Use an algorithm to find:
##sqrt(33) = [5;bar(12110)] = 5+1/(1+1/(2+1/(1+1/(10+1/(1+1/(2+…))))))##
##~~5.744562646538##
##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(12110)]## The first economical approximation for ##sqrt(33)## is then: ##sqrt(33) ~~ [5;121] = 5+1/(1+1/(2+1/1)) = 23/4 = 5.75## The next is: ##sqrt(33) ~~ [5;12110121] = 1057/184 ~~ 5.7445652174## Actually ##sqrt(33)## is closer to: ##sqrt(33) ~~ 5.744562646538##