Is 0.99999 repeating equal to 1?

Question

0.999999999999999999.... == 1

Is this correct?

I think it is. Am I wrong?

If 1/3 + 1/3 + 1/3 = 1, and 1/3 = .3333333, and .333333... + .333333... + .333333... = .999999..., then isn't .9999999 = 1? Prove me wrong, please.

Is it not true that in order for one number to be greater than another, there must be a number in between? What number is between .9999999 repeating and 1?

Answer 1

YES, it is. That repeating decimal (where the 9s do go on forever, as noted by the ellipsis) is EXACTLY equal to 1.

Not "approximately".
Not "only if you round up".
It is EXACTLY equal to 1!

You can mathematically prove it too:

Let N = 0.999...
Then 10N = 9.999..., and 10N - N = 9.999... - 0.999...,
which means 9N = 9 so N = 1.

Also, if you've done work with sequences and series, you can prove that the infinite geometric sequence 9/10 + 9/100 + 9/1000 + ... converges to 1.

These pages explain it in more depth:
http://mathforum.org/dr.math/faq/faq.0.9999.html
http://mathforum.org/library/drmath/view/55746.html
http://mathforum.org/library/drmath/view/55748.html
http://mathforum.org/library/drmath/view/57174.html

Also search Yahoo Answers too, because this one comes up a lot.

EDIT: I like your 1/3 + 1/3 + 1/3 proof! I never thought of that one.

EDIT: Yes, big fan 2006, the normal convention is to have a bar over the repeating part, but writing the implied repeating part plus an ellipsis (...), which the person did, is an acceptable form too. And you're still wrong. See the links.

EDIT: I just realized that Puggy, the top math contributer, even mentions the "Yes, 0.999... = 1" thing in his profile. LOL

Answer 2

I hate to feed into another one of these questions (they come up regularly here it seems), but this time a few people seem quite convinced of the non-equality, so I'll try to pitch in. I doubt that I'll be able to convince anyone, but it's worth a shot.

Shawty: 1/3 is indeed equal to .3333repeating. If you simply do the long division you may be able to convince yourself that dividing 1 by 3 results in infinitely many 3's after the decimal point. Otherwise, your problem is a common one--misunderstanding the mathematical concept of infinity. I'll address this further down. Your rebuke to kyle's answer doesn't hold, as 9.999repeating-x IS possible: just remember what x was defined as (.999repeating), and it's easy to subtract.

Bucky: Your refutation of converging limits is admirable, but the thing to remember here is that when we say .999 repeats forever, we really do mean infinitely; we do not mean that .999999 has a number of digits that approaches infinity, but rather than it has infinitely many digits (all of which are 9's).

As mentioned before, when you consider 3N, a 7 does not appear at the "end" of the digit string, because there is no end: the nines just keep on coming. But suppose for a moment that there are infinitely many 9's, followed by a seven. Then the rest of your work goes to show that .99999repeating = .99999repeating. Okay, that's fine, but we've also shown that .99999repeating=1, using valid mathematical operations. Your last comment (before edits) seems to indicate you think that any manipulation of .999repeating should end up with it becoming 1. This is not the case. Think of y=x*(x+1). Using a bit of manipulation, we can see that y=x^2+x. But doing some more manipulations, we can likewise see that y=(x+1)*x. There's no problem here though, they're simply different ways of expressing the same thing.

Can you agree that 1/3=.333repeating? If so, then do your final manipulation backwards: Let N=1, so N/3=.333repeating, so N=.9999repeating.

Basically I think your roadblock here is that you consider .9999repeating to be a series of 9's that grows without limit. Technically though, the 9's are there, and they're infinite. They don't really expand outward, and that's why a 7 is never found at the end of 3N, as mentioned above. In your definition of convergence, notice the "as the number of terms increases"; the number of 9's in N is not increasing, they're there already.

rathan d: yes, 1-1=0, and 1-.99999repeating=0 as well.

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I rather like Nisovin's (the question asker) approach in the second add. details. On the real number line (and hopefully we can agree that N is a real number), there are several properties. One is that for any non-equivalent numbers a and b, there is a number between them. In fact, there should be infinitely many numbers between them. Also, every number has a decimal representation. So if there is a difference between .999repeating and 1, someone should be able to give me a decimal expansion that is between them.

And if anyone didn't read the links given in geezah's answer, please do at least peruse them. They might provide a new way to look at this problem.

