0

If $\frac{p_r}{q_r}$ be the $r^{\text{th}}$ convergent of the continued fraction of $\frac{\sqrt{5}+1}{2}$ then prove that $p_{n+1}=p_{n}+p_{n-1}$ and $p_{2n}=p_{2n-1}+p_{2n-2}$.

Attempt:

I have written the continued fraction of $\frac{\sqrt{5}+1}{2}$ which is equal to $1+\frac{1}{1+}\frac{1}{1+}\frac{1}{1+}\cdots$. But how to get the above two relations. Please help.

user1942348
  • 3,871
  • The second thing you ask seems to be a case of the first where one substitutes $2n-1$ for the $n$ in the first one. Did you intend that? – coffeemath Feb 29 '16 at 17:49
  • @coffeemath Yes It is correct. But how to prove the first? – user1942348 Feb 29 '16 at 17:50
  • Maybe look at how each convergent is related to the previous one in a general continued fraction (check wiki or a number theory book on that). Things should be simple in this case because of all the "denominators" being 1. – coffeemath Feb 29 '16 at 17:53
  • @coffeemath I am very happy if you help me to give a step for solving. – user1942348 Feb 29 '16 at 18:11
  • 1
    It's impossible to help you learn without more context. Have you worked with continued fractions before this example? Do you know the general method for computing numerators of convergents? Or is this the first time you're seeing it and it's intended for you to work it out by hand (which is quite possible)? – Greg Martin Feb 29 '16 at 18:14
  • user1942348 As Greg Martin points out in his comment, you should say in your question what course this question is for. If it is a number theory course, mention the textbook, and at least whether you read anything about continued fractions. As it is, your question will likely get closed... – coffeemath Feb 29 '16 at 18:37

1 Answers1

0

It's quite standard

$\displaystyle 1=\frac{1}{1}=\frac{p_{0}}{q_{0}}$

$\displaystyle 1+\frac{1}{1}=\frac{2}{1}=\frac{p_{1}}{q_{1}}$

$\displaystyle 1+\frac{1}{\frac{p_{n}}{q_{n}}}=\frac{p_{n}+q_{n}}{p_{n}}=\frac{p_{n+1}}{q_{n+1}}$

Note that if $\gcd(p_{n},q_{n}) = 1 \implies \gcd(p_{n},p_{n}+q_{n}) = 1$

Equating numerators and denominators then eliminate $q_{n}$. The result follows at once.

Put $m=2n-1$, then $p_{m+1}=p_{m}+p_{m-1} \implies p_{2n}=p_{2n-1}+p_{2n-2}$

Ng Chung Tak
  • 18,990