Question

How do you find the exact square root of 39?

Answer 1

`39 = 3 * 13` has no square factors, so its square root cannot be simplified. It is an irrational number, so cannot be represented by a fraction or by a terminating or repeating decimal expansion.

Explanation:

If you would like to prove that `sqrt(39)` is irrational, you can prove it in a similar way to `sqrt(15)` as shown here:

How do you prove that square root 15 is irrational?

You can find a succession of rational approximations for `sqrt(39)` using a Newton Raphson type method.

Typically you would start with an approximation `a_0` and iterate using a formula like:

`a_(i+1) = (a_i^2 + n) / (2a_i)`

where `n = 39` is the number you are trying to approximate the square root of.

I prefer to split `a_i` into a fraction `p_i/q_i` and iterate using the formulas:

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

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

If the resulting `p_(i+1)` and `q_(i+1)` have a common factor, then divide both by that factor before the next iteration...

So in our case, let `n = 39`, `p_0 = 6` and `q_0 = 1`. We use an initial approximation `sqrt(39) ~~ 6/1` since `6^2 = 36`.

Then:

`p_1 = p_0^2 + n q_0^2 = 6^2 + 39*1^2 = 36+39 = 75`

`q_1 = 2 p_0 q_0 = 2 * 6 * 1 = 12`

Now both of these are divisible by `3`, so divide both by `3` to get:

`p_(1a) = 25`

`q_(1a) = 4`

If we stopped here we would get `sqrt(39) ~~ 25/4 = 6.25`

Next iteration:

`p_2 = p_(1a)^2 + n q_(1a)^2 = 25^2 + 39*4^2 = 625 + 624 = 1249`

`q_2 = 2 p_(1a) q_(1a) = 2 * 25 * 4 = 200`

If we stopped here we would get `sqrt(39) ~~ 1249/200 = 6.245`

Next iteration:

`p_3 = p_2^2 + n q_2^2 = 1249^2 + 39*200^2 = 1560001 + 1560000 = 3120001`

`q_3 = 2 * 1249 * 200 = 499600`

If we stop here, we get the approximation:

`sqrt(39) ~~ 3120001 / 499600 ~~ 6.2449979983987`

Actually:

`sqrt(39) ~~ 6.24499799839839820584`