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Loser66

  • one year ago

Show that gcd (even, odd) = odd Please, help

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  1. Loser66
    • one year ago
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    @zzr0ck3r

  2. zzr0ck3r
    • one year ago
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    if it was even, then an even number divides an odd number, yikes!

  3. Loser66
    • one year ago
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    I need algebraic proof, not logic one. Please.

  4. Loser66
    • one year ago
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    even = 2k odd = 2m +1 gcd (2k, 2m +1) = d, and prove d is odd.

  5. Empty
    • one year ago
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    This troubles me

  6. zzr0ck3r
    • one year ago
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    Suppose by way of contradiction that \(c=\gcd(2k, 2k_0+1)\) where \(c=2k_2\) is even. Then \(2k_0+1=2(2k_2)\)

  7. Loser66
    • one year ago
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    This is what I have: d| 2k d| 2m +1 hence d| 2k + 2m +1 d| 2(k+m) +1 then ... stuck.

  8. zzr0ck3r
    • one year ago
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    Then \(2k_0+1=k(2k_2)\) for osme \(k\).

  9. Loser66
    • one year ago
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    From where, you have \(2k_0 +1 = 2 (2k_2)\)?

  10. zzr0ck3r
    • one year ago
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    that should say \[2k_0+1=k(2k_2)\]

  11. Loser66
    • one year ago
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    Use other letter, please, it confuses me when the fist element is k, and this k.

  12. Loser66
    • one year ago
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    Got you.!! thanks a lot

  13. zzr0ck3r
    • one year ago
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    lol ok PF Suppose \(a\) is even, \(b\) is odd. Then \(b=2k_1+1\) for some \(k_1\in \mathbb{Z}\). Now suppose \(c=\gcd(a,b)\) and \(c\) is even. Then \(c=2k_2\) for some \(k_2\in \mathbb{Z}\). Since \(c\) is the GCD of \(a\) and \(b\) it must divide both. So \(c\) divdes \(b\) and thus \(b=c*k_3\) where \(k_3\in \mathbb{Z}\) So \(2k_1+1=k_3(2k_2)\) a contradiction

  14. zzr0ck3r
    • one year ago
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    an even number cant divide an odd number evenly is actually what you prove.

  15. zzr0ck3r
    • one year ago
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    np

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