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# hmw2s - 3 If every real number was algebraic then R = A ∩...

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18.100B, Fall 2002 Problem set 2, Solutions (1) Rudin, Chapter 2, Problem 2 An algebraic equation (1) a 0 z n + a 1 z n 1 + · · · + a n 1 z + a n = 0 in which at least one of the coeﬃcients a j is not zero, can have at most n complex solutions. For each integer N consider the set A N = z C ; z satisfies (1) for some integers a 0 , a 1 , . . . , a n , n with 1 ≤ | a 0 | + | a 1 | + · · · + | a n | + n N Now, A N is finite, since there are only finitely many equation here and each has only finitely many solutions. Furthermore, the set of algebraic numbers is ± A = A N , N =1 i.e. every algebraic number is in one of the A N ’s. Thus A is a countable union of finite sets, so is countable (it is not finite since the integers are clearly algebraic, as z = n satisfies z n = 0) . (2) Rudin, Chapter 2, Problem
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Unformatted text preview: 3 If every real number was algebraic then, R = A ∩ R would be countable. We have shown in class that R is not countable, so R ±⊂ A and hence there must be a non-algebraic real number; indeed there must be an uncountably inﬁnite set of them. (3) Rudin, Chapter 2, Problem 4 We have shown in class that the set of rational numbers, Q ⊂ R is countable. Since R is uncountable it cannot be equal to Q so there must exist irrational real numbers. 1...
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