Adding 1/3 to 2/3 gives 1 EXACTLY. But expanding the two fractions and then adding gives 0.99999 Where is the flaw in this reasoning?
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It is the same flaw which states $\frac{1}{3}=0.3333...$ – evil999man Jul 09 '15 at 15:37
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1$0.99999\ldots=1$ exactly. – Jul 09 '15 at 15:42
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This question has been asked before under the guise of "why does .999... = 1?". – Paul Sundheim Jul 09 '15 at 15:43
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The flaw is that you assumed the sequence of digits in $0.99999\ldots$ terminates at some point. It doesn't. The sequence $0.99999\ldots$ is $$ \frac 9 {10} + \frac 9 {100} + \frac 9 {1000} + \frac 9 {10000} + \cdots $$ and by the usual method of finding the sum of an infinite geometric series, that adds up to $1$, exactly.
Notice that $0.33333$, terminating after five digits, is not exactly $1/3$ either. If the sequence of $3$s terminates at some point, then it's less than $1/3$. You can make it as close as you want to $1/3$ by making the number of $3$s big enough.
(I think various slight variations on the theme of this question have been posted here a number of times.)
