What is #sqrt(83)#?
1 Answer
Mar 29, 2016
Explanation:
Since
It is not a whole number and cannot be represented in the form
In decimal form it is approximately
It can be expressed as a simple continued fraction
#sqrt(83) = [9;bar(9, 18)] = 9+1/(9+1/(18+1/(9+1/(18+1/(9+1/(18+...))))))#
You can use this continued fraction to calculate approximations for
For example,
For more accurate approximations, truncate later:
#sqrt(83) ~~ [9;9,18] = 9+1/(9+1/18) = 9+18/163 ~~ 9.11043#