How do you find the square root of 7?

Since `7` is a prime number, it has no square factors and its square root cannot be simplified.

It is an irrational number, so cannot be exactly represented by `p/q` for any integers `p, q`.

We can however find good rational approximations to `sqrt(7)`.

First note that:

`8^2 = 64 = 63+1 = 7*3^2 + 1`

This is in Pell's equation form:

`p^2 = n q^2 + 1`

with `n = 7`, `p = 8` and `q = 3`.

This means that `8/3` is an economical approximation for `sqrt(7)` and it also means that we can use `8/3` to derive the continued fraction expansion of `sqrt(7)`:

`8/3 = 2 + 1/(1+1/(1+1/1))`

and hence we can deduce:

`sqrt(7) = [2;bar(1,1,1,4)] = 2 + 1/(1+1/(1+1/(1+1/(4+1/(1+1/(1+1/(1+1/(4+...))))))))`

The next economical approximation is given by truncating the continued fraction expansion just before the next `4`, i.e.

`sqrt(7) ~~ [2;1,1,1,4,1,1,1] = 2 + 1/(1+1/(1+1/(1+1/(4+1/(1+1/(1+1/1)))))) = 127/48 = 2.6458bar(3)`

This is also a solution of Pell's equation for `7`, since we find:

`127^2 = 16129 = 16128+1 = 7*48^2+1`

If you want more accuracy, truncate just before the next `4` or the one after.

By expanding the repeating part of the continued fraction for `sqrt(7)` we can derive a generalised continued fraction:

`sqrt(7) = 21/8+(7/64)/(21/4+(7/64)/(21/4+(7/64)/(21/4+(7/64)/(21/4+...))))`

Using a calculator, we find:

`sqrt(7) ~~ 2.645751311`