Thread: Strong induction on Fibonacci numbers

Hi,

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

Let be the Fibonacci number. Prove that for all m ≥ 1 and all n, .

Can someone help me with it?

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?

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 with and with .

Thank you very much!

Originally Posted by TheFangel

Hi,

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

Let be the Fibonacci number. Prove that for all m ≥ 1 and all n, .

Can someone help me with it?

Hi The Fangel,

P(k)



P(k+1)



Proof



and



then it follows









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

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.

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!