# How do you find the square root of 2000?

Dec 29, 2015

sqrt(2000) = 20 sqrt(5) = 20 [2;bar(4)] ~~ 44.7

#### Explanation:

If $a , b \ge 0$ then $\sqrt{a b} = \sqrt{a} \sqrt{b}$

So:

$\sqrt{2000} = \sqrt{400 \cdot 5} = \sqrt{400} \cdot \sqrt{5} = 20 \sqrt{5}$

Since $5 = {2}^{2} + 1$ is of the form ${n}^{2} + 1$, $\sqrt{5}$ has a simple expansion as a continued fraction:

sqrt(5) = [2;bar(4)] = 2 + 1/(4+1/(4+1/(4+1/(4+...))))

According to how accurate an approximation we want we can terminate this continued fraction at more or fewer terms.

For example:

sqrt(5) ~~ [2;4,4] = 2+1/(4+1/4) = 2 + 4/17 = 38/17

So:

$\sqrt{2000} = 20 \sqrt{5} \approx 20 \cdot \frac{38}{17} \approx 44.71$

Actually:

$\sqrt{2000} \approx 44.72135954999579392818$

As another way of calculating the successive approximations provided by the continued fraction, consider the sequence:

$0 , 1 , 4 , 17 , 72 , 305 , \ldots$

where ${a}_{1} = 0$, ${a}_{2} = 1$, ${a}_{i + 2} = {a}_{i} + 4 {a}_{i + 1}$

This is similar to the Fibonacci sequence, except the rule is ${a}_{i + 2} = {a}_{i} + \boldsymbol{4} {a}_{i + 1}$ instead of ${a}_{i + 2} = {a}_{i} + {a}_{i + 1}$.

This is strongly related to the continued fraction:

[4;bar(4)] = 4+1/(4+1/(4+1/(4+1/(4+...))))

The ratio between successive terms of the sequence tends to $2 + \sqrt{5}$ (somewhat faster than the Fibonacci sequence does to $\frac{1}{2} + \frac{\sqrt{5}}{2}$)

For example, we can find an approximation for $\sqrt{5}$ in:

$\frac{305}{72} - 2 = \frac{161}{72}$

Hence $\sqrt{2000} \approx 20 \cdot \frac{161}{72} = \frac{3220}{72} = 44.7 \dot{2}$