Can anyone prove that 0.999 repeating to infinity is exactly equal to one?

Answer 1

Yes, several proofs were given 1 day ago when the same question was asked:

http://answers.yahoo.com/question/index;...

See also my response to that question.

Furthermore, the people who say it is not the same as 1 are wrong. If 0.99999... has a value, then it is 1. Saying that it is not quite 1 is a conceptual error -- if we were to truncate at some point, i.e., at 0.999999, then that value is not quite 1. But the recurring decimal 0.9999999... is not the same as any of its truncations. Notice that none of its truncations are equal, so if it were the same as one of its truncations, how would you be able to decide which one?

The DEFINITION of 0.9999... is the sum of the series 9/10^1 + 9/10^2+9/10^3+..., an infinite series. Using the theory of such series, the value is 1. People who say 0.99999... is less than 1 would also think that pi=3.14, but it is not. It has infinitely many digits, and if you give me any number n, I could in principal give you the first n digits of pi. The same concept holds for 0.9999...

Answer 2

No. And helpful as wikipedia can be, I wouldn't call it an authoritative source.

Firstly, a qualification - when referring to 0.999..., I don't consider it the same as the fraction 1/3 multiplied by 3 with decimal numbers. So what you're specifying is a standalone number, 0.999... with an infinite number of 9's repeating after the decimal point.

(I believe calculators have confused the issue somewhat with their tendency to produce 'recurring' results on relatively simple operations, the results of which can justifiably be rounded)

The common proof shown tends to include lines something like this:

n = 0.999....
10n = 9.999...

Let me interject here for a second. We have 10n being inifinitely close to 10. Not stated as equal to 10. Now continue:

10n-n = 9.
or
10n-n=9n=9.

I'm sorry, but you've just contradicted yourself. If 10n = 9.999..., then 9n = 8.999...

You say that's what you're proving wrong? Well, ok then, let's say 9n=9, like you say.

What, then, is 9-n?

8n? 7.999... ?

8?

or 8.000... 001 ?

While n may be infinitely close to 1, it's also an infinitely small amount less than 1. Not a specific decimal amount, I'm talking infinity here.

To be honest I'm not sure you can meaningfully multiply a number containing recurring digits without rounding or simplifying it to some extent first anyway. Though many people do.

(Just quickly, to those who think in terms of: 1/3 = 0.333..., so 3 times that equals 1, let me ask: what's the decimal for 2/3? That's right, it's 0.666... 667. It's not double the decimal for 1/3 at all. Hence 0.999... has nothing to do with thirds; 3 x 1/3 = 1.)

In mathematics, infinite limits are described such that a number 'approaches' x. It doesn't reach x. Never. You can define the value x and use it in subsequent calculations because, in practical terms, it's so close (infinitely!) that your results will not be skewed to any noticeable degree. But it's still not equal.

mathbear, I do not think that Pi = 3.14. And while your series is correct, and the limit of the sum of that series is 1, the sum is not. The sum is infinitely close to 1.

mathematician: firstly, the limit is 1. Secondly, if your theory is correct, there are THREE ways to write 1: 1, 0.999..., and 1.000...001. The fact is, 0.999... is an infinitely small amount less than 1, while 1.000...001 is an infinitely small amount more than 1.

Neither is EXACTLY equal to 1.

Answer 3

No, it is an irrational number that you have just defined as not being equal to 1.

[edit]

All those "proofs" below and on other sites are crap. 0.999... is an IRRATIONAL number because it is NOT of the form 'a/b' where 'a' and 'b' are integers. The proofs cited below indicate that the only possible value for 'a' and 'b' is a=b (e.g.,=1). However, in our decimal system, 1/1 is indicated using the symbol '1'. Also 0.999... is NOT a repeating decimal. Repeating decimals do take on the form 'a/b', but again 9/9 (or 1/1) is universally indicated as '1' in our decimal system.

The only thing that these proofs really mean is that the notation 0.999...(infinite 9's) is an acceptable way of indicating the value 1. But representing the value 1 with an infinite series of digits is impractical. The practical way to indicate a value of 1 is using the notation '1'. Since impractical is not the same as practical, they are not equal.

Answer 4

You've got some good proofs and some misguided answers here, so I'm adding in my two cents in the hopes of clarifying. It is true that the repeating decimal .9999.... is equal to 1. The proof using n and 10n is a good proof. It may also help to think of it this way: .9999... is the sum of an infinite series, .9 + .09 + .009 + .0009 + ...
Of course, if you stop adding at any point along the way, the sum will be less than one by some small amount, but the point of an *infinite* series is that you *never* stop adding. The infinite sum is *actually* equal to one.

Answer 5

To all those who say these are not the same: please don't answer questions where you don't know the answer! There are TWO ways of writing the number 1 as a decimal. One is 1.0000.....; the other is .9999....

The problem is to figure out what the symbol .99... means. It *means* the limit of a sequence. Which sequence? The sequence whose first term is .9; whose second term is .99; whose third term is .999; etc. As you go further out in this sequence, is there some number that the terms of the sequence are getting closer and closer to? Yes. That number is what the symbol .999... means. It is also clear that that number is also 1.

Answer 6

I'm sure we had this question like 3 days ago (might have been this one?) If it isn't it's usually a good idea to search first before asking (although not having tried this I can't speak for the usefullness of the YA search engine)

Answer 7

Chuck Norris counted to infinity, twice.

Answer 8

No, because it isn't. It's a very little off. It's approximately 1.

Yeti, the second line of your equation is incorrect. 9n=9*.9999....

Answer 9

Yes, see the wikipedia entry on this for more details.

Answer 10

it's not equal, .999 always need a little more