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Using the rules for addition of fractions, we know that $1/3+2/3=1$. However, we also know that $1/3=0.333\dots$ up to infinity, and $2/3=0.666\dots$ up to infinity, and adding those two decimals we get $0.999\dots$ up to infinity. Now, according to the concept of limits, the above value approaches $1$, but is not exactly equal to $1$. Why does this discrepancy occur?

Joe
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DEB SANKAR ROY
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    $0.99999999999 \dots$ is not approaching $1 $, it is $1$ – Sine of the Time Aug 12 '23 at 08:50
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    The issue is thinking $0.\dot{9}$ and $1$ are not the same thing. This article discusses it quite well: https://en.m.wikipedia.org/wiki/0.999... – Chris Lewis Aug 12 '23 at 08:50
  • How@Sine of the Time? – DEB SANKAR ROY Aug 12 '23 at 08:53
  • Note that $0.999\dots$ is a number, and it doesn't make sense to say that a number "approaches" another number. What we can say that is that the sequence $(0.9,0.99,0.999,0.9999,\dots)$ approaches $1$. Since $0.999\dots$ is, by definition, the limit of the aforementioned sequence, we see that $0.999\ldots = 1$. – Joe Aug 12 '23 at 09:26
  • Note that $1-0.999\ldots = 0.000\ldots = 0$ and also $0.999\ldots = 9\cdot\frac{1}{9} = 1$. – Abezhiko Aug 12 '23 at 09:45

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One way is $\frac13+\frac23=\frac{1+2}3=1$, the other is $$\sum_{k=1}^\infty 3\cdot 10^{-k}+\sum_{k=1}^\infty 6\cdot 10^{-k}=\sum_{k=1}^\infty 9\cdot 10^{-k}=\lim_{n\to\infty}\sum_{k=1}^n 9\cdot 10^{-k}=\lim_{n\to\infty}9\cdot \frac{10^{-n-1}-10^{-1}}{10^{-1}-1}=\\=\lim_{n\to\infty}1-10^{-n}=1$$

I don't see a problem or a discrepancy.