# How do you find the exact square root of 41?

Oct 24, 2015

$\sqrt{41}$ is an irrational number. It cannot be expressed as one integer divided by another or by a terminating or repeating decimal.

$\sqrt{41} \approx 6.40312423743284868648$

#### Explanation:

$\sqrt{41}$ can be represented as a repeating continued fraction:

sqrt(41) = [6;bar(2,2,12)] = 6+1/(2+1/(2+1/(12+1/(2+1/(2+...)))))

We can use this expression to get rational approximations for $\sqrt{41}$

For example,

sqrt(41) ~~ [6;2,2] = 6+1/(2+1/2) = 6+2/5 = 6.4

Alternatively, use a Newton Raphson type approach:

To find approximations for $\sqrt{n}$, start with a reasonable approximation ${a}_{0}$ and iterate using the formula:

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

So if we started with $n = 41$, ${a}_{0} = 6$

${a}_{1} = \frac{{a}_{0}^{2} + n}{2 {a}_{0}} = \frac{{6}^{2} + 41}{12} = \frac{77}{12} = 6.41 \dot{6}$

${a}_{2} = \frac{{a}_{1}^{2} + n}{2 {a}_{1}} = \frac{{\left(\frac{77}{12}\right)}^{2} + 41}{2 \cdot \left(\frac{77}{12}\right)}$

$= \frac{{77}^{2} + 41 \cdot {12}^{2}}{2 \cdot 77 \cdot 12} = \frac{5929 + 41 \cdot 144}{1848}$

$= \frac{5929 + 5904}{1848} = \frac{11833}{1848} \approx 6.40314$