This preview shows page 1. Sign up to view the full content.
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...
View
Full
Document
This note was uploaded on 03/18/2012 for the course MAT MAT 25 taught by Professor Stevenklee during the Fall '10 term at UC Davis.
 Fall '10
 StevenKlee
 Calculus

Click to edit the document details