# What is the square root of 67?

Oct 12, 2016

$67$ is a prime, and cannot be factored......

#### Explanation:

.........and thus ${67}^{\frac{1}{2}}$ $=$ $\pm \sqrt{67}$.

Feb 19, 2017

$\sqrt{67} \approx \frac{34313}{4192} \approx 8.185353$

#### Explanation:

$67$ is a prime number, so in particular has no square factors. So its square root is irrational and not simplifiable.

There are several methods you can use to find rational approximations.

Here's a method based on the Babylonian method...

To find the square root of a number $n$, choose an initial approximation ${p}_{0} / {q}_{0}$ where ${p}_{0} , {q}_{0}$ are integers.

Then apply the following formulas repeatedly to get better approximations:

$\left\{\begin{matrix}{p}_{i + 1} = {p}_{i}^{2} + n {q}_{i}^{2} \\ {q}_{i + 1} = 2 {p}_{i} {q}_{i}\end{matrix}\right.$

In our example, let $n = 67$, ${p}_{0} = 8$ and ${q}_{0} = 1$, since ${8}^{2} = 64$ is quite close to $67$. Then:

$\left\{\begin{matrix}{p}_{1} = {p}_{0}^{2} + n {q}_{0}^{2} = {8}^{2} + 67 \cdot {1}^{2} = 64 + 67 = 131 \\ {q}_{1} = 2 {p}_{0} {q}_{0} = 2 \cdot 8 \cdot 1 = 16\end{matrix}\right.$

$\left\{\begin{matrix}{p}_{2} = {p}_{1}^{2} + n {q}_{1}^{2} = {131}^{2} + 67 \cdot {16}^{2} = 17161 + 17152 = 34313 \\ {q}_{2} = 2 {p}_{1} {q}_{1} = 2 \cdot 131 \cdot 16 = 4192\end{matrix}\right.$

If we stop here, we get:

$\sqrt{67} \approx \frac{34313}{4192} \approx 8.185353$

which is accurate to $6$ decimal places.