Frequently Questioned Answers: 0.999… = 1

Having looked at two common questions in probability that are often challenged, let’s turn to the realm of numbers. Non-terminating decimals are inherently problematic, and one particular example causes difficulty for many, even after they fully accept the mathematics of it. Our FAQ page on this topic, at 0.9999… = 1, is very brief, and doesn’t begin to cover the discussions we’ve had. Repeated challenges force us to look for different approaches and deeper explanations.

Proving and believing

The first question we got, within a couple months of the start of the service in 1994, came from Joe, who could give a good proof of the fact, but, like many, wasn’t satisfied:

Point nine repeating equals one?

I have had a question about this for years but I don't remember finding a satisfactory answer.

When I learned how to convert repeating decimals to fractions, we were given the following example:
             _                   _
   Let n = .99      so  10n = 9.99

Subtracting the first equation from the second yields:

   9n = 9 since the repeating decimals subtract out
                                              _
which gives us n = 1,  but we know that n = .99  so
     _
   .99 = 1   

The problem I have is that I can't logically believe this is true, and I don't see an error with the math, so what am I missing or forgetting to resolve this?

(If memory serves, he also said that there are several other ways of proving that .9999...  = 1 but I don't remember them)

There are two issues: how to prove it convincingly, and how to really believe it. We’ll deal later with the things that stand in the way of believing it. Doctor Steve first quickly suggested one of the other approaches to a proof:

I think it is true and you did a beautiful job presenting it. 

If it was not equal to one then there would be a number between it and 1. What number would that be?

That’s just a hint at an idea we’ll see more fully developed later.

For now, Molly (Doctor Sydney) offered additional ideas:

Hello there!  Thanks for writing Dr. Math.  You asked an excellent question, and I liked Steve's first response to you, but I thought I might add two things.

Another way to think about this is this:  Would you agree that 1/3 = .33333...?  .3333....is the way to write 1/3 using decimals.  If you multiply both sides of the above equation by three you get 1 = .99999...., right?

This is particularly convincing to many people. If we accept that \(0.333… = \frac{1}{3}\) (which we’ll come back to later), then \(3\times 0.333… = 3\times \frac{1}{3} = 1\). But the left-hand side is clearly 0.999… . (We’ll see an issue some people have with that, though, too!)

I think the problem you are having, though, is BELIEVING it is true, right?  I admit, depending on how you look at it, it can seem false. After all, how can 2 different numbers be equal?  The thing is, these 2 numbers AREN'T different.  I think saying 1 = .9999... may seem contradictory to us because we aren't realizing that .999.... is a repeating decimal that really does go on forever.  Obviously saying 1 = .9 is false, as is saying 1 = .99, 1 = .999, 1 = .9999, etc.  But we aren't dealing with finite decimals here.  So, you might think of .9999.... as another name for 1, just as .333... is another name for 1/3.  

What Steve said really should clinch it for you, but I thought I'd just add these thoughts anyway.  Hope it helps.

Molly has said two important things here.

First, the very idea of a decimal that continues forever trips us up, as in our minds we always picture it stopping somewhere. That will pop up again and again.

Second, there is nothing wrong with having two names for the same number; 1, 1.0, 1/1, 3/3, and so on are all representations of the same number. We just need to become aware that 0.99… is yet another way to write that number. It turns out, in fact, that any terminating decimal, like 0.123, can also be represented by a non-terminating decimal (0.122999… in this case).

Higher authority … and geometric series

A 1996 question came from an anonymous teacher looking for support:

Repeating Decimals

I am a high school teacher who is having a running argument with some students about repeating decimals. I claim that 0.9 repeating is equal to 1. One of my proofs involves the lack of numbers between these two numbers, much like any other two identical numbers. Another good proof is the pattern that comes from 1/9 = 0.1 repeating, 2/9 = 0.2 repeating, ... 8/9 = 0.8 repeating, and therefore 9/9 = 0.9 repeating which of course = 1 (9/9). 

