$$F = \left \{ (x,y,z) \in R^{3}: x+y+z = 1\right \}$$ with addition and multiplication defined as: $$(x,y,z) \boxplus (u,v,w) = (x+u-1, y+v, z+w)$$ $$a \boxdot (x,y,z) = (ax - a + 1, ay, az)$$ $$\forall (x,y,z), \space (u,v,w) \in F; \space \forall a \in \mathbb{R}$$
Show that F, with this operations, is a vector space under the real numbers.
I tried to check if the defined addition respects the commutative axiom. $$(x,y,z) \boxplus (u,v,w) = (x+u-1, y+v, z+w) = (u+x-1, v+y, w+z) = (u,v,w) \boxplus (x,y,z) $$
I'm not sure if this is correct.