Q: Is 0.9999… repeating really equal to 1?

Mathematician: Usually, in math, there are lots of ways of writing the same thing. For instance:

\frac{1}{4} = 0.25 = \frac{1}{\frac{1}{1/4}} = \frac{73}{292} = (\int_{0}^{\infty} \frac{\sin(x)}{\pi x} dx)^2
 
As it so happens, 0.9999… repeating is just another way of writing one. A slick way to see this is to use:

0.9999... = (9*0.9999...) / 9 = ((10-1) 0.9999...) / 9
 
= (10*0.9999... - 0.9999...) / 9
 
= (9.9999... - 0.9999...) / 9
 
= (9 + 0.9999... - 0.9999...) / 9 = 9 / 9 = 1
 

One.

Another approach, that makes it a bit clearer what is going on, is to consider limits. Let’s define:

p_{1} = 0.9
p_{2} = 0.99
p_{3} = 0.999
p_{4} = 0.9999

and so on.

Now, our number 0.9999... is bigger than p_{n} for every n, since our number has an infinite number of 9’s, whereas p_{n} always has a finite number, so we can write:

p_{n} < 0.9999... \le 1 for all n.

Taking 1 and subtracting all parts of the equation from it gives:

1-p_{n} > 1-0.9999... \ge 0

Then, we observe that:

1 - p_{n} = 1 - 0.99...999 = 0.00...001 = \frac{1}{10^n}
and hence

\frac{1}{10^n} > 1-0.9999... \ge 0.

But we can make the left hand side into as small a positive number as we like by making n sufficiently large. That implies that 1-0.9999… must be smaller than every positive number. At the same time though, it must also be at least as big as zero, since 0.9999… is clearly not bigger than 1. Hence, the only possibility is that

1-0.9999... = 0

and therefore that

0.9999... = 1.

What we see here is that 0.9999… is closer to 1 than any real number (since we showed that 1-0.9999… must be smaller than every positive number). This is intuitive given the infinitely repeating 9’s. But since there aren’t any numbers “between” 1 and all the real numbers less than 1, that means that 0.9999… can’t help but being exactly 1.

Update: As one commenter pointed out, I am assuming in this article certain things about 0.9999…. In particular, I am assuming that you already believe that it is a real number (or, if you like, that it has a few properties that we generally assume that numbers have). If you don’t believe this about 0.9999… or would like to see a discussion of these issues, you can check this out.

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90 Responses to Q: Is 0.9999… repeating really equal to 1?

  1. mathman says:

    lol. You need to know what each “number” is before you can take the difference. Of course the difference is zero, because you say that they are the same before you subtract, assuming you are subtracting “numbers”

  2. George says:

    While the mathematical argument presented here is correct, I have to say that limit and/or mathematical infinity introduced created confusions for many. What I wanted to say is that math infinity is really a math trick. It cannot and should not be real in the real physical world. Today this infinity plagues physics and its calculations. Maybe we need something new…

  3. EMINEM says:

    go to this https://www.youtube.com/watch?v=wsOXvQn3JuE site and get a mathematical proof that 0.999… doesnot equal to 1….

  4. . says:

    EMINEM, the video was published on April fool’s day for a reason!

  5. Josh says:

    The simplest explanation I’ve seen for this is:

    1/3 = 0.333…
    (1/3)3=1 = (0.333…)3=0.999…
    1 = 0.999…

    I don’t see how anyone can come to any conclusion other than 1=0.999… after understanding this.

  6. mathman says:

    1 / 3 = ??
    Use long division. There are 2 parts: the quotient and the remainder. For every iteration.
    First iteration is 0 R 1/3
    Second is 0.3 R 1/30
    Third is 0.33 R 1/300
    Etc etc.

    Where does your remainder go, Josh?
    You don’t account for it anywhere.

    The remainder is never zero.

    How many 9’s are there in your 0.999… ?

  7. Josh says:

    Hey mathman,

    There are infinitely many 9s in my “0.999…”. If it were anything other than infinitely many, then it would be less than 1.

