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Thread: Check if ive done right please binary operations

  1. #1
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    Check if ive done right please binary operations

    An operation * is defined on a set of numbers S by x * y = x + y - 2(x^2)(y^2)


    in z+, q and q+

    for z + its not a binary operation because you can get a negative answer say x = 3 and y = 4

    for q it is a binary operation

    for q+ it is a binary operation
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by adam_leeds View Post
    An operation * is defined on a set of numbers S by x * y = x + y - 2(x^2)(y^2)


    in z+, q and q+

    for z + its not a binary operation because you can get a negative answer say x = 3 and y = 4

    for q it is a binary operation

    for q+ it is a binary operation
    You are correct to note that $\displaystyle *:S^2\mapsto S$ so when $\displaystyle S=\mathbb{N}$ we ave that $\displaystyle *(3,4)\notin\mathbb{N}$. But, why doesn't that exact same example work when $\displaystyle S=\mathbb{Q}^+$?
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  3. #3
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    Quote Originally Posted by adam_leeds View Post
    An operation $\displaystyle *$ is defined on a set of numbers $\displaystyle S$ by $\displaystyle x * y = x + y - 2x^2y^2$


    in $\displaystyle \mathbb{Z}^+$, $\displaystyle \mathbb{Q}$ and $\displaystyle \mathbb{Q}^+$

    for $\displaystyle \mathbb{Z}^+$ its not a binary operation because you can get a negative answer say $\displaystyle x = 3$ and $\displaystyle y = 4$

    for $\displaystyle \mathbb{Q}$ it is a binary operation

    for $\displaystyle \mathbb{Q}^+$ it is a binary operation
    Since $\displaystyle (3,4)\in\mathbb{Q}^+\times\mathbb{Q}^+$, then clearly $\displaystyle \mathbb{Q}^+\times\mathbb{Q}^+$ is not a valid domain for the above function into $\displaystyle \mathbb{Q}^+$.

    However, $\displaystyle \mathbb{Q}\times\mathbb{Q}$ is a valid domain into $\displaystyle \mathbb{Q}$.
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by hatsoff View Post
    Since $\displaystyle (3,4)\in\mathbb{Q}^+\times\mathbb{Q}^+$, then clearly $\displaystyle \mathbb{Q}^+\times\mathbb{Q}^+$ is not a valid domain for the above function into $\displaystyle \mathbb{Q}^+$.

    However, $\displaystyle \mathbb{Q}\times\mathbb{Q}$ is a valid domain into $\displaystyle \mathbb{Q}$.
    I'm sorry. Did I say something incorrect?
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  5. #5
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    Quote Originally Posted by Drexel28 View Post
    I'm sorry. Did I say something incorrect?
    Oh no. It just takes me longer to type than you, apparently, so that I logged my response after yours even though we set out at approximately the same time.
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  6. #6
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    Sorry i dont understand why q+ is not a binary operation. I thought q meant fractions. Why are you using 3 and 4?
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  7. #7
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by adam_leeds View Post
    Sorry i dont understand why q+ is not a binary operation. I thought q meant fractions. Why are you using 3 and 4?
    3 and 4 are fractions:

    $\displaystyle 3 = \frac{3}{1}$,

    $\displaystyle 4 = \frac{4}{1}$.
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  8. #8
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    Quote Originally Posted by Swlabr View Post
    3 and 4 are fractions:

    $\displaystyle 3 = \frac{3}{1}$,

    $\displaystyle 4 = \frac{4}{1}$.
    woops
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