# What is the square root of 89?

Oct 1, 2015

The square root of $89$ is a number which when squared gives $89$.

$\sqrt{89} \approx 9.434$

#### Explanation:

Since $89$ is prime, $\sqrt{89}$ cannot be simplified.

You can approximate it using a Newton Raphson method.

I like to reformulate it a little as follows:

Let $n = 89$ be the number you want the square root of.

Choose ${p}_{0} = 19$, ${q}_{0} = 2$ so that ${p}_{0} / {q}_{0}$ is a reasonable rational approximation. I chose these particular values since $89$ is about halfway between ${9}^{2} = 81$ and ${10}^{2} = 100$.

Iterate using the formulas:

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

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

This will give a better rational approximation.

So:

${p}_{1} = {p}_{0}^{2} + n {q}_{0}^{2} = {19}^{2} + 89 \cdot {2}^{2} = 361 + 356 = 717$

${q}_{1} = 2 {p}_{0} {q}_{0} = 2 \cdot 19 \cdot 2 = 76$

So if we stopped here, we would get an approximation:

$\sqrt{89} \approx \frac{717}{76} \approx 9.434$

Let's go one more step:

${p}_{2} = {p}_{1}^{2} + n {q}_{1}^{2} = {717}^{2} + 89 \cdot {76}^{2} = 514089 + 514064 = 1028153$

${q}_{2} = 2 {p}_{1} {q}_{1} = 2 \cdot 717 \cdot 76 = 108984$

So we get an approximation:

$\sqrt{89} \approx \frac{1028153}{108984} \approx 9.43398113$

This Newton Raphson method converges fast.

$\textcolor{w h i t e}{}$
Actually, a rather good simple approximation for $\sqrt{89}$ is $\frac{500}{53}$, since ${500}^{2} = 250000$ and $89 \cdot {53}^{2} = 250001$

$\sqrt{89} \approx \frac{500}{53} \approx 9.43396$

If we apply one iteration step to this, we get a better approximation:

$\sqrt{89} \approx \frac{500001}{53000} \approx 9.4339811321$

$\textcolor{w h i t e}{}$
Footnote

All square roots of positive integers have repeating continued fraction expansions, which you can also use to give rational approximations.

However, in the case of $\sqrt{89}$ the continued fraction expansion is a little messy so not so nice to work with:

sqrt(89) = [9; bar(2, 3, 3, 2, 18)] = 9+1/(2+1/(3+1/(3+1/(2+1/(18+1/(2+1/(3+...)))))))

The approximation $\frac{500}{53}$ above is [9; 2, 3, 3, 2]