2. Let V be the set of all real numbers with the two operations O and O defined by u Ov=u+v - 2 and kOu = ku + (1+ k) (a) Compute 201 Solution : 201 = (b) Compute 083 Solution : 083 = (c) Verify Ariom 4: Solution : We have u( ) = %3D = u for every u = u in V; thus the number 1 plays role of the zero vector in V; so 0 = (d) Verify Axiom 5: Solution : For each u = u in V, we have u serves as the negative of u = ) = u + ( )- 2 = 0; thus the number u in V. u = (e) Verify Axiom 7: Solution : kO(uÐv) = k®(u+ v- 2) = %3D %3D but kOu O kOv %3D %3D so ko(uOv) = kOuOkOv (f)Verify Ariom 8 : Solution : (k +1)Ou = but kOu 1Ou = so (k +1)Ou= kOuIOu (g) Verify Ariom 9: Solution : kO(1®u) = k® %3D %3D but (kl)Ou = so, kO(1Ou) = (kl)®u.
2. Let V be the set of all real numbers with the two operations O and O defined by u Ov=u+v - 2 and kOu = ku + (1+ k) (a) Compute 201 Solution : 201 = (b) Compute 083 Solution : 083 = (c) Verify Ariom 4: Solution : We have u( ) = %3D = u for every u = u in V; thus the number 1 plays role of the zero vector in V; so 0 = (d) Verify Axiom 5: Solution : For each u = u in V, we have u serves as the negative of u = ) = u + ( )- 2 = 0; thus the number u in V. u = (e) Verify Axiom 7: Solution : kO(uÐv) = k®(u+ v- 2) = %3D %3D but kOu O kOv %3D %3D so ko(uOv) = kOuOkOv (f)Verify Ariom 8 : Solution : (k +1)Ou = but kOu 1Ou = so (k +1)Ou= kOuIOu (g) Verify Ariom 9: Solution : kO(1®u) = k® %3D %3D but (kl)Ou = so, kO(1Ou) = (kl)®u.
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 O defined by
u O v =u+ v - 2 and k&u = ku + (1 +k)
(a) Compute 2i
Solution : 24O13=
(b) Compute 083
Solution : 003 =
(c) Verify Ariom 4:
Solution : We have u( ) =
= u for every u = u in V; thus the number 1 plays
role of
the zero vector in V; so 0 =
(d) Verify Axiom 5:
Solution : For each u = u in V, we have
) = u + (
serves as the negative of u = u in V.
)- 2 = 0; thus the number
u =
(e) Verify Axiom 7:
Solution : k(uÐv) = kO(u + v – 2) =
but
kOu O kOv =
so ko(uOv) = kOuOkOv
(f) Verify Ariom 8 :
Solution : (k +1)Qu =
but
so
(k +1)Ou = kOu IOu
(g) Verify Ariom 9
Solution : k&(1Qu) = kO
but
(kl)Ou =
kO(18u) = (kl)Ou.
80,
*COPYRIGHT
64°F
rch
Expert Solution
Step 1
"Since you have posted a question with multisubparts, we will solve the first three subparts for
you. To get remaining subparts solved, please repost the complete question and mention the
subparts to be solved."
It is given that the binary operation in the set V of real numbers are and such that
and for all the vectors and the scalars k.
Step by step
Solved in 2 steps
Recommended textbooks for you
Advanced Engineering Mathematics
Advanced Math
ISBN:
9780470458365
Author:
Erwin Kreyszig
Publisher:
Wiley, John & Sons, Incorporated
Numerical Methods for Engineers
Advanced Math
ISBN:
9780073397924
Author:
Steven C. Chapra Dr., Raymond P. Canale
Publisher:
McGraw-Hill Education
Introductory Mathematics for Engineering Applicat…
Advanced Math
ISBN:
9781118141809
Author:
Nathan Klingbeil
Publisher:
WILEY
Advanced Engineering Mathematics
Advanced Math
ISBN:
9780470458365
Author:
Erwin Kreyszig
Publisher:
Wiley, John & Sons, Incorporated
Numerical Methods for Engineers
Advanced Math
ISBN:
9780073397924
Author:
Steven C. Chapra Dr., Raymond P. Canale
Publisher:
McGraw-Hill Education
Introductory Mathematics for Engineering Applicat…
Advanced Math
ISBN:
9781118141809
Author:
Nathan Klingbeil
Publisher:
WILEY
Mathematics For Machine Technology
Advanced Math
ISBN:
9781337798310
Author:
Peterson, John.
Publisher:
Cengage Learning,
