Mixing
Proving That Consecutive Fibonacci Numbers are Relatively Prime
$begingroup$
The Problem:
Prove that consecutive Fibonacci numbers are relatively prime.
I have seen proofs where people use induction and show that if $gcd(F_n, F_{n+1})=1$, then $gcd(F_{n+1}, F_{n+2})=1$ through the Fibonacci property.
My proof uses induction as well, but in a different way.
We will use induction to show that if such a base case exists that $F_{n}$ and $F_{n+1}$ are both divisible by some integral $k ge 2$, then all $F_{i}$ are divisible by $k$.
$$F_{n} equiv F_{n+1} equiv 0 pmod{k}$$
$$F_{n} equiv F_{n}+F_{n-1} equiv 0 pmod{k}$$
$$F_{n} equiv F_{n-1} equiv 0 pmod{k}$$
Since $F_{3}=2$ and $F_4=3$ are not divisible be any common integer $k ge 2$, our assumption is wrong. Hence, $F_{n}$ and $F_{n+1}$ are relatively prime.
proof-verification induction fibonacci-numbers coprime
edited Jan 3 at 23:11
José Carlos Santos
155k22124227
asked Jan 3 at 22:41
Shrey JoshiShrey Joshi
30713
$endgroup$
add a comment |
$begingroup$
The Problem:
Prove that consecutive Fibonacci numbers are relatively prime.
I have seen proofs where people use induction and show that if $gcd(F_n, F_{n+1})=1$, then $gcd(F_{n+1}, F_{n+2})=1$ through the Fibonacci property.
My proof uses induction as well, but in a different way.
We will use induction to show that if such a base case exists that $F_{n}$ and $F_{n+1}$ are both divisible by some integral $k ge 2$, then all $F_{i}$ are divisible by $k$.
$$F_{n} equiv F_{n+1} equiv 0 pmod{k}$$
$$F_{n} equiv F_{n}+F_{n-1} equiv 0 pmod{k}$$
$$F_{n} equiv F_{n-1} equiv 0 pmod{k}$$
Since $F_{3}=2$ and $F_4=3$ are not divisible be any common integer $k ge 2$, our assumption is wrong. Hence, $F_{n}$ and $F_{n+1}$ are relatively prime.
proof-verification induction fibonacci-numbers coprime
edited Jan 3 at 23:11
José Carlos Santos
155k22124227
asked Jan 3 at 22:41
Shrey JoshiShrey Joshi
30713
$endgroup$
add a comment |
$begingroup$
The Problem:
Prove that consecutive Fibonacci numbers are relatively prime.
I have seen proofs where people use induction and show that if $gcd(F_n, F_{n+1})=1$, then $gcd(F_{n+1}, F_{n+2})=1$ through the Fibonacci property.
My proof uses induction as well, but in a different way.
We will use induction to show that if such a base case exists that $F_{n}$ and $F_{n+1}$ are both divisible by some integral $k ge 2$, then all $F_{i}$ are divisible by $k$.
$$F_{n} equiv F_{n+1} equiv 0 pmod{k}$$
$$F_{n} equiv F_{n}+F_{n-1} equiv 0 pmod{k}$$
$$F_{n} equiv F_{n-1} equiv 0 pmod{k}$$
Since $F_{3}=2$ and $F_4=3$ are not divisible be any common integer $k ge 2$, our assumption is wrong. Hence, $F_{n}$ and $F_{n+1}$ are relatively prime.
proof-verification induction fibonacci-numbers coprime
edited Jan 3 at 23:11
José Carlos Santos
155k22124227
asked Jan 3 at 22:41
Shrey JoshiShrey Joshi
30713
$endgroup$
The Problem:
Prove that consecutive Fibonacci numbers are relatively prime.
I have seen proofs where people use induction and show that if $gcd(F_n, F_{n+1})=1$, then $gcd(F_{n+1}, F_{n+2})=1$ through the Fibonacci property.
My proof uses induction as well, but in a different way.
We will use induction to show that if such a base case exists that $F_{n}$ and $F_{n+1}$ are both divisible by some integral $k ge 2$, then all $F_{i}$ are divisible by $k$.
$$F_{n} equiv F_{n+1} equiv 0 pmod{k}$$
$$F_{n} equiv F_{n}+F_{n-1} equiv 0 pmod{k}$$
$$F_{n} equiv F_{n-1} equiv 0 pmod{k}$$
Since $F_{3}=2$ and $F_4=3$ are not divisible be any common integer $k ge 2$, our assumption is wrong. Hence, $F_{n}$ and $F_{n+1}$ are relatively prime.
