arises from not grasping infinity. 0.999… has an infinite number of nines. If we somehow remove a nine from this sequence, then we would still have an infinite number of nines. So there are an equal number of nines in 0.999… and 9.999…

The same thing happens here: consider two sets of numbers, A and B, where
$$A=\{0,1,2,3,4,…\}$$
and
$$B=\{1,2,3,4,…\}$$
Both sets are infinite. And actually, both sets have an equal number of elements. But A doesn’t contain 0, so it has one fewer element than B? Yes, but this reasoning only applies to finite sets. For infinite sets, it’s quite possible to have one element less and still have an equal number of elements. Indeed, consider the following correspondence:

$$0\leftrightarrow 1,~1\leftrightarrow 2,~2\leftrightarrow 3,~3\leftrightarrow 4,…,~n\leftrightarrow n+1,…$$
So for an element n in A, there exists a unique element in B that corresponds to n, namely n+1. This means, by definition actually, that both sets have the same number of elements.

Maybe we should just abandon our base 10 number system and move to a number system where every number does have a unique representation.
Tempting, but sadly this is not possible. The problem arises in every base! For example, in base 2, we have $1=0.111…$. There is no way around it.
The following forum members have contributed to this FAQ:
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111 replies
1. mathexam says:

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

2. Mark44 says:

[QUOTE=”Nemika, post: 5312734, member: 573538″]*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![/QUOTE]

[QUOTE=”WWGD, post: 5312760, member: 69719″]Please look up the definition of a finite number. There is a difference between an infinite _decimal expansion_ and an infinite number.[/QUOTE]WWGD is correct. Finite numbers are bounded; infinite numbers are unbounded.

3. WWGD says:

[QUOTE=”Nemika, post: 5312734, member: 573538″]*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![/QUOTE]

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

4. Mark44 says:

[QUOTE=”Ernest S Walton, post: 5276913, member: 560204″]So does this mean that .99999….. = .99999…[B]8[/B] ?[/QUOTE]
As already mentioned, the right side is meaningless for two reasons. On the left side the [I]ellipsis[/I] (…) means that the pattern you see repeats endlessly. On the right side, it is not known how many 9s there are in front of the final 8 digit.

[QUOTE=”Ernest S Walton, post: 5276931, member: 560204″]Okay… but it seems like to make .999 = 1 you also have to terminate the run of nines.
[/quote]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.
[QUOTE=”Ernest S Walton”]
Anyway I know this has been discussed several times so I’ll read up some more.[/QUOTE]

5. Ernest S Walton says:

[QUOTE=”pwsnafu, post: 5276918, member: 259282″]No, because the right hand side is nonsense. You can’t have a non-terminating run of nines and then terminate it.[/QUOTE]
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.

6. pwsnafu says:

[QUOTE=”Ernest S Walton, post: 5276913, member: 560204″]So does this mean that .99999….. = .99999…[B]8[/B] ?[/QUOTE]
No, because the right hand side is nonsense. You can’t have a non-terminating run of nines and then terminate it.

7. Ernest S Walton says:

So does this mean that .99999….. = .99999…[B]8[/B] ?

8. SlowThinker says:

[QUOTE=”Mark44, post: 5262295, member: 147785″]Why would the proof need to invoke surreal numbers? The proofs are all concerned with two representations of the same real number.[/QUOTE]
If the proof can be used to prove a false statement, then the proof must be wrong.
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,
[QUOTE][B]Can we define number systems such that 1=0.999… does not hold? [/B]Of course![/QUOTE]
seems to be the answer that people are looking for, together with “[B]Real numbers[/B] are defined in such a way that 0.999…=1”.

9. Mark44 says:

[QUOTE=”Mark44, post: 5262160, member: 147785″]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.[/QUOTE]
[quote=SlowThinker]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.
[/QUOTE]
Why would the proof need to invoke surreal numbers? The proofs are all concerned with two representations of the same real number.
[QUOTE=”Mark44, post: 5262160, member: 147785″]Please show me where.[/QUOTE]
[QUOTE=”HallsofIvy, post: 5260930, member: 331″]Now, turning to the problem at hand. 0.999… is, by definition of the decimal numeration system, ##0.9+ 0.09+ 0.009+ cdotcdotcdot## = ##0.9+ 0.9(0.1)+ 0.9(0.01)+ cdotcdotcdot## = ##0.9+ 0.9(.1)+ 0.9(.1^2)+ cdotcdotcdot##, [B]precisely a geometric series with a= 0.9 and r= 0.1[/B]. So the sum, and the value of 0.9999…, is $frac{0.9}{1- 0.1}= frac{0.9}{0.9}= 1$.[/QUOTE]
[quote=SlowThinker]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.
[/QUOTE]And that’s the set we’re concerned with here.