They still refuse to believe me, and have asked that I consult a higher authority. Can you help?

“Anon” gave, in essence, two of the explanations we saw above: that there is no number between 0.999… and 1 (because if you add anything to any digit, the result will be greater than 1), and that, just as 1/3 = 0.333…, 1/9 = 0.111…, and multiplication gives the result.

Doctor Tom started with a tongue-in-cheek answer:

Okay, I'm a "higher authority," and I pronounce that you're right!

Is that enough?

Of course, mathematics is not based on authority, but on proof; decreeing an answer has no power.

If not, there are other ways to think of it:

For example, if they believe that 1/3 = .3333..., then clearly 1/3 + 1/3 + 1/3 = 1 = .99999...

This, again, is a different perspective on Anon’s second idea.

Or, if they believe that .00000... is zero, then if .99999... is different from 1, it must clearly be less. So what is 1 - .99999...?  If it's not zero, its decimal expansion must have some non-zero term.  Wherever that term is, it is easy to show that the difference between 1 and .99999... is less than that.

This is a clearer statement of the first, expressed in terms of subtraction.

The next is a new and important idea:

Or, if they believe that

.99999... = 9/10 + 9/100 + 9/1000 + 9/10000 + ...,

then just sum up the geometric series, where the first term is 9/10, and the factor is 1/10.

If S = a + ar + ar^2 + ar^3 + ..., then

S = a/(1-r).  a = 9/10, (1-r) = 9/10, so S = 1.

Repeating decimals are geometric series in disguise. We’ll be looking later at the fact that infinite decimals in general represent infinite series, and a formal proof must take that definition into account. The formula given here for the sum of a geometric series is, in fact, derived by the same manipulations that Joe used in the first question above.

See this page for alternative versions of the approaches we’ve seen:

Getting 0.99999...

Isn’t it just approaching 1?

Skipping ahead to 2008, we have this question that focuses attention on what a non-terminating decimal means, which to my thinking is essential to a thorough answer:

0.99... Repeating Can't Really Equal One, Can It?

Doesn't .999 repeating simply get infinitely close to 1, but never quite get there?  Kind of as if it had an asymptote at 1?  It always gets closer and closer, but will never quite get there, because if it did then all the 9s would turn to 0s, and then it would be 1.  Each 9 that is added simply makes the number more precise, closer to 1.  But the number will ONLY become more PRECISE, it will never become, well, precise, only MORE precise.

I think this lies behind many objections to our claim, and it involves a fundamental misunderstanding of non-terminating decimals. On the other hand, what John says is almost exactly right! I answered:

Sure, if you ever STOP somewhere in writing out 0.999..., then it will not be equal to 1.  But that's not what a repeating decimal means.

Think about 0.333... . That equals 1/3, right? But something like

  0.33333333333333333333333333333333333333333333333333

does not equal 1/3, because I've left off all the digits after the 50th.  The value of any non-terminating decimal like 0.333... is defined to be, not the value when you stop somewhere, but the "limit" that you approach closer and closer as you take more digits into account.  So in fact what you say is true of ANY such number, even something like pi = 3.14159...; each digit you add gets closer to the actual value, giving you better and better approximations without ever actually reaching it.  But that actual value is still out there, beyond all those approximations; it is not one of them.

So an infinite decimal does not mean any of the numbers that you get when you stop it somewhere, but the number they are approaching.

So the value of 0.999... is not the value of

  0.99999999999999999999999999999999999999999999999999

but the value, namely 1, that you are approaching as you add more digits.  (In fact, if you could stop somewhere and get a value of 1, that would prove that 0.999... was NOT 1, since the next digit would make it greater!)

By the way, if 0.999... were not 1, how would you explain the fact that

  1 = 3 * 1/3 = 3 * 0.333... = 0.999... ?

We’ve already seen this last argument a couple times, haven’t we? I think part of its power is that it forces us to work with the entire representation, not one digit at a time, working with whatever understanding we have of the nature of infinite decimals as limits.

