What is the simplest radical form of #sqrt115#?

2 Answers
Jul 22, 2016

Answer:

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#

Jul 22, 2016

Answer:

#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)#

#color(white)()#
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#