2. Let V be the set of all real numbers with the two operations and defined by u v=u+v- 2 and ku = ku+ (1+ k) (a) Compute 201 Solution : 201 = (b) Compute 083 Solution : 083 = (c) Verify Ariom 4: Solution : We have u ) = role of %3! %3D = u for every u = u in V; thus the nu: the zero vector in V; so 0 = (d) Verify Axiom 5: Solution : For each u = u in V, we have %3D 0: thus the number
2. Let V be the set of all real numbers with the two operations and defined by u v=u+v- 2 and ku = ku+ (1+ k) (a) Compute 201 Solution : 201 = (b) Compute 083 Solution : 083 = (c) Verify Ariom 4: Solution : We have u ) = role of %3! %3D = u for every u = u in V; thus the nu: the zero vector in V; so 0 = (d) Verify Axiom 5: Solution : For each u = u in V, we have %3D 0: thus the number
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
Transcribed Image Text:2. Let V be the set of all real numbers with the two operations and defined by
u Ov=u+v – 2 and kOu = ku + (1+ k)
(a) Compute 201
Solution : 2Ð1 =
(b) Compute 083
Solution : 083 =
(c) Verify Axiom 4:
Solution : We have u( ) =
%3D
= u for every u = uin V; thus the number 1 plays
role of
the zero vector in V; so 0 =
(d) V erify Axiom 5:
Solution : For each u = u in V, we have
uÐ
serves as the negative of u = u in V.
) = u + ( )– 2 = 0; thus the number
%3D
(e) Verify Axiom 7 :
Solution : kO(uÐv) = k®(u+ v – 2) =
but
kOu O kOv
so ko(uÐv) = k®uOkOv
(f) Verify Axiom 8 :
Solution : (k +1)Ou =
%3D
but
so
(k +1)Qu= k&u 1Ou
(g) Verify Axiom 9 :
Solution : kO(1Ou) = k®
%3D
%3D
%3D
but
(kl)Ou =
k®(1Ou) = (kl)®u.
%3D
so,
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