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"if 2 is not equal to 5"


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The professor is preparing you for the next step where he/she says "and now assume 2 and 5 were any arbitrary value: all steps I did still hold true, so...": "When Tv1 = av1 and Tv2 = bv1: If a!=b then v1 and v2 are not colinear". Either that or for some other reason or for no reason whatsoever.

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I would assume the professor means we are not in a discrete arithmetic in which 5 = 2 (mod p) for some p. That p would be 3, as 5 = 2 (mod 3)

 

Edited by joigus
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23 minutes ago, joigus said:
Spoiler

I would assume the professor means we are not in a discrete arithmetic in which 5 = 2 (mod p) for some p. That p would be 3, as 5 = 2 (mod 3)

 

Spoiler

He only assumes that vector space is defined over a field. Quotient field mod 3 is as good for this as any.

+1

Edited by Genady
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Exactly. Just to correct myself. The p would be any integer multiple of 3, not necessarily 3.

And yes, fields is the relevant concept there. I forgot.

Sorry, I didn't realise this was under the Teasers and Puzzles section. I added a spoiler.

Edited by joigus
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