Can every transcendental number be expressed as an infinite continued fraction?

Question

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?

Answer 1

The infinite continued fractions are precisely the irrational numbers; you will find a proof here, along with a proof that the expansion is unique.

Answer 2

To express an irrational number $\alpha$ as a continued fraction, do this:

1. Put $\alpha_0 = \alpha$ and $n=0$.
2. Let $a_n = \lfloor\alpha_n\rfloor$, the greatest integer not exceeding $\alpha_n$.
3. Put $\alpha_{n+1} = {(\alpha_n - a_n)}^{-1}$.
4. Add 1 to $n$.
5. Go back to step 2.

The continued fraction expansion of $\alpha$ is then:

$$\alpha = a_0 + \cfrac{1}{ a_1 + \cfrac{1}{ a_2 + \cfrac{1}{ a_3 + \cdots}}}$$

For example, let's take $\alpha = \pi = 3.141529\ldots$. Then

1. $\alpha_0 = 3.141529\ldots$ and $a_0 = 3$.
2. $\alpha_1 = {(0.14159\ldots)}^{-1} \approx 7.06251$ and $a_1 = 7$.
3. $\alpha_2 = {(0.06251\ldots)}^{-1} \approx 15.9966$ and $a_2 = 15$.
4. $\alpha_3 = {(0.09966\ldots)}^{-1} \approx 1.00341$ and $a_3 = 1$.
5. …

And so we have

$$\pi = 3 + \cfrac{1}{ 7 + \cfrac{1}{ 15 + \cfrac{1}{ 1 + \cdots}}}$$

which is correct.

Note that since $a_n = \lfloor\alpha_n\rfloor$, we know that $\alpha_n-1 < a_n \le \alpha_n$, and therefore that $\alpha_n - a_n$ must be at least 0, but less than 1. If $\alpha_n - a_n$ is nonzero, its reciprocal, which is $\alpha_{n+1}$, must be strictly greater than 1. Then $a_{n+1}$ must be a positive integer, as required.

If you take $\alpha$ to be a rational number, the process will terminate when one of the $\alpha_n$ is zero. If $\alpha$ is an irrational number, the process can't terminate in this way, since if it had, you would be able to write $\alpha$ as a finite continued fraction, which you would be able to put into standard $\frac ab$ form, which is impossible. That is a (somewhat handwavy) proof that you were looking for.