Date: 07/22/97 at 12:32:32 From: Cat Fyle Subject: Modern Algebra I have a test tomorrow in Modern Algebra, but the test is going to consist of complex numbers. I have not had complex numbers, so I have no idea where to begin. Here is one of the questions: Let E be a subset of P such that i) 1 is in E ii) whenever n is in E, also n+1 is in E. Then E = P. Prove the above statement. This is the principle of Mathematical Induction for well-ordered positive integers: Let P denote the set of all positive integers such that S <= P and S <> P, then there exists an integer m in S such that m <= n for all n in S. I would appreciate any help you could provide. Thank you very much, CAT
Date: 07/22/97 at 12:58:26 From: Doctor Rob Subject: Re: Modern Algebra Assume that E <> P. Let S be the set of integers in P but not in E. Apply the Principle of Mathematical Induction. Then show that the m that it produces leads to a contradiction, that there is an element of E which is larger than m by showing m+1 is in E. Then the assumption that E <> P must be wrong, so E = P. -Doctor Rob, The Math Forum Check out our web site! http://mathforum.org/dr.math/
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