proof-verification induction fibonacci-numbers coprime
proof-verification induction fibonacci-numbers coprime
edited Jan 3 at 23:11
José Carlos Santos
155k22124227
asked Jan 3 at 22:41
Shrey JoshiShrey Joshi
30713
edited Jan 3 at 23:11
José Carlos Santos
155k22124227
asked Jan 3 at 22:41
Shrey JoshiShrey Joshi
30713
edited Jan 3 at 23:11
José Carlos Santos
155k22124227
edited Jan 3 at 23:11
José Carlos Santos
155k22124227
edited Jan 3 at 23:11
José Carlos Santos
155k22124227
155k22124227
asked Jan 3 at 22:41
Shrey JoshiShrey Joshi
30713
asked Jan 3 at 22:41
Shrey JoshiShrey Joshi
30713
asked Jan 3 at 22:41
Shrey JoshiShrey Joshi
30713
30713
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
It is a waste of time to use modular arithmetic for this. What you proved was that$$kmid F_nwedge kmid F_{n+1}implies kmid F_{n-1},$$which is indeed correct. But, after having done this, I think that your best option is to say that you are using the well-ordering principle: if there was some $ninmathbb N$ such that $F_n$ and $F_{n+1}$ are not coprime, you take the smallest such $n$. Of course, $n$ cannot be $1$, since $F_1$ and $F_2$ are coprime. So, $n-1inmathbb N$ and it follows from what you proved that $F_{n-1}$ and $F_n$ are coprime too. This contradicts the definition of $n$.
answered Jan 3 at 22:49
José Carlos SantosJosé Carlos Santos
155k22124227
$endgroup$
add a comment |
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1 Answer
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active
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1 Answer
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$begingroup$
It is a waste of time to use modular arithmetic for this. What you proved was that$$kmid F_nwedge kmid F_{n+1}implies kmid F_{n-1},$$which is indeed correct. But, after having done this, I think that your best option is to say that you are using the well-ordering principle: if there was some $ninmathbb N$ such that $F_n$ and $F_{n+1}$ are not coprime, you take the smallest such $n$. Of course, $n$ cannot be $1$, since $F_1$ and $F_2$ are coprime. So, $n-1inmathbb N$ and it follows from what you proved that $F_{n-1}$ and $F_n$ are coprime too. This contradicts the definition of $n$.
answered Jan 3 at 22:49
José Carlos SantosJosé Carlos Santos
155k22124227
$endgroup$
add a comment |
$begingroup$
It is a waste of time to use modular arithmetic for this. What you proved was that$$kmid F_nwedge kmid F_{n+1}implies kmid F_{n-1},$$which is indeed correct. But, after having done this, I think that your best option is to say that you are using the well-ordering principle: if there was some $ninmathbb N$ such that $F_n$ and $F_{n+1}$ are not coprime, you take the smallest such $n$. Of course, $n$ cannot be $1$, since $F_1$ and $F_2$ are coprime. So, $n-1inmathbb N$ and it follows from what you proved that $F_{n-1}$ and $F_n$ are coprime too. This contradicts the definition of $n$.
answered Jan 3 at 22:49
José Carlos SantosJosé Carlos Santos
155k22124227
$endgroup$
add a comment |
$begingroup$
It is a waste of time to use modular arithmetic for this. What you proved was that$$kmid F_nwedge kmid F_{n+1}implies kmid F_{n-1},$$which is indeed correct. But, after having done this, I think that your best option is to say that you are using the well-ordering principle: if there was some $ninmathbb N$ such that $F_n$ and $F_{n+1}$ are not coprime, you take the smallest such $n$. Of course, $n$ cannot be $1$, since $F_1$ and $F_2$ are coprime. So, $n-1inmathbb N$ and it follows from what you proved that $F_{n-1}$ and $F_n$ are coprime too. This contradicts the definition of $n$.
answered Jan 3 at 22:49
José Carlos SantosJosé Carlos Santos
155k22124227
$endgroup$
It is a waste of time to use modular arithmetic for this. What you proved was that$$kmid F_nwedge kmid F_{n+1}implies kmid F_{n-1},$$which is indeed correct. But, after having done this, I think that your best option is to say that you are using the well-ordering principle: if there was some $ninmathbb N$ such that $F_n$ and $F_{n+1}$ are not coprime, you take the smallest such $n$. Of course, $n$ cannot be $1$, since $F_1$ and $F_2$ are coprime. So, $n-1inmathbb N$ and it follows from what you proved that $F_{n-1}$ and $F_n$ are coprime too. This contradicts the definition of $n$.
answered Jan 3 at 22:49
José Carlos SantosJosé Carlos Santos
155k22124227
edited Jan 4 at 12:29
answered Jan 3 at 22:49
José Carlos SantosJosé Carlos Santos
155k22124227
answered Jan 3 at 22:49
José Carlos SantosJosé Carlos Santos
155k22124227
answered Jan 3 at 22:49
José Carlos SantosJosé Carlos Santos
155k22124227
155k22124227
add a comment |
add a comment |
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