Question

How do you simplify `sqrt(257)`?

Answer 1

`257` is prime, so `sqrt(257)` cannot be simplified.

Explanation:

If the radicand has a square factor, then you can move the square root of that factor outside of the square root.

For example, `sqrt(12) = sqrt(2^2*3) = sqrt(2^2)sqrt(3) = 2sqrt(3)`

If `n` has no square factor (e.g. `n` is prime), then `sqrt(n)` cannot be simplified in this way.

In your case you can note that `257` is `256+1` and `sqrt(256) = 16`. So `16` makes a very good first approximation to the square root for use with a method like Newton Raphson.

So you can put `a_0 = 16` and iterate using:

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

or suchlike to quickly find a very good approximation.

In our case `a_1 = 16+1/32 = 16.03125` while `sqrt(257) ~= 16.03122`