    And as far as 1/3, your “etc. etc.” is an infinitely repeating “0.333…”. I’m failing to see how anything you’ve presented has disproved what I said.

  8. Angel says:

    @mathman:

    The remainder does not need to be accounted for. The division 1/3 is never formally complete if there is a remainder. Therefore, one must keep iterating the operations in long division. As the number of iterations n approaches infinity, the remainder does approach zero. That is what matters here. The number of 3s in the decimal expansion of 1/3 is infinite, and so is the number of 9s in the figure 0.999…

  9. Anthony.R.Brown says:

    The Original author of the Proof below…used in this Thread 🙂

    http://www.mathisfunforum.com/viewtopic.php?id=6069

    It’s impossible for Infinite/Recurring 0.9 (0.999…) to ever equal 1

    Because of the Infinite/Recurring 0.1 (0.001…) Difference! 🙂

    Only by the using the + Calculation can it be possible (0.999…) + (0.001…) = 1

    Anthony.R.Brown

  10. Ángel Méndez Rivera says:

    Anthony R. Brown,

    Your argument is flawed for the simple reason that the “infinite difference” you describe equivalent to 0.000…1 is not a number. It doesn’t exist in mathematics. It can’t exist in mathematics. It is more of a problematic number than 0/0 or defining a left-identity of exponentiation Phi such that Phi^x=x. It is simply nonsense.

  11. Mathman says:

    Anthony,
    Being someone who has researched and understands both sides of this issue, you will be told it’s nonsense by mathematicians. It’s their way of putting themselves above you to put you down. The issue with your argument is that 0.001… is defined as 0.00111111111111…. You are having a problem defining what the idea in your head is. 0.000…1 is also not a number. Mathematicians for some reason cannot visualize repeated division the same as they can visualize it in addition. There quick to use the word nonsense, and then in the next breath say that by adding infinite numbers they can have a finite result.

  12. Anthony.R.Brown says:

    Hi Ángel Méndez Rivera

    Your try at a counter argument…saying (0.001…) does not exist in mathematics ? can’t exist in mathematics if (0.999…) exist’s which it does! 🙂
    The two are the related from the calculation 1 – (0.999…) = (0.001…) one cannot exist without the other!

    Anthony.R.Brown

  13. Anthony.R.Brown says:

    I had the reply below from Mathman…

    Anthony,
    Being someone who has researched and understands both sides of this issue, you will be told it’s nonsense by mathematicians. It’s their way of putting themselves above you to put you down. The issue with your argument is that 0.001… is defined as 0.00111111111111…. You are having a problem defining what the idea in your head is. 0.000…1 is also not a number. Mathematicians for some reason cannot visualize repeated division the same as they can visualize it in addition. There quick to use the word nonsense, and then in the next breath say that by adding infinite numbers they can have a finite result.

    Then when I wanted to reply the message above is not shown ? so I will Answer below…

    First of all what you are showing as 0.00111111111111…. is wrong! there is only a single (1) following the Infinite/recurring calculation so your example above should look like 0.00000000000001….
    The amount of zeros depends on how far one wants to show the calculation regarding decimal places!

    Example 0.9… has the infinite/recurring difference 0.1…
    Example 0.99… has the infinite/recurring difference 0.01…
    Example 0.99… has the infinite/recurring difference 0.001…

    So as you can see the infinite/recurring difference is always the same length as the infinite/recurring 0.9 🙂

    Anthony.

  14. Mathman says:

    0.001….. = 0.0011111111111111111…..
    Anthony do you know what you are taking about?

  15. Anthony.R.Brown says:

    Hi Mathman

    Yes I fully know what I am talking about but you appear not to ?

    As I pointed out in your example ? = 0.0011111111111111111….. it is wrong there can only be a single (1) otherwise when you do the calculation…

    0.0011111111111111111….. + 0.0000000000000000009….. it will equal something like = 0.0011111201111111111… ?

    Where as the True calculation (with both sides the 0.9… and the 0.1… the same amount of decimal places)

    Calculates 0.9999999999…. + 0.0000000001…. = 1

    Regards Anthony.

