How do you find the square root of 33?

`33=3*11` has no square factors, so `sqrt(33)` cannot be simplified.

It is an irrational number a little less than `6`, since `6^2 = 36`.

To find a rational approximation, I will find a continued fraction expansion for `sqrt(33)` then truncate it.

`color(white)()`
To find the simple continued fraction expansion of `sqrt(n)`, use the following algorithm:

`m_0 = 0`
`d_0 = 1`
`a_0 = floor(sqrt(n))`

`m_(i+1) = d_i a_i - m_i`

`d_(i+1) = (n - m_(i+1)^2)/d_i`

`a_(i+1) = floor((a_0 + m_(i+1)) / d_(i+1))`

Stop when `a_i = 2a_0`, marking the end of the repeating part of the continued fraction.

The continued fraction expansion is then:

`[a_0; a_1, a_2, a_3,...]= a_0 + 1/(a_1 + 1/(a_2 + 1/(a_3 + ...)))`

`color(white)()`
In our example, `n = 33` and `floor(sqrt(n)) = 5`, since `5^2 = 25 < 33 < 36 = 6^2`.

So:

`{ (m_0 = 0), (d_0 = 1), (a_0 = floor(sqrt(33)) = color(blue)(5)) :}`

`{ (m_1 = d_0 a_0 - m_0 = 5), (d_1 = (n - m_1^2)/d_0 = (33-5^2)/1 = 8), (a_1 = floor((a_0 + m_1)/d_1) = floor((5+5)/8) = color(blue)(1)) :}`

`{ (m_2 = d_1 a_1 - m_1 = 8 - 5 = 3), (d_2 = (n - m_2^2)/d_1 = (33-9)/8 = 3), (a_2 = floor((a_0 + m_2)/d_2) = floor((5+3)/3) = color(blue)(2)) :}`

`{ (m_3 = d_2 a_2 - m_2 = 6 - 3 = 3), (d_3 = (n - m_3^2)/d_2 = (33-9)/3 = 8), (a_3 = floor((a_0 + m_3)/d_3) = floor((5+3)/8) = color(blue)(1)) :}`

`{ (m_4 = d_3 a_3 - m_3 = 8-3=5), (d_4 = (n - m_4^2)/d_3 = (33-25)/8 = 1), (a_4 = floor((a_0 + m_4)/d_4) = floor((5+5)/1) = color(blue)(10)) :}`

Having reached a value `color(blue)(10)` which is twice the first value `color(blue)(5)`, this is the end of the repeating pattern of the continued fraction, and we have:

`sqrt(33) = [5;bar(1,2,1,10)]`

The first economical approximation for `sqrt(33)` is then:

`sqrt(33) ~~ [5;1,2,1] = 5+1/(1+1/(2+1/1)) = 23/4 = 5.75`

The next is:

`sqrt(33) ~~ [5;1,2,1,10,1,2,1] = 1057/184 ~~ 5.7445652174`

Actually `sqrt(33)` is closer to:

`sqrt(33) ~~ 5.744562646538`