(d) Show that Axiom 5 holds by producing an ordered pair- u such that u + (-u) = 0 for u = (u1, u2). Solution: If u = (u1, u2), then - u= ( satisfies u + (- u) = ( , ) = 0; thus Axiom 5 holds. %3D %3D (e) Find two vector space axioms that fail to hold. Solution: Axioms 7 fail to hold. For example: ) = ( ku +kv = (ku1, ku2) + (kv1, kv2) = ( k(u + v) + ku +kv, k 1. Axioms 8 fail to hold. For example: (k+1)(u1, u2) =( ku + lu = (ku1, ku2) + (lu1, lu2) = ( k(u + v) = k( %3D (k+l)u
(d) Show that Axiom 5 holds by producing an ordered pair- u such that u + (-u) = 0 for u = (u1, u2). Solution: If u = (u1, u2), then - u= ( satisfies u + (- u) = ( , ) = 0; thus Axiom 5 holds. %3D %3D (e) Find two vector space axioms that fail to hold. Solution: Axioms 7 fail to hold. For example: ) = ( ku +kv = (ku1, ku2) + (kv1, kv2) = ( k(u + v) + ku +kv, k 1. Axioms 8 fail to hold. For example: (k+1)(u1, u2) =( ku + lu = (ku1, ku2) + (lu1, lu2) = ( k(u + v) = k( %3D (k+l)u
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter1: Fundamental Concepts Of Algebra
Section1.1: Real Numbers
Problem 38E
Related questions
Question
Please solve question ( d, e)
Transcribed Image Text:Fill up the blanks
Let V be the set of all ordered pairs of real numbers, and consider the following addition and
scalar multiplication operations on u (u1, u2) and v = (v1, v2):
u + v = (u1 + v1 - 1, u2 + v2 - 1), ku = (ku1, ku2)
(a) Compute u + v and ku for u = (1, 2), v = (– 1,3) and k = 2.
Solution:
u + v =
ku =
(b) Show that (0, 0) # 0.
Solution : We have (u1, u2) + (0,0) = (
u in V.
); thus u + (0,0) u for every
) = 0.
(c) Show that (
Solution: We have (u1, u2) + (
thus u + (
)3 (u1, u2);
) = (u1
)= u for every u in V.
, u2
u such that u+ (- u) = 0
(d) Show that Axiom 5 holds by producing an ordered pair
for u = (u1, u2).
Solution: If u = (u1, u2), then
satisfies u + (- u) = (
u=
) = 0;
thus Axiom 5 holds.
(e) Find two vector space axioms that fail to hold.
Solution: Axioms 7 fail to hold. For example:
k(u + v) = k(
) = (
ku +kv = (ku1, ku2) + (kvi, kv2) = (
k(u + v) + ku +kv, k#1.
Axioms 8 fail to hold. For example:
(k + 1)u = (k+1)(u1, u2) = (
ku + lu = (ku1, ku2) + (lu1, lu2) = (
(k +l)u + ku + lu.
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