Is #sqrt3# rational or irrational?
2 Answers
Irrational
Explanation:
Suppose
#x = 1+1/(1+1/(1+x))#
#color(white)(x) = 1+(1+x)/(2+x)#
#color(white)(x) = (3+2x)/(2+x)#
Multiplying both ends by
#x^2+2x = 3+2x#
Subtracting
#x^2 = 3#
and hence:
#x = sqrt(3)#
So we have found:
#sqrt(3) = 1+1/(1+1/(1+sqrt(3)))#
#color(white)(sqrt(3)) = 1+1/(1+1/(2+1/(1+1/(2+1/(1+1/(2+...)))))#
Since this continued fraction does not terminate,
Bonus
Here's a fun method to calculate rational approximations to
First note that:
#7^2 = 49 = 48+1 = 3(4^2)+1#
Hence an efficient approximation to
Consider the quadratic with zeros
#(x-7-4sqrt(3))(x-7+4sqrt(3)) = (x-7)^2-48#
#color(white)((x-7-4sqrt(3))(x-7+4sqrt(3))) = x^2-14x+49-48#
#color(white)((x-7-4sqrt(3))(x-7+4sqrt(3))) = x^2-14x+1#
Now if
Define a related sequence of integers recursively by:
#{ (a_0 = 0), (a_1 = 1), (a_(n+2) = 14a_(n+1)-a_n) :}#
Then the ratio between successive terms will tend to
The first few terms are:
#0, 1, 14, 195, 2716, 37829, 526890#
Let's stop there to find:
#7+4sqrt(3) ~~ 526890/37829#
So:
#sqrt(3) ~~ 1/4(526890/37829-7) = 1/4(262087/37829) = 262087/151316 ~~ 1.7320508076#
Irrational
Explanation:
#6/1# #18/3# #30/5#
We cannot express
NOTE: