# How do you approximate -sqrt30?

Jul 7, 2015

Let's forget the $-$sign for now. It will be put before the answer later. We do $\sqrt{30}$.

#### Explanation:

Now $25$ is the nearest square below $30$, and $36$ is the first higher square.
So the answer must be between $5 \mathmr{and} 6$.

Since $30$ is also roughly halfway between $25 \mathmr{and} 36$, the answer should also be in that neighbourhood.
My first approximation would be $5.5$ which gives a square of $30.25$, a little bit high.

The calculator answer is $5.477 \ldots$ which will be rounded to $5.5$

Jul 7, 2015

Use Newton Raphson method to find a sequence of approximations for $\sqrt{30}$:

$\frac{11}{2} = 5.5$

$\frac{241}{44} \cong 5.4772$

$\frac{116161}{21208} \cong 5.477225575$

#### Explanation:

The Newton-Raphson method, specialised to finding square roots boils down to choosing a reasonable first approximation ${a}_{0}$, then iterating using the formula:

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

where $n$ is the number you are approximating the square root of.

In our case let us choose ${a}_{0} = 5$, $n = 30$.

Then:

${a}_{1} = \frac{{a}_{0}^{2} + n}{2 {a}_{0}}$

$= \frac{{5}^{2} + 30}{2 \cdot 5} = \frac{25 + 30}{10} = \frac{55}{10} = \frac{11}{2} = \textcolor{red}{5.5}$

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

$= \frac{{\left(\frac{11}{2}\right)}^{2} + 30}{2 \cdot \frac{11}{2}} = \frac{\frac{121}{4} + \frac{120}{4}}{11} = \frac{241}{44} \cong \textcolor{red}{5.4772}$

${a}_{3} = \frac{{a}_{2}^{2} + n}{2 {a}_{2}}$

$= \frac{{\left(\frac{241}{44}\right)}^{2} + 30}{2 \cdot \frac{241}{44}}$

$= \frac{\frac{58081}{{44}^{2}} + \frac{58080}{{44}^{2}}}{\frac{482}{44}}$

$= \frac{116161}{21208} \cong \textcolor{red}{5.477225575}$

Notice that the number of significant digits roughly doubles on every iteration.

Actually, if we want to use rational approximations,

it is nicer to write ${a}_{0} = {p}_{0} / {q}_{0}$ and iterate using the formulas:

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

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

optionally dividing ${p}_{i}$ and ${q}_{i}$ by any common factor.

That avoids some of the messy fractions.