Answer 3

Before you are ready to accept a proof in mathematics, you need to have a clear understanding of what the terms involved mean. What I want you to focus on is, what *is* .9999...? Or any decimal expansion, like 3.1415... or whatever. A number doesn't change. It's not 0.9 one moment and then 0.99 the next, as if some outside person is changing it by contemplating it. It's stuck somewhere on the number line, no matter how you describe it. And if I ask, what is the difference between 1 and 0.9999...., it's going to have to be a fixed number too, not some "infinitely small" thing. There has to be an answer; you can always subtract numbers. And it's easy to see that the difference, if positive, would have to be smaller than any other positive number. (its decimal expansion would start with more zeros 0.0000...) If not positive, it must be zero. Can it be that there is a positive number smaller than any other positive number? No, then it would be smaller than itself (or half itself). Contradiction. Here's another perspective: do you believe that 1/2 + 1/4 + 1/8 + 1/16 + ...=1? (Classic Achilles & the Hare problem.) That's the same as saying 0.11111.... in binary is equal to 1. Your problem is the same idea base ten.

Answer 4

I've never seen so many incorrect answers. Only Geezah, G M, and tsr21 are correct.

It IS equal to 1.

Proof
Let x = 0.9999....

Multiply both sides by 10:
10(0.99999.....) = 10x
9.9999999..... = 10x

Subtract x from both sides:
9.9999999..... - x = 10x - x
9 = 9x

Divide both sides by 9
x = 1
0.9999.... = 1

QED

Answerer below me:
What are you talking about? 1 + 2 = 3, not 4. There was no manipulation in my proof. This proof has been done for years. Have you ever taken a course in proof? If not, why did you even answer? This isn't an opinionated question.

edit: I just realized Geezah already gave this proof. I guess we both realized the obviousness of this. Give him the best answer.

edit2: To answer your other question, there is no number between 0.999...repeating and 1 because they are EQUAL.

edit: To Bucky:
Here's the problem with your counterexample:
x = 0.99999....repeating infinitely
3x = 3(0.99999....)
3x = 2.9999.....7
That step is incorrect. There are an infinite number of 9s after the decimal so the decimal would never terminate after multiplying it by a number (similar example: 3*∞ = ∞, it doesn't change to a finite number). Therefore the only things you can multiply by are multiples of 10 since they will not change that property of the decimal.

Answer 5

No! Everyone keeps writing about geometric series and convergent numbers. It is great that you have an BS and MS in mathematics. Try looking up the definition of converge. .999..... gets infinitely close to 1, but it never equals 1. 1 is the only number that equals 1 without rounding.

Your proof:

Let N = 0.999...
Then 10N = 9.999..., and 10N - N = 9.999... - 0.999...,
which means 9N = 9 so N = 1.

Oh yeah, this is GREAT!

Now lets try it with a different number:

Let N = 0.999...
Then 3N = 2.99999...7, and 3N - N = 2.999...7 - 0.999...,
which means 2N = 1.999....8 so N = .999.....

Let N = 1...
Then 3N = 3, and 3N - N = 3 - 1,
which means 2N = 2 so N = 1.

The problem with this theory is that unless you are working with 10 or 100 and so on, you get a number that is not .99999.... across the board. Somewhere along the line when you subtract your .999... you will not end with a whole number. The only way for this to be true is if it works for all numbers. IT DOES NOT!

They converge, but they are not equal.

edit: To Kyle

You make a good point, incorrect, but a good point. In my example, my numbers can continue to infinity, but decimal for decimal (no matter how far out they get) my number will always end in a 7 and N will always end in a 9. It doesn't work out to a perfect 1. You are correct, multiples of 10 do not change the property of the decimal. If you are worried about changing the property of the number, then this entire question is off. The property changed when 1/3 became .33333......

Your theory is wrong because it does not work for all possible numbers. Only 1 can work for every number, therefore only 1 can equal 1.

Lets go back to the definition of converge:
to approach a limit as the number of terms increases without limit

The limit is 1, they will infinitely approach this, but never equal it. KEYWORD = APPROACH!

Answer 6

No it is not at all equal to 1 and is not approximately 1. It is very much less than 1. Proof:
For k is a non 1 value, ε exists, greater than k and less than 1.
Eg. For 0.9 exist, 0.99 exist.
For 0.9999 recurring to exist, 0.9999 recuring & an extra 9 must also exist.

Answer 7

Why does 0.99999999999 (repeating) equal 1? This was adddressed in great detail about a year ago, right here in answers.yahoo The link is below .

Answer 8

Yes. It's a convergent geometric series, with a very well known formula to evaluate the sum a/(1-r) where a is the first term and r is the common ratio. In this case, a = .9 and r = .1
Thus we get the sum as .9 / (1-.1) = .9/.9 = 1.

Answer 9

definitely they both are not the same then is 1-1 and 1- .9999999 are the same ?

Answer 10

yes by geometric series

.99999 repeating
= 9(.111111 repeating)
=9( sum n=1 to inf (1/10^n))
= 9(10/9-1)
= 9(1/9)
= 1