Question

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

Answer 1

`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`