timo Posted October 21 Posted October 21 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.
Genady Posted October 21 Author Posted October 21 This part is correct: 9 minutes ago, timo said: for some other reason 😉 He has a mathematical reason to say "if".
Genady Posted October 22 Author Posted October 22 Hint. Here he is asking, "5 can be equal to 2? 5 can't be equal to 2?"
joigus Posted October 22 Posted October 22 (edited) 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) Edited October 22 by joigus spoiler added 1
Genady Posted October 22 Author Posted October 22 (edited) 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 October 22 by Genady
joigus Posted October 22 Posted October 22 (edited) Spoiler 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 October 22 by joigus spoiler added
Genady Posted October 22 Author Posted October 22 19 minutes ago, joigus said: I added a spoiler. Me too.
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