# How do you find the repeating decimal 0.82 with 82 repeated as a fraction?

##### 1 Answer

#### Explanation:

In case you have not encountered it, let me introduce you to some notation:

A repeating decimal can be represented using a bar placed over the repeating pattern of decimals.

So instead of writing

If we multiply

#(100-1) 0.bar(82) = 100 * 0.bar(82) - 1 * 0.bar(82)#

#color(white)((100-1) 0.bar(82)) = 82.bar(82) - 0.bar(82)#

#color(white)((100-1) 0.bar(82)) = 82#

Notice that the

Next divide both ends by

#0.bar(82) = 82/(100-1) = 82/99#

**Alternative method**

An alternative method recognises that:

#0.bar(82) = 0.82 + 0.0082 + 0.000082 +...#

is a geometric series, with initial term

This has sum given by the formula:

#s_oo = a/(1-r) = 0.82/(1-1/100) = 0.82/(99/100) = (0.82*100)/99 = 82/99#