How would i go around proving a(u+v)+ u= (a+1)u+av using axioms? I started with distributivity with respect to vector addition, associativity of addition, commutativity of addition, distributivity with respect to field addition. I think i am missing one axiom.
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1$\begingroup$ What are these objects? If you mean $a$ is a field element and $u,v$ are in some vector space, then this is usually false since the LHS is $au+av$ and the RHS is $au+av+u$, so unless $u=\mathbf{0}$ you're SOL. $\endgroup$– Adam HughesOct 11, 2014 at 20:25
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$\begingroup$ Now you are right $\endgroup$– Ri-LiOct 11, 2014 at 20:34
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$\begingroup$ @AdamHughes changed it now $\endgroup$– BOBOct 11, 2014 at 20:34
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$\begingroup$ Sure i haven't missed one. $\endgroup$– BOBOct 11, 2014 at 20:35
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1 Answer
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distributive law says $(a+1)\mathbf{u}=a\mathbf{u}+1\cdot \mathbf{u}$. Identity law says that $1\cdot \mathbf{u}=\mathbf{u}$.
As such you see that
$$(a+1)\mathbf{u}+a\mathbf{v}=(a\mathbf{u}+\mathbf{u})+a\mathbf{v}$$ $$=a\mathbf{u}+(\mathbf{u}+a\mathbf{v}) = a\mathbf{u}+(a\mathbf{v}+\mathbf{u})$$ $$=(a\mathbf{u}+a\mathbf{v})+\mathbf{u} =a(\mathbf{u}+\mathbf{v})+\mathbf{u}$$
answered Oct 11, 2014 at 20:38