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Math Help - [SOLVED] Relations

  1. #31
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    Yes indeed.
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  2. #32
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    Prove that card(ℕℕ) = \aleph_0.

    Since we have already shown f(m,n) is equal to the product of the prime numbers, f(m,n)= (not sure how to finish it but I think this setup should go somewhere).....
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  3. #33
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by dwsmith View Post
    Prove that card(ℕℕ) = \aleph_0.

    Since we have already shown f(m,n) is equal to the product of the prime numbers, f(m,n)= (not sure how to finish it but I think this setup should go somewhere).....
    You've just proved that m,n)\mapsto 2^{m-1}(2n-1)" alt="f:\mathbb{N}\times\mathbb{N}\to\mathbb{N}m,n)\mapsto 2^{m-1}(2n-1)" /> is a bijection. So, by definition \text{card }\mathbb{N}\times\mathbb{N}=\text{card }\mathbb{N}=\aleph_0
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  4. #34
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    Quote Originally Posted by Drexel28 View Post
    You've just proved that m,n)\mapsto 2^{m-1}(2n-1)" alt="f:\mathbb{N}\times\mathbb{N}\to\mathbb{N}m,n)\mapsto 2^{m-1}(2n-1)" /> is a bijection. So, by definition \text{card }\mathbb{N}\times\mathbb{N}=\text{card }\mathbb{N}=\aleph_0
    Oh, boy, I had a lot fun watching Drexel Math Show.
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