10. SlowThinker says:

[QUOTE=”Mark44, post: 5262160, member: 147785″]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.[/QUOTE]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.
[QUOTE=”Mark44, post: 5262160, member: 147785″]Please show me where.[/QUOTE]
[QUOTE=”HallsofIvy, post: 5260930, member: 331″]Now, turning to the problem at hand. 0.999… is, by definition of the decimal numeration system, ##0.9+ 0.09+ 0.009+ cdotcdotcdot## = ##0.9+ 0.9(0.1)+ 0.9(0.01)+ cdotcdotcdot## = ##0.9+ 0.9(.1)+ 0.9(.1^2)+ cdotcdotcdot##, [B]precisely a geometric series with a= 0.9 and r= 0.1[/B]. So the sum, and the value of 0.9999…, is $frac{0.9}{1- 0.1}= frac{0.9}{0.9}= 1$.[/QUOTE]
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.

11. Mark44 says:

[QUOTE=”Algr, post: 5262024, member: 122860″]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.
[/quote]How is this assumption insulting. Just by ordinary division you can show that the 3 digit keeps repeating.
[QUOTE=Algr]And why assume that .999… and 9.999… have different numbers of nines when this would never happen for any finite number?[/QUOTE]
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.
[QUOTE=”SlowThinker, post: 5262131, member: 572662″]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.
[/quote]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.
[QUOTE=SlowThinker]
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.[/QUOTE]Please show me where.

12. FactChecker says:

[QUOTE=”SlowThinker, post: 5262131, member: 572662″]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.[/QUOTE]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.

13. SlowThinker says:

[QUOTE=”FactChecker, post: 5262069, member: 500115″]If they are different, how different are they? Suppose you say that they are different by more than 0.0001. Then 1 – 0.99999 is closer and the more 9’s you add, the closer it gets. So you can not say there is any difference.[/QUOTE]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.

14. FactChecker says:

If they are different, how different are they? Suppose you say that they are different by more than 0.0001. Then 1 – 0.99999 is closer and the more 9’s you add, the closer it gets. So you can not say there is any difference.

15. Greg Bernhardt says:

[QUOTE=”HallsofIvy, post: 5260930, member: 331″]While there are several [B]arguments[/B] here that people might accept as proofs, they all, as stated, are non-rigorous, involving some basic, unstated assumptions.[/QUOTE]
Here is the previous FAQ
[URL]https://www.physicsforums.com/insights/is-there-a-rigorous-proof-of-1-0-999/[/URL]

16. HallsofIvy says:

While there are several [b]arguments[/b] 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 [b]rigorous[/b] proof using “geometric series”. It is easily shown that the sum $sum_{i= 0}^n ar^i$ is $$frac{a(1- r^n}{1- r}$$. To do that, let $S_n= sum_{i= 0}^n ar^i= a+ ar+ ar^2+ cdotcdotcdot+ ar^n)$. Since that is a [b]finite[/b] sum, the usual properties of arithmetic hold and we can write $S_n- a= ar+ ar^2+ cdotcdotcdot+ ar^n= r(a+ ar+ cdotcdotcdot+ ar^{n-1})$. The quantity in the last parentheses is [b]almost[/b] $S_n$ itself. It is only missing the last term, $ar^n$. Restore that by adding $ar^{n+1}$ to both sides: $S_n- a+ ar^n= r(a+ ar+ cdotcdotcdot+ ar^{n-1})+ ar^{n+ 1}$. Taking that last term inside the parentheses we have $S_n- a+ ar^n= r(a+ ar+ cdotcdotcdot+ ar^{n-1}+ ar^n)= rS_n$. We can write that as $rS_n- S_n= (r- 1)S)_n= ar^n- a= a(r^n- 1)$ and dividing both sides by r- 1, $S_n= frac{a(r^n- 1)}{r- 1}$ which, for r< 1 can be written more as $S_n= frac{a(1- r^n)}{1- r}$The geometric series, $sum_{i= 0}^infty ar^i$, is, by definition, the limit of the "partial sums", $S_n= sum_{i= 0}^n ar^i$, as n goes to infinity. As long as $|r|< 1$, $lim_{ntoinfty} r^n= 0$ so that limit is easily calculated as $sum_{i= 0}^infty ar^i= frac{a}{1- r}$.Now, turning to the problem at hand. 0.999... is, by definition of the decimal numeration system, $0.9+ 0.09+ 0.009+ cdotcdotcdot= 0.9+ 0.9(0.1)+ 0.9(0.01)+ cdotcdotcdot= 0.9+ 0.9(.1)+ 0.9(.1^2)+ cdotcdotcdot$, precisely a geometric series with a= 0.9 and r= 0.1. So the sum, and the value of 0.9999..., is $frac{0.9}{1- 0.1}= frac{0.9}{0.9}= 1$.

17. Kegan says:

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

18. Nemika says:

*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!

19. Algr says:

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.  And why assume that .999… and 9.999… have different numbers of nines when this would never happen for any finite number?