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Let's start considering a simple fractions like $\dfrac {1}{2}$ and $\dfrac {1}{3}$.

If I choose to represent those fraction using decimal representation, I get, respectively, $0.5$ and $0.3333\overline{3}$ (a repeating decimal).

That is where my question begins.

If I multiply either $\dfrac {1}{2}$ or $0.5$ by $2$, I end up with $1$, as well as, if I multiply $\dfrac {1}{3}$ by $3$.

Nonetheless, if I decide to multiply $0.3333\overline{3}$ by $3$, I will not get $1$, but instead, $0.9999\overline{9}$

What am I missing here?

*Note that my question is different than the question Adding repeating decimals

Community
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carlosrafaelgn
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2 Answers2

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Hint: compute the difference between $1$ and $0.9\bar9$. How much is that ? What do you conclude ?

Tom-Tom
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  • Sorry, but I did not get the point properly :( Isn't the difference between $1$ and $0.9\overline{9}$ equals to $0.000000...0001$ (infinite zeroes followed by one)? I understand that if I have infinite zeroes, than this $1$ will never be reached, therefore, giving me a practical value of 0. So, if the difference between $1$ and the number is $0$, than the number is $1$. But why?! Why do we write $0.9\overline{9}$, an endless amount of nines?! – carlosrafaelgn Sep 17 '14 at 13:12
  • You have understood. – Tom-Tom Sep 17 '14 at 13:17
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    Every digit has a place: there's a tenth's place, a hundredth's place, a thousandth's place, and so forth. If you have infinitely many zeroes, then all of those places have a 0 in them, so there's nowhere to put a 1. The difference is all zeroes. –  Sep 17 '14 at 13:18
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$0.9999$ repeating is equal to $1$. There is some crucial thing you should understand: Every finite decimal of the form $0.999$ is $not$ equal to $0.999$ repeating. The latter represents the number one and no other number. Watch this video and the others related to the construction of reals by the professor Francis Su: https://www.youtube.com/watch?v=sqEyWLGvvdw

LearningMath
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