What is the square root of 8 to the nearest integer?

1 Answer
Apr 22, 2018

Answer:

#sqrt(8) ~~ 3#

Explanation:

Note that:

#2^2 = 4 < 8 < 9 = 3^2#

Hence the (positive) square root of #8# is somewhere between #2# and #3#. Since #8# is much closer to #9 = 3^2# than #4 = 2^2#, we can deduce that the closest integer to the square root is #3#.

We can use this proximity of the square root of #8# to #3# to derive an efficient method for finding approximations.

Consider a quadratic with zeros #3+sqrt(8)# and #3-sqrt(8)#:

#(x-3-sqrt(8))(x-3+sqrt(8)) = (x-3)^2-8 = x^2-6x+1#

From this quadratic, we can define a sequence of integers recursively as follows:

#{ (a_0 = 0), (a_1 = 1), (a_(n+2) = 6a_(n+1)-a_n) :}#

The first few terms are:

#0, 1, 6, 35, 204, 1189, 6930,...#

The ratio between successive terms will tend very quickly towards #3+sqrt(8)#.

So:

#sqrt(8) ~~ 6930/1189-3 = 3363/1189 ~~ 2.828427#