How can you simplify #sqrt(6)# ?

Question

HIX Tutor · Accepted Answer

undefined To simplify √6, first, we check if there are any perfect squares that can be factored out of 6. Since 6 = 2 * 3, there are no perfect squares that can be factored out. Therefore, √6 is already simplified.

Abigail Allen · Answer

Here are three methods...

Note that #6 = 2*3# has no square factors, so #sqrt(6)# cannot be simplified. It is an irrational number somewhere between #sqrt(4) = 2# and #sqrt(9) = 3# We can find rational approximations using any one of a number of methods (there are at least 25 different methods). I will present a few. In each of these examples I will use #sqrt(6) ~~ 5/2# as an initial approximation, since that lies between #2=sqrt(4)# and #3=sqrt(9)#. Babylonian method Given a rational approximation #p_i/q_i# to the square root of #n#, a better approximation is given by #p_(i+1)/q_(i+1)# where: #{ (p_(i+1) = p_i^2+n q_i^2), (q_(i+1) = 2 p_i q_i) :}# Using #n=6#, #p_0 = 5# and #q_0 = 2# we find: #{ (p_1 = p_0^2+n q_0^2 = color(blue)(5)^2+color(blue)(6)(color(blue)(2)^2) = 25+24 = 49), (q_1 = 2 p_0 q_0 = 2(color(blue)(5))(color(blue)(2)) = 20) :}# #{ (p_2 = p_1^2+n q_1^2 = color(blue)(49)^2 + color(blue)(6)(color(blue)(20)^2) = 2401+2400 = 4801), (q_2 = 2 p_1 q_1 = 2(color(blue)(49))(color(blue)(20)) = 1960) :}# Given that we started with an efficient approximation (which we did), the number of correct significant digits of each successive approximation will be approximately the total number of digits in the numerator and denominator. So: #sqrt(6) ~~ 4801/1960 ~~ 2.4494898# A calculator tells me: #sqrt(6) ~~ 2.44948974278317809819# So we were almost good to #8# significant digits as expected. Generalised continued fraction #sqrt(a^2+b) = a+b/(2a+b/(2a+b/(2a+b/(2a+...))))# So putting #a=5/2# and #b=6-(5/2)^2 = 24/4-25/4 = -1/4# we find: #sqrt(6) = 5/2-(1/4)/(5-(1/4)/(5-(1/4)/(5-(1/4)/(5-...))))# We can truncate this early to find: #sqrt(6) ~~ 5/2-(1/4)/(5-(1/4)/5)# #color(white)(sqrt(6)) = 5/2-(1/4)/(99/20)# #color(white)(sqrt(6)) = 5/2-5/99# #color(white)(sqrt(6)) = 5/2-5/99# #color(white)(sqrt(6)) ~~ 2.44949# For more accuracy, use more terms. Limiting ratio of integer sequence Given that: #2sqrt(6) ~~ 5# Consider the quadratic with zeros #5+2sqrt(6)# and #5-2sqrt(6)#... #(x-5-2sqrt(6))(x-5+2sqrt(6)) = (x-5)^2-4*6# #color(white)((x-5-2sqrt(6))(x-5+2sqrt(6))) = x^2-10x+1# So the zeros satisfy: #x^2=10x-1# Define a sequence recursively based on this quadratic: #a_0 = 0# #a_1 = 1# #a_(n+2) = 10a_(n+1)-a_n# Then the ratio between successive terms will tend to #5+2sqrt(6)# The first few terms are: #0, 1, 10, 99, 980, 9701, 96030, 950599# Hence: #sqrt(6) ~~ 1/2(950599/96030-5) = 1/2(470449/96030) = 470449/192060 ~~ 2.44948974279#