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Math Help - Tower Law problem

  1. #1
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    Tower Law problem

    Show [\mathbb{Q}(\alpha):\mathbb{Q}]=2 where \alpha = \sqrt{4+2\sqrt{3}}

    This is what i have.

    [\mathbb{Q}(\sqrt{4+2\sqrt{3}}):\mathbb{Q}]=[\mathbb{Q}(\sqrt{4+2\sqrt{3}}):\mathbb{Q}\sqrt{3}] [\mathbb{Q}\sqrt{3}:\mathbb{Q}]

    [\mathbb{Q}(\sqrt{4+2\sqrt{3}}):\mathbb{Q}] = 2*2 = 4
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  2. #2
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    Quote Originally Posted by frankdent1 View Post
    Show [\mathbb{Q}(\alpha):\mathbb{Q}]=2 where \alpha = \sqrt{4+2\sqrt{3}}

    This is what i have.

    [\mathbb{Q}(\sqrt{4+2\sqrt{3}}):\mathbb{Q}]=[\mathbb{Q}(\sqrt{4+2\sqrt{3}}):\mathbb{Q}\sqrt{3}] [\mathbb{Q}\sqrt{3}:\mathbb{Q}]

    [\mathbb{Q}(\sqrt{4+2\sqrt{3}}):\mathbb{Q}] = 2*2 = 4
    Hint: \sqrt{4+2\sqrt{3}}=\sqrt{3}+ 1.
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