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Math Help - Subring of R

  1. #1
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    Subring of R

    Let S be the set of all elements of the form a+(b*cubed root of 2)+(c*cubed root of 4) with a,b,c rationals. Show that S is a subring of R.

    My book does not require multiplicative identity.

    I think I can handle showing the zero element is there, and closure under addition, but I'm having trouble with closure under multiplication and additive identity. Thanks.
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by zhupolongjoe View Post
    Let S be the set of all elements of the form a+(b*cubed root of 2)+(c*cubed root of 4) with a,b,c rationals. Show that S is a subring of R.

    My book does not require multiplicative identity.

    I think I can handle showing the zero element is there, and closure under addition, but I'm having trouble with closure under multiplication and additive identity. Thanks.
    I'm assuming that this says S=\left\{a+b\sqrt[3]{2}+c\sqrt[3]{4}:a,b,c\in\mathbb{Q}\right\}. Closure under mult. is just a lot of work. But, what do you mean by additive identity? Isn't the zero element the additive identity?
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  3. #3
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    Yes, that is the way it looks. Sorry, I meat additive inverse, not additive identity.
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by zhupolongjoe View Post
    Yes, that is the way it looks. Sorry, I meat additive inverse, not additive identity.
    a+b\sqrt[3]{2}+c\sqrt[3]{4}+-a+-b\sqrt[3]{2}+-c\sqrt[3]{4}=0?
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  5. #5
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    I didn't think it would be that easy lol, but thanks...the multiplication is just troublesome.
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  6. #6
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by zhupolongjoe View Post
    I didn't think it would be that easy lol, but thanks...the multiplication is just troublesome.
    Wolfram|Alpha
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