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Math Help - [SOLVED] Binary Operations

  1. #1
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    [SOLVED] Binary Operations

    Suppose that two binary operations, denoted by \oplus and \odot, are defined on a nonempty set S, and that the following conditions are satisfied \forall x,y,z\in S.

    (1) x\oplus y and x\odot y are in S
    (2) x\oplus (y\oplus z)=(x\oplus y) \oplus z and x\odot (y\odot z)=(x\odot y) \odot z
    (3) x\oplus y=y\oplus x

    Also, \forall x\in S and \forall n\in \mathbb{Z}^+, the elements nx and x^n are defined recursively as follows:

    1x=x^1=x
    if kx and x^k have been defined, then (k+1)x=kx\oplus x and x^{k+1}=x^k\odot x

    Which of the following must be true?

    (i) (x\odot y)^n=x^n\odot y^n \forall x,y\in S and \forall n\in\mathbb{Z}^+

    (ii) n(x\oplus y)=nx\oplus ny \forall x,y\in S and \forall n\in\mathbb{Z}^+

    (ii) x^m\odot x^n=x^{m+n} \forall x\in S and \forall m,n\in\mathbb{Z}^+
    This is one obviously true.

    I am struggling with proving or disproving 1 and 2
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  2. #2
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    Quote Originally Posted by dwsmith View Post
    Suppose that two binary operations, denoted by \oplus and \odot, are defined on a nonempty set S, and that the following conditions are satisfied \forall x,y,z\in S.

    (1) x\oplus y and x\odot y are in S
    (2) x\oplus (y\oplus z)=(x\oplus y) \oplus z and x\odot (y\odot z)=(x\odot y) \odot z
    (3) x\oplus y=y\oplus x

    Also, \forall x\in S and \forall n\in \mathbb{Z}^+, the elements nx and x^n are defined recursively as follows:

    1x=x^1=x
    if kx and x^k have been defined, then (k+1)x=kx\oplus x and x^{k+1}=x^k\odot x

    Which of the following must be true?

    (i) (x\odot y)^n=x^n\odot y^n \forall x,y\in S and \forall n\in\mathbb{Z}^+

    (ii) n(x\oplus y)=nx\oplus ny \forall x,y\in S and \forall n\in\mathbb{Z}^+

    (ii) x^m\odot x^n=x^{m+n} \forall x\in S and \forall m,n\in\mathbb{Z}^+
    This is one obviously true.

    I am struggling with proving or disproving 1 and 2
    I think (i) is false, but I don't think my demonstration constitutes a rigorous proof.

    We can rewrite

    (x\odot y)^n=\underbrace{(x\odot y)\odot (x\odot y) \odot ... \odot (x\odot y)}_{n \text{ instances of }(x\odot y)}

    Since \odot is not commutative, we cannot rearrange terms.

    I think (ii) is true and can be proven from the definition of multiplication and conditions (2) and (3).

    We can rewrite

    n(x\oplus y)=\underbrace{(x\oplus y) \oplus (x\oplus y) \oplus \cdots \oplus (x\oplus y)}_{n \text{ instances of }(x\oplus y)}

    Now with commutativity and associativity we can simply rearrange terms to get

    n(x\oplus y)=\underbrace{x \oplus x \oplus \cdots \oplus x}_{n \text{ instances of } x} \oplus \underbrace{y \oplus y \oplus \cdots \oplus y}_{n \text{ instances of } y}

     = nx\oplus ny
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  3. #3
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    Why does n(x\oplus y) mean (x\oplus y)\oplus (x\oplus y)....? I don't see how you deciphered that from what was giving.
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  4. #4
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    Quote Originally Posted by dwsmith View Post
    Why does n(x\oplus y) mean (x\oplus y)\oplus (x\oplus y)....? I don't see how you deciphered that from what was giving.
    Well 1x = x

    2x = 1x \oplus x = x \oplus x

    3x = 2x \oplus x = (x \oplus x) \oplus x = x \oplus x \oplus x

    ...
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  5. #5
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    nx=(n-1+1)x=(n-1)x\oplus x=(n-2+1)x\oplus x=(n-2)x\oplus x \oplus x.....
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  6. #6
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    Quote Originally Posted by dwsmith View Post
    nx=(n-1+1)x=(n-1)x\oplus x=(n-2+1)x\oplus x=(n-2)x\oplus x \oplus x.....
    This relies on

    (n_1+n_2)x = n_1x \oplus n_2x

    which I suppose is obvious since it's just the \oplus version of (iii), nevertheless it still requires proof.

    The way I wrote it down, weak induction could be used for a formal proof.

    I'm curious as to your obvious proof of (iii), since I would have used

    x^n = \underbrace{x \odot x \odot \cdots \odot x}_{n \text{ instances of } x}

    from which (iii) follows trivially.
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  7. #7
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    x^{k+1}=x^k\odot x

    This is why 3 is obvious.
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