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?

  • $\begingroup$ Yes. The infinite continued fractions with $0$ integer part are precisely the irrational numbers in $(0,1)$. $\endgroup$ Commented Feb 11, 2013 at 3:16
  • $\begingroup$ how do you prove it? $\endgroup$
    – user61779
    Commented Feb 11, 2013 at 3:21

4 Answers 4


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

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    $\begingroup$ The link you posed does not exists anymore, please posed then content if possible. $\endgroup$
    – Peter Teoh
    Commented Jun 29, 2016 at 2:35
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    $\begingroup$ @Peter: Apparently they rearranged their site, but the page still exists, and I've updated the link. $\endgroup$ Commented Jun 29, 2016 at 2:50
  • $\begingroup$ Thank you. I have not read a proof that an infinite continued fraction with only positive integer convergents are irrational. One corollary, of course, is that an infinite continued fraction with opnly rational convergents would be irrational. The converse is easy to prove. A finite continued fraction with rational convergents would in effect be a finite sequence of arithmetic operations among the rationals, so the result must be rational. $\endgroup$ Commented Feb 15 at 0:30
  • $\begingroup$ If one or more convergents are allowed to be irrational, anything goes. Lambert proved that the tangent of $x$ radians has an infinite continued fraction representation. In particular, an infinitude of convergents are powers of $x$. Set $x=\pi/4$, and the infinite continued fraction is equal to 1. (This also establishes the irrationality of $\pi$.) $\endgroup$ Commented Feb 15 at 0:33

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 $a_3 = 1$.

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.

  • $\begingroup$ If the transcendental number is nonreal, then its continued fraction representation must have at least one nonreal convergent. All infinite continued fractions with real (let alone rational) convergents are real numbers. $\endgroup$ Commented Feb 15 at 0:36

Any arbitrary complex number (and I spect, any number in a normed division algebra) can be expressed as an infinite continued fraction.

Irrational numbers have a unique continued fraction expression consisting solely of integers. See here.


As such, real transcendental numbers (being irrational) all have a unique continued fraction expression with integer convergents.

An infinite continued fraction with only rational convergents must be irrational, per that same article. As such, nonzero rational numbers can only be represented by an infinite continued fraction if at least one convergent is irrational.

An infinite continued fraction expression with real convergents must be real. This means that for nonreal transcendental numbers, any infinite continued fraction expression must have at least one nonreal convergent.


The set of non-computable numbers is a strict subset of the transcendental numbers. If infinite continued fractions are computable, then I believe that they would not include all the transcendental numbers. Chaitin's constant is transcendental, but cannot be represented by an infinite continued fraction. Chaitin's constant cannot be represented through an algorithmic process.

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    $\begingroup$ Where is there any restriction that the continued fractions be computable? (Specifically, the sequence of $a_i$s is not required to be computable.) $\endgroup$ Commented Jan 3, 2019 at 17:53

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