# Continued Fraction

If the continued fractional representation of an irrational number $\alpha$ is given by [1,1,1,…], I can compute that $\alpha = \frac{1+\sqrt{5}}{2}$ by solving the equation $\alpha = 1+ \frac{1}{\alpha}$ (and noting that $\alpha$ is positive).

But this seems a bit informal to me.

Is there a more formal way to show that [1,1,1,…] = $\frac{1+\sqrt{5}}{2}$?

Thanks.

#### Solutions Collecting From Web of "Continued Fraction"

The only other thing you really need to show if you want to be precise is that the sequence of partial fractions given by $a_1 = [1]$, $a_2 = [1,1]$, $a_3 = [1,1,1]$, etc. does tend to a limit (it suffices to show that the sequence $\{a_n\}_{n\in\mathbb{N}}$ is bounded above by something and increasing eventually). Then your calculation shows that $\alpha$ is the unique positive solution, and hence must be equal to the infinite continued fraction (which is formally the limit of the partial fractions you get when you stop after $n$ 1’s: $[1,1,1\ldots] := \lim_{n\to\infty} a_n$).

You could prove, by induction, that $[1,1,\dots,1]=f_{n+1}/f_n$, where $f_n$ is the $n$th Fibonacci number, and then prove (using, say, the Binet formula for $f_n$) that $\lim_{n\to\infty}(f_{n+1}/f_n)=(1+\sqrt5)/2$.

Actually, it’s quite visual representation. Let’s take your fraction and write it as it should be written
$$\alpha = 1 + \frac 1{1 + \frac 1{1+ \frac 1{1+\frac 1{1+ \frac 1{1 + \ldots}}}}}$$
and compare it to the equation
$$\alpha = 1+\frac 1{\alpha}$$
and substitute $\alpha$ which is in denominator with itself, and you’ll get
$$\alpha = 1 + \frac 1{1 + \frac 1{\alpha}}$$
if you continue that substitution, you’ll get
$$\alpha = 1 + \frac 1{1 + \frac 1{1+ \frac 1{1+\frac 1{1+ \frac 1{1 + \ldots}}}}}$$
Solve the quadratic $x^2-x-1$ in two ways. Using the quadratic formula gives you $(1+\sqrt{5})/2$ (and the other root of course), and using continued fractions gives you $[1,1,1,\cdots]$.
You want to show that the continued fraction expansion of $\alpha=\frac{1+\sqrt5}{2}$ is as stated. So just use the standard process:
The largest integer less than $\alpha$ is easily seen to be $1$. So one sees that the first number is $1$, and obtain: $\alpha=1+\frac{1}{\frac{\sqrt5+1}{2}}$. Now, by induction, one finds that every number in the expansion is $1$. As to the convergence of this expansion, it is just in the theory of such fractions.