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There have been a few comments regarding this in the shoutbox recently, so I thought I'd start a thread to discuss the topic.
The question of whether 0.999..=1 is an age old problem, which I'm sure everyone encounters sometime in their life/education. The simple answer to the question is that, yes, they are the same, as they are simply two different decimal expansions which represent the same number. One can think of a decimal expansion as an infinite sum. For example, take the sum 0.999..=9/10+9/100+9/1000+... . This is a geometric series whose sum is one. Further, consider the decimal expansion for 1: 1=1+0/10+0/100+0/1000+... . This is also a geometric series whose sum is one.
I think that once we look at the problem in terms of infinite series, it is not surprising that two infinite series add up to the same finite value.
Any questions or comments are welcome.
~neutralino
If you haven't found something strange during the day, it hasn't been much of a day - John A. Wheeler.
The Following User Says Thank You to neutralino For This Useful Post:
But infinity never ends, you'll always have a .9 never ending.
Yes, that is what 0.999.. means. Think of it another way: the number 0.9999 is quite close to 1, but then the number 0.9999999999 is even closer, and 0.999999999999999999999999999 is even closer than that. Now consider the number 0.99... with infinitely many nines. This is the closest possible number to 1, but since we are dealing with real numbers I should be able to find another number closer to 1 still *. But, by definition I cannot, since there are infinitely many 9's. Thus 0.99.. must be equal to 1.
* There exists a real number between any two real numbers. To show this consider r<s. Then r<(r+s)/2<s.
~neutralino
If you haven't found something strange during the day, it hasn't been much of a day - John A. Wheeler.
Now 1/9~.1111and if you keep carrying this out never stoping you'll always have another.1 to get it a bit closer, but being infinite it's never going to end and at no point because there is no end point to infinity will it ever =.111... It will always be a very very close approximation.
I had a convo with Lloyd way back when about this, and we put it into particulate terms if I remember right. Would it be fair to infer that .999... can be representative of 1 composite particle consisting of an infinite number of point particles; and 1 representative of a 1 point particle? This way both would equal 1, but of a different type: the composite, infinite; and the point, absolute.
I've always considered there to be an extremely fine line between infinite and absolute, and for this reason I feel there will always remain debateable points as to whether or not the larger is larger than the largest. It might be the cause for the quantum fluctuation in there somewhere.
"To show this consider r<s. Then r<(r+s)/2<s."
If r is 1 and s is 2, the result is 3/2, but I don't see how that implies 4/2. The infinite set taken as one set I understand though.
Special thanks to you, Neutralino, for your help before.
![=2\left[\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\cdots \right]](/forum/latex/img/0f9d013509ab517db4c5e921197aaf8e-1.gif "=2\left[\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\cdots \right]")
![=2\left[\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\cdots\right]](/forum/latex/img/74026e5d28a2fe2e1de2de3e6710ab5b-1.gif "=2\left[\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\cdots\right]")
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