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Question Number 61840 by psyche last updated on 10/Jun/19
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)∈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,completetheproof
Commented by arcana last updated on 10/Jun/19
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)]
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