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FieldAxioms - a Indeed-a =-a 0 by A3 =-a a b by =-a a b by...

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Math 25: Advanced Calculus Fall 2010 Examples using the Field Axioms We proved the following Lemmas in class. You are free to use them on any homework assignment you wish, but you must write out the statement of the lemma that you are using. Lemma 1: If F is a field, then 0 · x = 0 for all elements x F . Proof: We compute 0 = 0 · x + ( - (0 · x )) by A4 = (0 + 0) · x + ( - (0 · x )) by A3 = (0 · x + 0 · x ) + ( - (0 · x )) by AM1 = 0 · x + (0 · x + ( - (0 · x )) by A2 = 0 · x + 0 by A4 = 0 · x by A3 . Lemma 2: For any a F , the number - a is the unique number for which a +( - a ) = 0. Proof: Suppose that a + b = 0 . ( * ) We will show that b =
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Unformatted text preview: a . Indeed,-a =-a + 0 by A3 =-a + ( a + b ) by (*) = (-a + a ) + b by A2 = 0 + b by A4 = b by A3 . ± Lemma 3: For any a ∈ F , we have (-1) · a =-a . Proof: We compute 0 = 0 · a by Lemma 1 = (1 + (-1)) · a by A4 = 1 · a + (-1) · a by AM1 = a + (-1) · a by M3 . By Lemma 2, we know that (-a ) is the unique element of F with the property that a + (-a ) = 0. Since we have shown that a + (-1) · a = 0, it must be the case that (-a ) = (-1) · a . ± 1...
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