Is a number transcendental if its continued fraction is non-periodic?

Question

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).

Answer 1

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.