  16. Joshua says:

    The flaw in your logic is thinking that there is ever space for a 0.0…01. The number 0.999… repeating forever by definition has no room to add anything to it. The 9’s never stop to allow the insertion of any variation of 0.0001. And if there is no room to ever add anything to 0.999… infinitely repeating, then there is no space in between 0.999… infinitely repeating and 1, and therefore they are equal. If the 9’s stopped repeating at any point, you would be correct, but they do not stop repeating, so you are incorrect.

  17. Mathman says:

    The ellipsis “…” mean “repeating”. I know what you are trying to say but you are using symbols incorrectly.
    1/1000… would be a more accurate to convey your idea.

  18. Anthony.R.Brown says:

    Hi Joshua

    Your Quote:”The number 0.999… repeating forever by definition has no room to add anything to it”

    A.R.B

    The same as “The number 0.001… repeating forever by definition has no room to add anything to it”

    They both run side by side! 🙂

    0.99999999999999999999999999999999999999999999999999999999….
    0.0000000000000000000000000000000000000000000000000001….

    It’s only when one want’s to make the Value (1) that the + Calculation is needed it’s the only way!

    ( 0.9999999999….) + (0.0000000001….) = 1

  19. Joshua says:

    Sorry Anthony, you have not wrapped your head around what 0.999… repeating for infinity means. It’s the infinite part that’s really important. I promise you, you are wrong, and once you understand what infinitely repeating means, you’ll get it.

  20. Ángel Méndez Rivera says:

    Anthony, no. They don’t run side by side. The number you describe as the difference between 1 and 0.999… doesn’t exist, because by definition, it is impossible to have a 1 after an infinite amount of zeros. It just is. There is no such calculation as the one you performed to be made because the number you describe doesn’t exist.

  21. Mathman says:

    There is an undefined infinitesimal space between 0.999… and 1. Just because it’s undefined doesn’t mean it doesn’t exist.

  22. Ángel says:

    Mathman,

    1. If it is undefined, then it does not exist. One implies the other.
    2. No, there is no infinitesimal space between 0.999… = 1, and this can be proven.

    0.999… = 9/10 + 9/100 + 9/1000 + … this is an infinite geometric series, which is evaluated and in fact defined by the formula a/(1 – r), so the sum 0.999.. = (9/10)/(1 – 1/10) = (9/10)/(9/10) = 1. As for the formula a/(1 – r), it can be easily derived and proven

  23. Anthony.R.Brown says:

    ( 1.1 ) x 0.9 = 0.99 ” One Decimal Place = ” 0.01 < 1

    ( 1.11 ) x 0.9 = 0.999 " Two Decimal Place's = " 0.001 < 1

    ( 1.111 ) x 0.9 = 0.9999 " Three Decimal Place's = " 0.0001 < 1

    ( 1 / 0.9 ) x 0.9 = 0.9… " Infinite Decimal Place's = " 0.1… < 1 "

  24. Joshua says:

    Mathman,

    You would be correct if the repeating 0.999… ever stopped, no matter how long it was. But the number that we are talking about literally never stops, and therefore does not leave any space. There is absolutley no number that would fit in between it and 1.

    Most of the confusion over this matter seems to be about the nature of infinity, and how it applies to infinitely repeating decimals. The importance of wrapping our heads around what “infinitely repeating” means cannot be overstated.

  25. Ángel Méndez Rivera says:

    Anthony,

    (1.1)(0.9) = 0.99

    (1.11)(0.9) = 0.999

    (1.111)(0.9) = 0.9999

    (1.111…)(0.9) = 0.999…. = (1/0.9)(0.9), because 1.1111… = 1/0.9. But any number divided by itself equals 1, so (0.9)/(0.9) = 1, therefore 1 = 0.9999….

  26. Anthony.R.Brown says:

    Quote: “Joshua says:
    You would be correct if the repeating 0.999… ever stopped, no matter how long it was. But the number that we are talking about literally never stops, and therefore does not leave any space. There is absolutley no number that would fit in between it and 1.”

