Originally posted by: Kyteland
We're having a debate at work. Is 1=0.99999..... repeating. I say that this holds but one of my coworkers claims that multiplication breaks down for an infinitely repeating number.
x = 0.9999...
10x = 9.9999...
10x - x = 9.9999... - 0.9999...
9x = 9
x = 1.
What do you think?
Originally posted by: SilentRunning
Silverpig what is the fraction which is equal to 0.999999........ and don't say 1/1 because dividing 1 by 1 equals 1 with a remainder 0.
If you can represent 0.999999..... with a fractional equivalent then I will concede that 0.99999.... is not an irrational number.
Originally posted by: SilentRunning
OK, part of the proof
For -1<r<1 , the sum converges as n approaches infinity ,in which case
Sure seems to be stating an approximation to me. Converge, does not equate to equal to.
They are stating that r^(n+1) approaches zero as n approaches infinity
That is all for now, must do something constructive.
Originally posted by: MadRat
Hence, objective values cannot be used to define subjective ones.
Originally posted by: silverpig
Originally posted by: SilentRunning
OK, part of the proof
For -1<r<1 , the sum converges as n approaches infinity ,in which case
Sure seems to be stating an approximation to me. Converge, does not equate to equal to.
They are stating that r^(n+1) approaches zero as n approaches infinity
That is all for now, must do something constructive.
That's all fine and dandy but where in 0.999... do you see a sum? How can a number approach anything? What does a single number converge to?
All rational numbers are repeating decimals.
2/1 = 2.000000000000000000...
5/4 = 1.250000000000000000...
Do you agree?
Originally posted by: Alphathree33
...
Again, properties of the real number system are not approximate. They are axioms. These things work by definition, not by proof or by punching a lot of numbers on your calculator and scratching your head.
You are correct with your thinking here.Originally posted by: Rainsford
I would say yes because that proof makes sense. However, could someone explain something to me, I still do not understand this. Hopefully I can make this clear enough.
.9 is 9/10, right? So it could be considered as you moving 9/10 of the distance between two points (say, 0 and 1), right?
Now .09 is 9/100 which could also be thought of as 9/10 * 1/10. When you add .09 to .9 this could be thought of as moving 9/10 of the remaining 1/10 of the distance, right?
So for each 9 you add to the end, you can think of it as moving 9/10 of the remaining distance. So the distance between you and the second point (1) gets smaller and smaller with each 9 added on to the end. However, it does not seem like you would ever actually reach the second point because you are only moving a fraction of the remaining distance each time. No matter how small, that distance would still exist, right? Or am I missing something?
lets say you have n number of 9's, as n approaches infinity the value approaches 1, but we are conserned with when n equals infintiy.Originally posted by: Rainsford
I would say yes because that proof makes sense. However, could someone explain something to me, I still do not understand this. Hopefully I can make this clear enough.
.9 is 9/10, right? So it could be considered as you moving 9/10 of the distance between two points (say, 0 and 1), right?
Now .09 is 9/100 which could also be thought of as 9/10 * 1/10. When you add .09 to .9 this could be thought of as moving 9/10 of the remaining 1/10 of the distance, right?
So for each 9 you add to the end, you can think of it as moving 9/10 of the remaining distance. So the distance between you and the second point (1) gets smaller and smaller with each 9 added on to the end. However, it does not seem like you would ever actually reach the second point because you are only moving a fraction of the remaining distance each time. No matter how small, that distance would still exist, right? Or am I missing something?
Originally posted by: Woodchuck2000
Any how do you intend to define .99 reccuring without using some kind of series? As the floating point precision with which you define the number increases, it tends towards one.incorrect. as stated earlier a single number is in fact, a single number and nothing more. single numbers, such as 0.999..., tend toward nothing. only series of numbers, or sums, can trend toward something, or have a limit.
As far as your 'proof' goes, I'd like to start by challenging
which makes no sense whatsoever.Proof: 0.9999... = Sum 9/10^n
(n=1 -> Infinity)
followed by
Where you've introduced an undefined variable m and done mysterious things with it.= lim sum 9/10^n
(m -> Infinity) (n=1 -> m)
I'll challenge the rest once you've explained the first premises.
BTW You're going to have to explain your notation a little as it's mathematical in no sense of the word.
Does m -> infinity mean 'as m tends towards infinity' or are you defining a range?
