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#