TITLE: Finding the additive inverse in a vector space with unusual operations
QUESTION [5 upvotes]: Let $V=\mathbb{R}$. For $u,v\in V$ and $a\in\mathbb{R}$, define vector addition by $$u\boxplus v:=u+v+2$$
  and scalar multiplication by
  $$a\boxdot u:=au+2a−2.$$
  It can be shown that $(V,\boxplus,\boxdot)$ is a vector space over the scalar field $\mathbb{R}$. Find the following...

I have found the sum, scalar multiple and the zero vector. I don't know where the x is coming from, let alone how to find the additive inverse. Can someone assist me? Thanks.

REPLY [2 votes]: The additive inverse is defined when the identity element for the set is known like here +(u,v) = u + v + 2 so, we find +(u,x) = u where in x must be in Vspace so u+x+2 = u we get x = -2 hence, it is in Vspace.
So now identity element is -2 so now finding additive inverse goes this way
+(u,y) = -2 where y is additive inverse for u so by by performing the operation we get y = -4 - u .