# What is the square root of 48569834 ?

Apr 4, 2017

The square root is irrational.

#### Explanation:

${6969}^{2} = 48566961$ and
${6970}^{2} = 48580900$.
Since 48,569,384 lies between two consecutive perfect squares, its square root is irrational.
If you need an approximation for it every computer comes with a calculator that will provide one. Alternatively, use Google.

[Of course, every positive number has two square roots. The principle square root is positive, and the other square root is negative.]

Apr 4, 2017

$\sqrt{48569384} = 6969 + \frac{2423}{13938 + \frac{2423}{13938 + \frac{2423}{13938 + \ldots}}}$

#### Explanation:

To attempt to find the square root of $48569384$, we can first split off pairs of digits starting from the right to get:

$48 \text{|"56"|"93"|} 84$

Note that $48 < 49 = {7}^{2}$

Hence the square root is a little under $7000$

We can try using a method based on the Babylonian method to get a better approximation.

Starting with ${p}_{0} / {q}_{0} = \frac{7000}{1}$, to find better approximations for $\sqrt{n}$ where $n = 48569384$, apply the following formulae:

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

So:

$\left\{\begin{matrix}{p}_{1} = {p}_{0}^{2} + n {q}_{0}^{2} = {7000}^{2} + 48569384 \cdot {1}^{2} = 97569384 \\ {q}_{1} = 2 {p}_{0} {q}_{0} = 2 \cdot 7000 \cdot 1 = 14000\end{matrix}\right.$

Let's see where this has got us after just one step:

${p}_{1} / {q}_{1} = \frac{97569384}{14000} \approx 6969.2$

Try:

${6969}^{2} = 48566961 < 48569384$

${6970}^{2} = 48580900 > 48569384$

So the square root is somewhere strictly between $6969$ and $6970$. It cannot be rational, so it's an irrational number.

Note that in general:

$\sqrt{n} = a + \frac{b}{2 a + \frac{b}{2 a + \frac{b}{2 a + \ldots}}}$

where $0 < a < \sqrt{n}$ and $b = n - {a}^{2}$

Now:

$48569384 - {6969}^{2} = 2423$

and:

$2 \cdot 6969 = 13938$

So:

$\sqrt{48569384} = 6969 + \frac{2423}{13938 + \frac{2423}{13938 + \frac{2423}{13938 + \ldots}}}$