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So this is my first question and unfortunately it's quite a silly one. But I still hope I can get an answer because it drives me crazy! So here's my problem:
I know that $\frac{1}{3}=0.\overline{3}$ and $\frac{2}{3}=0.\overline{6}$. This makes totally sense to me since we divide a whole (the number one) in three equal parts. And so $\frac{2}{3}-\frac{1}{3}=0.\overline{6}-0.\overline{3}=0.\overline{3}$. So to get the next higher third we just add $\frac{1}{3}$ which means to every digit in $0.\overline{3}$ we add the digit 3 and we do so for an infinite number of digits. But that would mean that $\frac{3}{3}=0.\overline{9}$ but as I always learned it is $\frac{3}{3}=1$. Now here comes the point that really puzzles me. It is $1.0-0.\overline{3}=0.\overline{6}$ but it is also $0.\overline{9}-0.\overline{3}=0.\overline{6}$ since we just subtract the digit 3 from every digit behind zero for infinite digits. So maybe this is common knowledge and I was just asleep in math when we talked about that but is $0.\overline{9}=1$? I mean are these mathematically the exact same numbers? That we don't just round up and ignore the error of $0.\overline{0}1$, but that they are literally the exact number= And if so does that also mean that $6.\overline{9}=7$, $1.1\overline{9}=1.2$ and is $0.\overline{0}1=0$? Please help me because I feel like I lose my whole understanding of numbers...

Tony Fly
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  • Check out: http://math.stackexchange.com/questions/11/does-99999-1 – The_Sympathizer Aug 03 '13 at 21:30
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    Yes, 1 = 0.999...; there's an extensive discussion here. Regarding another of your questions, the symbol $0.000000...01$ is not meaningful. –  Aug 03 '13 at 21:31
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    You must be new here. – Ali Caglayan Aug 03 '13 at 21:32
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    Also, a comment: $0.\bar{0}1$ is not a valid decimal of a real. To see this, just ask yourself: "what would be the (negative) power of 10 corresponding to the place in which the '1' occupies? What if I tack another '1' on the end to make $0.\bar{0}11$? What is it's power of 10? What does this tell me about if one can make sense of $0.\bar{0}1$ as a decimal representation of a number?" – The_Sympathizer Aug 03 '13 at 21:44
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    Tanks a lot!! The fact that $0.\overline{0}1$ is not a valid decimal and that every two numbers that aren't the same have to have at least one number in between solved my life crisis. Btw I'm totally amazed how fast questions get answered here - and yes I am new here. – Tony Fly Aug 03 '13 at 21:51

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