What is the square root of 89?

Since `89` is prime, `sqrt(89)` cannot be simplified.

You can approximate it using a Newton Raphson method.

I like to reformulate it a little as follows:

Let `n = 89` be the number you want the square root of.

Choose `p_0 = 19`, `q_0 = 2` so that `p_0/q_0` is a reasonable rational approximation. I chose these particular values since `89` is about halfway between `9^2 = 81` and `10^2 = 100`.

Iterate using the formulas:

`p_(i+1) = p_i^2 + n q_i^2`

`q_(i+1) = 2 p_i q_i`

This will give a better rational approximation.

So:

`p_1 = p_0^2 + n q_0^2 = 19^2 + 89 * 2^2 = 361+356 = 717`

`q_1 = 2 p_0 q_0 = 2 * 19 * 2 = 76`

So if we stopped here, we would get an approximation:

`sqrt(89) ~~ 717/76 ~~ 9.434`

Let's go one more step:

`p_2 = p_1^2 + n q_1^2 = 717^2 + 89 * 76^2 = 514089 + 514064 = 1028153`

`q_2 = 2 p_1 q_1 = 2 * 717 * 76 = 108984`

So we get an approximation:

`sqrt(89) ~~ 1028153/108984 ~~ 9.43398113`

This Newton Raphson method converges fast.

`color(white)()`
Actually, a rather good simple approximation for `sqrt(89)` is `500/53`, since `500^2 = 250000` and `89 * 53^2 = 250001`

`sqrt(89) ~~ 500/53 ~~ 9.43396`

If we apply one iteration step to this, we get a better approximation:

`sqrt(89) ~~ 500001 / 53000 ~~ 9.4339811321`

`color(white)()`
Footnote

All square roots of positive integers have repeating continued fraction expansions, which you can also use to give rational approximations.

However, in the case of `sqrt(89)` the continued fraction expansion is a little messy so not so nice to work with:

`sqrt(89) = [9; bar(2, 3, 3, 2, 18)] = 9+1/(2+1/(3+1/(3+1/(2+1/(18+1/(2+1/(3+...)))))))`

The approximation `500/53` above is `[9; 2, 3, 3, 2]`