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999equals1

Why 1 Equals 0.999… — Explanations & Rigorous Proofs

October 17, 2015/111 Comments/in Mathematics FAQs/by Multiple_Authors
📖Read Time: 2 minutes
📊Readability: Accessible (Clear & approachable)
🔖Core Topics: numbernumbersproofinfinitedecimal

Table of Contents

  • Why do people say 1 and 0.999… are equal? Aren’t they two different numbers?
  • Intuitive proofs that 1 = 0.999…
    • Proof #1 — no number between
    • Proof #2 — simple algebra
    • Proof #3 — shrinking difference
    • Proof #4 — using 1/3
  • Common questions and objections
    • “But every number has a unique representation!”
    • “0.999… gets closer to 1 but never reaches it.”
    • Can we build number systems where 1 ≠ 0.999… ?
    • “In Proof #2, 9.999… seems to have one fewer nine than 0.999…”
    • Further reading
    • Base independence
    • More Related Articles

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

No — 1 and 0.999… really are the same number, although that can feel counterintuitive at first. Below are several accessible (non‑rigorous) proofs, followed by answers to common objections. I preserved the original voice and meaning while improving structure, clarity, and HTML semantics for better readability and SEO.

Intuitive proofs that 1 = 0.999…

Proof #1 — no number between

For any two unequal real numbers there is always another number between them (picture a number line). So, if 0.999… and 1 were different, there would be a number strictly between them. But there is no decimal number that is greater than 0.999… and less than 1. Therefore 0.999… = 1.

Proof #2 — simple algebra

Let x = 0.999…
10x = 9.999…
Subtract: 10x − x = 9.999… − 0.999… → 9x = 9 → x = 1.

Proof #3 — shrinking difference

If 0.999… is a number, how far below 1 can it be? It’s larger than 0.9999, so it is less than 0.0001 below 1. It’s also larger than 0.9999999, so it is less than 0.0000001 below 1. Continuing indefinitely shows the difference is smaller than 10⁻ⁿ for every n, so the difference must be 0. If two numbers differ by 0, they are the same, so 0.999… = 1.

Proof #4 — using 1/3

1/3 = 0.333… . Multiply both sides by 3: 1 = 3×(1/3) = 3×(0.333…) = 0.999… .

All four arguments are correct as intuitive proofs. (For a fully rigorous proof you need to work with limits or the completeness of the real numbers; the next post in the original thread gives a formal version.)

Common questions and objections

“But every number has a unique representation!”

That is a common misconception. Numbers can have multiple representations: for example, 1/3 = 2/6 = 3/9 = 0.333… . Similarly, 1 and 0.999… are two different decimal representations of the same real number. Accepting this avoids unnecessary complications in the number system.

“0.999… gets closer to 1 but never reaches it.”

That phrasing treats 0.999… as a process rather than a number. Mathematicians treat 0.999… as a completed infinite decimal, i.e. a number in the same sense as 2 or 3. Saying “it gets close but never reaches” is therefore misleading.

Can we build number systems where 1 ≠ 0.999… ?

Yes, you can define nonstandard or extended number systems where representations behave differently, but those systems are less useful for everyday analysis and calculus because they don’t match the usual completeness and limit properties of the real numbers.

“In Proof #2, 9.999… seems to have one fewer nine than 0.999…”

This confusion comes from misunderstanding infinity. Both 0.999… and 9.999… have infinitely many nines; removing one nine from an infinite sequence still leaves an infinite sequence. A helpful analogy is the infinite sets A = {0,1,2,3,…} and B = {1,2,3,4,…}: both are countably infinite and can be put into one‑to‑one correspondence (0↔1, 1↔2, 2↔3, …). The same intuition about infinity applies to the decimal expansions.

Further reading

Read this supportive article: Is there a rigorous proof of 1 = 0.999…?

Base independence

This phenomenon is not unique to base 10. For example, in base 2 we have 1 = 0.111… in the same way; every positional base has analogous non‑unique representations for some numbers.

The following forum members contributed to the original FAQ: AlephZero, Fredrik, micromass, tiny-tim, vela.

Click for forum comments

Multiple_Authors

This article was authored by several Physics Forums members with PhDs in physics or mathematics.

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111 replies
« Older Comments
  1. Isaac0427 says:
    May 19, 2016 at 11:29 pm

    “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.

