# What is the simplest radical form of sqrt115?

Jul 22, 2016

There is no simpler form

#### Explanation:

With radicals you try to factorize the argument, and see if there are any squares that can be 'taken out from under the root'.

Example: $\sqrt{125} = \sqrt{5 \times 5 \times 5} = \sqrt{{5}^{2}} \times \sqrt{5} = 5 \sqrt{5}$

In this case, no such luck:
$\sqrt{115} = \sqrt{5 \times 23} = \sqrt{5} \times \sqrt{23}$

Jul 22, 2016

$\sqrt{115}$ is already in simplest form.

#### Explanation:

The prime factorisation of $115$ is:

$115 = 5 \cdot 23$

Since there are no square factors, it is not possible to simplify the square root. It is possible to express it as a product, but that does not count as simpler:

$\sqrt{115} = \sqrt{5} \cdot \sqrt{23}$

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

In common with any irrational square root of a rational number, $\sqrt{115}$ has a repeating continued fraction expansion:

sqrt(115) = [10;bar(1,2,1,1,1,1,1,2,1,20)]

$= 10 + \frac{1}{1 + \frac{1}{2 + \frac{1}{1 + \frac{1}{1 + \frac{1}{1 + \frac{1}{1 + \frac{1}{1 + \frac{1}{2 + \frac{1}{1 + \frac{1}{20 + \frac{1}{1 + \ldots}}}}}}}}}}}$

You can truncate the continued fraction expansion early to give rational approximations for $\sqrt{115}$.

For example:

sqrt(115) ~~ [10;1,2,1,1,1,1,1,2,1]

$= 10 + \frac{1}{1 + \frac{1}{2 + \frac{1}{1 + \frac{1}{1 + \frac{1}{1 + \frac{1}{1 + \frac{1}{1 + \frac{1}{2 + \frac{1}{1}}}}}}}}}$

$= \frac{1126}{105}$

In fact, by truncating just before the end of the repeating section of the continued fraction, we have found the simplest rational approximation for $\sqrt{115}$ that satisfies Pell's equation.

That is:

$115 \cdot {105}^{2} = 1267875$

${1126}^{2} = 1267876$

only differ by $1$.

This makes $\frac{1126}{105} \approx 10.7 \overline{238095}$ an efficient approximation for $\sqrt{115} \approx 10.7238052947636$