Does .999 repeating exist?

I will attempt, however erroneously it may be, that the value of .999... doesn't exist at all!

1/9 = .111...

(1/9)*(9/1) = 1

.999...=.111...*9

.999...=(1/9)*(9/9)=1

This is yet another occurence where the English Language fails us. I'm not sure that it can be explained perfectly in any language. We can agree that .999... is definitely .111...*9. But it is actually 1. So .999... does not exist, because infinity is a concept, not a value or measure.

You have not proven .999... doesn't exist.
You have proven .999... is equal to 1.

Now, if you want to really say .999... doesn't exist, then, since you proved .999... and 1 are identical, then I guess you'll also have to say 1 does not exist.

-Lou
:)

Here is some proofs why it equals to 1 (maybe the poll should've been does it equal to 1). you see, .9 repeating can exist and equal to 1 at the same time. but for clearance sakes, if you believe that .9 repeating is not one, vote it exists. otherwise (if you believe that it does not exist or u think it exists, vote for not exist).

just a note that i haven't been able to find one website where it says that .9 repeating does not equal to 1!

.(9) (just gonna stand for .9 repeating to make my life easier) has some "very special" properties, such as it has no square roots, and (.(9)+1)/2 is undefined (not something over 0 -- take the word undefined in the face value).

but it is precisely as I have told you, these two properties does NOT make any sense! for example, no one said you can't divide by 2 (or for that matter, any number that's not 0!) if the number is (1.(9))! or you can't take the sqrt(x) if x is .(9)!

as my proof indicated, x<(x+1)/2<1 if x<1

in terms of an algebraic sense, lets look at two graphs:
y=x, y=(x+1)/2

you have to agree that by solving for the point of intersection,
it will be (1,1).

these two equations are linear functions, which means a line [with no holes].

now how could that be? didn't we say that (.(9)+1)/2 is undefined?

my point is, saying (.(9)+1)/2 is undefined is saying that my original proof that has algebraically proved that x<(x+1)/2<1 if x<1 is wrong!

if you look over my original proof, there is absolutely nothing wrong with what I did or my reasoning, except you claim that 0.(9) does not fit this case.

do you understand of the proof by contradition?

you assume something is true, show that a contradiction exists, and therefore proving your assumption to be false.

what you are doing here however, is claiming something to be true and show that it exempts from properties of real numbers! that is pretty absurd to me.

I did a search for this topic, and came up with every website saying (actually i only visited around 4) .(9)=1

here is their proof:

1/3 = .(3)
2/3 = .(6)
so 1/3+2/3=.(3)+.(6)
so 1=.(9)

in case you claim that 2/3 is not .(6) because you typically see .(6) as .666666...6667 on a calculator. i want to say that as you told me your self, .(6) is a conceptual number, and .666...6667 is a close approxmation to .(6).

in case you are still not convinced,

1/3+1/3+1/3=.(3)+.(3)+.(3)
3*(1/3)=3*0.(3)
1=.(9)

now the only way you are getting out of this proof is to say 1/3 is not .(3) is that what you are saying?

and by the way, you seem to have believe that .(9) is not 1. can you offer any proof of any other source that says the same thing (like a professional website, hopefully university website)

this site was one of the 4-5 sites i visited, it tells a very interesting story :
http://reflectiveeclectic.org/tales_of_truth.htm

this is a sweet proof:
http://mathforum.org/dr.math/faq/faq.0.9999.html



here is a post about limits and such: they all came from the same website i gave above



quote:
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Rather than saying "giving infinity a value," it's perhaps a bit
clearer to say, "giving the concept of a limit of an infinite sequence
of numbers a value."

.9 is not 1; neither is .999, nor .9999999999. In fact if you stop the
expansion of 9s at any finite point, the fraction you have (like .9999
= 9999/10000) is never equal to 1. But each time you add a 9, the
error is less. In fact, with each 9, the error is ten times smaller.

You can show (using calculus or other methods) that with a large
enough number of 9s in the expansion, you can get arbitrarily close to
1, and here's the key:

THERE IS NO OTHER NUMBER THAT THE SEQUENCE GETS ARBITRARILY CLOSE TO.

Thus, if you are going to assign a value to .9999... (going on
forever), the only sensible value is 1.

There is nothing special about .999... The idea that 1/3 = .3333...
is the same. None of .3, .33, .333333, etc. is exactly equal to 1/3,
but with each 3 added, the fraction is closer than the previous
approximation. In addition, 1/3 is the ONLY number that the series
gets arbitrarily close to.

