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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.

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  • $\begingroup$ For the inverse $x$ of $u$, we want $u+x+2=-2$. $\endgroup$ Commented Feb 11, 2014 at 1:25
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    $\begingroup$ Excuse me: what $\;x\;$ ?? $\endgroup$
    – DonAntonio
    Commented Mar 10, 2014 at 4:52

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This vector space is nice, never encountered such an example.

If you found that the zero vector is $-2$, the additive inverse is not too far away. You actually don't need to care about scalar multiplication for this part.

The problem of finding the additive inverse is to find a function $i:V\to V$ such that for every $u\in V$, you have $u\boxplus i(u)=e$, where $e$ is the zero vector of $(V,\boxplus,\boxdot)$.

So the first step is to find the zero vector, which you claim you had. Just to be sure, I show here how it is found: for all $u\in V$, we want $u\boxplus e=u$. This means for all $u\in \mathbb R$, we have $u+e+2=u$. We immediatly get $e=-2$, by choosing any $u$ (for instance it suffices to look at $u=0$).

Now we want to find the function $i$ which gives us the additive inverse of any vector of $V$. Given $u$, we need to find $i(u)$ such that $u\boxplus i(u)=e$. This last equation translates $u+i(u)+2=-2$. we solve the equation for $i(u)$, and we get $i(u)=-4-u$, for all $u\in V$. So this shows that the additive inverse in $V$ is given by the formula $i(u)=-4-u$. Notice that for instance $i(e)=-4-(-2)=-4+2=-2=e$, which is normal: zero is always its own inverse.

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  • $\begingroup$ What exactly is u'? $\endgroup$ Commented Feb 11, 2014 at 1:30
  • $\begingroup$ the additive inverse of $u$. $\endgroup$
    – Denis
    Commented Feb 11, 2014 at 1:30
  • $\begingroup$ Can you explain a little more? I'm completely lost. $\endgroup$ Commented Feb 11, 2014 at 1:32
  • $\begingroup$ ok I update the answer with more details $\endgroup$
    – Denis
    Commented Feb 11, 2014 at 1:34
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    $\begingroup$ @KnowledgeGeek: You state in the Question that you found the "zero vector". The zero vector is the same as the additive identity, which is -2. Thus the additive inverse should provide a "vector" that "added" to $u$ gives back $-2$ (in terms of this weirdly defined operations). $\endgroup$
    – hardmath
    Commented Mar 10, 2014 at 5:55
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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 .

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