    A.R.B

    Absolutely Correct!!! So now you and I agree it’s impossible for 0.999…to ever stop and become equal to 1 🙂

    So how do we make 0.999… = 1 and that’s where we have the answer!…

    ( 0.999…) + (0.001…) = 1

    {When great minds think alike Mountains can be moved!} 🙂

  27. Joshua says:

    No, Anthony, you still don’t understand. The fact that it never stops is the very thing that makes it equal to 1. Again, I come back to encouraging you to take the time to actually understand what an infinitely repeating decimal is.

  28. Anthony.R.Brown says:

    Hi Joshua

    What you don’t seem to understand is that infinite/recurring 0.999… is a Calculation on a Number! and the number Starts as 0.9 (It’s not something that is just plucked from the sky from nothing!)

    Your biggest mistake! is “The fact that it never stops is the very thing that makes it equal to 1” ? a Number that Starts < 1 can never be made to equal 1 unless the + Calculation is used!

  29. Joshua says:

    Anthony, the number 0.999… with infinite 9s is not a calculation. It is a static number. The 9s are not constantly being generated, they exist (all infinity of them) at once. It does not “start” at 0.9. It exists as 0.999… with an infinite number of 9s. You don’t have to work up to it, it just is.

    It is clear that you still do not understand what an infinitely repeating decimal is. You do not understand the basic premise of what we are talking about. Therefore, you can not speak to it in an intelligent or useful manner.

    I’m not trying to be mean or anything, either. I appreciate people pointing out when I am wrong so I can learn what is correct. You clearly don’t understand the basics of this topic, therefore you are wrong, and it is an opportunity for you to learn something new and broaden your knowledge base.

  30. Angel says:

    “What you don’t seem to understand is that recurring 0.999… is a calculation on a number!”

    This is wrong. 0.999… is a number. It isn’t a process, it already is a number. It doesn’t “start” out as 0.9. It already is and exists as 0.999…, it doesn’t start anywhere or end anywhere. It isn’t a changing or moving thing. It just is a number.

  31. Anthony.R.Brown says:

    So Joshua & Angel…

    Are both wrong! if they think (the number 0.999… with infinite 9s is not a calculation) ???

    Because all the so called Proofs ? are based on some kind’s of Calculations!!!

    Just to back that up…a Google search for 0.999… proofs shows (About 367,000 results) 21/09/2017

    And they are all some kind’s of Calculations! 🙂

    One fine example is shown below…with many Calculations!

    https://en.wikipedia.org/wiki/0.999

    Anthony.R.Brown

  32. Mathman says:

    Ángel,
    a/(1-r) comes from a limit. The LIMIT of the infinite series is 1. The definition is flawed and contradicts the very theory of limits.

    It may be impossible to have a 1 AFTER an “infinite amount”. That is because “infinite amount” doesn’t exist. Stop trying to quantify infinity. It is not impossible to describe the idea, however: 1/1000…

    The very definition of limits show there is ALWAYS AN EPSILON.
    n=1: 1-0.9= 0.1
    n=2: 1-0.99 = 0.01
    For ALL values n, there is a difference, or, contradictory to Anthony, there is always a space. If you travserse 90% of the remaining distance , ad infinitum, there will always be 10% of the last distance ahead of you. You’ll never complete the journey of say, 1 mile, going 90% each time.

    I argue that “0.999…” is not a number at all.
    Σ (n=1 to N) 9/10^n as N goes to infinity, implies there is always a rational number being added to the previous total. The number 1 has nothing being addd to it. Therefore they aren’t the same.
    Also, ‘something’ that always has another addend can’t be a number. It’s contradictory because it always has another number being added to it.

  33. Mathman says:

    Thus: 1-0.999(n-times) = 1/10^n
    You may let n get as arbitrarily large as you like.
    You cannot have “infinite amount of 9’s” though.
    You can have the LIMIT as the number of 9’s gets arbitrarily large. This is a number. “Endless 9’s” really is nonsense when it comes to a number which should really be finite.
    So then, for ALL n in N, that is, for every integer, there is a difference which is smaller than the last and never 0.
    Learn what arbitrarily large means, and how limits really work, and what infinite means, and how infinity is applied in calculus. Infinity and limits go hand in hand. You don’t plop infinity wherever you feel.

