1
$\begingroup$

Define the Fibonacci numbers by $F_0=1$, $F_1=1$, and for each $n\ge2$, $F_n=F_{n-1}+F_{n-2}$. I need to prove that no Fibonacci number is twice that of another Fibonacci number (except for $F_2=2=2\times 1=2F_1$). I tried to use induction but I can't seem to figure it out. Thanks!

$\endgroup$
2
$\begingroup$

Sequence of Fibonacci numbers is, except for first two values, increasing. From it follows that $F_{n+1}=F_n+F_{n-1} > 2F_{n-1}$. So given value exceeds twice its value very quickly, only way to get $F_n=2F_m$ would be for two consecutive values. But that means $F_{n+1}=2F_n=F_n+F_{n-1}$ which implies $F_{n-1}=F_{n}$ which is true only for $n=1$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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