Originally posted by: DoNotDisturb
Originally posted by: bleeb
They are not the same because you can NEVER reach infinity. And the only way for 0.9999... to converge to 1 is if you reach infinity, but there is no way to reach infinity. THERFORE 0.9999.... != 1. (hhaha i love it)
that is true. anything that is considered "infinite" is theoretical, the LIMIT approaches 1 as n -> infinity, HOWEVER, there is theoretically no "end point", thus you cannot say it's exactly equal to 1. it gets closer and closer to 1, but will never equal it because infinity is not attainable. if you don't know why its theoretical, read on infinity. if you look at the proof in that dr.math site, n -> infinity.
Originally posted by: Woodchuck2000
As proofs go, it's badly phrased in my opinion...
As i stated earlier, I've seen a much better proof of the proposition, I was just playing devil's advocate because josphII was bullsh*tting...
Originally posted by: TuxDave
Originally posted by: FrustratedUser
Originally posted by: TuxDave
Originally posted by: McPhreak
Originally posted by: TuxDave
That proof looks good to me. I believe 0.9999... = 1. For those who don't believe still, show me 1-0.999... doesn't equal 0.
And if you say it equals 0.000000.....001... there's an infinite number of zeros, so that '1' never really exists, so technically 1-0.999... = 0
Does that mean 0.99999.....98 = 0.999999999.....99?
Ahh.. but there's a freaky difference. I say numbers like 0.9999....98 doesn't exist because we define there to be an infinite number of nines... and here with 0.9999...98 we say, there's a last nine, but there isn't.. there's infinite!
Well, that means 0.999999.... is infinitely smaller than 1.
Hence my latter question, is there such thing as a number that's infinitely smaller than 1. Because that implies that there exists a largest number smaller than 1. And I say no... such a number does not exist...
<edit>
just as there is no smallest number greater than 1. So since nothing can go between 0.999... and 1, AAND we say that there is no such largest number smaller than 1, we must conclude that they are equal.
Originally posted by: FrustratedUser
Originally posted by: halik
.9999 infinatelly repeating is EXACTLY equal to 1...sheesh people
I must be stupid but an infinitely long string of 0.999999.... still need to be added to 0.0000......001 to be = to 1.
Originally posted by: kenleung
it sounds like an asymptotic (sp?) number to me... well for a function that is but i can find the similarities
graph f(x)=1/(x-1). x approaches 1 but it never reaches 1
apply the same logic to this question, .999999.... isn't 1
Originally posted by: SilentRunning
My point about e=mc^2 is that that is not an actual equation, It is an approximation of the actual equation.
And you seem to be missing the point
1/3 = 0.3333........ + 10^(-n) as n approaches infinity
to say that as n approaches infinity 10^(-n) = zero is wrong, it approaches zero.
Nothing you can say will change that fact. So 0.33333...... is an approximation of 1/3
True, I'm amazed that 117 people are confident that it is wrong.Originally posted by: spidey07
Yeah,
Funny how you can perform so many proofs using series (supports the fact the the two are the same number), then how the definition of the real number system supports it, and so on.
The only real argument is some folks just don't get it.
Originally posted by: Yomicron
lets say you have n number of 9's, as n approaches infinity the value approaches 1, but we are conserned with when n equals infintiy.Originally posted by: Rainsford
I would say yes because that proof makes sense. However, could someone explain something to me, I still do not understand this. Hopefully I can make this clear enough.
.9 is 9/10, right? So it could be considered as you moving 9/10 of the distance between two points (say, 0 and 1), right?
Now .09 is 9/100 which could also be thought of as 9/10 * 1/10. When you add .09 to .9 this could be thought of as moving 9/10 of the remaining 1/10 of the distance, right?
So for each 9 you add to the end, you can think of it as moving 9/10 of the remaining distance. So the distance between you and the second point (1) gets smaller and smaller with each 9 added on to the end. However, it does not seem like you would ever actually reach the second point because you are only moving a fraction of the remaining distance each time. No matter how small, that distance would still exist, right? Or am I missing something?
if you were to just keep adding 9s you would never reach 1, because you are looking at a finite number of 9s
Originally posted by: silverpig
Originally posted by: MadRat
Hence, objective values cannot be used to define subjective ones.
What, pray tell, is an objective value, and what is a subjective value?