  34. Anthony.R.Brown says:

    LIMIT Cannot be used with Infinite/Recurring 0.9 (0.999…) because you will then Contradict the term (Infinite/Recurring) by forcing the .9’s to cut off at some point depending on where you want the LIMIT to be, regarding decimal places calculated,this is much the same as the cut off point used when Infinite/Recurring 0.9 (0.999…) is forced at some point to become equal to 1

    So back to…

    It’s impossible for Infinite/Recurring 0.9 (0.999…) to ever equal 1 (Unless you use a LIMIT) ? 🙁
    And the Infinite/Recurring 0.1 (0.001…) Difference! (Also does not us a LIMIT)

    Both are LIMITless 🙂

    Only by the using the + Calculation can it be possible for (0.999…) + (0.001…) = 1

  35. Ángel Méndez Rivera says:

    No, we do not get to choose where the limit is. The limit is a number larger than every element of the sequence of partial evaluations of a function, which means that the limit is larger than the sequence itself. In other words, the term of the sequence which equals the limit is itself located at an infinite position of the sequence. Any finite evaluation yields less than the limit. However, 0.99… has an infinite number of 9s, so it exactly equals the limit. We’re not cutting it off to equal 1, you’re cutting it off to equal less than 1, but you’re Not actually using the sequence of infinite nines, so you’re argument is wrong.

    There is no such a thing as limitless in mathematics, and 0.000…1 is nonsense and does not exist.

  36. Ángel Méndez Rivera says:

    Mathman,

    Read my previous response to your argument. I already explained why your description of the definition of a limit is wrong and why the idea of there always being an epsilon is irrelevant when actually evaluating the limit of a function as the argument approaches c.

  37. Anthony.R.Brown says:

    Quote: “Ángel Méndez Rivera says:
    September 22, 2017 at 1:50 pm
    No, we do not get to choose where the limit is.

    A.R.B
    Yes! you do by using it in the first place!

    Quote: “Ángel Méndez Rivera says:
    “However, 0.99… has an infinite number of 9s, so it exactly equals the limit. ”

    A.R.B
    How can an infinite number of 9s, exactly equals the limit (You have set!) which is 1 ? which is an Integer number > 0.99…

    Quote: “Ángel Méndez Rivera says:
    There is no such a thing as limitless in mathematics, and 0.000…1 is nonsense and does not exist.

    A.R.B
    Infinite/Recurring 0.9 (0.999…)
    Infinite/Recurring 0.1 (0.001…)
    Are both (Individually!) LIMITless they only have a LIMIT when added! together to equal 1
    (0.999…) + (0.001…) = 1

  38. The Physicist The Physicist says:

    @Anthony.R.Brown
    They’re not using the word “limit” the way any reasonable English speaking person uses the word limit, they mean it in the mathematical sense. You can read about what a mathematical limit is here.

  39. Anthony.R.Brown says:

    The Physicist says: Quote:”They’re not using the word “limit” the way any reasonable English speaking person uses the word limit, they mean it in the mathematical sense. You can read about what a mathematical limit is here.”

    A.R.B

    The word (LIMIT) however you define it ? or look it up on some Math sites ? is being used in a way… and the same way ? so as to make a Calculated Number,that starts life less than 1 somehow become equal to 1 ? 🙁

    A better way to look at the problem is…

    Infinite/Recurring 0.9 (0.999…) = Universe! 🙂

    And if we place a Wall at the End of the Universe! ? there will still be Universe! behind the Wall…

    0.9 (0.999…) Universe! { Wall = 1} 0.9 (0.999…) Universe!…

  40. Ángel Méndez Rivera says:

    Anthony, no. There is no such a thing as a limitless number. Please, study the topic you’re talking on before you use terminology. The Physicist already told you where you can study the idea of what a limit is. Take the suggestion and study it.

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