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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?

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Yes. The infinite continued fractions with $0$ integer part are precisely the irrational numbers in $(0,1)$. –  Brian M. Scott Feb 11 '13 at 3:16
    
how do you prove it? –  user61779 Feb 11 '13 at 3:21
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2 Answers

up vote 3 down vote accepted

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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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.

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