Question

What is the square root of 26?

Answer 1

You can only have an approximation: 5.09901951...

Explanation:

The square root of a number `x` is a number `y` such that `y^2=x`. So, we're looking for a number `y` such that `y^2=26`. Since `5^2=25` and `6^2=36`, the square root of `26` is between `5` and `6`. Since there is no algorithm to compute it exactly, you can only have an approximation. A possible way is the following: we know that `sqrt(26)` is between `5` and `6`. So, since `5^2=25` and `5.1^2=26.01`, `sqrt(26)` must be between `5` and `5.1`.

Iterating this process gives you all the decimal digits you need.

Answer 2

`sqrt(26)` does not simplify, but you can calculate an approximation efficiently using Newton Raphson method as:

`sqrt(26) ~~ 54100801 / 10610040 ~~ 5.099019513592786`

Explanation:

`26 = 2 * 13` has no square factors, so `sqrt(26)` cannot be simplified.

If you want to calculate an approximation by hand, then I would recommend a form of Newton Raphson method, starting with first approximation `a_0 = 5`.

To iterate you can use the formula:

`a_(i+1) = (a_i^2 + n)/(2a_i)`

where `n = 26` is the number you are approximating the square root of.

Personally, I like to deal with these approximations as rational approximations in the form `p_i/q_i = a_i` where `p_i` and `q_i` are integers as follows:

`n = 26`
`p_0 = 5`
`q_0 = 1`

Iterate using:

`p_(i+1) = p_i^2 + n q_i^2`
`q_(i+1) = 2 p_i q_i`

So:

`p_1 = 5^2 + 26*1^2 = 25+26 = 51`
`q_1 = 2*5*1 = 10`

`p_2 = 51^2 + 26*10^2 = 2601 + 2600 = 5201`
`q_2 = 2*51*10 = 1020`

`p_3 = 5201^2 + 26*1020^2 = 27050401 + 27050400 = 54100801`
`q_3 = 2*5201*1020 = 10610040`

Stop when you think you have enough significant digits (typically about the number of significant digits of `p_i` + the number of significant digits of `q_i`).

`sqrt(26) ~~ 54100801 / 10610040 ~~ 5.099019513592786`

Actually `sqrt(26) ~~ 5.099019513592785`