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Topic: 0.999... (Read 9747 times) 

Grimbal
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Re: 0.999...
« Reply #150 on: Feb 26^{th}, 2013, 1:22am » 
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on Feb 25^{th}, 2013, 10:41pm, Pavel wrote:So for every a(i) of S there must exist some number number [Epsilon](a(n))>0 such that for any n=1,2,3,... 1[Epsilon]=a(n) thus 1=a(n)+[Epsilon] since [Epsilon] is a positive real number 1>a(n) 
 Epsilon is not a positive real number. In this case, Epsilon is a function of a(n). You should write Epsilon[n]. on Feb 25^{th}, 2013, 10:41pm, Pavel wrote:1>a(i) for all a(i) belonging to S therefore .999...<1 
 You cannot draw that conclusion. You can have a(i)<1 and lim a(i)=1. on Feb 25^{th}, 2013, 10:41pm, Pavel wrote:.999..... is a concept not a number and its unfair to compare it to the number 1 and not the concept of 1 (thus most proofs above). 
 I disagree. 0.999... is a number. Its value is sum _{k=1,2,...} (9*10^{k}).

« Last Edit: Feb 26^{th}, 2013, 1:31am by Grimbal » 
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rmsgrey
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Re: 0.999...
« Reply #151 on: Feb 26^{th}, 2013, 4:11am » 
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on Feb 25^{th}, 2013, 10:41pm, Pavel wrote:1>a(i) for all a(i) belonging to S therefore .999...<1 
 We also have: .999...>a(i) for all (finite) i therefore? .999...<.999... Something is wrong with the logic there.


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peoplepower
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Re: 0.999...
« Reply #152 on: Feb 26^{th}, 2013, 4:44am » 
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Something is wrong with the logic elsewhere too. on Feb 25^{th}, 2013, 10:41pm, Pavel wrote:.999..... is a concept not a number and its unfair to compare it to the number 1 and not the concept of 1 (thus most proofs above). 
 There are multiple meanings that can be and often are attached to the single symbol 1. The meanings of 1 as a natural number, rational number, and real number are all different even though in some sense it has the same value, which is grounds for confusion. When we are asked to compare the real number 0.999... with 1, we need to ask which meaning fits best from the context we are given. Of course, one is to choose the real number 1. Quote: the discussion i believe is not about limits or convergence but one of a number that looks like this: 0.9999999999999999999999999999 but there is always one bigger but with the trait that it MUST start 0.9..... , and so never quiet 1. 
 The assumption inherent in the problem is that we choose to work in some ordercompletion the rational numbers (like the real numbers). Thus, by definition really, the supremum of S exists taking the value 0.999... Quote:the series S contains infinitely many elements and it does converge to 1 but there isn't a single number in there like 1. 
 We are working in an ordered field. Likeness is based on distance rather than some properties of the decimal representation of the number.

« Last Edit: Feb 26^{th}, 2013, 4:44am by peoplepower » 
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riddler358
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Re: 0.999...
« Reply #153 on: May 4^{th}, 2016, 11:15pm » 
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note: i didn't read all of the answers if we agree that 0,(3) = 1/3 and we agree that 3 * 0,(3) = 0,(9) we substitute and get 0,(9) = 3 * 1/3 then we probably should conclude that 0,(9) = 1


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Grimbal
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Re: 0.999...
« Reply #154 on: May 25^{th}, 2016, 8:34am » 
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And that's perfectly correct. The problem most people face, I think, is that they have the intuition that if two numbers are written differently, then they must be different.


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Grimbal
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Re: 0.999...
« Reply #156 on: Sep 28^{th}, 2016, 8:39am » 
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one != 1 by commutativity: 1=one !


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towr
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Re: 0.999...
« Reply #157 on: Sep 28^{th}, 2016, 8:51am » 
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I wonder if there's a programming language with ! as factorial operator and where it binds stronger than than unequal (or just doesn't have != as unequal operator, and uses = as equality operator) $: 1!=1 > True


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