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

Apr 22, 2018

$\sqrt{8} \approx 3$

#### 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 $3 - \sqrt{8}$:

$\left(x - 3 - \sqrt{8}\right) \left(x - 3 + \sqrt{8}\right) = {\left(x - 3\right)}^{2} - 8 = {x}^{2} - 6 x + 1$

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

$\left\{\begin{matrix}{a}_{0} = 0 \\ {a}_{1} = 1 \\ {a}_{n + 2} = 6 {a}_{n + 1} - {a}_{n}\end{matrix}\right.$

The first few terms are:

$0 , 1 , 6 , 35 , 204 , 1189 , 6930 , \ldots$

The ratio between successive terms will tend very quickly towards $3 + \sqrt{8}$.

So:

$\sqrt{8} \approx \frac{6930}{1189} - 3 = \frac{3363}{1189} \approx 2.828427$