changing repeating decimal to fractions

Anne Davis

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Joined
Jan 28, 2009
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1
Please change the following repeating decimals to fractions:

1.19 with the 9 repeating 0.183 with the 83 repeating
 
0.183 with the 83 repeating

\(\displaystyle .18383838383.....\)

\(\displaystyle x=.1838383...\)
\(\displaystyle 10x=1.838383...\)
\(\displaystyle 1000x=183.838383...\)

Subtract 1000x-10x:

\(\displaystyle 1000x=183.838383...\)
\(\displaystyle 10x=1.838383...\)
------------------------------------------------
\(\displaystyle 990x=182\)

\(\displaystyle x=\frac{182}{990}\)

See what I done?. This is a standard technique with these.

Let's do .278278278278.............

x=.278278278
1000x=278.278278...

Subtract:

999x=278

\(\displaystyle \frac{278}{999}\)

You have to multiply by something like 10, 100, 1000, etc so that when you subtract you get an integer when you subtract them, then solve for x.
 
Hello, Anne!

galactus is absolutely correct!
I enjoy formatting these problems . . .


Change the repeating decimal to a fraction.

\(\displaystyle (a)\;\;1.1999\overline{9}\hdots\)

\(\displaystyle \text{Let }\,x\:=\:1.1999\hdots\)

\(\displaystyle \begin{array}{cccccc}\text{Multiply by 100:} & 100x &=& 119.999\hdots \\ \text{Multiply by 10:} & 10x &=&\;\; 11.999\hdots \end{array}\)

\(\displaystyle \text{Subtract: }\:90x \:=\:108\quad\Rightarrow\quad x \:=\:\frac{108}{90} \:=\:\frac{6}{5}\)




\(\displaystyle (b)\;\;0.18383\overline{83}\hdots\)

\(\displaystyle \text{Let }\,x \:=\:0.1838383\hdots\)

\(\displaystyle \begin{array}{cccccc}\text{Multiply by 1000:} & 1000x &=& 183.838383 \hdots \\ \text{Multiply by 10:} & \;\;10x &=& \quad1.838383\hdots \end{array}\)

\(\displaystyle \text{Subtract: }\:990x \:=\:182 \quad\Rightarrow\quad x \:=\:\frac{182}{990} \:=\:\frac{91}{495}\)

 
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