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GoldenPho
 Post subject: Sets and Relation BASIC
PostPosted: Sun, 22 Apr 2012 21:11:44 UTC 
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Hi

Let set A={1,2},

and relation R={ (1,2), (2,1),(2,2)}, would this be asymmetric? if not why??

my answer: i think yes, because (2,1) in R and (1,2) in R then (2,2) in R... but i dont know if it is the right thinking.

Or....do you also need (1,1) in R since (1,2) in R and (2,1) in R
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Shadow
 Post subject: Re: Sets and Relation BASIC
PostPosted: Mon, 23 Apr 2012 03:56:37 UTC 
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GoldenPho wrote:
Hi

Let set A={1,2},

and relation R={ (1,2), (2,1),(2,2)}, would this be asymmetric? if not why??

my answer: i think yes, because (2,1) in R and (1,2) in R then (2,2) in R... but i dont know if it is the right thinking.

Or....do you also need (1,1) in R since (1,2) in R and (2,1) in R


Are you sure you mean asymmetric and not anti-symmetric?

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outermeasure
 Post subject: Re: Sets and Relation BASIC
PostPosted: Mon, 23 Apr 2012 04:03:31 UTC 
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Shadow wrote:
GoldenPho wrote:
Hi

Let set A={1,2},

and relation R={ (1,2), (2,1),(2,2)}, would this be asymmetric? if not why??

my answer: i think yes, because (2,1) in R and (1,2) in R then (2,2) in R... but i dont know if it is the right thinking.

Or....do you also need (1,1) in R since (1,2) in R and (2,1) in R


Are you sure you mean asymmetric and not anti-symmetric?


Either way it is no, not yes... ;)

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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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Shadow
 Post subject: Re: Sets and Relation BASIC
PostPosted: Mon, 23 Apr 2012 04:04:13 UTC 
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outermeasure wrote:
Shadow wrote:
GoldenPho wrote:
Hi

Let set A={1,2},

and relation R={ (1,2), (2,1),(2,2)}, would this be asymmetric? if not why??

my answer: i think yes, because (2,1) in R and (1,2) in R then (2,2) in R... but i dont know if it is the right thinking.

Or....do you also need (1,1) in R since (1,2) in R and (2,1) in R


Are you sure you mean asymmetric and not anti-symmetric?


Either way it is no, not yes... ;)


True, but at least we can fix terminology, as "asymmetric" is not a generally useful thing to know. :-)

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