Is this set a vector space and if not, why? Determine if the following is a vector space under the given
operations. If not, list some of the axioms that fail to hold.
The set of all triples of real numbers $(x, y, z)$ with the operations
$$(x, y, z) + (x'
, y'
, z'
) = (x + x
'
, y + y
'
, z + z
'
)$$ 
and
$$λ(x, y, z) = (λx, y, z)$$
 A: no it is not. and there is two of the axioms that fail to hold:
1) Associativity of addition:
this axiom requires : u+(v+w) = (u+v)+w so if we assume that u = v = (x,y,z) and w = (x',y',z')
then based on the space property, for the left side of equation we will have:
(x,y,z) + ((x,y,z)+(x',y',z')) = (x,y,z) + ((x+x',y+y',z+z')) = (2x+x',2y+y',2z+z') 
and for the right side of equation we will have:
((x,y,z) + (x,y,z))+(x',y',z') = 2(x,y,z) + (x',y',z') = (2x+x',y+y',z+z')
so:
(x,y,z) + ((x,y,z)+(x',y',z')) is not equal to ((x,y,z) + (x,y,z))+(x',y',z')
and this axiom fails to hold.
2)Inverse elements of addition:
For every v ∈ V, there exists an element −v ∈ V, called the additive inverse of v, such that v + (−v) = 0.
but in this space if we assume v = (x,y,z) then:
v + (-v) = (x,y,z) + (-1(x,y,z)) = (x,y,z)+(-x,y,z) = (0,2y,2z) which is not equal to 0.
A: I suppose scalers come from $\mathbb{R}$.
A counterexample for not being a vector space over $\mathbb{R}$ : 
$v=(3,4,5)$ , $\lambda_{1}= 1$ , $\lambda_{2}= 2$
Then, $(\lambda_{1} + \lambda_{2} )v= (1+2)(3,4,5)=3(3,4,5)= (9,4,5 )$
But On the other side : $\lambda_{1}v + \lambda_{2}v= 1 ( 3,4,5) +2 (3,4,5)= ( 3,4,5) + ( 6,4,5)= ( 9,8,10)$
Namely, $(\lambda_{1} + \lambda_{2} )v\neq\lambda_{1}v + \lambda_{2}v$
