# What is the square root of 60?

Oct 1, 2015

$\sqrt{60} = 2 \sqrt{15} \approx \frac{1921}{248}$

#### Explanation:

$60 = {2}^{2} \cdot 3 \cdot 5$ has a square factor ${2}^{2}$

So we can simplify $\sqrt{60}$ using $\sqrt{a b} = \sqrt{a} \sqrt{b}$ as follows:

$\sqrt{60} = \sqrt{{2}^{2} \cdot 15} = \sqrt{{2}^{2}} \sqrt{15} = 2 \sqrt{15}$

It is not possible to simplify $\sqrt{15}$ further, but you can find rational approximations for it using a Newton Raphson type method.

Let $n = 15$, ${p}_{0} = 4$, ${q}_{0} = 1$ and iterate using the formulae:

${p}_{i + 1} = {p}_{i}^{2} + n {q}_{i}^{2}$

${q}_{i + 1} = 2 {p}_{i} {q}_{i}$

At each iteration, ${p}_{i} / {q}_{i}$ is a rational approximation for $\sqrt{n}$

So:

${p}_{1} = {p}_{0}^{2} + n {q}_{0}^{2} = {4}^{2} + 15 \cdot {1}^{2} = 16 + 15 = 31$

${q}_{1} = 2 {p}_{0} {q}_{0} = 2 \cdot 4 \cdot 1 = 8$

Then:

${p}_{2} = {p}_{1}^{2} + n {q}_{1}^{2} = {31}^{2} + 15 \cdot {8}^{2} = 961 + 960 = 1291$

${q}_{2} = 2 {p}_{1} {q}_{1} = 2 \cdot 31 \cdot 8 = 496$

We could go further to get a better approximation, but stop here to get:

$\sqrt{15} \approx \frac{1291}{496}$

So

$\sqrt{60} = 2 \sqrt{15} \approx 2 \cdot \frac{1291}{496} = \frac{1291}{248}$