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I was reading Penrose’s The road to reality. In chapter 3 (section 2) he introduces us to irrational numbers and how they can be expressed as an infinite continued fraction. I noticed that in the examples he offers, the first two (those being $\sqrt{2}$ and $7 - \sqrt{3})$ have periodic infinite continued fractions, these being: $$\sqrt{2} = 1 + \frac{1}{2 + \frac{1}{2 + \frac{1}{2 + ...}}}$$ $$7 - \sqrt{3} = 5 + \frac{1}{3 + \frac{1}{1 + \frac{1}{2 + \frac{1}{1 + \frac{1}{2 + …}}}}}$$ These numbers are also algebraic. However with the case of transcendental numbers (such as $\pi$ or $e$) their continued fractions are non-periodic. Is this the case with any transcendental number? My intuition says it must be so, otherwise you can construct an equation (with integer coefficients) that satisfies the infinite continued fraction (like the case of the golden ratio).

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    $\begingroup$ In the title you asked if nonperiodic continued fraction implies transcendental, whereas in the body you asked if transcendental implies nonperiodic continued fraction !? $\endgroup$ – J. W. Tanner Jul 5 at 21:32
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    $\begingroup$ I think the answer was (correct) to the question in the original title, though the comment to the answer mitigates that $\endgroup$ – J. W. Tanner Jul 5 at 21:38
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No. If a real has a periodic continued fraction, then it satisfies a quadratic equation over $\Bbb Q$. Therefore $\sqrt[3]2$ has a non-periodic continued fraction.

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    $\begingroup$ And conversely, a positive real number has a periodic continued fraction if and only if it's a quadratic irrational. (Where "periodic" means "eventually periodic" and "continued fraction" means "simple continued fraction".) $\endgroup$ – bof Jul 5 at 20:00

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