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Math Help - [SOLVED] Relations

  1. #16
    MHF Contributor Drexel28's Avatar
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    [QUOTE=dwsmith;499962][quote=Drexel28;499961]
    Quote Originally Posted by dwsmith View Post
    Well if it equals zero, the left is odd and the right is odd.

    I am confused on how that proves it is an injection though.
    We may conclude that m=m' right? And so f(m,n)=f(m',n')\implies 2^m(2n-1)=2^{m'}(2n-1)=2^m(2n'-1)\implies 2n-1=2n'-1...soo
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  2. #17
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    [quote=Drexel28;499963][quote=dwsmith;499962]
    Quote Originally Posted by Drexel28 View Post

    We may conclude that m=m' right? And so f(m,n)=f(m',n')\implies 2^m(2n-1)=2^{m'}(2n-1)=2^m(2n'-1)\implies 2n-1=2n'-1...soo
    n=n'
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  3. #18
    MHF Contributor Drexel28's Avatar
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    [QUOTE=dwsmith;499964][quote=Drexel28;499963]
    Quote Originally Posted by dwsmith View Post

    n=n'
    So!!!

    m=m'\text{ and }n=n' and so...
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  4. #19
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    [quote=Drexel28;499965][quote=dwsmith;499964]
    Quote Originally Posted by Drexel28 View Post

    So!!!

    m=m'\text{ and }n=n' and so...
    f is injective if for all a and b in A, if f(a) = f(b), then a = b; that is, f(a) = f(b) implies a = b.
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  5. #20
    MHF Contributor Drexel28's Avatar
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    [QUOTE=dwsmith;499966][quote=Drexel28;499965]
    Quote Originally Posted by dwsmith View Post

    f is injective if for all a and b in A, if f(a) = f(b), then a = b; that is, f(a) = f(b) implies a = b.
    And m=m'\text{ and }n=n'\implies (m,n)=(m',n')...Ta-da!
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  6. #21
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    Prove that f is a surjection. This is actually a consequence of the Fundalmental Theorem of Arithmetic, Theorem 8.16.

    I don't know what the fact that a number is prime or the product of prime numbers has to be with solving the surjection.
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  7. #22
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by dwsmith View Post
    Prove that f is a surjection. This is actually a consequence of the Fundalmental Theorem of Arithmetic, Theorem 8.16.

    I don't know what the fact that a number is prime or the product of prime numbers has to be with solving the surjection.
    You know that if n\in\mathbb{N} that it may be represented as the product of primes n=p_1^{\alpha_1}\cdots p_n^{\alpha_n}, right? Well, except for 2 every prime is odd. So, we could for the sake of convenience write every integer as n=2^{\beta}\cdot p_1^{\alpha_1}\cdots p_n^{\alpha_n} where \beta may be zero of course. But, as previously mentioned since every other prime is odd and the product of odd numbers is odd it follows that p_1^{\alpha_1}\cdots p_n^{\alpha_n}\text{ is odd}\implies p_1^{\alpha_1}\cdots p_n^{\alpha_n}=2m+1 for some m\in\mathbb{N}, right? So now what?
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  8. #23
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    Quote Originally Posted by Drexel28 View Post
    You know that if n\in\mathbb{N} that it may be represented as the product of primes n=p_1^{\alpha_1}\cdots p_n^{\alpha_n}, right? Well, except for 2 every prime is odd. So, we could for the sake of convenience write every integer as n=2^{\beta}\cdot p_1^{\alpha_1}\cdots p_n^{\alpha_n} where \beta may be zero of course. But, as previously mentioned since every other prime is odd and the product of odd numbers is odd it follows that p_1^{\alpha_1}\cdots p_n^{\alpha_n}\text{ is odd}\implies p_1^{\alpha_1}\cdots p_n^{\alpha_n}=2m+1 for some m\in\mathbb{N}, right? So now what?
    Well 2m+1 is both the right and left side minus the 2^{\beta}
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  9. #24
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by dwsmith View Post
    Well 2m+1 is both the right and left side minus the 2^{\beta}
    So, every number may be written in the form \pm2^{\beta}(2m+1)..so
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  10. #25
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    Quote Originally Posted by Drexel28 View Post
    So, every number may be written in the form \pm2^{\beta}(2m+1)..so
    Well that product is even unless \beta=0
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  11. #26
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by dwsmith View Post
    Well that product is even unless \beta=0
    Does it matter? Isn't that f(\beta,m)?
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  12. #27
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    Quote Originally Posted by Drexel28 View Post
    Does it matter? Isn't that f(\beta,m)?
    Yeah it is f(b,m)
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  13. #28
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by dwsmith View Post
    Yeah it is f(b,m)
    And a surjection is what?
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  14. #29
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    When f(x,y)=z
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  15. #30
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by dwsmith View Post
    When f(x,y)=z
    And haven't we done that?
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