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Thread: Addition of relatively prime numbers

  1. #1
    Senior Member I-Think's Avatar
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    Addition of relatively prime numbers

    RTP: If $\displaystyle gcd(a,b)=1$, then $\displaystyle gcd(a, (a+b))=1$, where a,b are positive integers

    Proof
    Note $\displaystyle 1=as+bt$
    Assume $\displaystyle gcd(a, (a+b))=k$
    Then
    $\displaystyle am+(a+b)n=a(m+n)+bn=k*1$
    $\displaystyle a(m+n)+bn=aks+bkt$
    Thus
    $\displaystyle m+n=ks,n=kt$
    So
    $\displaystyle m+kt=ks,m=ks-kt$
    So
    $\displaystyle a(ks-kt)+bkt=aks+bkt$
    $\displaystyle -akt=0$
    This is impossible, so our assumption is wrong
    Thus our statement is true

    Questions
    1. Is this proof valid?
    2. Can I generalize to :If gcd(a,b)=k, then gcd(a, (a+b))=k
    3. Is there a simpler non-contradiction, non-FTA proof?
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  2. #2
    Senior Member I-Think's Avatar
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    This thread is a repost. Sorry. Marking it as solved and requesting that it be removed please
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  3. #3
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by I-Think View Post
    RTP: If $\displaystyle gcd(a,b)=1$, then $\displaystyle gcd(a, (a+b))=1$, where a,b are positive integers

    Proof
    Note $\displaystyle 1=as+bt$
    Assume $\displaystyle gcd(a, (a+b))=k$
    Then
    $\displaystyle am+(a+b)n=a(m+n)+bn=k*1$
    $\displaystyle a(m+n)+bn=aks+bkt$
    Thus
    $\displaystyle m+n=ks,n=kt$
    So
    $\displaystyle m+kt=ks,m=ks-kt$
    So
    $\displaystyle a(ks-kt)+bkt=aks+bkt$
    $\displaystyle -akt=0$
    This is impossible, so our assumption is wrong
    Thus our statement is true

    Questions
    1. Is this proof valid?
    2. Can I generalize to :If gcd(a,b)=k, then gcd(a, (a+b))=k
    3. Is there a simpler non-contradiction, non-FTA proof?
    Quote Originally Posted by I-Think View Post
    This thread is a repost. Sorry. Marking it as solved and requesting that it be removed please
    I can't find your other thread. Your proof looks fine from what I see. Yes, you're theorem generalizes. One merely needs note that if $\displaystyle \text{gcd}(a,a+b)\mid \text{gcd}(a,b)$ since $\displaystyle z\mid a\text{ and }z\mid a+b\implies z\mid a\text{ and }z\mid b$. Also, $\displaystyle \text{gcd}(a,b)\mid \text{gcd}(a,b)$ since $\displaystyle z\mid a\text{ and }z\mid b\implies z\mid a\text{ and }z\mid a+b$
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