# Infinite repeated decimals in a number

I don't now how to convert infinite periodic decimal number $$x=3,1(42)=3.1424242...$$ to a fraction $$\frac{a}{b}$$ a,b are integers. Need to find a,b THANks

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Equation $$x = 3.1\overline{42}$$ first multiply by $10$ we get

$$10x = 31.\overline{42}$$ now multiply by $100$ we have

$$1000x = 3142.\overline{42}$$ from third equation subtract second we get

$$990x = 3111$$

$$x = \frac{3111}{990}=\frac{1037}{330}$$

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I think you mean $\frac{1037}{330}$ – Mike Nov 15 '12 at 0:29
Note it is "multiplying" and "multiply". – Pedro Tamaroff Nov 15 '12 at 0:47

There are several ways to do this. One way would be first to focus on the repeating part. So let $$x = 0.042424242\ldots = 0.0\overline{42}.$$

Then you have $100x = 4.2424242\ldots = 4.2\overline{42}$. So $$99x = 100x - x = 4.2424242\ldots - 0.042424242 = 4.2.$$ So $$x = \frac{4.2}{99} = \frac{42}{990}.$$

Now then you can probably find $3.1424242\ldots = 3.1 + x = \ldots$

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Any repeating decimal can be written as a convergent geometric series. Here we have $$3.1424242\ldots=3.1+{42\over 10^3}+{42\over 10^5}+{42\over 10^7}+\cdots$$ $$=3.1+{42\over 10^3}\Bigl(1+{1\over 100}+{1\over 100^2}+{1\over 100^3}+\cdots \Bigr)$$ $$=3.1+{42\over 10^3}{1\over 1-{1\over 100}}=3.1+{42\over 990}$$

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$$0.0\overline{42} = 42 \cdot 0.0\overline{01} = \frac{42}{99} \cdot 0.0\overline{99} = \frac{42}{99} \cdot 0.1$$

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