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 Post subject: Sequence problem
PostPosted: Fri, 11 Feb 2011 19:25:22 UTC 
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Is there a monotonic increasing sequence a_n such that a_n+a_m=a_{nm}?

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 Post subject: Re: Sequence problem
PostPosted: Fri, 11 Feb 2011 19:37:38 UTC 
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Shadow wrote:
Is there a monotonic increasing sequence a_n such that a_n+a_m=a_{nm}?


How about a_n = \log(n)?

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 Post subject: Re: Sequence problem
PostPosted: Fri, 11 Feb 2011 22:30:08 UTC 
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robbwrr wrote:
Shadow wrote:
Is there a monotonic increasing sequence a_n such that a_n+a_m=a_{nm}?


How about a_n = \log(n)?


What if the indexing set is \mathbb{Z}?

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\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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 Post subject: Re: Sequence problem
PostPosted: Wed, 29 Feb 2012 17:16:05 UTC 
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Shadow wrote:
Is there a monotonic increasing sequence a_n such that a_n+a_m=a_{nm}?


Maybe I'm missing something, but would a_n=0 satisfy the conditions, as well as outermeasure's extension?


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 Post subject: Re: Sequence problem
PostPosted: Wed, 29 Feb 2012 17:18:40 UTC 
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DJJazzyDan wrote:
Shadow wrote:
Is there a monotonic increasing sequence a_n such that a_n+a_m=a_{nm}?


Maybe I'm missing something, but would a_n=0 satisfy the conditions, as well as outermeasure's extension?


Monotone increasing.

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 Post subject: Re: Sequence problem
PostPosted: Wed, 29 Feb 2012 19:31:18 UTC 
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Shadow wrote:
Monotone increasing.


My mistake, confused with nondecreasing.
Answer and proof:
Spoiler:
There cannot be such a sequence when the indexing set is \mathbb{Z}
Proof: Consider there was such a sequence a_n
Then for some n > 1, say 2
a_1+a_n=a_{1n}=a_n which implies a_1=0
and
a_0+a_n=a_{0n}=a_0 which implies a_n=0

Since the sequence is monotone increasing, a_{i+1}>a_{i} which is a contradiction.


There is a similar proof, I believe, by considering the index -2,-1, and 2


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