Fibonacci numbers. how to prove? [duplicate]
This question already has an answer here:
Fibonacci modular results $ F_nmid F_{kn},,$ $, gcd(F_n,F_m) = F_{gcd(n,m)}$
5 answers
We are give Fibonacci numbers.{fi
| i ∈ N}, where f0 = 0, f1 = 1, fn+2 = fn +fn+1, n∈ N. How to proof with mathematical induction that if n divides by m, then fn divides by fm?
I am having trouble with thinking, what should be the transition.
induction divisibility fibonacci-numbers
asked Dec 27 '18 at 23:28
Edvards Zakovskis
234
marked as duplicate by anomaly, Crostul, amWhy, Bill Dubuque divisibility
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Dec 27 '18 at 23:57
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
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This question already has an answer here:
Fibonacci modular results $ F_nmid F_{kn},,$ $, gcd(F_n,F_m) = F_{gcd(n,m)}$
5 answers
We are give Fibonacci numbers.{fi
| i ∈ N}, where f0 = 0, f1 = 1, fn+2 = fn +fn+1, n∈ N. How to proof with mathematical induction that if n divides by m, then fn divides by fm?
I am having trouble with thinking, what should be the transition.
induction divisibility fibonacci-numbers
asked Dec 27 '18 at 23:28
Edvards Zakovskis
234
marked as duplicate by anomaly, Crostul, amWhy, Bill Dubuque divisibility
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Dec 27 '18 at 23:57
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
Hi & welcome to MSE. Please show us what you've tried so far & are having difficulty with. Thanks.
– John Omielan
Dec 27 '18 at 23:29
add a comment |
This question already has an answer here:
Fibonacci modular results $ F_nmid F_{kn},,$ $, gcd(F_n,F_m) = F_{gcd(n,m)}$
5 answers
We are give Fibonacci numbers.{fi
| i ∈ N}, where f0 = 0, f1 = 1, fn+2 = fn +fn+1, n∈ N. How to proof with mathematical induction that if n divides by m, then fn divides by fm?
I am having trouble with thinking, what should be the transition.
induction divisibility fibonacci-numbers
asked Dec 27 '18 at 23:28
Edvards Zakovskis
234
This question already has an answer here:
Fibonacci modular results $ F_nmid F_{kn},,$ $, gcd(F_n,F_m) = F_{gcd(n,m)}$
5 answers
We are give Fibonacci numbers.{fi
| i ∈ N}, where f0 = 0, f1 = 1, fn+2 = fn +fn+1, n∈ N. How to proof with mathematical induction that if n divides by m, then fn divides by fm?
I am having trouble with thinking, what should be the transition.
This question already has an answer here:
Fibonacci modular results $ F_nmid F_{kn},,$ $, gcd(F_n,F_m) = F_{gcd(n,m)}$
5 answers
induction divisibility fibonacci-numbers
induction divisibility fibonacci-numbers
asked Dec 27 '18 at 23:28
Edvards Zakovskis
234
asked Dec 27 '18 at 23:28
Edvards Zakovskis
234
edited Dec 27 '18 at 23:36
asked Dec 27 '18 at 23:28
Edvards Zakovskis
234
asked Dec 27 '18 at 23:28
Edvards Zakovskis
234
asked Dec 27 '18 at 23:28
Edvards Zakovskis
234
234
marked as duplicate by anomaly, Crostul, amWhy, Bill Dubuque divisibility
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Dec 27 '18 at 23:57
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Dec 27 '18 at 23:57
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
Hi & welcome to MSE. Please show us what you've tried so far & are having difficulty with. Thanks.
– John Omielan
Dec 27 '18 at 23:29
add a comment |
Hi & welcome to MSE. Please show us what you've tried so far & are having difficulty with. Thanks.
– John Omielan
Dec 27 '18 at 23:29
Hi & welcome to MSE. Please show us what you've tried so far & are having difficulty with. Thanks.
– John Omielan
Dec 27 '18 at 23:29
Hi & welcome to MSE. Please show us what you've tried so far & are having difficulty with. Thanks.
– John Omielan
Dec 27 '18 at 23:29
add a comment |
1 Answer
1
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oldest
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First prove by induction on $m$ that
$F_{m+n}=F_{m+1}F_n+F_mF_{n+1}-F_m F_n$
Then put $m=kn$ and find that if $F_n|F_{kn}$ then $F_n|F_{(k+1)n}$.
answered Dec 27 '18 at 23:48
Oscar Lanzi
12.1k12036
+1 then -1. The battle is on!
– Oscar Lanzi
Dec 27 '18 at 23:58
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
First prove by induction on $m$ that
$F_{m+n}=F_{m+1}F_n+F_mF_{n+1}-F_m F_n$
Then put $m=kn$ and find that if $F_n|F_{kn}$ then $F_n|F_{(k+1)n}$.
answered Dec 27 '18 at 23:48
Oscar Lanzi
12.1k12036
+1 then -1. The battle is on!
– Oscar Lanzi
Dec 27 '18 at 23:58
add a comment |
First prove by induction on $m$ that
$F_{m+n}=F_{m+1}F_n+F_mF_{n+1}-F_m F_n$
Then put $m=kn$ and find that if $F_n|F_{kn}$ then $F_n|F_{(k+1)n}$.
answered Dec 27 '18 at 23:48
Oscar Lanzi
12.1k12036
+1 then -1. The battle is on!
– Oscar Lanzi
Dec 27 '18 at 23:58
add a comment |
First prove by induction on $m$ that
$F_{m+n}=F_{m+1}F_n+F_mF_{n+1}-F_m F_n$
Then put $m=kn$ and find that if $F_n|F_{kn}$ then $F_n|F_{(k+1)n}$.
answered Dec 27 '18 at 23:48
Oscar Lanzi
12.1k12036
First prove by induction on $m$ that
$F_{m+n}=F_{m+1}F_n+F_mF_{n+1}-F_m F_n$
Then put $m=kn$ and find that if $F_n|F_{kn}$ then $F_n|F_{(k+1)n}$.
answered Dec 27 '18 at 23:48
Oscar Lanzi
12.1k12036
answered Dec 27 '18 at 23:48
Oscar Lanzi
12.1k12036
answered Dec 27 '18 at 23:48
Oscar Lanzi
12.1k12036
answered Dec 27 '18 at 23:48
Oscar Lanzi
12.1k12036
12.1k12036
+1 then -1. The battle is on!
– Oscar Lanzi
Dec 27 '18 at 23:58
add a comment |
+1 then -1. The battle is on!
– Oscar Lanzi
Dec 27 '18 at 23:58
+1 then -1. The battle is on!
– Oscar Lanzi
Dec 27 '18 at 23:58
+1 then -1. The battle is on!
– Oscar Lanzi
Dec 27 '18 at 23:58
add a comment |
Hi & welcome to MSE. Please show us what you've tried so far & are having difficulty with. Thanks.
– John Omielan
Dec 27 '18 at 23:29