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Suppose u₁ and u2 both have the property of being an additive inverse of v. That is, for this arbitra particular vector v, u₁ + v = 0 and u₂ + v= 0. Supply a reason for each line of the string of equ proving that u₁ = u₂, thus proving that additive inverses are unique [and therefore -1 v is the ad inverse of v]. U₁=U₁+0 = U₁ + (u₂ + v) = U₁ + (v + U₂) (U₁ + v) + U₂ = = 0 + U₂ Select an answer [Select an answer Select an answer Select an answer Select an answer Soloct on answor

Suppose u₁ and u2 both have the property of being an additive inverse of v. That is, for this arbitra particular vector v, u₁ + v = 0 and u₂ + v= 0. Supply a reason for each line of the string of equ proving that u₁ = u₂, thus proving that additive inverses are unique [and therefore -1 v is the ad inverse of v]. U₁=U₁+0 = U₁ + (u₂ + v) = U₁ + (v + U₂) (U₁ + v) + U₂ = = 0 + U₂ Select an answer [Select an answer Select an answer Select an answer Select an answer Soloct on answor

Algebra and Trigonometry (6th Edition)
Algebra and Trigonometry (6th Edition)
6th Edition
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
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Suppose ₁ and ₂ both have the property of being an additive inverse of v. That is, for this arbitrary but particular vector v, u₁ + v = 0 and u₂ + v = 0. Supply a reason for each line of the string of equalities proving that u₁ = u₂, thus proving that additive inverses are unique [and therefore -1 v is the additive inverse of v]. Select an answer = U₁ + (U₂ + v) Select an answer = U₁ + (v + U₂) [Select an answer = (U₁ + v) + U₂ Select an answer = 0 + U₂ Select an answer Select an answer = U2 U₁ = U₁ + 0
Transcribed Image Text:Suppose ₁ and ₂ both have the property of being an additive inverse of v. That is, for this arbitrary but particular vector v, u₁ + v = 0 and u₂ + v = 0. Supply a reason for each line of the string of equalities proving that u₁ = u₂, thus proving that additive inverses are unique [and therefore -1 v is the additive inverse of v]. Select an answer = U₁ + (U₂ + v) Select an answer = U₁ + (v + U₂) [Select an answer = (U₁ + v) + U₂ Select an answer = 0 + U₂ Select an answer Select an answer = U2 U₁ = U₁ + 0
Suppose u₁ and u₂ both have the property of being an additive inverse of v. That is, for this arbitrary but particular vector v, u₁ + v = 0 and u₂ + v= 0. Supply a reason for each line of the string of equalities proving that u₁ = U₂, thus proving that additive inverses are unique [and therefore -1 v is the additive inverse of v]. U₁ = U₁ + 0 = U₁ + (U₂ + v) = U₁ + (v + U₂) (U₁ + v) + U₂ = 0 + U₂ = U12 = Check Answer ✓ Select an answer assumption of the claim vector space property 1: closure under addition vector space property 2: vector addition is commutative vector space property 3: vector addition is associative vector space property 4: property of the additive identity vector space property 5: existence of additive inverses vector space property 6: closure under scalar multiplication vector space property 7: property of the multiplicative identity, 1 vector space property 8: distribution of scalars over vectors vector space property 9: distribution of vectors over scalars vector space property 10: associativity of scalar multiplication
Transcribed Image Text:Suppose u₁ and u₂ both have the property of being an additive inverse of v. That is, for this arbitrary but particular vector v, u₁ + v = 0 and u₂ + v= 0. Supply a reason for each line of the string of equalities proving that u₁ = U₂, thus proving that additive inverses are unique [and therefore -1 v is the additive inverse of v]. U₁ = U₁ + 0 = U₁ + (U₂ + v) = U₁ + (v + U₂) (U₁ + v) + U₂ = 0 + U₂ = U12 = Check Answer ✓ Select an answer assumption of the claim vector space property 1: closure under addition vector space property 2: vector addition is commutative vector space property 3: vector addition is associative vector space property 4: property of the additive identity vector space property 5: existence of additive inverses vector space property 6: closure under scalar multiplication vector space property 7: property of the multiplicative identity, 1 vector space property 8: distribution of scalars over vectors vector space property 9: distribution of vectors over scalars vector space property 10: associativity of scalar multiplication
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