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  2. Kegan says:
    May 19, 2016 at 11:29 pm

    “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

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  3. Mark44 says:
    May 19, 2016 at 11:29 pm

    “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.

    Log in to Reply
  4. Mark44 says:
    May 19, 2016 at 11:29 pm

    “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.

    Log in to Reply
  5. Kegan says:
    May 19, 2016 at 11:29 pm

    “and no quantum physics in high school”
    i really want to go to california institute of technology

    Log in to Reply
  6. Kegan says:
    May 19, 2016 at 11:29 pm

    “what bothers me they don’t have physics in the 6th grade”
    and no quantum physics in high school

    Log in to Reply
  7. Kegan says:
    May 19, 2016 at 11:29 pm

    “yeah im 11 i like physics and math”

    what bothers me they don’t have physics in the 6th grade

    Log in to Reply
  8. Kegan says:
    May 19, 2016 at 11:29 pm

    “Gauss name is commonly mentioned in Physics. Still, between 10 and 20. Right?”

    yeah im 11 i like physics and math

    Log in to Reply
  9. Kegan says:
    May 19, 2016 at 11:29 pm

    “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

    Log in to Reply
  10. WWGD says:
    May 19, 2016 at 11:29 pm

    “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?

    Log in to Reply
  11. Kegan says:
    May 19, 2016 at 11:29 pm

    “[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

    Log in to Reply
  12. WWGD says:
    May 19, 2016 at 11:29 pm

    “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.

    Log in to Reply
  13. Kegan says:
    May 19, 2016 at 11:29 pm

    “oh ok and i have a question for you how old do you think i am ,seriously”

    And who is Gauss

    Log in to Reply
  14. Kegan says:
    May 19, 2016 at 11:29 pm

    “Close: What Would Gauss Do.”

    oh ok and i have a question for you how old do you think i am ,seriously

    Log in to Reply
  15. WWGD says:
    May 19, 2016 at 11:29 pm

    “and i really like your username does it stand for” what would god do””
    Close: What Would Gauss Do.

    Log in to Reply
  16. Kegan says:
    May 19, 2016 at 11:29 pm

    “yeah I Know i just forgot the….”

    and i really like your username does it stand for” what would god do”

    Log in to Reply
  17. Kegan says:
    May 19, 2016 at 11:29 pm

    “[USER=584906]@Kegan[/USER] It is not .999 , it is .9999….. with an infinite string of 9s.”

    yeah I Know i just forgot the….

    Log in to Reply
  18. Kegan says:
    May 19, 2016 at 11:29 pm

    “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 ?

    Log in to Reply
  19. Isaac0427 says:
    May 19, 2016 at 11:29 pm

    “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.

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  20. jbriggs444 says:
    May 19, 2016 at 11:29 pm

    “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.

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  21. WWGD says:
    May 19, 2016 at 11:29 pm

    “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.

    Log in to Reply
  22. micromass says:
    May 19, 2016 at 11:29 pm

    “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.

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  23. MrAnchovy says:
    May 19, 2016 at 11:29 pm

    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]

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  24. WWGD says:
    May 19, 2016 at 11:29 pm

    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.

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  25. MrAnchovy says:
    May 19, 2016 at 11:29 pm

    “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.

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  26. micromass says:
    May 19, 2016 at 11:29 pm

    “Instead of thinking of .999… as a number”

    No. 0.99999…. is a number.

    Log in to Reply
  27. Isaac0427 says:
    May 19, 2016 at 11:29 pm

    “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?

    Log in to Reply
  28. WWGD says:
    May 19, 2016 at 11:29 pm

    “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.

    Log in to Reply
  29. Gjmdp says:
    May 19, 2016 at 11:29 pm

    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…)

    Log in to Reply
  30. HallsofIvy says:
    May 19, 2016 at 11:29 pm

    “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.

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  31. FactChecker says:
    May 19, 2016 at 11:29 pm

    “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.

    Log in to Reply
  32. gmax137 says:
    May 19, 2016 at 11:29 pm

    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?

    Log in to Reply
  33. pwsnafu says:
    May 19, 2016 at 11:29 pm

    “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”.

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  34. jbriggs444 says:
    May 19, 2016 at 11:29 pm

    “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.

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  35. Isaac0427 says:
    May 19, 2016 at 11:29 pm

    “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.

    Log in to Reply
  36. lavinia says:
    May 19, 2016 at 11:29 pm

    “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.

    Log in to Reply
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