# How do you find the square root of 203?

Aug 31, 2016

Use a Newton Raphson method to find:

$\sqrt{203} \approx \frac{6497}{456} \approx 14.247807$

#### Explanation:

$203 = 7 \cdot 29$ has no square factors. So its square root cannot be simplified.

It is an irrational number between $14$ and $15$, since:

${14}^{2} = 196 < 203 < 225 = {15}^{2}$

As such, it cannot be represented in the form $\frac{p}{q}$ for integers $p , q$.

We can find rational approximations using a Newton Raphson method:

To approximate the square root of $n$, start with an approximation ${a}_{0}$ then iterate using the formula:

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

I prefer to re-formulate this slightly using separate integers ${p}_{i} , {q}_{i}$ where ${p}_{i} / {q}_{i} = {a}_{i}$ and formulae:

${p}_{i + 1} = {p}_{i}^{2} + n {q}_{i}^{2}$

${q}_{i + 1} = 2 {p}_{i} {q}_{i}$

If the resulting pair of integers have a common factor, then divide by that before the next iteration.

So for our example, let $n = 203$, ${p}_{0} = 29$ and ${q}_{0} = 2$ (i.e. ${a}_{0} = 14.5$) ...

$\left\{\begin{matrix}{p}_{0} = 29 \\ {q}_{0} = 2\end{matrix}\right.$

{ (p_1 = p_0^2+ n q_0^2 = 29^2 + 203*2^2 = 841+812 = 1653), (q_1 = 2 p_0 q_0 = 2*29*2 = 116) :}

Note that both ${p}_{1}$ and ${q}_{1}$ are divisible by $29$, so divide both by that:

$\left\{\begin{matrix}{p}_{1 a} = \frac{1653}{29} = 57 \\ {q}_{1 a} = \frac{116}{29} = 4\end{matrix}\right.$

Next iteration:

{ (p_2 = p_(1a)^2 + n q_(1a)^2 = 57^2+203*4^2 = 3249+3248 = 6497), (q_2 = 2 p_(1a) q_(1a) = 2*57*4 = 456) :}

If we stop here, we get:

$\sqrt{203} \approx \frac{6497}{456} \approx 14.247807$

Each iteration roughly doubles the number of significant figures in the approximation.