# Can an infinite continued fraction over the irrationals converge to an integer?

Let $$\eta_i$$,$$\zeta_i$$ be irrational real numbers, not necessarily distinct. $$\eta_0$$ may also be zero. I have two questions. My first question:

Can we express $$1$$ as $$\eta_0+\dfrac1{\eta_1+\frac1{\eta_2+\frac1\ddots}}$$? If so, can we also determine $$\eta_j$$?

And my second question:

Can we express $$1$$ as $$\eta_0+\dfrac{\zeta_0}{\eta_1+\frac{\zeta_1}{\eta_2+\frac{\zeta_2}\ddots}}$$? If so, can we also determine $$\eta_j,\zeta_j$$?

I am not sure how to answer this because I do not know exactly what qualifies as 'expressing $$1$$ as...' Surely, as $$\zeta_i \to0$$ and $$\eta_0\to1$$, the second c.f. converges to $$1$$. But this is not exactly 'representing' $$1$$ as a c.f. in the same sense as actually finding what distinct values of $$\eta_i,\zeta_i$$, when evaluated in the c.f., are equal to $$1$$. This ambiguity makes my second question a bit weak. My first question, however, does not have this ambiguity, as it is a simple continued fraction. The problem then is that I have made no progress on it.

• @saulspatz Those simple continued fractions have integer elements, so that doesn't apply to this question. – Robert Israel Dec 8 '19 at 5:05
• @RobertIsrael I misread the question. Thanks. – saulspatz Dec 8 '19 at 12:27

If $$r = a_0 + \dfrac{1}{a_1 + \dfrac{1}{a_2 + \dfrac{1}{a_3 + \ldots}}}$$ is a simple continued fraction, where $$r$$ is a positive irrational number and $$a_i$$ are positive integers, then
$$1 = a_0/r + \dfrac{1}{a_1 r + \dfrac{1}{a_2/r + \dfrac{1}{a_3 r + \ldots}}}$$
expresses $$1$$ as a continued fraction with irrational elements $$\eta_i = \cases{a_i/r & if i is even\\ a_i r & if i is odd}$$