Does .99999… = 1?

Question

I'm told by smart people that 0.999... = 1 and I believe them but is there a proof that explains why?

Answer 1

What does it mean when you refer to $.99999...$? Symbols don't mean anything in particular until you've defined what you mean by them.

In this case the definition is that you're taking the limit of $.9$, $.99$, $.999$, $.9999$, etc. What does it mean to say that limit is $1$? Well, it means that no matter how small a number $x$ you pick, I can show you a point in that sequence such that all further numbers in the sequence are within distance $x$ of $1$. But certainly whatever number you chose your number is bigger than $10^{-k}$ for some $k$. So I can just pick my point to be the $k$th spot in the sequence.

A more intuitive way of explaining the above argument is that the reason $.99999... = 1$ is that their difference is zero. So let's subtract $1.0000... -.99999... = .00000... = 0$. That is,

$1.0 -.9 = .1$

$1.00-.99 = .01$

$1.000-.999=.001$,

$...$

$1.000... -.99999... = .000... = 0$

Answer 2

Suppose this was not the case, i.e. 0.9999... != 1. Then 0.9999... < 1 (I hope we agree on that). But between two distinct real numbers, there's always another one (say x) in between, hence 0.9999... < x < 1.

The decimal representation of x must have a digit somewhere that is not 9 (otherwise x == 0.9999...). But that means it's actually smaller – x < 0.9999..., contradicting the definition of x.

Thus, the assumption that there's a number between 0.9999... and 1 is false, hence they're equal.

Answer 3

What I really don't like about all the above answers, is the underlying assumption that 1/3=0.3333..., How do you know that?. It seems to me like assuming the something which is already known.

A proof I really like is:

0.9999... × 10 = 9.999999...

0.9999... × (9+1) = 9.999999...  [by distribution rule:]

0.9999... × 9 + 0.99999.... × 1 = 9.99999....

0.9999... × 9 = 9.9999....-0.9999.... = 9

0.9999... × 9 = 9

0.9999... = 1

The only things I need to assume is, that 9.999.... - 0.9999... = 9 and that 0.999... x 10 = 9.999.... These seems to me intuitive enough to take for granted.

The proof is from an old highschool level math book of the Open University in Israel.

Answer 4

Assuming:

1. infinite decimals are series where the terms are the digits divided by the proper power of the base
2. the infinite geometric series a + a * r + a * r^2 + a * r^3 + ... has sum a/(1 - r) as long as |r|<1

0.99999... = 9/10 + 9/10^2 + 9/10^3 + ...

This is the infinite geometric series with first term a = 9/10 and common ratio r = 1/10, so it has sum (9/10) / (1 - 1/10) = (9/10) / (9/10) = 1.

Answer 5

Okay I burned a lot of reputation points (at least for me) on MathOverflow to gain clarity on how to give some intuition into this problem, so hopefully this answer will be at least be somewhat illuminating.

To gain a deeper understanding of what is going on, first we need to answer the question, "What is a number?"

There are a lot of ways to define numbers, but in general numbers are thought of as symbols that represent sets.

This is easy for things like the natural numbers. So 10 would correspond to the set with ten things -- like a bag of ten stones. Pretty straight forward.

The tricky part is that when we consider ten a subset of the real numbers, we actually redefine it. This is not emphasized even in higher mathematics classes, like real analysis; it just happens when we define the real numbers.

So what is 10 when constructed in the real numbers? Well, at least with the Dedekind cut version of the real numbers, all real numbers correspond to a set with an infinite amount of elements. This makes 10 under the hood look drastically different, although in practice it operates exactly the same.

So let's return to the question: Why is 10 the same as 9.99999? Because the real numbers have this completely surprising quality, where there is no next real number. So when you have two real numbers that are as close together as possible, they are the same. I can't think of any physical object that has this quality, but it's how the real numbers work (makes "real" seem ironic).

With integers (bag of stones version) this is not the same. When you have two integers as close to each other as possible they are still different, and they are distance one apart.

Put another way, 10 bag of stones are not the same as 9.9999999 but 10 the natural number, where natural numbers are a subset of the real numbers is.

The bottom line is that the real numbers have these tricky edge cases that are hard to understand intuitively. Don't worry, your intuition is not really failing you. :)

I didn't feel confident answering until I got this Terence Tao link: http://www.google.com/buzz/114134834346472219368/RarPutThCJv/In-the-foundations-of-mathematics-the-standard.

Answer 6

x = 0.999...
10x = 9.999...
10x - x = 9.999... - 0.999... = 9
-> 9x = 9
-> x = 1 thus, 0.999... = 1

Answer 7

.999... = 1 because .999... is a concise symbolic representation of "the limit of some variable as it approaches one." Therefore, .999... = 1 for the same reason the limit of x as x approaches 1 equals 1.

Answer 8

You can visualise it by thinking about it in infinitesimals. The more 9's you have on the end of 0.999, the closer you get to 1. When you add an infinite number of 9's to the decimal expansion, you are infinitely close to 1 (or an infinitesimal distance away).

