# Strong induction on Fibonacci numbers

• May 26th 2010, 03:06 PM
TheFangel
Strong induction on Fibonacci numbers
Hi,

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

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

Can someone help me with it?
• May 26th 2010, 03:24 PM
wonderboy1953
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?
• May 26th 2010, 03:51 PM
TheFangel
Quote:

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 $F_n = F_{n-1} + F_{n-2}$ with $F_0 = 0$ and $F_1 = 1$ with $n \ge 2$.

Thank you very much!
• May 27th 2010, 01:55 AM
Quote:

Originally Posted by TheFangel
Hi,

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

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

Can someone help me with it?

Hi The Fangel,

P(k)

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

P(k+1)

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

Proof

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

and

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

then it follows

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

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

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

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

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

$F_{m+k+1}=F_{m-1}F_{k+1}+F_mF_{k+2}+0$
• May 27th 2010, 06:52 AM
wonderboy1953
Looked it up in my book. Here's how it describes it (reworded of course):

First it describes it as a lemma, http://www.mathhelpforum.com/math-he...ad948943-1.gif 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 http://www.mathhelpforum.com/math-he...14270780-1.gif. Now since http://www.mathhelpforum.com/math-he...fc6783d1-1.gif, we must verify that http://www.mathhelpforum.com/math-he...36d1ee07-1.gif which must be true since it's a defining relationship for the Fibonacci numbers.

When n = 2, then we need to verify that http://www.mathhelpforum.com/math-he...5f4a6d90-1.gif. Now since http://www.mathhelpforum.com/math-he...fc6783d1-1.gif and http://www.mathhelpforum.com/math-he...0a7d8b13-1.gif, then we need to verify that http://www.mathhelpforum.com/math-he...b819ac72-1.gif which is true based on the following:

http://www.mathhelpforum.com/math-he...cfce1549-1.gif

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

http://www.mathhelpforum.com/math-he...6cbab4e8-1.gif and http://www.mathhelpforum.com/math-he...652652a9-1.gif which is the
induction hypothesis. From out of the next series of equations:

http://www.mathhelpforum.com/math-he...f1eb9ce3-1.gif
= http://www.mathhelpforum.com/math-he...84ff0bdf-1.gif
= http://www.mathhelpforum.com/math-he...648ef0de-1.gif
= http://www.mathhelpforum.com/math-he...af562370-1.gif
= http://www.mathhelpforum.com/math-he...0a4c5864-1.gif
= http://www.mathhelpforum.com/math-he...cf232561-1.gif

we have http://www.mathhelpforum.com/math-he...b8f2dba8-1.gif which is the statement for n = k + 1 which completes the proof.
• May 27th 2010, 11:00 AM
TheFangel
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.