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Does .99999… = 1?

$\frac{1}{3} + \frac{1}{3} + \frac{1}{3} = \frac{3}{3} = 1$

but

$0.\overline{3} + 0.\overline{3} +0.\overline{3} = 0.\overline{9}$

Does that mean that $0.\overline{9} = 1$ ? Any proof?

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    $0.\overline{9}$ _is_ one. You've just proven it.2012-11-15
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    I'm not so sure that $0.\overline{9}$ *is* $1$. But, under the equivalence relation $=$, it's safe to say that $0.\overline{9}=1$. (I'm reminded of "[The Treachery of Images](http://en.wikipedia.org/wiki/The_Treachery_of_Images)")2012-11-15
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    @DouglasS.Stones : The question of whether $0.\bar9$ "is" $1$ is analogous to whether Albert Einstein "is" the author of the photon theory of light. We can imagine an alternate history in which the theory was proposed by someone else, but in the model we assume by default, the different labels "Albert Einstein" and "author of the photon theory of light" refer to the same individual. For $0.\bar9$, the only question is whether you're talking about the "real number system", or instead another number system that you haven't bothered to mention or define which would it from "being" $1$.2012-11-15
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    @NieldeBeaudrap: Under [the Cauchy sequence construction of the real numbers](http://en.wikipedia.org/wiki/Construction_of_the_real_numbers#Construction_from_Cauchy_sequences), what we write as $0.\overline{9}$ would naturally be defined as the Cauchy sequence $(0,0.9,0.99,\ldots)$, whereas what we write as $1$ would naturally be defined as $(1,1,1,\ldots)$. So, in this context, $0.\overline{9}$ is not $1$ (since the two sequences are not equal), but, at the same time, $0.\overline{9}=1$, where $=$ is an equivalence relation.2012-11-15
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    @DouglasS.Stones: you forgot the part where the Cauchy construction is equivalence classes of sequences, under the relation of differences of sequences converging to zero. As the two sequences are equivalent under that relation, the real numbers represented by those sequences are identical. The same goes for any construction of the real numbers. You might as well complain that 1/2 is "not necessarily" 2/4 in the rational numbers.2012-11-15
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    @DouglasS.Stone: it doesn't really add anything to the discussion to note that = is an equivalence relation. It is true, but more to the point, it is that equivalence relation which allows one to substitute one expression for another. Properly used, equality *is* identity, in mathematics. To contemplate whether $0.\bar9$ *is* $1$, given that $0.\bar9$ *equals* $1$, is to introduce a distinction which doesn't exist (or at least, to make it evident that you need to define your terms clearly so that others can figure out what it is that you actually mean).2012-11-15

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