# Vector Space and Addition Inverse

Determine whether the set $\mathbb{R}^2$, with the operations $(x_1,y_1)+(x_2,y_2)=(x_1x_2,y_1y_2)$ and $c(x_1,y_1)=(cx_1,cy_1)$, is a vector space. If it is, verify each vector space axiom; if not, state all vector space axioms that fail.

I know this set is not a vector space under the specified operations since axioms like distributive multiplication do not hold true, but my problem is that the book uses the below argument to deduce that it's not a vector space:

"The set is not a vector space. It fails because $(1,1)$ is the additive identity, so $(0,0)$ has no additive inverse."

If we pick vector $u=(0,0)$ to be in the set $\mathbb{R}^2$, then and according to the book's solution, $u$ has no additive inverse. Why is this the case? I mean isn't $(0,0)$ in this question the additive inverse of all vectors in the set $\mathbb{R}^2$ including $u$? Is this related to

"The additive inverse of $u$ is unique. That is, if $u+v=0$, then $v =-u$."

And what is meant by "unique" in the previous statement?

I think (although I might be wrong) that you're confused with what the word "additive" refers to. In the context of this question, "additive" refers not to the usual addition of vectors, but to the operation defined in this question. In a vector space $V$, the definition of the inverse element (which, by the way, has to be given only after the identity element has been defined) states:

For an element $v\in V$, its inverse element is the element $v'\in V$ such that $v+v'=v'+v=\mathrm{Id}$.

In this example we've already established that $(1,1)\in\mathbb{R}^2$ would have to be the identity element. But then for any $v=(x,y)\in\mathbb{R}^2$ we have $$(0,0)+(x,y)=(0x,0y)=(0,0)\neq(1,1)=\mathrm{Id}.$$ This computation shows two things for us:

• $(0,0)$ does not have an additive inverse with respect to this operation because adding any element to it never results in the identity element;
• The claim that "$(0,0)$ in this question is the additive inverse of all vectors in the set $\mathbb{R}^2$" is not true for the same reason.

EDIT. I forgot to answer your last question. The word "unique" means that there's only one such element: if $v'$ and $v''$ are two additive inverses of the same element $v\in V$, then necessarily $v'=v''$. This uniqueness property is not required in the axioms of a vector space, but it's easily deduced from them, so the bottom line is that it always holds in a vector space. I don't feel like this property is directly related to your task here, since you're asked to verify the axioms, not other properties of vector spaces.

An additive inverse'' of some $u$ is another element such that $u + v$ equals the identity. The key point is that $(0,0)$ is no longer the identity.

It looks like your book is using the symbol '$0$' to refer to the identity, which might be confusing you. In the current setting '$0$' is the element $(1,1)$, and not $(0,0)$.

So even though there exist several elements $v$ such that $(0,0) + v = (0,0)$, the element $(0,0)$ does not have an inverse since there is no $v$ such that $$(0,0) + v = (1,1).$$

For the additive inverse of some $u$ to be unique, there would have to be one and only one $v$ such that $u + v$ were equal to the identity. But the problem here is that there are no such $v$, not that there are more than one.

By the definition of addition given above, $(0,0) + (x,y) = (1,1)$ implies that $(0,0) = (1,1)$. In other words, there is no pair $u \in \mathbb{R}^2$ such that $(0,0) + u = (1,1)$, thus $(0,0)$ has no additive inverse.