Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
1  
See my answer here which deals with the very same problem. –  Pedro Tamaroff Feb 27 '13 at 15:50

5 Answers 5

up vote 5 down vote accepted

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: $[1,1,1\ldots] := \lim_{n\to\infty} a_n$).

share|improve this answer
    
Thank you Stahl - appreciate the clear Real Analysis explanation. –  Conan Wong Feb 27 '13 at 7:25

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 $\lim_{n\to\infty}(f_{n+1}/f_n)=(1+\sqrt5)/2$.

share|improve this answer
    
Thank you Gerry - your explanation ties in nicely with the theory on continued fractions which I'm currently studying. –  Conan Wong Feb 27 '13 at 7:26

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 $$ \alpha = 1 + \frac 1{1 + \frac 1{1+ \frac 1{1+\frac 1{1+ \frac 1{1 + \ldots}}}}} $$ which is your fraction again.

share|improve this answer
1  
I think the key here is, that it needs justification for the possibility of the replacement of the infinite sub-continued fraction as pointed out by @Stahl –  Gottfried Helms Feb 26 '13 at 23:37
3  
I think this is the argument the OP already has but is worried about. But thanks for the lovely typesetting nonetheless. :) –  Fixee Feb 26 '13 at 23:38
    
Thanks Kaster. Gottfried and Fixee are correct. Your argument is how I got my equation $\alpha=1+\frac{1}{\alpha}$ in the first place, but I'm concerned about the justification for doing this. Nevertheless, as Fixee points out, beautiful continued fraction you have here :-) –  Conan Wong Feb 26 '13 at 23:42
    
@ConanWong I confused the meanings of formal and informal. lol –  Kaster Feb 27 '13 at 0:05

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 $[1,1,1,\cdots]$.

share|improve this answer
    
Thank you Fixee - this is a nice approach, very clear. –  Conan Wong Feb 27 '13 at 7:28
    
You're welcome. This is the cutest (and most elementary) solution I could think of. Good luck. –  Fixee Mar 1 '13 at 6:22

Since no one alluded to this point, I think it is appropriate to mention it.
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.
P.S. This approach is to go backwards: once you know what the solution should look like, you check that it is, and this is quite familiar, right?
P.S.D. If one thinks of this answer as quite frivolous, it is understandable: just tell me. Thanks.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.