How do you find the square root of 33?

Use an algorithm to find:

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