What is $\sqrt{65}$ $sqrt65$ in simplified radical form?

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Answer 1 · 2016-06-19T21:52:13

$\sqrt{65}$ $sqrt(65)$ cannot be simplified.

The prime factorisation of $65$ $65$ is:

$65 = 5 \cdot 13$ $65 = 5 * 13$

Since this has no square factors $\sqrt{65}$ $sqrt(65)$ cannot be simplified.

Bonus

$65 = 64 + 1 = 8^{2} + 1$ $65 = 64+1 = 8^2+1$

is in the form $n^{2} + 1$ $n^2+1$ .

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:

$sqrt(65) ~~ 8.0622577483$

What is sqrt65√65sqrt65 in simplified radical form?