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Unformatted text preview: f . That leaves the question: Is f one to one? The squares are a hint the the answer is no. To prove this is actually the case, we need to ﬁnd a vector r s t u v for which ( * ) has at least two solutions. To simplify things, let us decrease the number of variables, and see if we can do this with a = b = 0. (Why not? Try simple things ﬁrst. There is a square in each variable, so keeping one “active” should suﬃce.) We ﬁnd f c = c 2 21 /september/ 2005; 22:09 5...
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This note was uploaded on 10/03/2010 for the course MATH 380 taught by Professor Gladue during the Fall '08 term at Roger Williams.
 Fall '08
 Gladue
 Calculus

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