Why 1 = 0.999…: Four Simple Proofs Explained
Yes, 1 and 0.999… are exactly the same real number, not merely two numbers that happen to be extremely close. This surprises many people because 0.999… looks like a process of approaching 1 rather than a fixed value, but treated as a complete infinite decimal, it equals 1 exactly. Several independent, intuitive arguments confirm this, and a fully rigorous version relies on the completeness property of the real numbers.
Table of Contents
Key Takeaways
- 1 and 0.999… are identical values, not two numbers separated by an infinitely small gap.
- Algebraic manipulation of x = 0.999… leads directly to x = 1 in three steps.
- Real numbers can have more than one decimal representation, as with 1/3 = 0.333… = 2/6.
- The same pattern holds in other number bases, such as 1 = 0.111… in base 2.
- No rigorous number system used in standard calculus treats 0.999… as less than 1.
Four intuitive proofs that 1 = 0.999…
Proof 1: There is no number between them
Between any two distinct real numbers, another real number always exists somewhere on the number line. If 0.999… and 1 were different values, a number would have to exist strictly between them. No such decimal number exists that is greater than 0.999… and less than 1, so 0.999… and 1 cannot be different numbers.
Proof 2: Simple algebra
Let x = 0.999…
10x = 9.999…
Subtracting the first equation from the second gives 10x − x = 9.999… − 0.999…, which simplifies to 9x = 9, so x = 1.
Proof 3: The shrinking difference
0.999… is larger than 0.9999, so the gap between 0.999… and 1 is smaller than 0.0001. It is also larger than 0.9999999, making the gap smaller than 0.0000001. Extending this pattern indefinitely shows the difference between 0.999… and 1 is smaller than 10⁻ⁿ for every possible value of n. A gap that is smaller than every positive number must equal zero, and two numbers with a difference of zero are the same number.
Proof 4: Using 1/3
1/3 equals 0.333… as a repeating decimal. Multiplying both sides of that equation by 3 gives 1 = 3 × (1/3) = 3 × 0.333… = 0.999…, confirming the same result from a different starting point.
Each of these four arguments is a valid intuitive demonstration rather than a formal mathematical proof. A fully rigorous proof requires working with limits or the completeness property of the real numbers.
Common objections, answered
Doesn’t every number have a unique decimal representation?
No, that assumption is a common misconception. Many real numbers have more than one valid decimal representation. For example, 1/3 equals 2/6 and also 3/9, and all three fractions equal the repeating decimal 0.333…. In the same way, “1” and “0.999…” are simply two different representations of one identical real number.
Doesn’t 0.999… just get closer to 1 without ever reaching it?
That description treats 0.999… as an ongoing process instead of a fixed number. Mathematicians define 0.999… as a completed infinite decimal, a specific number in exactly the same sense that 2 or 3 are specific numbers. Describing it as something that “approaches but never reaches” 1 mischaracterizes what the notation actually represents.
Can a number system be built where 1 does not equal 0.999…?
Nonstandard or extended number systems can define representations that behave differently from the standard real numbers. These systems are generally less useful for everyday analysis and calculus, however, because they do not preserve the usual completeness and limit properties that make the real number system work consistently.
In Proof 2, doesn’t 9.999… have one fewer nine than 0.999…?
This objection comes from a misunderstanding of how infinite sequences work. Both 0.999… and 9.999… contain infinitely many nines, and removing one nine from an infinite sequence still leaves an infinite sequence. The infinite sets A = {0, 1, 2, 3, …} and B = {1, 2, 3, 4, …} illustrate the same idea: both sets are countably infinite and can be matched one-to-one (0 with 1, 1 with 2, 2 with 3, and so on), even though B appears to be “missing” the element 0.
Does this only happen in base 10?
No, this phenomenon appears in every positional numbering system, not only base 10. In base 2 (binary), for instance, 1 equals 0.111… in exactly the same way that 1 equals 0.999… in base 10. Every positional base has analogous cases of non-unique representations for certain numbers.
Frequently Asked Questions
Is 0.999… exactly equal to 1, or just extremely close to it?
0.999… is exactly equal to 1. It is not an approximation or a value infinitesimally smaller than 1. Treated as a completed infinite decimal, it represents precisely the same real number as 1, confirmed by multiple independent algebraic and logical arguments.
Why does this feel wrong even though it’s mathematically correct?
The confusion usually comes from picturing 0.999… as a number that is still “in the process” of adding more nines, rather than as a fixed, completed value. Once 0.999… is understood as a specific number rather than a sequence in motion, the equality with 1 follows naturally from basic algebra.
Is the algebra proof (10x − x = 9) considered rigorous by mathematicians?
The algebraic proof is a valid and convincing intuitive argument, but it is not considered fully rigorous on its own. A rigorous treatment requires defining 0.999… formally using limits or the completeness property of the real numbers.
Does 1 = 0.999… apply in number bases other than base 10?
Yes. The same pattern of non-unique decimal representation occurs in every positional numbering system. In base 2, for example, 1 equals 0.111… for the same underlying reason that 1 equals 0.999… in base 10.
Can I read a more formal, rigorous version of this proof?
Yes, a more formal treatment using limits and the completeness of the real numbers is available in Is there a rigorous proof of 1 = 0.999…? on Physics Forums Insights.
This FAQ draws on discussion originally contributed by forum members AlephZero, Fredrik, micromass, tiny-tim, and vela. The original discussion thread is available at Why do people say that 1 and .999… are equal? on Physics Forums.
