Concept guide
Real vector space: auditing axiom instances in ℝ³ and proving one distributive law
In ℝ³ take u=(1,−2,3), v=(4,0,−1), a=2, b=−3. (1) Confirm the dimensions agree; (2) check exactly the five axiom instances a(u+v)=au+av, (a+b)u=au+bu, (ab)u=a(bu), 1u=u and u+(−u)=0; (3) prove a(u+v)=au+av for every real a and all u, v∈ℝ³; (4) use a modified scalar multiplication on ℝ² to show that successful instances do not establish an axiom.