Question

What is the square root of 82?

Answer 1

`10>sqrt82>9` , `sqrt82 ~~9.0554`

Explanation:

`x_"n+1"=1/2(x_"n"+S/x_"n") ->sqrtS` for `n -> oo`
S is the number of which you're aproxximating its sqaure root. In this case `S=82`

Heres what this means and how it is used:
First, take a guess, what might the square root of 82 be?
the square root of 81 is 9, so it must be sligthly higher than 9 right?

Our guess will be `x_"0"`, let's say 9.2, `x_"0" = 9.2`

Inserting 9.2 as "x" in the formula will give us `x_"0+1"=x_"1"`
This will be the next number we put into the equation. This is because we started with a guess of 9.2 = `x_"0"`, this gave us a number `x_"1"`, inserting this number will give us `x_"2"`, which will give us `x_"3"` and so on, always giving us the next number when we insert the previous. The right side of the equation indicated with "`->`" means that when "n" gets bigger and bigger, the number also gets closer and closer to the square root of S, in this case 82.

Let's say we did the same calculation 100 times! Then we would have `x_"100"` . This number would be very close to the square root of S.

Enough talking, let's do some actual calculations!

We start with our guess `x_"0"=9.2`
#x_"1" = 1/2(9.2+82/9.2)~~ 9.05652#

Now do the same with the new number: #x_"2"=1/2(9.05652+82/9.05652)~~ 9.05549#

Let's do it one last time: #x_"3"=1/2(9.05549+82/9.05546)~~ 9.0554#

That means `sqrt82~~9.0554`

And there you have it!

Sorry if all my talking was annoying. I tried to explain it in-depth and in a simple way, which is always nice if you're not very familiar with a certain field in mathematics. I don't see why some people has to be so posh when explaining mathematics :)

Answer 2

`sqrt(82) = 9+1/(18+1/(18+1/(18+1/(18+...)))) ~~ 9.0553851381374`

Explanation:

The prime factorisation of `82` is:

`82 = 2*41`

Since there are no square factors, `sqrt(82)` cannot be simplified. It is an irrational number a little larger than `9`.

However, note that `82=81+1 = 9^2+1`.

Since this is of the form `n^2+1`, the square root has a very regular form as a continued fraction:

`sqrt(82) = [9;bar(18)] = 9+1/(18+1/(18+1/(18+1/(18+...))))`

More generally:

`sqrt(n^2+1) = [n;bar(2n)] = n+1/(2n+1/(2n+1/(2n+1/(2n+...))))`

More generally still:

`sqrt(n^2+m) = n+m/(2n+m/(2n+m/(2n+m/(2n+...))))`

In any case, we can use the continued fraction to get rational approximations to `sqrt(82)` by truncating.

For example:

`sqrt(82) ~~ [9;18] = 9+1/18 = 163/18 = 9.0bar(5)`

`sqrt(82) ~~ [9;18,18] = 9+1/(18+1/18) = 2943/325 = 9.05bar(538461)`

`sqrt(82) ~~ [9;18,18,18] = 9+1/(18+1/(18+1/18)) = 53137/5868 ~~ 9.05538513974`

A calculator tells me that:

`sqrt(82) ~~ 9.0553851381374`

So you can see that our approximations are accurate to just about as many significant digits as the total number of digits in the quotient.