do irregular multiplication of vector space hold while taking an additive inverse? I have been doing some university work and have a question related to vector space
Question: Determine whether the set $V = {(x, 1) ∶ x \in \mathbb{R}}$ equipped with the operations $(x_1, 1) + (x_2, 1) = (x_1 + x_2, 1)$ and $c(x, 1) = (cx, 1)$ is a vector space or not?
Now when we take additive inverse of a vector let's say $v_1 = (x_1,1)$
Should it be like $-v_1$ as in scalar operation that will result in $-v_1 = (-x_1,-1)$
Or will it involve the irregular vector multiplication given above which will result is $-1(v_1) = (-x1,1)$?
If latter is the case, can someone explain why don't we use the given operation of multiplication while we will still use irregular addition and does that mean additive inverse in both cases will not satisfy?
Thank you!
 A: The second entry isn’t changed by addition or multiplication, so you can ignore it and see that the resulting structure acts like the real numbers. This means that it acts like a vector space in terms of addition and scalar multiplication.
As for subtraction, you can think of it in one of two ways:
Multiplication by $-1$: Given the multiplication rule provided, the second case is true.
Inverse of addition: $-x$ is defined by $x+(-x)=0$
The addition defined doesn’t affect the second entry so there’s no need to change it. This means that the negative of an element has nothing to do with the it. The number that negates the vector is the number that negates the first entry.
Therefore, $-(x,1)=(-x,1)$
A: The additive inverses cannot be of the form $(-x,-1)$, since the $-1$ part does not exist in your vector space. All elements of $V$ are of the form $(x,1)$, for some $x\in\mathbb R$. We need to determine what the additive identity is before we can find additive inverses. To be clear, the additive identity may not be what you expect (e.g. $(0,0)$).
So, what element in $V$ has the property that when you add it to $(x,1)$, you get back $(x,1)$. Well, $(0,1)$ of course! Since
$$(x,1)+(0,1)=(x+0,1)=(x,1).$$
So $(0,1)$ is the additive identity.
Now, what is the additive inverse of $(x,1)$? It must be $(-x,1)$, since
$$(x,1)+(-x,1)=(x-x,1)=(0,1).$$
