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Meanwhile, every number larger than 1 will be larger than any decimal of the form 0.999...9 for any finite number of nines. Therefore, 0.999... cannot be identified with any number larger than 1, either. Because 0.999... cannot be bigger than 1 or smaller than 1, it must equal 1 if it is to be any real number at all. [1] [2]
All the arguments for 0.999... equals 1 are flawed... A common argument is that since 1/3 = 0.333… then we can simply multiply both sides by 3 to get 1 = 0.999… This argument requires that we start by accepting that 1/3 equals 0.333…
OK... I learned about 0.(9) being the same as 1 in high school too... But now I have this question. 1 = 0.99999... right? 2 = 1.99999... every ok so far. does this ...
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(Another way of looking at this is by considering the fact that the sequence 0.9, 0.09, 0.009, 0.0009, ... converges to zero, but its sum is equal to 1.) Redquark 00:11, 26 October 2006 (UTC) Ok, for the record I do understand that 0.999... = 1, so I don't have a problem with that. My problem is how the picture represents this concept.
The correct notation is (0.9; 0.99; 0,999; ...). Q: If it is possible to construct number systems in which 0.999... is less than 1, shouldn't we be talking about those instead of focusing so much on the real numbers? Aren't people justified in believing that 0.999... is less than one when other number systems can show this explicitly?
If (0.999...) is a real number, then it MUST be equal to 1, since there is no such thing as "the number prior to 1" by definition of the reals. Then, let's all agree ...
I don't feel strongly about "exactly", so I'm going to take this issue and run with it. Among non-mathematicians, real numbers often are decimals, in which case the equality of nu