And this isn't a rigorous proof, just an aid to visualisation of the result.

Answer 9

If you take two real numbers x and y then there per definition of the real number z for which x < z < y or x > z > y is true.

For x = 0.99999... and y = 1 you can't find a z and therefore 0.99999... = 1.

Answer 10

The argument against this is that 0.99999999... is "somewhat" less than 1. How much exactly?

      1 - 0.999999... = ε              (0)

If the above is true, the following also must be true:

9 × (1 - 0.999999...) = ε × 9

Let's calculate:

0.999... ×
9        =
───────────
8.1
  81
   81
     .
      .
       .

───────────
8.999...

Thus:

     9 - 8.999999... = 9ε              (1)

But:

         8.999999... = 8 + 0.99999...  (2)

Indeed:

8.00000000... +
0.99999999... =
────────────────
8.99999999...

Now let's see what we can deduce from (0), (1) and (2).

9 - 8.999999... = 9ε                      because of (2)
9 - 8.999999... = 9 - (8 + 0.99999...) =  because of (1)
                = 9 -  8 - (1 - ε)        because of (0)
                =   1    -  1 + ε         
                =               ε.

Thus:

9ε = ε

8ε = 0

ε = 0

1 - 0.999999... = ε = 0

Quod erat demostrandum. Pardon my unicode.

Answer 11

Here's a collection of proofs:

http://premium.uploadit.org/AlphaNumeric/Recurring.pdf

Answer 12

Indeed this is true. The underlying reason is that decimal numbers are not unique representations of the reals. (Technically, there does not exist a bijection between the set of all decimal numbers and the reals.)

Here's a very simple proof:

1 / 3 = 0.333... (by long division)

=> 0.333 * 3 = 0.999... (multiplying each digit by 3)

But then we already know 0.333... * 3 = 1

Therefore 0.999... = 1

Answer 13

The proof I've always seen was that if 1/3 == 0.333... then 0.333... x 3 must be equal to 1, but at the same time, calculating it on a digit level gives 0.999...

Answer 14

Given (by long division):
$\frac{1}{3} = 0.\bar{3}$

Multiply by 3:
$3\times \left( \frac{1}{3} \right) = \left( 0.\bar{3} \right) \times 3$

Therefore:
$\frac{3}{3} = 0.\bar{9}$

QED.

Answer 15

well the simplest is 3x(1/3) = 1 where 1/3 = 0.33·

Answer 16

This should have been a comment; but I post as an answer anyway. I just want to remark that in spite of all the criticisms, all the statements in the most popular proof in the list above are true.

That is, in the real number system, it is indeed true that:

1/3 = 0.333333...

It is again true that:

0.33333.... * 3 = 0.99999....

Etc..

I would still stand by the superiority of Noah Snyder's answer, however. That's the one addressing the subtleties.

Answer 17

For any digit n = 0,1,2,3,4,5,6,7,8,9 the fraction

n/9 = 0.nnnnnnnn... (periodically).

Therefore 9/9 is equal to 0.999999999.... but also since 9 divided by 9 equals one,

9/9 = 0.9999999.... = 1

Answer 18

If you want to go the calculus route, then basically you're gonna follow this pattern and use mathy magic and limits to follow it up to its logical conclusion:

0.9 0.99 0.999 0.9999

And so on. This pattern is a sum of 9*10^-i from i=1 to i=infinity. Every iteration, the difference between the pattern and 1 is a bit less. So at i=infinity, the difference is so small that it is literally zero.

Answer 19

I do not think that 0.999... = 1 unless we're defining the equality operator in some strange way. Assuming you allow N decimals in the 0.999... term, you can take N as large as you want and it will still be less than 1. There is no N for which 0.999... is not less than 1. Sure, if you take the limit as N approaches infinity then you get 1, but "lim 0.999... as N approaches infinity" is not the same as "0.999..." by itself. By itself, there is always some infinitesimal difference.

I think the whole point of writing 0.999... is to indicate that it is NOT equal to 1 and to indicate that there is some infinitesimal difference. Otherwise you would just write 1.

My opinion is that most people say 0.999... = 1 because that's what they were taught in college and they accepted it without really thinking about it. All the arguments I've heard or read so far are just contrived ways of making the equality hold, but there is always some subtle flaw that takes a long time to point out, that the proponents will not accept anyway (for reasons I will not get into here).

From a comment I made:

10x - x != 9. 10x - x would be 8.9999...1. However infinite the extent of 9s is in x, if we multiply it by 10, the nines are shifted left by one position and a zero inserted at the "last" place, and then when you subtract the other number there is a nine subtracted from a zero at the far right. Otherwise we'd have to give 0.999.. some unusual properties like automatically increasing the number of nines when it is multiplied. It would not be just an ordinary number. Maybe that's the problem. 0.999... might just not be an ordinary type number as some people are using it.

The current winning response as of 7/22/10 treats 0.999... as being equivalent to lim (0.999...) as the number of decimals goes to infinity. But the sequence and the limit of the sequence are not the same thing.