Question and Answers Forum

All Questions      Topic List

Others Questions

Previous in All Question      Next in All Question      

Previous in Others      Next in Others      

Question Number 61840 by psyche last updated on 10/Jun/19

consider the triple of real numbers (x,y,z) defined by the addittion (x,y,z)+(x′,y′,z′)=(x+x′,y+y′,z+z′) and scalar multiplication by 𝛂(x,y,z)=(0,0,0). Show that all axioms for a vector space are satisfied except axiom 8.

considerthetripleofrealnumbers(x,y,z)definedbytheaddittion(x,y,z)+(x′,y′,z′)=(x+x′,y+y′,z+z′)andscalarmultiplicationbyα(x,y,z)=(0,0,0).Showthatallaxiomsforavectorspacearesatisfiedexceptaxiom8.

Answered by arcana last updated on 10/Jun/19

if ∀(x,y,z)∈R^3 1∙(x,y,z)=(0,0,0)=(x,y,z) iff x=0, y=0,z=0

if∀(x,y,z)∈R31⋅(x,y,z)=(0,0,0)=(x,y,z)iffx=0,y=0,z=0

Commented by psyche last updated on 10/Jun/19

please, complete the proof

please,completetheproof

Commented by arcana last updated on 10/Jun/19

axiom 1,2,3,4 define (R^3 ,“+”) is a structure of group with sum usual in R^3 we need a field K because define ∙ operation. elements in K are scalars, elements inR^3 are vectors if R=K (field) ∙:R×R^3 →R^3 (α,a)→α∙a axiom5 α,β∈R,(x,y,z)∈R^3 α+β∈R ⇒(α+β)∙(x,y,z)=(0,0,0) def. “∙” β∙(x,y,z)=(0,0,0) ; β∙(x,y,z)=(0,0,0) ⇒(α+β)∙(x,y,z)=α∙(x,y,z)+β∙(x,y,z) axiom 6 α∈R,(x,y,z),(x′,y′,z′)∈R^3 α∙[(x,y,z)+(x′,y′,z′)]=α∙(x+x′,y+y′,z+z′) (x+x′,y+y′,z+z′)∈R^3 ⇒α∙(x+x′,y+y′,z+z′)=(0,0,0) α∙[(x,y,z)+(x′,y′,z′)]α∙(x,y,z)=(0,0,0) ; α∙(x′,y′,z′)=(0,0,0) ⇒α∙[(x,y,z)+(x′,y′,z′)]=α∙(x,y,z)+α∙(x′,y′,z′) axiom 7 α,β∈R,(x,y,z)∈R^3 .αβ∈R (αβ)∙(x,y,z)=(0,0,0) β∙(x,y,z)=(0,0,0) ⇒α∙[β∙(x,y,z)]=α∙(0,0,0)=(0,0,0) ⇒(αβ)∙(x,y,z)=α∙[β∙(x,y,z)]

axiom1,2,3,4define(R3,‘‘+″)isastructureofgroupwithsumusualinR3weneedafieldKbecausedefine⋅operation.elementsinKarescalars,elementsinR3arevectorsifR=K(field)⋅:R×R3→R3(α,a)→α⋅aaxiom5α,β∈R,(x,y,z)∈R3α+β∈R⇒(α+β)⋅(x,y,z)=(0,0,0)def.‘‘⋅″β⋅(x,y,z)=(0,0,0);β⋅(x,y,z)=(0,0,0)⇒(α+β)⋅(x,y,z)=α⋅(x,y,z)+β⋅(x,y,z)axiom6α∈R,(x,y,z),(x′,y′,z′)∈R3α⋅[(x,y,z)+(x′,y′,z′)]=α⋅(x+x′,y+y′,z+z′)(x+x′,y+y′,z+z′)∈R3⇒α⋅(x+x′,y+y′,z+z′)=(0,0,0)α⋅[(x,y,z)+(x′,y′,z′)]α⋅(x,y,z)=(0,0,0);α⋅(x′,y′,z′)=(0,0,0)⇒α⋅[(x,y,z)+(x′,y′,z′)]=α⋅(x,y,z)+α⋅(x′,y′,z′)axiom7α,β∈R,(x,y,z)∈R3.αβ∈R(αβ)⋅(x,y,z)=(0,0,0)β⋅(x,y,z)=(0,0,0)⇒α⋅[β⋅(x,y,z)]=α⋅(0,0,0)=(0,0,0)⇒(αβ)⋅(x,y,z)=α⋅[β⋅(x,y,z)]

Terms of Service

Privacy Policy

Contact: info@tinkutara.com