Divide the radicand into groups of two, starting
from the decimal point and going in either
direction. For rational radicands, add pairs of
zeroes to the right of the decimal point as
needed. If the number is irrational, you already
have an infinite sequence of pairs. Now that the
radicand is marked off in pairs, forget the
decimal point (except in the answer). Call each
new digit of the approximation D, and the
current best approximation of the square root R.
Find the largest integer whose square is less
than or equal to the leftmost group. Since
groups are at most two digits long, the integer
is one digit long: in this case, the leftmost group
is 2, so we begin with D = 1, R = 1.
2 -1 = 1, append the next pair of digits, giving
100.
Find the largest digit D such that D*(20*1+D) is
less than or equal to 100.
(HINT: estimate D by dividing 20*1 into 100 -
this approach might overestimate.)
The result is 4*(20*1+4) = 4*24 = 96, so D = 4, R
= 1.4.
100 - 96 = 4, append next pair of digits, result is
400.
400/(20*14) = 1 + fraction, 1*(20*14+1) =
%3D
1*281, so D = 1, R = 1.41.
400 - 281 = 119, appending next pair yields
11900.
11900/(20*141)
= 4 + fraction, 4*(20*141+4) =
4*2824 = 11296, so D = 4, R = 1.414, etc. etc.
What justifies this?
Suppose we know R, the best N-digit
approximation from this algorithm. Then the
next approximation will be 10*R + D, and D must
be the largest digit that satisfies
10R + D)? = (100R? + 20RD + D² )
(= 100R? + D(20R+ D) < radicand
