# Thread: Strong induction on Fibonacci numbers

1. ## Strong induction on Fibonacci numbers

Hi,

today I'm really stuck with strong inductions...

Let $\displaystyle F_n$ be the $\displaystyle n^{th}$ Fibonacci number. Prove that for all m ≥ 1 and all n, $\displaystyle F_{m+n} = F_{m-1} F_n + F_m F_{n+1}$.

Can someone help me with it?

2. ## Response

Recently got a book by Posamentier on Fibonacci numbers that I believe discusses strong induction on them. I'll look it up when I get home and let you know what I find out.

Question for you. Have you tried Googling for an answer?

3. Originally Posted by wonderboy1953
Question for you. Have you tried Googling for an answer?
Yes I tried but this particular proof is not explained. I can only find the strong induction proof of the fibonacci numbers defined as $\displaystyle F_n = F_{n-1} + F_{n-2}$ with $\displaystyle F_0 = 0$ and $\displaystyle F_1 = 1$ with $\displaystyle n \ge 2$.

Thank you very much!

4. Originally Posted by TheFangel
Hi,

today I'm really stuck with strong inductions...

Let $\displaystyle F_n$ be the $\displaystyle n^{th}$ Fibonacci number. Prove that for all m ≥ 1 and all n, $\displaystyle F_{m+n} = F_{m-1} F_n + F_m F_{n+1}$.

Can someone help me with it?
Hi The Fangel,

P(k)

$\displaystyle F_{m+k}=F_{m-1}F_k+F_mF_{k+1}$

P(k+1)

$\displaystyle F_{m+k+1}=F_{m-1}F_{k+1}+F_mF_{k+2}$

Proof

$\displaystyle F_{k+1}=F_k+F_{k-1}\ \Rightarrow\$$\displaystyle F_k=F_{k+1}-F_{k-1}$

and

$\displaystyle F_{k+2}=F_{k+1}+F_k\ \Rightarrow\ F_{k+1}=F_{k+2}-F_k$

then it follows

$\displaystyle F_{m+k+1}=F_{m+k}+F_{m+k-1}$

$\displaystyle =F_{m-1}F_k+F_mF_{k+1}+F_{m+k-1}$

$\displaystyle =F_{m-1}F_{k+1}-F_{m-1}F_{k-1}+F_mF_{k+2}-F_mF_k+F_{m+k-1}$

$\displaystyle =F_{m-1}F_{k+1}+F_mF_{k+2}+\left(F_{m+k-1}-F_mF_k-F_{m-1}F_{k-1}\right)$

as the part in brackets is zero based on P(k), then

$\displaystyle F_{m+k+1}=F_{m-1}F_{k+1}+F_mF_{k+2}+0$

Looked it up in my book. Here's how it describes it (reworded of course):

First it describes it as a lemma, which needs to be proven (using the statements for n = k - 1 and for n = k to prove the statement for n = k + 1 so you'll be needing two base cases instead of just one meaning we have to check for n = 1 and n = 2).

When n = 1, we have to verify that . Now since , we must verify that which must be true since it's a defining relationship for the Fibonacci numbers.

When n = 2, then we need to verify that . Now since and , then we need to verify that which is true based on the following:

Next we assume the statement holds for n = k - 1 and n = k or

and which is the
induction hypothesis. From out of the next series of equations:

=
=
=
=
=

we have which is the statement for n = k + 1 which completes the proof.

6. Thank you very much to both of you!

In the end I chose the one of wonderboy because it shows in a clearer way the strong induction mechanism, even though both are very valid.

Thanks again for your time!