Question

What is the simplest radical form of `sqrt115`?

Answer 1

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: `sqrt125=sqrt(5xx5xx5)=sqrt(5^2)xxsqrt5=5sqrt5`

In this case, no such luck:
`sqrt115=sqrt(5xx23)=sqrt5xxsqrt23`

Answer 2

`sqrt(115)` is already in simplest form.

Explanation:

The prime factorisation of `115` is:

`115 = 5*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)*sqrt(23)`

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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 + 1/(1+1/(2+1/(1+1/(1+1/(1+1/(1+1/(1+1/(2+1/(1+1/(20+1/(1+...)))))))))))`

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 + 1/(1+1/(2+1/(1+1/(1+1/(1+1/(1+1/(1+1/(2+1/1))))))))`

`=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*105^2 = 1267875`

`1126^2 = 1267876`

only differ by `1`.

This makes `1126/105 ~~ 10.7bar(238095)` an efficient approximation for `sqrt(115) ~~ 10.7238052947636`