John responded:

Touche.

For a similar answer, see:

How can .999999.... equal 1?

Non-terminating decimals are limits

We’ll close with this question from another doubter, earlier in 2008:

Introduction to Infinity, Limits, and Why 0.999... Equals One

I have read a good deal of the FAQs and other questions sent to you concerning this problem, but I'm still not convinced that 0.999... equals 1. 

The problem is that whenever you try to explain it, you tend to get back to the point of 0.999 vs. 0.999... you say that since it is constantly getting bigger, and gets infinitely close to 1, it must be 1.  But if 0.999... ever actually reached 1, that would bring an end to infinity as quickly as 0.999 would.  In order for 0.999... to reach 1, there would have to be some point, somewhere along the number line, where a finite quantity was added that would cause it to reach 1, and then it would stop increasing.

Plus, whether it progresses infinitely or not, it still doesn't reach one, because it adds a smaller value each time.  It's not like I'm adding 1 a bunch of times, and saying that I'll never reach 100 because I don't want to write out that many 1's.

Infinity louses up math pretty much anywhere you use it, and it seems to me that all you're saying is "Well, 0.999... is really really close to 1, and since it would be a lot harder to work with 0.999... than 1, we might as well just call it 1 and save everyone the trouble."  I have no problem with that, as I have as much trouble grasping the concept of infinity as everyone else on the planet, but when you're not working with it formally or towards a purpose, I see no reason not to admit that 0.999... is not actually 1.

Another thing I see wrong with this, is that if 0.999... is equal to one, what's to stop pi from being equal to 4, or, at the very least, 3.2?  Pi is an infinitely long number, so it keeps adding up decimals.  It should eventually get as close to 3.2 as 0.999... will ever get to 1.

Doctor Rick answered, diving more deeply than we yet have into the concept of limits. He first commented on the idea of “constantly getting bigger”:

We should be clear in our language.  The number 0.999... does NOT "constantly get bigger."  It is a NUMBER, and a number just is what it is; it doesn't change its value.

This is important: A non-terminating decimal is a representation of a number, not a process. Before we can talk about whether it equals 1, we have to be clear on what we mean by such a representation in the first place. As I said above, it doesn’t represent any of the numbers you get if you stop along the way. It also is not the entire sequence:

What you're trying to describe is something that mathematicians formalize as a LIMIT.  You will learn all about this when you get to calculus.  The idea is that we have a sequence of finite decimals:

  0.9, 0.99, 0.999, 0.9999, 0.99999, ...

As we keep adding 9's at the end, the numbers get closer and closer to 1--but you're right, they never actually reach 1.

The problem with your arguments is that 0.999... is NOT this sequence (because the sequence is not one number, but many), nor is it any of the numbers in that sequence (because each of them has a finite number of 9's).

When we say that 0.999… = 1, we absolutely do not mean that any of the terms in the sequence equals 1.

Infinity is indeed a difficult concept to grasp. The theory of limits is the formalism that was developed to deal with infinite processes.  It does so essentially by NOT dealing with infinity itself, but with sequences of finite quantities, each of which we know how to work with.

If you try to think of 0.999... (or any other repeating decimal, such as 0.333...) in the same way we deal with finite decimals like 0.125, we run into a problem.  While we can work out something like 0.125 = 1/10 + 2/100 + 5/1000, we can't do the same for 0.333... = 3/10 + 3/100 + 3/1000 + ... --because we can't do an infinite number of additions!  As you suggest, this isn't the same as just having 100 additions and getting tired before we're done; conceptually there is nothing different about doing 2 additions, or 100, or 1 billion.  But an infinite number of additions truly CAN'T be done.

To put this another way, before we can talk about infinite decimals, we have to define what it means to add an infinite number of terms – or rather, what we will replace such a sum with, since actually adding is impossible.

