How do you find the square root of 33?

1 Answer
Aug 30, 2016

Answer:

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#

Explanation:

#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#