# More 0.999~ vs 1

1. Jun 24, 2006

### BSMSMSTMSPHD

Here's how I've always though of it (with details left out for the sake of brevity):

If two real numbers are different, then you can find another real number between them (their average, for example). So, assume .999... and 1 are different numbers. Can you find a number between them? Since you cannot, the numbers are therefore the same.

2. Jul 6, 2006

### dansydney

On the side a bit, is 1/3 equal to 0.3333 because the 3's are infinite, the same with 2/3 the 6's are infinite. Are 1/3 and 0.333(repeater) really the exact same number? 1/3 is one third of something whereas 0.333 is 0.333 of something and nomatter how many 3's are added it will not give the same result as 1/3 because you cannot use infinite? Not sure if you will understand what im trying to say but feel free to make me look stupid. Cheers dansydney

3. Jul 6, 2006

### HallsofIvy

Okay, I'll feel free! "it will not give the same result as 1/3 because you cannot use infinite" is incorrect. You don't have to "use infinite" because an "infinite number of digits" arises only in the base 10 representation of a number and has nothing to do with the number itself.
Once again, the definition of a base 10 representation of a number is that 0.a1a2...an... is the limit of the sequence 0.a1, 0.a1a2, ..., 0.a1a2...an,... (not the sequence itself- that's the confusion that leads to such nonsense).
It's easy to prove that the limit of such a sequence exists and is a real number (it's obviously an increasing sequence and has upper bound 0.a1+ 0.1). It's even easier to prove that 0.333... which, by definition, means the limit of the sequence 0.3, 0.33, 0.333, 0.3333, ... is the limit of the sequence of partial sums of the geometric series
$$\Sum_{n=0}^\infty 0.3(0.1)^n$$
and so has sum $\frac{0.3}{1-0.1}= \frac{1}{3}$.

Similarly, 0.9999... means the sum of the geometric series
$$\Sum_{n=0}^\infty 0.9(0.1)^n$$
and so has sum $\frac{0.9}{1-0.9}= 1$.

4. Jul 6, 2006

### HallsofIvy

Can you prove that there is no number between them? The fact that I can't do something hardly proves that it can't be done! (I've proven that many times!)

5. Jul 6, 2006

### dansydney

I sort of get what your saying. Im only a yr11 student so I havent covered all the limits topics thus far.

6. Jul 7, 2006

### Robokapp

Imagine the fraction...(x^2-x)/(x-1)=0
It can be simplified in x(x-1)/(x-1)=0

Find the limit as x -> 1

lim- = 1
lim+ = 1
lim exists therefore at x=1
What is the value of the function at x=1? There isn't any bc 1 is not a ponit of continuity (i wish i knew how to write fancy the lim as x-> a of f(x)=f(a) proves that point x, f(x) belongs to f(x) and it is a ponit of continuity)

So...0.99999999999 is part of the function and one/infinity of a unit to the right, 1 is NOT a point that belongs to the graph. Sorry to break your heart.

Also think this way. To be equal, in the true equality meaning, it must have the exact same characteristic...like a triangle. Equality in two traingles means all sides, angles, perimeter, area...whatever is identical...basically if you don't label them you can't tell the difference.

Well I can tell the difference between 1 and 0.999999... which no one here can even write fully...I know this is not math but it's 6:30 am lol.

7. Jul 7, 2006

### HallsofIvy

I have absolutely no idea what this means. What does it mean to say that "0.99999999999 is part of the function" and what is "one/infinity of a unit"?

I certainly agree that "1 is NOT a point that belongs to the graph."- 1 is not a point at all! If you mean that there is no point with x-coordinate 1 on the graph of f(x)= (x2-1)(x-1) (I assume that's what you meant. You never did say what f was.), that's true. f(x) = x+1 for x not equal to 1 and is not defined at x=1. But I don't see what that has to do with whether 0.9999...= 1. There is no point with x= 0.999... on the graph.

8. Jul 7, 2006

### arildno

Robokapp:
I have no idea, either!
I thought the real number "1" was an equivalence class of (a subset of) rational sequences with respect to a particular equivalence relation, but obviously, I'm mistaken.

9. Jul 7, 2006

### neophysique

I think there's something illogical about comparing a number that
is not complete (and which can never be complete) like 0.999... to
a number that is completed, like 1. It's no different than someone
claiming to know the biggest number in the infinite set of all integers.
And wouldn't make any less sense if someone also claimed that number
to be 1.

It's like there's a 1 meter track. A runner (the infinite sequencer)
is programmed to keep shrinking and shrinking forever as he runs
that 1 meter track. He by definition can never finish the track,
though it started only as 1 meter long when he was "fully" sized.
For all practical purposes, the track has became infinitely long
to the runner and not 1 meter.

So how does one compare a track that is .999... meters long (one
that can never be completed) with a 1 meter long track ran
by a "normal" runner?

10. Jul 7, 2006

### matt grime

I think there is something illogical about using the word complete without defining what it means for a number to be complete (or incomplete).

