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Math Help - Proof by contradiction

  1. #1
    Member
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    Unhappy Proof by contradiction

    If u and v have different parities( one is even and one is odd), then ( a,b,c) is primitive.

    Fact: If (a, b, c) is not primitive, then there exists a prime number p and natural numbers a`, b`, c` , such that a=pa`, b=pb` , c=pc`

    Fact 2: if p is a prime number, and m,n are elements of N such tha p|mn , then p|m or p|n
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  2. #2
    Senior Member
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    Atlanta, GA
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    Pythagorean Theorem

    Well, for starters, you ought to specify that we're referring to the equations (a,b,c)=(u^2-v^2,2uv,u^2+v^2). Also, this theorem is false: u and v must not only have opposite parities, they must be relatively prime. Example: 6 and 3 have opposite parity, yet (27,36,45) is NOT primitive. The correct theorem should read: "If (u,v)=1, then (a,b,c) is primitive." Moving on...

    To prove by contradiction, we start with the premise " u and v have no common divisors AND there exists a prime p that divides a,b,c." We will show this premise to be impossible:

    Lemma: if (u,v)=1, u and v must have opposite parities, one even one odd. p|b so p|2uv, so p=2 or p|u or p|v. If p=2 then since u and v have opposite parities, so p|u or p|v. WLOG, suppose p|u. Now, p|a so p|u^2-v^2. Since p|u, p|v^2 so p|v. Thus, (u,v)\neq1, and we have our contradiction. Therefore our premise is false.
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