# What is the square root of 3 ?

##### 2 Answers
Nov 3, 2016

$\sqrt{3} = 1.732$

#### Explanation:

3 is not a perfect square, so does not have an exact square root.

$\sqrt{3}$ is an irrational number. The answer is an infinite, non-recurring decimal.

Using a calculator gives:

$\sqrt{3} = 1.7320508 \ldots . .$

Give your answer rounded to a reasonable number of decimal places.

Mar 19, 2017

The square root of $3$ is an irrational number which we can approximate as:

$\sqrt{3} \approx \frac{18817}{10864} \approx 1.7320508$

#### Explanation:

"The" square root of $3$ is an irrational number. It is not expressible in the form $\frac{p}{q}$ for integers $p , q$ and its decimal expansion neither repeats nor terminates.

It can be expressed by a (non terminating) continued fraction:

sqrt(3) = [1;bar(1,2)] = 1+1/(1+1/(2+1/(1+1/(2+1/(1+1/(2+1/(1+...)))))))

This positive square root is also known as the principal square root of $3$. The number $- \sqrt{3}$ is also a square root of $3$.

We can get rational approximations to $\sqrt{3}$ by truncating the continued fraction early.

For example:

sqrt(3) ~~ [1;1,2,1] = 1+1/(1+1/(2+1/1)) = 7/4 = 1.75

sqrt(3) ~~ [1;1,2,1,2,1] = 1+1/(1+1/(2+1/(1+1/(2+1/1)))) = 26/15 = 1.7bar(3)

sqrt(3) ~~ [1;1,2,1,2,1,2,1] = 1+1/(1+1/(2+1/(1+1/(2+1/(1+1/(2+1/1)))))) = 97/56 = 1.732bar(142857)

$\textcolor{w h i t e}{}$
Another method to find approximations to $\sqrt{n}$ begins with a rational approximation ${p}_{0} / {q}_{0}$ then repeatedly applies the formulae:

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

Let us start with $n = 3$, ${p}_{0} = 2$, ${q}_{0} = 1$.

Then:

$\left\{\begin{matrix}{p}_{1} = {p}_{0}^{2} + n {q}_{0}^{2} = {2}^{2} + 3 \cdot {1}^{2} = 4 + 3 = 7 \\ {q}_{1} = 2 {p}_{0} {q}_{0} = 2 \cdot 2 \cdot 1 = 4\end{matrix}\right.$

$\left\{\begin{matrix}{p}_{2} = {p}_{1}^{2} + n {q}_{1}^{2} = {7}^{2} + 3 \cdot {4}^{2} = 49 + 48 = 97 \\ {q}_{2} = 2 {p}_{1} {q}_{1} = 2 \cdot 7 \cdot 4 = 56\end{matrix}\right.$

$\left\{\begin{matrix}{p}_{3} = {p}_{2}^{2} + n {q}_{2}^{2} = {97}^{2} + 3 \cdot {56}^{2} = 9409 + 9408 = 18817 \\ {q}_{3} = 2 {p}_{2} {q}_{2} = 2 \cdot 97 \cdot 56 = 10864\end{matrix}\right.$

Stopping here, we get:

$\sqrt{3} \approx \frac{18817}{10864} \approx 1.7320508$

As you may notice, this method is exponentially faster than the continued fraction method, effectively doubling the number of terms each time.