This article was authored by several Physics Forums members with PhDs in physics or mathematics.
“yeah that does make more sense so what you saying is that .999… to infinity approaches the number 1 so close it is consideribly the same number is that what you are saying ?”
No, .999… is NOT a sequence. It is a number. The … means that the value of the number is equal to the limit of the sequence (.9, .99, .999, …). If it helps, you can think of it as the term of the sequence in which n=infinity. The value of this term, by definition, will be the limit of the sequence as n approaches infinity. If you do not understand this, look up limits.
“No, .999… doesn’t approach 1 — it is exactly equal to 1.
The limit of the sequence {.9, .99, .999, …} is 1, which means the farther you go in the sequence, the closer a number in the sequence is to 1.
BTW, it’s redundant to write “to infinity” after .999… The dots (an ellipsis) already means that the 9 digits repeat endlessly.”
Ok that makes sense the more 9’s you have the closer it is to the number 1 ,and i’m only eleven and language arts is not my best subject
“what bothers me they don’t have physics in the 6th grade”
You can’t understand physics without having at least a competency in algebra, or better yet, calculus.
“and no quantum physics in high school”
But they do have regular (or Newtonian) physics in high school. I took that when I was a senior at my high school.
“yeah that does make more sense so what you saying is that .999… to infinity approaches the number 1 so close it is consideribly the same number is that what you are saying ?”
No, .999… doesn’t approach 1 — it is exactly equal to 1.
The limit of the sequence {.9, .99, .999, …} is 1, which means the farther you go in the sequence, the closer a number in the sequence is to 1.
BTW, it’s redundant to write “to infinity” after .999… The dots (an ellipsis) already means that the 9 digits repeat endlessly.
“and no quantum physics in high school”
i really want to go to california institute of technology
“what bothers me they don’t have physics in the 6th grade”
and no quantum physics in high school
“yeah im 11 i like physics and math”
what bothers me they don’t have physics in the 6th grade
“Gauss name is commonly mentioned in Physics. Still, between 10 and 20. Right?”
yeah im 11 i like physics and math
“oh ok im more of a quantum Phyisics guy and how old do you think i am please let me know”
thanks for the link that was really cool i did not know that
“oh ok im more of a quantum Phyisics guy and how old do you think i am please let me know”
Gauss name is commonly mentioned in Physics. Still, between 10 and 20. Right?
“[URL]https://en.wikipedia.org/wiki/Carl_Friedrich_Gauss[/URL]
Maybe the best Mathematician of all times. Hope he is not pissed I am using his name.”
oh ok im more of a quantum Phyisics guy and how old do you think i am please let me know
“And who is Gauss”
[URL]https://en.wikipedia.org/wiki/Carl_Friedrich_Gauss[/URL]
Maybe the best Mathematician of all times. Hope he is not pissed I am using his name. I assume you are young , since you’re interested in Math and don’t know about Gauss. Maybe between 10 and 20.
“oh ok and i have a question for you how old do you think i am ,seriously”
And who is Gauss
“Close: What Would Gauss Do.”
oh ok and i have a question for you how old do you think i am ,seriously
“and i really like your username does it stand for” what would god do””
Close: What Would Gauss Do.
“yeah I Know i just forgot the….”
and i really like your username does it stand for” what would god do”
“[USER=584906]@Kegan[/USER] It is not .999 , it is .9999….. with an infinite string of 9s.”
yeah I Know i just forgot the….
“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?”
yeah that does make more sense so what you saying is that .999… to infinity approaches the number 1 so close it is consideribly the same number is that what you are saying ?
“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.”
That was the point. I understand it is a number, but the value of the number can be thought of the limit of that sequence. IMO, this is easier to comprehend, but as you said, it’s a mater of opinion.
“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.”
I was going to write that one could consider the distinction between numerals or formulas on the one hand and numbers on the other. It seems that the distinction you want to make is between exemplars of an equivalence class and the class itself.
6 of one, ##frac{dozen}{2}## of the other.
“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.
“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.
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 [s]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.[/s]
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.
“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.
“Instead of thinking of .999… as a number”
No. 0.99999…. is a number.
“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?
“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”
[USER=584906]@Kegan[/USER] It is not .999 , it is .9999….. with an infinite string of 9s.
The easiest proof for me is that:
0.9999999…=0.3333333…*3
0.3333333…=1/3
0.9999999…=(1/3)*3=1
Then 0.9999999…=1
Look that also, 0.999999…8=0.99999999(=1):
1.999999999…8=0.6666666666…*3
0.66666666…=2/3
1.999999…8=(2/3)*3=2=1+1=1+0.9999999…=1.999999…
Then 1.999999…8=1.999999999…
1.999999…8 -1= 1.99999… -1
So 0.999999…8=0.999999(=1…)
“Going back to post #2
Maybe (probably) I’m being dense, but didn’t you just add [itex]ar^n[/itex] to the left and [itex]ar^{n+ 1}[/itex] to the right?”
Yes, that was a typo. It should have been [itex]ar^{n+1}[/itex] on both sides.
“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.
Going back to post #2
”
…
Restore that by adding [itex]ar^{n+1}[/itex] to both sides:
[itex]S_n- a+ ar^n= r(a+ ar+ cdotcdotcdot+ ar^{n-1})+ ar^{n+ 1}[/itex]
…”
Maybe (probably) I’m being dense, but didn’t you just add [itex]ar^n[/itex] to the left and [itex]ar^{n+ 1}[/itex] to the right?
“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”.
“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.
“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.
“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.