What is #sqrt65# in simplified radical form?

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Answer 1 · 2016-06-19T21:52:13

#sqrt(65)# cannot be simplified.

The prime factorisation of #65# is:

#65 = 5 * 13#

Since this has no square factors #sqrt(65)# cannot be simplified.

Bonus

#65 = 64+1 = 8^2+1#

is in the form #n^2+1#.

The square roots of such numbers have a simple form of continued fraction expansion:

#sqrt(n^2+1) = [n;bar(2n)] = n+1/(2n+1/(2n+1/(2n+1/(2n+1/(2n+1/(2n+...))))))#

So in our example:

#sqrt(65) = [8;bar(16)] = 8+1/(16+1/(16+1/(16+1/(16+1/(16+1/(16+...))))))#

You can get rational approximations of #sqrt(65)# to any desired accuracy by truncating the continued fraction expansion early.

For example:

#sqrt(65) ~~ [8;16] = 8+1/16 = 8.0625#

#sqrt(65) ~~ [8;16,16] = 8+1/(16+1/16) = 8+16/257 ~~ 8.0622568#

The actual value is more like:

#sqrt(65) ~~ 8.0622577483#