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anonymous

  • 4 years ago

"suppose \(c|(a+b)\) where \(a,b,c\in\mathbb{Z}\). then \(c|(pa+qb)\) where \(p,q\in\mathbb{Z}\)." can you really make such assumption!?

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  1. anonymous
    • 4 years ago
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    "for SOME integers \(p\) and \(q\)" i meant to add

  2. anonymous
    • 4 years ago
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    i don't seem to get why x.x

  3. JamesJ
    • 4 years ago
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    Choose p=q=1

  4. anonymous
    • 4 years ago
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    No we can't make such assumption.

  5. anonymous
    • 4 years ago
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    like a=2, b=4, c=3 c|(a+b) -> 3|6 which is true but c|(2a+3b) -> 3|16 is false :(

  6. JamesJ
    • 4 years ago
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    \[ c | (a+b) \implies c | (1\cdot a + 1\cdot b) \]

  7. anonymous
    • 4 years ago
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    \(3|(4+5)\) but 3 does not divide \((4\times 3+5 \times 5) \)

  8. anonymous
    • 4 years ago
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    ooooh

  9. JamesJ
    • 4 years ago
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    so if the question is there exist at least one pair p,q such your statement is true, then yes.

  10. anonymous
    • 4 years ago
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    But this question is asking for for all \(p,q \in \mathbb{Z}\), hence incorrect.

  11. anonymous
    • 4 years ago
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    all or any*

  12. JamesJ
    • 4 years ago
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    ffm, pre-algebra says immediately below "for SOME p and q ..."

  13. anonymous
    • 4 years ago
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    Hey, EDIT feature is a must!!!!

  14. anonymous
    • 4 years ago
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    I have a tendency not to read any comment/answer before trying it on my own.

  15. JamesJ
    • 4 years ago
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    no kidding

  16. anonymous
    • 4 years ago
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    "for SOME integers p and q" that's true.

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