# How do you simplify sqrt(257)?

Jul 29, 2015

$257$ 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, $\sqrt{12} = \sqrt{{2}^{2} \cdot 3} = \sqrt{{2}^{2}} \sqrt{3} = 2 \sqrt{3}$

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:

${a}_{i + 1} = {a}_{i} + \frac{n - {a}_{i}^{2}}{2 {a}_{i}}$

or suchlike to quickly find a very good approximation.

In our case ${a}_{1} = 16 + \frac{1}{32} = 16.03125$ while $\sqrt{257} \cong 16.03122$