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Exercise 4.2

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Proof. No. Both $x^{m}$ and $- x^{m} + 1$ are in the set, but $x^{m} + (- x^{m} + 1) = 1$ has degree 0, therefore the set is not closed under addition. □

celiopassos

2017-10-06 00:00

Exercise 4.2

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