- #1
KevB
- 11
- 0
It's my understanding that algebraic numbers are the roots of polynomials with rational (or equivalently integer) coefficients. I know all surds have a simple repeating continued fraction representation
Is it also the case that all simple repeating continued fractions are algebraic numbers?
e.g. [tex]\sqrt{}2= [ 1; 2, 2, 2, 2, ...] = [1,2],[0][/tex]
[tex]\sqrt{}3 = [ 1; 1, 2, 1, 2,...] = [1, 1, 2],[0,0][/tex]
[tex] \ \ \varphi = [1; 1, 1, 1, 1, ... ] = [1],[0] = golden \ ratio
[/tex]
While many transcendental numbers, like e, have interesting continued fractions, but the pattern isn't a simple repeat.
e.g.
[tex]e \ \ = [2; 1, 2, 1, 1, 4, 1, 1, 6, 1, ...] = [2; 1, 2, 1],[0, 2, 0] = [ 1, 0, 1],[0, 2, 0] [/tex]
[tex]e^{(1/n)}_{}= [1, n-1, 1, 1, 3n-1, 1, 1, 5n-1, ...] = [1, (n-1), 1],[0, 2n, 0][/tex]
Is it also the case that all simple repeating continued fractions are algebraic numbers?
e.g. [tex]\sqrt{}2= [ 1; 2, 2, 2, 2, ...] = [1,2],[0][/tex]
[tex]\sqrt{}3 = [ 1; 1, 2, 1, 2,...] = [1, 1, 2],[0,0][/tex]
[tex] \ \ \varphi = [1; 1, 1, 1, 1, ... ] = [1],[0] = golden \ ratio
[/tex]
While many transcendental numbers, like e, have interesting continued fractions, but the pattern isn't a simple repeat.
e.g.
[tex]e \ \ = [2; 1, 2, 1, 1, 4, 1, 1, 6, 1, ...] = [2; 1, 2, 1],[0, 2, 0] = [ 1, 0, 1],[0, 2, 0] [/tex]
[tex]e^{(1/n)}_{}= [1, n-1, 1, 1, 3n-1, 1, 1, 5n-1, ...] = [1, (n-1), 1],[0, 2n, 0][/tex]