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Let a be the number of real numbers between 0 and 1

Let b be the number of all integers.

Can I say b=2a+1, since for every real number n where 0<n<1, I can take n and remove the 0. in the beginning, making it an integer, and have a positive and negative of that integer? I will also have 0 in the integer set which cannot be reached through above method, hence the +1.

I think the logic is pretty sound, are there any mistakes in this argument?

what is sleep
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    This fails, since an integer cannot have infinite length. There are uncountably many real numbers between 0 and 1 and countably many integers. – Daniel R Dec 19 '13 at 19:36
  • Can you refer me to an article that says integer cannot have infinite length? I have never seen it. – what is sleep Dec 19 '13 at 19:37
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    Yes, your logic is pretty unsound. Both these numbers are infinite, and infinities are nothing like finite numbers. The infinity of the real numbers between $0$ and $1$ is much much much larger than that of the integers. Nothing that you can write as a simple equation like that. – Asaf Karagila Dec 19 '13 at 19:37
  • If you search this site or Wikipedia for Cantor+diagonal you will find much discussion of this – Ross Millikan Dec 19 '13 at 19:39
  • @AsafKaragila Can you elaborate. I can see both a and b are infinite, and there can be relations between different infinities, can you point out where the logic is unsound, and why are there more real numbers between 0 and 1 than integers. – what is sleep Dec 19 '13 at 19:43
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    Because the integers are only those numbers which can be written as a finite sum of $1$, and equivalently written by finitely many digits. Numbers like $\frac13$ don't have a finite decimal expansion to begin with, so removing the $0$ first will not turn to an integer. – Asaf Karagila Dec 19 '13 at 19:46
  • @AsafKaragila I get it now. If I remove the 0. before an irrational number between 0 and 1, I will get infinity which is not an integer. – what is sleep Dec 19 '13 at 19:53

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