11. Jul 7, 2006

### StatusX

The key point about 0.999... is that it is defined as:

$$0.999... = \lim_{n \rightarrow \infty} \sum_{k=1}^n 9 \frac{1}{10^k}$$

When people say things like "it can never be complete," I can only imagine that they are trying to visualize a number with many many 9's. This is not what 0.999... means. It is true that for any number n, the above sum is less than 1. It is also true that as n gets larger and larger, the numbers get closer and closer to 1 (and not to any other number), and this is what we mean when we say the limit of this series is 1. In fact, infinity is only really defined in terms of limits. When someone says:

$$\sum_{n=1}^\infty a_n = L$$

They are really just using shorthand for:

$$\lim_{N \rightarrow \infty} \sum_{n=1}^N a_n = L$$

In turn, an infinite limit like:

$$\lim_{n \rightarrow \infty} a_n = L$$

just means that for any e>0, no matter how small, there is some N (which will in general be larger and larger as e gets closer to 0) such that for all n>N, |an-L|<e, that is, the terms get arbitrarily close to L. Note that every step in the definition of infinite limits and sums involves normal, finite numbers. You never have to stretch your mind thinking about infinity if you just remember that it is usually just a shorthand for these straighforward definitions.

Last edited: Jul 7, 2006
12. Jul 7, 2006

### Robokapp

$$\lim_{n \rightarrow\ "1"} \ (x^2-x)/(x-1)$$

Okay, I'll try to write it the fancy way. Question is...is 0.9999999 = 1
Well f(x)=x(x-1)/(x-1) so the graph looks just like f(x)=x but it's having a hole at x=1 Correct?

Well point x=0.99999 does belong to the graph. Point x=1 does NOT.
Point 0.99999... is where the $$\lim_{n \rightarrow\ "1-"}$$ occurs. It's a real point, (0.(9), 0.(9)) on the graph. Point for x=1 does not have a y-value. big difference between the two!

Last edited: Jul 7, 2006
13. Jul 7, 2006

### Hurkyl

Staff Emeritus
As you've pointed out before,

$$\lim_{x \rightarrow 1^-} f(x) = 1$$

Now, for what value of x are you claiming that f(x) = 1?

14. Jul 7, 2006

### matt grime

no that is not the question. 0.9999999 is not equal to 1. 0.99... is equal to one (as base 10 representations of real numbers).

15. Jul 7, 2006

### HallsofIvy

I wanted to start this "I think there's something illogical about" your post but Matt Grime beat me to it!

Well, yes there is a difference! There is no such number. There IS such a number as 0.9999... and apparently even you agree to that. Do you insist that there's a "biggest number" lest than 1? If not, what such numbers are larger than 0.9999....?

No, it's not the same. You are making the same mistake I mentioned before: 0.9999... is not the sequence 0.9, 0.99, 0.999, ..., (which you could think of as "never ending"), it is the limit of that sequence and as far as 'never ending' is concerned, no different from the numbers 2 or 3 or any other number.

Last edited by a moderator: Jul 8, 2006
16. Jul 7, 2006

### Robokapp

I'm so confused I don't even know if I'm lost or not but...What i am claiming is that my equation is a y=x scenario with a hole for x=1. basically i created a straight line with the domain and range of {R|x or y not = 1}

so the 0.999 is there, fine and well but 1 is not. I'm not really sure what the question of the topic is since now I see some stuff I didn't read...but if you're talking about approximation, my approach would be that since you can't write all those 9s in there, you got to round down or up. rounding is closer to up into 0 than down into a zero...so one 9 somewhere down the lnie turns into a 0...cascading the others into 0s until the units turn into a 1.

Edit: hurkyl you're post 6999 :D write something special for 7000 and mention my name for noticing you hehe.

17. Jul 7, 2006

### Integral

Staff Emeritus
Unfortunately your function says NOTHING about the equaltiy of 1 and .999.... . You have stated but not proven that .999... is "there"... What ever that means.

edit:
OK I guess I will post this...Again...http://home.comcast.net/~Integral50/Math/proof2a.pdf" [Broken]

Last edited by a moderator: May 2, 2017
18. Jul 7, 2006

### MalayInd

I will say hat 0.999... is equal to 1 if the the trailing dots indicate that the number of repeating nines is greater than any natural number.
1=9/9
Now divide 9 by 9 by taking the quotient 0 rather than 1
You will get 9 as the remainder, then we follow it by zero and thus our quotient builds to 0.999... (a never ending repitition)
I would like to have your comments on this.

Keep Smiling
Malay

19. Jul 8, 2006

### Integral

Staff Emeritus
Yes ( ... ) is called an ellipsis, it denotes the repetition of a pattern endlessly. So .999... means repete the pattern 999 endlessly.

I have seen the 9/9 division before, it is yet another Demonstration of the equality.

Last edited: Jul 8, 2006
20. Jul 8, 2006

### shmoe

Once you've proven that doing division like this will end up giving you a decimal representation of your number of course.

The same will go for the proof using 3*(1/3)=3*(0.333...) etc. While swell that this seems to convince alot of people, it's often overlooked why we are justified performing operations on non-terminating decimals like this.

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