There is a lot of questions about sum and product of irrationals here, so I hope you'll bear with me.

Simple continued fraction is a very convenient way to represent any number since every real number has unique representation and for every rational number it's finite, while for every irrational it's infinite.

$$ x \in R, x \notin Q \Leftrightarrow x=a_0+\dfrac{1}{a_1+\dfrac{1}{a_2+1/...}}=[a_0;a_1,a_2,a_3,...,a_n,...] $$

So what I ask is this - can we define the sum and product of two irrational numbers as the sum of the Cauchy sequences of their simple continued fractions convergents? The sum and product of two Cauchy sequences are defined as sequences of sums and products of their respective terms.

Unlike general Cauchy sequences (of which there can be infinitely many for a given irrational number) the sequence of simple continued fraction convergents is supposed to be unique for every real number. There are recurrence relations defining every convergent of a simple continued fraction.

It may be possible then to gain a better understanding on the problem of irrationality of sums and products of irrationals like $e+\pi$ for example.

Maybe you can recommend some references on the topic? I'm sure it's been thoroughly investigated.


2 Answers 2


See: Rieger, “A new approach to the real numbers (motivated by continued fractions),” Abhandlungen der Braunschweigischen Wissenschaftlichen Gesellschaft, vol. 33, pp. 205–217, 1982.

The author gives a construction of the reals using continued fractions. A survey of this construction appears in this article.

I'm not sure how this helps though in studying the (ir)rationality of, e.g., $\pi + e$.

  • $\begingroup$ Simple continued fractions, unlike the decimal expansion, give a straightforward way to determine the rationality of a number. So if we find the continued fraction for $e+\pi$ we may be able to prove it's rational if it's CF terminates (but not the other way around, of course). $\endgroup$
    – Yuriy S
    Sep 17, 2015 at 10:40
  • $\begingroup$ One should expect it to be very hard to compute the continued fraction of such numbers. $\endgroup$ Sep 17, 2015 at 11:28

Note that you can take any sequences $a_n\to\alpha$ and $b_n\to\beta$ in order to obtain a sequence converging to $\alpha+\beta$ (or $\alpha\cdot\beta$); you don't need continued fractions in order to obtain a representant sequence for $\alpha+\beta$.

Now in your approach any real number $\alpha$, rational or irrational, has a certain unique "normal form": its continued fraction expansion. The theory of continued fractions then allows you to construct approximants $a_n={p_n\over q_n}\in{\mathbb Q}$ for $\alpha$, and similarly for $\beta$. This then allows to compute approximants ${r_n\over s_n}:=a_n+b_n$ for $\alpha+\beta$. Up to this point everything is fine. But the whole setup only would be useful if you had a simple algorithm that finds the definitive continued fraction expansion of $\alpha+\beta$, taking as input the sequence $\left({r_n\over s_n}\right)_{n\geq0}$. Think of $\alpha:=\pi+{3\over4}$, $\beta:=7-\pi$. At which point would this algorithm decide that $\alpha+\beta$ is in fact rational?


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