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Loser66
 one year ago
Show that gcd (even, odd) = odd
Please, help
Loser66
 one year ago
Show that gcd (even, odd) = odd Please, help

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zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2if it was even, then an even number divides an odd number, yikes!

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1I need algebraic proof, not logic one. Please.

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1even = 2k odd = 2m +1 gcd (2k, 2m +1) = d, and prove d is odd.

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2Suppose by way of contradiction that \(c=\gcd(2k, 2k_0+1)\) where \(c=2k_2\) is even. Then \(2k_0+1=2(2k_2)\)

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1This is what I have: d 2k d 2m +1 hence d 2k + 2m +1 d 2(k+m) +1 then ... stuck.

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2Then \(2k_0+1=k(2k_2)\) for osme \(k\).

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1From where, you have \(2k_0 +1 = 2 (2k_2)\)?

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2that should say \[2k_0+1=k(2k_2)\]

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1Use other letter, please, it confuses me when the fist element is k, and this k.

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1Got you.!! thanks a lot

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2lol 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

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2an even number cant divide an odd number evenly is actually what you prove.
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