Yet we NEED repeating decimals!  Do you recall what motivated their introduction?  When we do a division such as 1 divided by 3, the division process never ends; we never get a remainder of 0.  If we just cut off the process after some finite number of digits (no matter how great that number), the result would NOT be EXACTLY equal to the fraction 1/3.  If we're going to have a way to write any fraction as a decimal, we need a way to represent the idea of a division process that never ends.  And that way is repeating decimals.

When we divide 1 by 3, we get 0.333… . We don’t get a terminating decimal (whose meaning is clear); instead, we get this odd infinite thing whose meaning we then have to decide. And we define that meaning so that its value will be what it has to be, namely 1/3.

So what value do we assign to a repeating decimal such as 0.333..., if we can't treat it the same as a terminating decimal?  As I have said, it has to be a NUMBER--in this case, it has to equal 1/3 exactly.  The only number available is the LIMIT of the sequence 0.3, 0.33, 0.333, 0.3333, ...; that is, the number that this sequence APPROACHES as the terms go on without end.  As we have said, none of those terms is exactly equal to the limit (1/3); that's not the point.   The point is that the terms get closer and closer to 1/3.  Furthermore, there is no other number that the terms get closer and closer to.  Therefore the limit of the sequence is 1/3; it is the only number uniquely associated with the sequence, so (by definition) we say that it is the value of the repeating decimal 0.333...

So we define the value of a non-terminating decimal to be the limit of the sequence of partial decimals – the number they are approaching but never actually reach. Until we make this definition, 0.999… or 0.333…, or 3.14159… is meaningless.

How about the question of why pi doesn’t end up equaling 4?

Here's what is wrong with your thinking this time: Once we know that the tenths digit of the decimal expansion of pi is 1, we know that pi can't be greater than 3.2.  And once we have found that the hundredths digit is 4, we know that pi can't be greater than 3.15. Each digit we learn puts limits on the range of possible values of pi--precisely because of one of your observations above: that we're adding a smaller value each time in the sequence

  pi = 3 + 1/10 + 4/100 + ...

So Devan’s recognition that the partial decimals never reach the goal is exactly right: It’s what allows the limit definition to work for both 0.999… and 3.14159… .

I appreciate your thinking about this topic.  You are asking the right kind of questions, questions that lead on into the field of calculus.  Please don't stop!

If you'd like to continue discussion of this topic, may I ask you to tell me your thoughts on repeating decimals in general.  Do you think that 0.333..., for example, is exactly equal to 1/3, and the only problem is with 0.999...; or do you have a problem with ALL repeating decimals?  I don't recall anyone broadening out the argument this way; everyone seems to be concerned about 0.999... = 1 specifically.  How about you?

Devan answered:

Okay, I see where my flaw is, and I think I get the limit thing, but I'm going to make one more desperate attempt to save face before I give up: where does this leave Zeno's runner, who had to cross an infinite number of halfway points before he could reach the finish line?  Of course, we all know that that would never stop a real runner, but if Zeno really did take, say, two minutes to cross each new halfway point, he never would reach the finish line.  It really is the same problems as repeating decimals; which, by the way, I do (or did) have a problem in general with.

Doctor Rick responded to that last sentence:

I'm glad to hear it!  Many of those who write to us about this topic (and there are a lot of them) have no problem with 0.333... = 1/3. This tells me that their problem with 0.999... = 1 has more to do with an unfounded feeling that there can't be two decimal representations of the same number, and less to do with the real mathematical issue of what an infinite sequence of digits can mean.  You have been thinking well, and there is no need to save face.

Yes, Devan is doing well!

He also commented on Zeno’s paradox of the runner; you can read that yourself, and also see this page (again by Doctor Rick):

Why the Motionless Runner Parodox Fails

For other answers involving the limit concept, see:

Other Ways to See That 0.999... = 1

The Infamous .999... = 1

0.999... and Infinity

As you can see, the debate over 0.999… never ends …

1 thought on “Frequently Questioned Answers: 0.999… = 1”

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