Question

How do you simplify `sqrt65`?

Answer 1

`65 = 5*13` has no square factors, so `sqrt(65)` is the simplest form.

Explanation:

If a radicand (the part under the root sign) of a square root has a square factor, then it can be simplified:

`sqrt(a^2b) = abs(a) sqrt(b)`

or if you know that `a >= 0`, more simply:

`sqrt(a^2b) = a sqrt(b)`

For example, `sqrt(24) = sqrt(2^2*6) = 2sqrt(6)`

In our example, we find `65 = 5 * 13` has no square factors, so cannot be simplified in this way.

If you like, you can reexpress it:

`sqrt(65) = sqrt(5)sqrt(13)`

but that is not (as far as I know) considered 'simpler'.

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Note that `sqrt(65)` is an irrational number. That is, it cannot be expressed as a fraction `p/q` for integers `p` and `q`. As a result, its decimal expansion does not terminate or recur.

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Bonus

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

is of the form `n^2 + 1` with `n = 8`.

As a result, the square root can be expressed as a very simple continued fraction ...

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

You can use this to give you good approximations for `sqrt(65)`, by truncating the continued fraction after a few terms.

For example,

`[8; 16] = 8+1/16 = 129/16 = 8.0625`

`[8; 16, 16] = 8+1/(16+1/16) = 8+16/257 = 2072/257 ~~ 8.0622568`

Actually `sqrt(65)` is closer to `8.06225774829854965236`