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= Problem 5  =
  
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Clearly&nbsp;<math>\mathbb{Q}(\alpha)\subset \mathbb{Q}(2^{\frac{1}{3}})</math>. But also,
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<math>2\alpha - \alpha = 16-17 \cdot 2^{\frac{1}{3}}</math>
  
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So <math>\mathbb{Q}(2^{\frac{1}{3}}) \subset \mathbb{Q}(\alpha)</math> and <math>\mathbb{Q}(2^{\frac{1}{3}}) = \mathbb{Q}(\alpha)</math>.
  
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Thus <math>3=|\mathbb{Q}(2^{\frac{1}{3}}):\mathbb{Q}| = |\mathbb{Q}(\alpha):\mathbb{Q}|</math>.
  
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Latest revision as of 04:11, 3 July 2013


Problem 5

Clearly $ \mathbb{Q}(\alpha)\subset \mathbb{Q}(2^{\frac{1}{3}}) $. But also,

$ 2\alpha - \alpha = 16-17 \cdot 2^{\frac{1}{3}} $

So $ \mathbb{Q}(2^{\frac{1}{3}}) \subset \mathbb{Q}(\alpha) $ and $ \mathbb{Q}(2^{\frac{1}{3}}) = \mathbb{Q}(\alpha) $.

Thus $ 3=|\mathbb{Q}(2^{\frac{1}{3}}):\mathbb{Q}| = |\mathbb{Q}(\alpha):\mathbb{Q}| $.


Back to NinjaSharksSet5

Alumni Liaison

Ph.D. 2007, working on developing cool imaging technologies for digital cameras, camera phones, and video surveillance cameras.

Buyue Zhang