# How do you find the square root of 26/89?

Feb 13, 2017

$\sqrt{\frac{26}{89}} = \frac{1}{2} + \frac{\frac{15}{356}}{1 + \frac{\frac{15}{356}}{1 + \frac{\frac{15}{356}}{1 + \ldots}}} \approx 0.5405$

#### Explanation:

Note that $26 = 2 \cdot 13$ and $89$ (which is prime) have no common factors and no square factors.

So $\sqrt{\frac{26}{89}}$ is an irrational number with no simpler form.

We can find rational approximations to it.

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Generalised continued fraction method

First here's a little theory...

Suppose:

$\sqrt{n} = a + \frac{b}{2 a + \frac{b}{2 a + \frac{b}{2 a + \ldots}}}$

Then:

$a + \frac{b}{a + \sqrt{n}} = a + \frac{b}{a + \textcolor{b l u e}{a + \frac{b}{2 a + \frac{b}{2 a + \frac{b}{2 a + \ldots}}}}} = \sqrt{n}$

Multiplying both ends by $\left(a + \sqrt{n}\right)$ we get:

${a}^{2} + \textcolor{red}{\cancel{\textcolor{b l a c k}{a \sqrt{n}}}} + b = \textcolor{red}{\cancel{\textcolor{b l a c k}{a \sqrt{n}}}} + n$

Subtracting ${a}^{2} + a \sqrt{n}$ from both sides we find:

$b = n - {a}^{2}$

So if we want a generalised continued fraction to help us approximate a square root $\sqrt{n}$ then pick $a$ such that ${a}^{2} < n$ and derive $b = n - {a}^{2}$. If ${a}^{2}$ is close to $n$ then $b$ will be smaller and the fraction will converge faster.

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Application

Let $n = \frac{26}{89}$ and $a = \frac{1}{2}$

Then:

$b = n - {a}^{2} = \frac{26}{89} - \frac{1}{4} = \frac{104 - 89}{356} = \frac{15}{356}$

So:

$\sqrt{\frac{26}{89}} = \frac{1}{2} + \frac{\frac{15}{356}}{1 + \frac{\frac{15}{356}}{1 + \frac{\frac{15}{356}}{1 + \ldots}}}$

We can truncate this to give rational approximations.

For example:

$\sqrt{\frac{26}{89}} \approx \frac{1}{2} + \frac{\frac{15}{356}}{1 + \frac{\frac{15}{356}}{1 + \frac{15}{356}}} = \frac{74273}{137416} \approx 0.54050$

Having found this, we can see that putting $a = \frac{54}{100} = \frac{27}{50}$ might be a better first approximation.

It gives:

$b = \frac{26}{89} - {\left(\frac{27}{50}\right)}^{2} = \frac{119}{222500}$

So:

$\sqrt{\frac{26}{89}} = \frac{27}{50} + \frac{\frac{119}{222500}}{\frac{27}{25} + \frac{\frac{119}{222500}}{\frac{27}{25} + \frac{\frac{119}{222500}}{\frac{27}{25} + \ldots}}}$

Just a couple of steps of this give us:

$\sqrt{\frac{26}{89}} \approx \frac{27}{50} + \frac{\frac{119}{222500}}{\frac{27}{25}} = \frac{129881}{240300} \approx 0.540495$

which is correct to $6$ decimal places.

This continued fraction expansion will give us approximately $3$ more decimal places for each additional step we include.