This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Homework 9 Solutions 1. ( + 1 i + 2 j + 3 k ) 2 = ( 2 2 1 2 2 2 3 ) + 2 1 i + 2 2 j + 2 3 k Hence, ( + 1 i + 2 j + 3 k ) 2 = 1 2 2 1 2 2 2 3 = 1 and 1 = 2 = 3 = 0 2 1 + 2 2 + 2 3 = 1 (and = 0). Thus there are infinitely many solutions (1 corresponding to each point on the unit sphere). 2. Call the proposed field R , to simplify matters. additive identity = 0 0 0 0 R commutative under addition since R is. closed under addition since a + c b + d b d a + c R addition is associative  clear multiplicative identity = 1 0 0 1 R for A R , det ( A ) = a 2 + b 2 = 0 a = b = 0 thus every element has an inverse except the additive identity (as desired). a b b a c d d c = ac bd ad + bc ad bc ac bd = c d d c a b b a So commuta tive under multiplication....
View
Full
Document
This note was uploaded on 04/19/2011 for the course MATH 21373 taught by Professor Johnson during the Spring '11 term at Carnegie Mellon.
 Spring '11
 Johnson
 Algebra

Click to edit the document details