Verifying a Vector Space Let $\vec{V}$ be the set of all pairs of real numbers of the form $(1,x)$ with the following operations: $(1,x) \bigoplus (1,y) = (1, x + y)$ , $ k \bigotimes  (1,y) = (1, ky)$. Determine whether $\vec{V}$ is a vector space. 
I kind of just want to talk about what's on my mind and have someone verify if I am on the right track. So I know that in order for $\vec{V}$ to be a vector space it must satisfy all of the vector space axioms. In the $\bigotimes$ operation I can see that the distributive property already holds and if $k$ is $1$ then by the property of real numbers we have a scalar identity and closure under multiplication. In the $\bigoplus$ I can see that closure under addition holds since the result is $(1,x+y)$ and also there would exist an additive identity since our additive inverse would be a scalar of $-1$ for either $x$ or $y$. Other than the commutative property which I don't know how to prove did I get everything and say that $\vec{V}$ is a vector space?
 A: Let's have proper definitions. Let $K$ be a field. $\vec{V}  := \{ (1,x) \big| x \in K\}$. For $k,x \in K$, $k \otimes (1,x) := (1, kx)$.
All you have to do is to chose a list of axioms for vector spaces and check that you fulfill them all. Just make a list and check them one by one.
1. $(\vec{V}, \oplus)$ is a commutative group:
$\oplus$ is associative. Let $u,v,w \in \vec{V}$, say $u=(1,x), \, v=(1,y), \textrm{ and } w=(1,z)$, one has $$u \oplus( v\oplus w) = (1, x+(y+z)) = (1, (x+y)+z) = (u\oplus v) \oplus w $$
$(1,0)$ is a neutral element for $\oplus$
Every element of $\vec{V}$ has an inverse. What is the inverse of $(1,x)$ ? 
For all $(u,v) \in \vec{V}$, $u \oplus v = v\oplus u$. Say $u=(1, x)$ and $v=(1,y)$, one has
$$u \oplus v = (1, x + y) = (1, y + x) = v \oplus u$$
2. $\otimes$ is a scalar multiplication:
1 is neutral. Let $u \in \vec{V}$, say $u = (1, x)$, then $$1 \otimes u = (1, 1 \times x) = (1, x) = u$$
Associativity of multiplication. Let $k,l \in K$ and $u \in \vec{V}$. Say $u=(1,x)$, then 
$$(kl)\otimes u = (1, (kl)x) = (1(k(lx)) = k \otimes (l \otimes x) $$
Distributivity of scalar sums. Let $k,l \in K$ and $u \in \vec{V}$. Say $u=(1,x)$, then
$$(k+l)\otimes u = (1, (k+l)x) = (1, kx + lx) = (1, kx) \oplus(1, lx) = k \otimes u \oplus l \otimes u$$ 
Distributivity of vector sums. Let $k \in K$ and $u,v \in \vec{V}$, say $u=(1,x)$ and $v=(1,y)$. One has
$$ k\otimes (u \oplus v) = k \otimes (1,x+y) = (1,kx + ky) = (1,kx) \oplus (1,ky) = k \otimes u \oplus k \otimes v $$
Aaand you're done!
A: As long as $V$ contains a zero vector (i guess it's $(1,0)$), I guess you're right. 
It's clear that $\oplus$ is commutative,  as $(1,x)\oplus (1,y)=(1,x+y)=(1,y+x)=(1,y)\oplus (1,x)$.
