114 Groups and Rings 200809
Suggested solutions to exercise set 10
May 6, 2009
38.2 This is not a homomorphism since (1 + 1) = 4 = 2 = (1) + (1).
38.4 This is not a homomorphism since (1) = 1 = 1 = (1).
38.6 This is not a homomorphism since
([ 0 1 ][ 0 0
114 Groups and Rings 200809
Suggested solutions to exercise set 9
April 27, 2009
35.8 Let a(x) = 2x5 3x3 + 2x + 1. Then
a(3) = 2 (3)5 3 (3)3 + 2 (3) + 1 = 410,
which, by the Remainder Theorem, is the remainder when a(x) is divided by
x + 3 = x (3).
35.10
114 Groups and Rings 200809
Suggested solutions to exercise set 6
April 8, 2009
28.2 Observe that for any x, y Dp we have
x > y x (x + y) > y (x + y) y > x
(here we are using Theorem 28.2(d) to add (x + y) to both sides of the
inequality).
Since 0 > c, th