How do you find the square root of 4783?

As `4783` is not a perfect square, to find the square root of `4783`, we should do a special long division, where we pair, the numbers in two, starting from decimal point in either direction.

Here first pair is `47` and the number whose square is just less than it is `6`, so we write `36` below `47`. The difference is `11` and now we bring down next two digits `83`. As a divisor we first write double of `6` i.e. `12` and then find a number `x` so that `12x` (here `x` stands for single digit in units place) multiplied by `x` is just less than the number, here `1183`. We find for `x=9`, we have `129xx9=1161` and get the difference as `22`.

Now as we still have a remainder of `22`, we bring `00` after decimal point. Also put decimal after `69` and this makes it `6900`.

Now recall we had brought as divisor `6xx2=12`, but this time we have `69` so we make the divisor as `138x` and identify an `x` so that `138x` multiplied by `x` is just less than `2200`. This number is just `1`, as making it `2` will make the product `1382xx2=2764>2200`.

We continue to do this till we have desired accuracy.

`color(white)(xx)6color(white)(xx)9color(white)(xx).1color(white)(xx)5color(white)(xx)9`
`ul6|bar(47)color(white)(.)bar(83).color(white)(.)bar(00)color(white)(.)bar(00)color(white)(.)bar(00)color(white)(.)bar(00)`
`color(white)(X)ul(36)color(white)(X)darr`
`color(red)(12)9|color(white)(X)11color(white)(.)83`
`color(white)(xxxxx)ul(11color(white)(.)61)`
`color(white)(x)color(red)(138)1|color(white)(X)22color(white)(.)00`
`color(white)(xxxxxxx)ul(13color(white)(.)81)`
`color(white)(xx)color(red)(1382)5|color(white)(.)8color(white)(.)19color(white)(.)00`
`color(white)(xxxxxxxx)ul(6color(white)(.)91color(white)(.)25)`
`color(white)(xx)color(red)(13830)9|color(white)()1color(white)(.)27color(white)(.)75color(white)(.)00`
`color(white)(xxxxxxxx)ul(1color(white)(.)24color(white)(.)47color(white)(.)81)`
`color(white)(xxxxxxxxx)color(white)(.)32color(white)(.)71color(white)(.)9`

Hence `sqrt4783~~69.159`

Here's a method using generalised continued fractions.

First the theory...

Suppose we can find numbers `a` and `b` such that:

`sqrt(n) = a+b/(2a+b/(2a+b/(2a+b/(2a+...))))`

Then we have:

`a+b/(a+sqrt(n)) = a+b/(a+color(blue)(a+b/(2a+b/(2a+b/(2a+b/(2a+...)))))) = sqrt(n)`

Multiplying both ends by `(a+sqrt(n))` we get:

`a^2+color(red)(cancel(color(black)(asqrt(n))))+b = color(red)(cancel(color(black)(asqrt(n))))+n`

Hence:

`b = n-a^2`

In our example, first note that `4783 < 4900 = 70^2`, so let's see what we get when we square `69`:

`69^2 = (70-1)^2 = 70^2-2*70+1 = 4900-140+1 = 4761`

So putting `n=4783` and `a=69` we get:

`b = n-a^2 = 4783-4761 = 22`

So:

`sqrt(4783) = 69+22/(138+22/(138+22/(138+...)))`

We can truncate after any number of terms to get a rational approximation for `sqrt(4783)`:

`sqrt(4783) ~~ 69`

`sqrt(4783) ~~ 69+22/138 = 4772/69 ~~ 69.1594`

`sqrt(4783) ~~ 69+22/(138+22/138) = 69+22/(9533/69) = 659295/9533 ~~ 69.1592363`

A calculator tells me:

`sqrt(4873) ~~ 69.1592365487`