And it doesn't limit itself to single repeated decimals. When we say:

1/7 = .142857142857142857...

none of the finite parts of the decimal is equal to 1/7; it's just
that the more you add, the closer you get to 1/7, and in addition, 1/7
is the UNIQUE number that they all get closer to.


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this next story is funny:

quote:
--------------------------------------------------------------------------------

you're _really_ thinking of

0.999...999

which is not the same thing. You're absolutely right that 0.999...999
is a little below 1, but 0.999999... doesn't fall short of 1 _until_
you stop expanding it. But you never stop expanding it, so it never
falls short of 1.

Suppose someone gives you $1000, but says: "Now, don't spend it all,
because I'm going to go off and find the largest integer, and after I
find it I'm going to want you to give me $1 back." How much money has
he really given you?

On the one hand, you might say: "He's given me $999, because he's
going to come back later and get $1."

But on the other hand, you might say: "He's given me $1000, because
he's _never_ going to come back!"

It's only when you realize that in this instance, 'later' is the same
as 'never', that you can see that you get to keep the whole $1000. In
the same way, it's only when you really understand that the expansion
of 0.999999... _never_ ends that you realize that it's not really 'a
little below 1' at all.
--------------------------------------------------------------------------------



this one makes sense:

quote:
--------------------------------------------------------------------------------
Another way to approach the question is to subtract.

1.0000000....
- .9999999....
--------------
0.0000000....

Sure looks equal to me. What about the "1" at the end, I hear you ask?
Well, I'll write it as soon as I finish writing infinitely many 0s.
Any decimal place you name (say, the four billion three hundred
twenty-eight million two hundred seven thousand four hundred
ninety-fifth) has a 0 in it. A number with a 0 in every decimal place
is certainly 0.

--------------------------------------------------------------------------------

Hey Bugz,
You forgot to vote Yes!
:D

I am sorry for my poor English. My point is that .999 repeating is the same as 1. My English, especially when it gets complicated, is not very well.

lol i voted for no. but as i said, i agree with ur point NotLKH

Don't calculators etc only show 2/3 as 0.666...67 because it rounds? It does the same for 5/9, ie 0.555...56, and I'm sure it does it for all fractions with a recurring number higher than 4.

yes it rounds. it also rounds 1/3=1.333... to 1.3333333
the former has an infinite number of 3s, latter has only 7 3s. thats a big difference

Quote:

---

Originally posted by bugzpodder
yes it rounds. it also rounds 1/3=0.333... to 0.3333333
the former has an infinite number of 3s, latter has only 7 3s. thats a big difference

---

Just fixed the error in your post for you.

How about this explanation in VB:

VB Code:

The_Power = 0
The_Number = 0
Do While 1 = 1
The_Power = The_Power +1
The_Number = The_Number + 9/(10^The_Power)
Loop

Now, you can hopefully see that The_Number represents 0.(9) If you stop the program ever, then the number won't = 0.(9), and so it won't ever = 1
However, as the program runs and runs and runs, The_number will get closer and closer to 1 (Disregarding any problems resulting in storing fractions in binary :p, and the variables being only 'Integer', or 'Double')
But, if the program could theoretcially run the loop ad infinitum then The_Number will equal its limit of 1. (acentote or something ??, i haven't really learnt it yet)

However, The_Number is still represented by 0.(9)
=> it exists
similar to saying 10^{log (n)} exists, even though it has the same value to a different number, and can be written using much less effort.

(Oh, and another inventive proof that 0.(9) = 1:
what is 1-0.(9)???
=0.(0)???)

i'm not trying to argue, but I proved(I think) that the two are exactly equal, not using the limited memory of VB, but real mathematics. Actually, depending on what type of variable you use, The_Number will be rounded to 1 much sooner than even a thousand places. If you can prove that .999...<>1, please do so.

I kinda like this:

Does the number 1 exist?

Does 0.(9) = 1?

If both are yes, then I'd say 0.(9) exists. ;)

Destined

yeah, that's sorta what i wasy trying to say:
Just 'cos there's a easier way of writing 0.(9) doens't mean it doesn't exist.

Oh, and i do believe that 0.(9) = 1, i was just pointing out the difference between physical and theoretical numbers.

P.S. you will note i said to disregard "the variables being only 'Integer', or 'Double' "