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This number 2.962962 can be rational

$$x=2.962962$$ $$10x=29.62962$$ $$100x=296.2962$$ $$1000x=2962.962$$ $$1000x-10x=\frac{990x}{990}=\frac{2933}{990}$$

why is this wrong? That way of getting the answer is how I was said to do it

Comment: $$1000x-x=\frac{999x}{999}=\frac{2960}{999}=?$$

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The last line is wrong. If the decimal part is supposed to be repeating, I imagine it should read $1000x - x = 999x = 2960$. –  Qiaochu Yuan Apr 2 '13 at 8:54
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Your solution makes very less sense. You have made many direct mistakes. Check it once again. –  lsp Apr 2 '13 at 8:55
    
@QiaochuYuan That can't be right because the answer has to be rational –  user1838781 Apr 2 '13 at 8:56
    
@user: Can you explain your reasoning in that last comment? –  Hurkyl Apr 2 '13 at 9:02
    
@Hurkyl it has to be a fraction = no decimals –  user1838781 Apr 2 '13 at 9:06

3 Answers 3

up vote 1 down vote accepted

$$ 1000x -x = 2962.962962\ldots - 2.962962\ldots = 2960. $$ The fractional part cancels out completely.

Then $$ x = \frac{2960}{999} = \frac{37\cdot80}{37\cdot27} = \frac{80}{27}. $$

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I assume you meant a repeating decimal like $2.\overline{962}$ (i.e. $2.9629629\ldots$) rather than one that terminates like you write.

Your work at the end is somewhat confused; I can't tell what you were trying to do. But the calculation of $1000x - 10x$ yields

$$ \begin{matrix} 2&9&\not 6^5&{}^1 2&.&9&\not6^5&{}^1 2&9&\not6^5&{}^12&... \\ & & 2&9&.&6&2&9&6&2&9&... \\\hline \\ 2 & 9 & 3 & 3 & . & 3 & 3 & 3 & 3 & 3 & 3 & \cdots \end{matrix} $$

(I hope that is how they still notate subtraction these days) and so you have

$$ 1000 x - 10 x = 2933 + \frac{1}{3} $$

and

$$ 1000 x - 10 x = 990 x $$

and so we've derived

$$ 990 x = 2933 + \frac{1}{3} $$


Of course, it would have been easier to compare $1000x$ to $x$....

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Hint: there's a very easy way to transform periodic into decimals without any calculation. Here's an example: $$0.123\overline{4567}=\frac{1234567-123}{9999000}.$$ In your case it's just $$2\frac{962}{999}.$$

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