Every infinite continued fraction is irrational. But can every number, in particular those that are not the root of a polynomial with rational coefficients, be expressed as a continued fraction?
[Math] Can every transcendental number be expressed as an infinite continued fraction
continued-fractionstranscendental-numbers
Related Solutions
I'm not entirely sure what you mean by 'as the base for a CRF representation'; continued fraction representations, at heart, don't use bases at all - just integers.
Assuming that what you mean is allowing the coefficients of a standard continued fraction representation to be elements of $\mathbb{Z}[\phi]$ rather than just $\mathbb{Z}$, then you run into trouble of a different source: in that case, there's no canonical CF representatation! The usual algorithm for generating a canonical continued fraction uses the floor operation $\lfloor x\rfloor$ (that is, 'the largest integer less than $x$') : $a_0 = \lfloor x\rfloor, a_1=\lfloor\frac{1}{x-a_0}\rfloor$, etc. But this operation can't be defined in $\mathbb{Z}[\phi]$ because the values in that ring are dense in the reals; there's no 'largest member less than $x$' for any $x$.
On the other hand, if you're interested in generalizing some of the properties of continued fractions to the ring of 'golden integers', there's another approach you could investigate:
As I'm sure you're already aware, the sequence of partial convergents to a continued fraction for some irrational number $x$ offers the best approximations to $x$, in a very canonical sense: each term $\frac{p}{q}$ in the sequence is the best approximation to $x$ of height less than $q$, where we define the height of a fraction $\frac{c}{d}$ in simplest terms to be just the denominator of the fraction. In essence, the height represents the simplicity of the number, and the best rational approximation property says that no 'simpler' number than any of the partial convergents of the CF can be a better approximation.
You could try and generalize this property by defining a canonical height for golden integers (for instance, $\mathrm{ht}(a+b\phi) = \max(|a|, |b|)$ or even just $\mathrm{ht}(a+b\phi) = |b|$) and then studying the sequence of 'convergents' to a number $x$ using this height and the best-approximation property; this is in some sense similar to the continued fraction approach in that both are looking at the best approximation to $x$ of the form $f(a,b)$ for particular functions $f$ and integers $a,b$ less than some specified bound (and studying how that approximation changes as the bound moves) — in the case of continued fractions then $f(a,b) = \frac{a}{b}$ while here it's $f(a,b) = a+b\phi$. It's very possible that the similarity of the problem means that these convergents have some canonical structure (similar to the way that the continued-fraction structure unites the best rational approximations) which you could use to define a continued-fraction equivalent for the representation by golden integers. Speaking personally, I'd certainly be curious to see whether anything interesting came out of it!
The question is still being researched, some results are known, some are conjectured.
For examples if continued fraction is periodic then it represents a quadratic irrational number
Theorem 8.11 (Lagrange) Every periodic regular continued fraction is a quadratic irrationality (i.e., $\frac{a+b\sqrt{c}}{d}$ for some integers $a,b,c,$ and $d$, where $b \ne0, c>1, d>0$ anf $c$ is square-free). The converse is also true: every quadratic irrationality has a periodic regular continued fraction.
from this book, page 94. Same result in Khinchin's famous book, page 48. To some extent - spread here.
It is conjectured that all infinite continued fractions with bounded terms that are not eventually periodic are transcendental (eventually periodic continued fractions correspond to quadratic irrationals).
and here is a good paper highlighting some of the latest developments.
And of course Liouville's theorem, quoting Khinchin's book, page 46
Liouville's theorem shows that algebraic numbers do not admit rational-fraction approximations of greater than a certain order of accuracy.
this includes the best rational approximations (generated by simple continued fractions) as well.
Best Answer
The infinite continued fractions are precisely the irrational numbers; you will find a proof here, along with a proof that the expansion is unique.