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Math Help - Closure

  1. #1
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    Smile Closure

    hi
    would you help me find out if those sets are closed under addition and scalar product or not.
    A_{1}=\left \{ (cosx,sinx)/x\in \mathbb{R} \right \}
    A_{2}=\left \{ (x,y)/x\in \mathbb{Z},y\in \mathbb{Z} \right \}
    for the second i think \mathbb{Z} is closed under addition,  \left (\forall x\in \mathbb{Z}  \right ),\left ( \forall y\in \mathbb{Z} \right ); \left (x+y  \right )\in \mathbb{Z}.am i right ?
    thanks for helping me.
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  2. #2
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    Quote Originally Posted by Raoh View Post
    hi
    would you help me find out if those sets are closed under addition and scalar product or not.
    A_{1}=\left \{ (cosx,sinx)/x\in \mathbb{R} \right \}
    A_{2}=\left \{ (x,y)/x\in \mathbb{Z},y\in \mathbb{Z} \right \}
    for the second i think \mathbb{Z} is closed under addition,  \left (\forall x\in \mathbb{Z} \right ),\left ( \forall y\in \mathbb{Z} \right ); \left (x+y \right )\in \mathbb{Z}.am i right ?
    thanks for helping me.
    Well A_1 is clearly not closed under multiplication by scalar: just remember that |\cos{x}|\leq{1}
    Try to do the second one by yourself

    Tonio
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  3. #3
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    Smile

    for the second one, \mathbb{Z} isn't closed under scalar product,but it's closed under addition. \left (\forall a\in \mathbb{Z}  \right ),\left (\exists \lambda \in \mathbb{R}  \right );\lambda a\notin \mathbb{Z},is it okay ?
    thanks a lot.
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  4. #4
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    Quote Originally Posted by Raoh View Post
    for the second one, \mathbb{Z} isn't closed under scalar product,but it's closed under addition. \left (\forall a\in \mathbb{Z} \right ),\left (\exists \lambda \in \mathbb{R} \right );\lambda a\notin \mathbb{Z},is it okay ?
    thanks a lot.

    Exactly. Anyway, none of them is a subspace.

    Tonio
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  5. #5
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    Smile

    thanks a lot.
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