# How can you calculate sqrt(2) in your head?

Sep 12, 2016

$\sqrt{2} \approx \frac{577}{408} \approx 1.4142$

#### Explanation:

One way of calculating rational approximations for $\sqrt{2}$ in your head uses a variation of a Newton Raphson method.

Your basic Newton Raphson method for finding approximations for the square root of a number $n$ is to choose an initial approximation ${a}_{0}$, then iterate using a formula like:

${a}_{i + 1} = \frac{{a}_{i}^{2} + n}{2 {a}_{i}}$

That's fine, but the fractions can get a bit messy and distracting.

So I prefer to split ${a}_{i} = {p}_{i} / {q}_{i}$ where ${p}_{i}$ and ${q}_{i}$ are integers, then use these formulae:

$\left\{\begin{matrix}{p}_{i + 1} = {p}_{i}^{2} + n {q}_{i}^{2} \\ {q}_{i + 1} = 2 {p}_{i} {q}_{i}\end{matrix}\right.$

So for $\sqrt{2}$, we have $n = 2$ and I might choose ${p}_{0} = 3$, ${q}_{0} = 2$ (i.e. my initial approximation is $\frac{3}{2}$).

Then:

$\left\{\begin{matrix}{p}_{1} = {p}_{0}^{2} + 2 {q}_{0}^{2} = {3}^{2} + \left(2 \cdot {2}^{2}\right) = 9 + 8 = 17 \\ {q}_{1} = 2 {p}_{0} {q}_{0} = 2 \cdot 3 \cdot 2 = 12\end{matrix}\right.$

So if we stopped after one iteration, our approximation would be $\frac{17}{12} = 1.41 \overline{6}$

Let's do another iteration to get more accuracy:

$\left\{\begin{matrix}{p}_{2} = {p}_{1}^{2} + 2 {q}_{1}^{2} = {17}^{2} + \left(2 \cdot {12}^{2}\right) = 289 + 288 = 577 \\ {q}_{2} = 2 {p}_{1} {q}_{1} = 2 \cdot 17 \cdot 12 = 408\end{matrix}\right.$

This is probably as many iterations as you want to do in your head, since you have to work with double the number of digits each time.

So it remains to long divide $\frac{577}{408}$ in your head to find:

$\sqrt{2} \approx \frac{577}{408} \approx 1.414216$

Not bad - it's actually closer to $1.414213562373$