pgrmdave Posted December 14, 2004 Report Posted December 14, 2004 for any number x, x = x - .0000...x, i.e. 1 = .9999....9 The idea goes like this: .333333...3 = 1/3 *3 = *3 .999999...9 = 3/3 = 1 But if this is true, it can be nested, (x - .000...x ) - .0000...(x-.000...x) and so on, to infinity. If taken to infinity, then you are able to subtact an infinite amount from any number and still have a number equal to the origional number. Am I wrong, or is this a big flaw with our number system.
Tim_Lou Posted December 14, 2004 Report Posted December 14, 2004 it isnt very well defined. its like saying what the answer of 0*infinity,which doesnt really exist.
Tim_Lou Posted December 14, 2004 Report Posted December 14, 2004 btw, what does it have to do with "1!" ?
pgrmdave Posted December 14, 2004 Author Report Posted December 14, 2004 btw, what does it have to do with "1!" ? 1 != 1 means "one doesn't equal one" in C++
Tim_Lou Posted December 14, 2004 Report Posted December 14, 2004 oh i see... i thought it was 1 factorial.
pgrmdave Posted December 14, 2004 Author Report Posted December 14, 2004 ah, I hadn't thought about people thinking that. I guess looking at it that way, the title doesn't make much sense.
sanctus Posted December 14, 2004 Report Posted December 14, 2004 I agree, it's like if you substract an infinity of times 0.Your reasoning works if you see the 3.3333333 with a finite number of 3 or .000000....x with a finite number of 0. But your first reasonig doesn't work because 0.33333 != 1/3 (I learned something I also thought it meant factorial, the oposite of your thinking I know this expression of C++, but never would have thought that somebody would use it like that).
pgrmdave Posted December 14, 2004 Author Report Posted December 14, 2004 how does 1/3 != .3333...3 ? At the end of infinity it should be perfectly exact.
Tormod Posted December 15, 2004 Report Posted December 15, 2004 how does 1/3 != .3333...3 ? At the end of infinity it should be perfectly exact. But there is no such thing. It's like asking for the last digit of pi. :)
pgrmdave Posted December 15, 2004 Author Report Posted December 15, 2004 I know that the end of infinity doesn't exist, but we still need to work with it.
pgrmdave Posted December 15, 2004 Author Report Posted December 15, 2004 Just like we work with imaginary numbers. the sqrt of -1 doesn't exist, but we still use it.
Tormod Posted December 15, 2004 Report Posted December 15, 2004 I know that the end of infinity doesn't exist, but we still need to work with it. LOL...this gets my vote for "cool quote of the month". :)
Tim_Lou Posted December 15, 2004 Report Posted December 15, 2004 actually, imaginary does exist in a way, it is very useful in math.it considers not only imaginary but polar geometry. anyway... infinitiy is not really a number, but a limit. A limit that a number approaches as it gets progressively big.
pgrmdave Posted December 15, 2004 Author Report Posted December 15, 2004 But what does this have to do with the original statement?
sanctus Posted December 15, 2004 Report Posted December 15, 2004 As Tim putted it 0.3333333...3333 --> 1/3, but it will always be different, you will never have equality this is the definition of limit.I stated that 0.3333!= 1/3, which is not amazing. To explain my self better your theory works only if you start with an infinite numbers of 3, if you write 0.333....33 it may be huge, but it's not infinite and the most important step in your demonstrationis then wrong, because0.3333333....333!=1/3
pgrmdave Posted December 15, 2004 Author Report Posted December 15, 2004 x = .99999...999910x = 9.99999...9999 -x = - .9999...9999x = 9x = 9/9 = 1 Does the same problem occur in this?
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