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mebigp
 Post subject: Induction proof
PostPosted: Mon, 20 Sep 2010 19:27:28 UTC 
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Given
(F_k)^2 = F_b F_{b+1} for b>=1

Prove by induction that if F_k is the kth Fibonacci number.

Base F(1)^2= F(1)*F(2)
1=1*1
1=1 True

Induction k=b

(F(k))^2=F(k)*F(k+1)

(F(k))^2=(F(k))^2+1


Last edited by mebigp on Wed, 22 Sep 2010 00:34:00 UTC, edited 9 times in total.

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Matt
 Post subject:
PostPosted: Mon, 20 Sep 2010 19:31:04 UTC 
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Please correct any typos in the problem statement.
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mebigp
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PostPosted: Mon, 20 Sep 2010 21:30:35 UTC 
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The typos were corrected
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Matt
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PostPosted: Mon, 20 Sep 2010 22:00:42 UTC 
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No, not all of them were fixed...
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mebigp
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PostPosted: Tue, 21 Sep 2010 00:48:20 UTC 
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I think I corrected them all
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outermeasure
 Post subject: Re: Induction proof
PostPosted: Tue, 21 Sep 2010 19:03:23 UTC 
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mebigp wrote:
Given
(F_k)^2 = F_b F_{b+1} for b>=1

Prove by induction that if F_k is the kth Fibonacci number.


How is k related to b?

mebigp wrote:
Base F(1)^2= F(1)*F(2)
1=1*1
1=1 True

Induction k=n

(F(k))^2=F(k)*F(k+1)

(F(k))^2=(F(k))^2+1


Huh? What exactly are you doing?

mebigp wrote:
The typos were corrected


Obviously not.

_________________
\begin{aligned} Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\ Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\ Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\ Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\ Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\ Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4 \end{aligned}
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mebigp
 Post subject:
PostPosted: Wed, 22 Sep 2010 00:35:55 UTC 
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Should of been let k=b
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Soroban
 Post subject: Re: Induction proof
PostPosted: Wed, 22 Sep 2010 16:05:35 UTC 
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Hello, mebigp!

I don't see any corrections . . .

Quote:
Given
(F_b)^2 = F_b*F_{b+1} for b >= 1
. . First of all, this is not true!

"The square of F_b is the product of F_b and the next Fibonacci number" ?
. . I don't think so!


Prove by induction that if F_k is the kth Fibonacci number.

Base: F(1)^2 = F(1)*F(2)
. . . . . . . . .1 = 1*1
. . . . . . . . .1 = 1 True


Induction k=b

(F_k)^2 = F_k*F_(k+1)

(F_k)^2 = (F_k)^2 + 1 . what?

A number = the number plus one ?

No one can reply . . . The problem is all wrong!
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kompik
 Post subject: Re: Induction proof
PostPosted: Thu, 23 Sep 2010 16:05:39 UTC 
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Soroban wrote:
No one can reply . . . The problem is all wrong!

My guess is that the intended problem could be something like:
Given that \sum_{k=0}^n {F_k}^2 = F_{n} F_{n+1}, F_0=0 and F_1=1, prove that F_n is the n-th Fibonacci number.

(It's not that hard to do this by induction.)
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