Algebraic and Transcendental irrational numbers

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SUMMARY

Algebraic numbers are defined as the roots of polynomials with rational coefficients, and all simple repeating continued fractions correspond to algebraic numbers. Examples include the golden ratio (φ) and square roots of integers, which exhibit periodic continued fraction expansions. In contrast, transcendental numbers like e have non-repeating continued fractions. Euler established that all simple repeating continued fractions represent irrational solutions to quadratic equations with rational coefficients.

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KevB
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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. \sqrt{}2= [ 1; 2, 2, 2, 2, ...] = [1,2],[0]

\sqrt{}3 = [ 1; 1, 2, 1, 2,...] = [1, 1, 2],[0,0]

\ \ \varphi = [1; 1, 1, 1, 1, ... ] = [1],[0] = golden \ ratio<br />

While many transcendental numbers, like e, have interesting continued fractions, but the pattern isn't a simple repeat.

e.g.
e \ \ = [2; 1, 2, 1, 1, 4, 1, 1, 6, 1, ...] = [2; 1, 2, 1],[0, 2, 0] = [ 1, 0, 1],[0, 2, 0]
e^{(1/n)}_{}= [1, n-1, 1, 1, 3n-1, 1, 1, 5n-1, ...] = [1, (n-1), 1],[0, 2n, 0]
 
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Hi KevB! :smile:

Check out wikipedia: http://en.wikipedia.org/wiki/Continued_fraction

The numbers with periodic continued fraction expansion are precisely the irrational solutions of quadratic equations with rational coefficients (rational solutions have finite continued fraction expansions as previously stated). The simplest examples are the golden ratio φ = [1; 1, 1, 1, 1, 1, ...] and √ 2 = [1; 2, 2, 2, 2, ...]; while √14 = [3;1,2,1,6,1,2,1,6...] and √42 = [6;2,12,2,12,2,12...]. All irrational square roots of integers have a special form for the period; a symmetrical string, like the empty string (for √ 2) or 1,2,1 (for √14), followed by the double of the leading integer.

So the simple repeating continued fractions that you're talking about are exactly the irrational solutions to quadratic equations. So they are indeed all algebraic. Euler was the first one to prove this.

Also, this would imply that the continued fraction of \sqrt[3]{2} is not a nice repeating continued fraction...
 
micromass said:
Also, this would imply that the continued fraction of \sqrt[3]{2} is not a nice repeating continued fraction...

Nonetheless, it does have a rapidly converging http://en.wikipedia.org/wiki/Generalized_continued_fraction" , namely

\frac{5}{4} + \frac{2.5}{252 - \frac{8}{759 - \frac{35}{\ldots}}}

Also you might be interested in http://en.wikipedia.org/wiki/Gauss%27s_continued_fraction" , which is a very wide class.
 
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