# How do you simplify sqrt41?

Jun 11, 2017

$\sqrt{41} \approx 6.4031242374$ is an irrational number which cannot be simplified.

#### Explanation:

$41$ is a prime number, so has no square factors.

As a result its square root cannot be simplified. It is an irrational number.

We find:

${6}^{2} = 36 < 41 < 49 = {7}^{2}$

So:

$6 < \sqrt{41} < 7$

To get a good approximation for $\sqrt{41}$ note that:

${64}^{2} = 4096$

So:

$\sqrt{41} \approx \sqrt{40.96} = 6.4 = \frac{32}{5}$

In general:

$\sqrt{{a}^{2} + b} = a + \frac{b}{2 a + \frac{b}{2 a + \frac{b}{2 a + \ldots}}}$

Putting $a = \frac{32}{5}$ and $b = \frac{1}{25}$, we find:

$\sqrt{41} = \frac{32}{5} + \frac{\frac{1}{25}}{\frac{64}{5} + \frac{\frac{1}{25}}{\frac{64}{5} + \frac{\frac{1}{25}}{\frac{64}{5} + \ldots}}}$

We can get rational approximations for $\sqrt{41}$ by truncating this continued fraction.

For example:

$\sqrt{41} \approx \frac{32}{5} + \frac{\frac{1}{25}}{\frac{64}{5}} = \frac{2049}{320}$

$\sqrt{41} \approx \frac{32}{5} + \frac{\frac{1}{25}}{\frac{64}{5} + \frac{\frac{1}{25}}{\frac{64}{5}}} = \frac{131168}{20485} \approx 6.4031242374$