# How do you find a one-decimal place approximation for sqrt 10?

Oct 17, 2015

Use one step of a Newton Raphson method to find:

$\sqrt{10} \approx \frac{19}{6} \approx 3.2$

#### Explanation:

To find the square root of a positive number $n$, choose a first approximation ${a}_{0}$ then apply the following iteration step as many times as you like, to get the accuracy you want:

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

In our case, $n = 10$ and let ${a}_{0} = 3$, since ${3}^{2} = 9$ is quite close to what we want.

Then:

${a}_{1} = \frac{{a}_{0}^{2} + n}{2 {a}_{0}} = \frac{{3}^{2} + 10}{2 \cdot 3} = \frac{9 + 10}{6} = \frac{19}{6} = 3.1 \dot{6}$

So to one decimal place $\sqrt{10} \approx 3.2$

Oct 17, 2015

Truncate the continued fraction expansion for $\sqrt{10}$ to find:

sqrt(10) = [3;bar(6)] ~~ [3;6] = 3+1/6 = 3.1dot(6) ~~ 3.2

#### Explanation:

The continued fraction expansion for $\sqrt{{n}^{2} + 1}$ is

sqrt(n^2+1) = [n;bar(2n)] = n + 1/(2n+1/(2n+1/(2n+1/(2n+...))))

Now

$10 = {3}^{2} + 1$

So

sqrt(10) = [3;bar(6)] = 3+1/(6+1/(6+1/(6+1/(6+...))))

So truncating at [3;6] we find:

sqrt(10) ~~ [3;6] = 3 + 1/6 = 3.1dot(6) ~~ 3.2

If you want more accuracy you can include more terms, e.g.

sqrt(10) ~~ [3;6,6,6] = 3+1/(6+1/(6+1/6)) = 3+37/228 ~~ 3.16228