does 1= .9999....(repeating infinity)?

Answer 1

Yes. It is identically equal.

"1" is not a unique representation of what we mean by
one. It's not just true as an approximation, by rounding, or
in a limit. It is exactly true.

Proof:
x = 0.9999...
so
10x = 9.9999...
then
10x - x = 9
so 9x = 9 and x = 1

0.9999.... and 1 are two ways to write the same number.

Answer 2

I'll chime in with the educated chorus and agree that 0.999... is exactly equal to 1. Good proofs have been given, and there's no need for another, but I'll try:

N = 0.9999... = lim(n --> infinity) Sum(k=1 to n) 9/10^n
1-N = 1 - lim(n --> infinity) Sum(k=1 to n) 9/10^n
= lim(n --> infinity) 10^-n = 0

If 1-N = 0, then N = 1.

As for Tom's claim that mathematicians have been "refuted" somehow, he's badly mistaken, lying, or both. Consider Cantor's Diagonalization Argument, for one -- the 0/9 condition is based on the equality of numbers with repeating 9s to some other decimal representation.

It's true that hyperreal numbers can be considered, and that these arguments won't hold (in general) for the hyperreals. However:
* I believe (but I'd have to check) that 0.999... = 1 even in the hyperreals
* As real numbers, 0.9999... = 1 according to all of the mathematical community.

This isn't a disputed point. Any college-level student writing that 0.999... =/= 1 would be marked off; if the student was a math major and in a serious class (400-level), they'd probably have to have a talk with the prof to see if the class or major was right for them. This is a really basic concept. I learned this in my freshman year in high school; how so many people can not know is amazing (and depressing).

Answer 3

The answer is yes, but it's a very good problem to think through yourself.

What I want you to focus on is, what does .9999... *mean*? 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, as many have pointed out, that the difference, if positive, would have to be smaller than any other positive number. 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

What does .9999... mean? It means 9/10+9/100+9/1000+... . In algebra and arithmetic, this does not make sense. Addition is defined only for pairs of numbers; it can be extended to any finite set of numbers using the associative law, but the associative law does not allow one to add an infinite set of numbers.

In calculus and analysis, the only way to define 0.99999.... is as a limit. The decimal 0.9999... is an infinite series, which is the infinite sequence 9/10, 9/10+9/100, 9/10+9/100+9/1000, ... or 0.9, 0.99, 0.999, ... . Let e be a small number, and consider the neighborhood of numbers from 1-e to 1+e. 1-e will start out typically as a series of 9s but will eventually produce a non-9 digit, say at the nth place. Go n+1 places with 0.99999999...9 , and call this number y. Then y is between 1-e and 1+e. We have just proved that the limit of the sequence is 1. In that sense, we can say that 0.9999... = 1, although it takes an infinite number of operations to do this, and we can take the sum of this infinite series in the sense that we can find a limit for it.

Answer 5

The answer is wholeheartedly yes! It doesn't require rounding, it doesn't require any tricks, it just is!

First, is 1/3 = 0.3333...? You wouldn't deny that, or say it is an approximation, would you. No, 0.3333... is how we represent 1/3.

Now what's 1/3 + 1/3 + 1/3? It's 1.

And what is 0.3333... + 0.3333.... + 0.3333...? It's 0.9999...

So the two are equivalent.

Here's another way to look at it:
1/9 = 0.1111...
2/9 = 0.2222...
3/9 = 0.3333...
4/9 = 0.4444...
5/9 = 0.5555...
6/9 = 0.6666...
7/9 = 0.7777...
8/9 = 0.8888...
9/9 = 0.9999... but 9/9 = 1, so they are the same number.

Let's subtract them (1 - 0.9999...). If there were 4 nines, this would 1/10^4. But there are infinite number of nines, so the difference is 1/10^∞ which is 0. So the difference between them is 0 and they are therefore the same.

If you still don't believe, ask any mathematician... or check out the attached Wikipedia article.

Answer 6

The two are definitely not identical. Don't believe everything you read on Wikipedia. There was a huge debate on this where every single argument and proof presented by Math Phds and others was refuted.

A common misconception is that if these numbers are not the same, then one should be able to find a number in between. Well, one can find infinitely many numbers inbetween - see the discussion edits in Wikipedia.

Some of the proofs in favor appear to be very convincing but even those proofs were shown to be faulty. Those in favor pulled out all the stops: they talk about infinitesimals, the completeness principle and lots of other stuff. They were all refuted.

Whichever way you decide to look at this debate, it really makes no difference in practical mathematics.

firat_c: Really? Try to write 0.999 (repeating) - can you do this?

firat_c is an example of what happens to those who obtain Phds - they become so confused that they cannot think straight anymore.

Don't believe anything you read here unless you can prove it. Good luck!

Answer 7

I asked my Algebra teacher this very question when I was in 8th grade: if 1/3 = .3 repeating and 2/3 = .6 repeating, why does 1/3+2/3=1, but .3 repeating + .6 repeating = .9 repeating?

Her answer: if .9 repeating never ends (that it continues to infinity), how can you add a 1 to the end of it to make it 1.0? You can never find the end of it to add to it to make it 1.0. Therefore, it IS 1.

Answer 8

No. Here is why.

We would agree that 1 is not equal to .9, yes? Or even .99. Why? Because the second digit the .09 is inherantly smaller in value than the origional .9. The same is true for every digit you add on to that. Yes, you could stretch it out to infinity, but then the nines you are adding would also be infinitly small.

Answer 9

Actually, it does. There are a number of proofs for this:

1)
1/3 = 0.3 repeating
3(1/3) = 3(0.3 repeating)
3/3 = 0.9 repeating
1 = 0.9 repeating

2)
let a = 0.9 repeating
10a = 9.9 repeating
10a - a = 9.9 repeating - 0.9 repeating
9a = 9
a = 1

3)
I'd use the infinite series proof, but without sigma notation it won't be easy. You get the idea.

Hope this helps.

Answer 10

Yes it does. 1/3 = .3 repeating
2/3 = .6 repeating

3/3 = .9 repeating = 1