# How do you find the simplest radical form of 433?

Dec 16, 2017

Hmmm...

#### Explanation:

If you mean the simplest form of $\sqrt{433}$ then it is $\sqrt{433}$ since $433$ is a prime number.

If you mean the simplest expression involving a radical with value $433$, then you might choose $433 \sqrt{1}$ or possibly $\sqrt{187489}$ (since $433 \cdot 433 = 187489$).

Approximations

Since $433$ is a prime number, its square root cannot be simplified. In addition, it is not close to a square number or half way between two square numbers. So it is a little fiddly to get a rapidly convergent continued fraction expressing it.

Note that:

${20}^{2} = 400 < 433 < 441 = {21}^{2}$

So a reasonable approximation is somewhere between $20$ and $21$. Linearly interpolating, we find it is about $20 + \frac{33}{41} \approx 20.8 = \frac{104}{5}$

Then:

$433 - {\left(\frac{104}{5}\right)}^{2} = \frac{9}{25}$

Now:

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

So:

$\sqrt{433} = \frac{104}{5} + \frac{\frac{9}{25}}{\frac{208}{5} + \frac{\frac{9}{25}}{\frac{208}{5} + \frac{\frac{9}{25}}{\frac{208}{5} + \ldots}}}$

$\textcolor{w h i t e}{\sqrt{104}} = \frac{1}{5} \left(104 + \frac{9}{208 + \frac{9}{208 + \frac{9}{208 + \ldots}}}\right)$

Hence, if we define a sequence recursively by:

$\left\{\begin{matrix}{a}_{0} = 0 \\ {a}_{1} = 1 \\ {a}_{n + 2} = 208 {a}_{n + 1} + 9 {a}_{n}\end{matrix}\right.$

then the ratio between successive terms rapidly converges to $5 \sqrt{433} + 104$

The first few terms are:

$0 , 1 , 208 , 43273 , 9002656$

So:

$5 \sqrt{433} \approx \frac{9002656}{43273} - 104 = \frac{4502264}{43273}$

So:

$\sqrt{433} \approx \frac{4502264}{5 \cdot 43273} = \frac{4502264}{216365} \approx 20.808652046$