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amistre64

  • 2 years ago

so what we learned today .... if a doesnt divide b, but a divides bc; then a divides c.

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  1. amistre64
    • 2 years ago
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    we also learned that if a divides bc, then a divides bcn ..... 30 minutes of my life gone :/

  2. KingGeorge
    • 2 years ago
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    Believe it or not, that's not inherently obvious. Depending on what structures you're using, if a doesn't divide b, but a does divide bc, it doesn't necessarily divide c.

  3. amistre64
    • 2 years ago
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    number theory ... :)

  4. inkyvoyd
    • 2 years ago
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    I'ma delete this shortly afterward, but can anyone help me http://openstudy.com/study#/updates/5064b1f2e4b0583d5cd3facc Also, does that apply for matrix algebra?

  5. amistre64
    • 2 years ago
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    mattrix algebra has quirks in that its .. cant recall. ring, field, group ??

  6. KingGeorge
    • 2 years ago
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    Number theory is an easy example, but you can define a ring with zero divisors, and suddenly, if ab=0, then neither a nor b must be 0.

  7. inkyvoyd
    • 2 years ago
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    o.o show me the logic!

  8. amistre64
    • 2 years ago
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    teacher did show us at the begining of the year about her thesis work on abstract algebra and made us write up Zmod 12 tables for addition and multiplication

  9. KingGeorge
    • 2 years ago
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    Zmod12 is a nice example. 2*6=0=3*4

  10. inkyvoyd
    • 2 years ago
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    No that's not helpful I don't understand zmods lool

  11. KingGeorge
    • 2 years ago
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    2*6 mod 12=0 If you understand modular multiplication, it's what you would imagine Zmod12 would be.

  12. amistre64
    • 2 years ago
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    if a not| b; the (a,b)=1; therefore 1 = ax+by multiply each side by c c = acx+bcy a|acx a|bcy , it was given in the thrm that a|bc therefore a | (acx+bcy) a | c

  13. amistre64
    • 2 years ago
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    all the 0 mod 12s have no recipricals :)

  14. amistre64
    • 2 years ago
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    .... er, no multiplicative inversai

  15. inkyvoyd
    • 2 years ago
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    I look at wikipedia and see funny symbols. http://en.wikipedia.org/wiki/Modular_arithmetic I quit this party.

  16. amistre64
    • 2 years ago
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    There must be some way out of here, said the joker to the thief. There's too much confusion ... I can't get no relief

  17. KingGeorge
    • 2 years ago
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    Well, if it's coprime to 12 it has a multiplicative inverse....

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