Vector addition and translations I'm reading Feynman's lectures and I have a question regarding volume I chapter 11. 
In this chapter, Feynman says that not every group of numbers is a vector. He says that if we apply an operation to some vectors the result is a vector if it's invariant towards a rotation. So if I define an operation over, say, vectors $\vec a$ and $\vec b$, like $\vec a + \vec b$, and get $\vec c$, I can check if $\vec c$ is a vector by rotating it to get $\vec {c'}$ and then rotating $\vec a$ and $\vec b$, to get $\vec {a'}$ and $\vec {b'}$, and do the operation to get $\vec {c'}$ again. If both $\vec {c'}$ are the same, we're good to go. 
The thing that's bothering me is that considering just a rotation of the coordinate system the usual vector algebra is indeed invariant. But it's not if I translate it. What I find is that vector addition is not invariant to equations 11.2 in Feynman's text:
$x' = x - A$
$y' = y$
$z' = z$
If I add, for example $\vec a = (1,2)$ and $\vec b = (2,1)$ I get $ \vec a + \vec b = \vec c = (3,3)$. If I apply the above transformation to $\vec c$ I get:  
$\vec c' = (0,3)$
On the other hand, if I apply the transformation to $\vec a$ and $\vec b$ before adding, I get: $\vec a' = (-2,2)$ and $\vec b' = (-1,1)$. If I add them, my result is:
$\vec c' = (-3,3)$
As you can see, the transformation is not invariant towards vector addition.
Is there a reason why translation is not important Feynman's argument? In the first part of the lecture he seems to imply that both spatial and rotational symmetry are important. 
Thanks. 
 A: Strictly speaking,  vectors can't be translated.  Translation is not defined in vector spaces.  All vectors have their tails at the origin.  This is clear from the way we write vectors: $$\vec{A} = A_x\hat{x} +  A_y\hat{y}+ A_z\hat{z}$$
How do I translate that?  I can multiply by a scalar.  I can form dot and cross products.  I can calculate a magnitude.  I can rotate it.  But I can't translate it.  It's tail is implicitly fixed at the origin.
The fact that physicists can usefully translate vectors is a peculiarity of Euclidean space that is outside of the mathematical nature of vectors.  What we are doing without knowing it is defining a vector space at every point in space so that we can define vectors anywhere.  But then we need a rule that tells how to move a vector from one vector space to another.  The rule for Euclidean space is so simple that we usually don't mention it: the components at the new location are the same as the components of the old location.  But all this is outside of the mathematics of vector spaces.
A: Ah, I think You must misunderstand Mr.Feynman. 
He says "Do these form a vector? “Well,” we might say, “they are three numbers, and every three numbers form a vector.” No, not every three numbers form a vector! In order for it to be a vector, not only must there be three numbers, but these must be associated with a coordinate system in such a way that if we turn the coordinate system, the three numbers “revolve” on each other, get “mixed up” in each other"
For any vectors, there exists a (maybe standard) Orthonormal basis, and the vectors can be expressed using it, e.g. in Cartesian coordinate system, a vector $\vec{a}$ can be expressed as $\vec{a}=a_x\hat{x}+a_y\hat{y}+a_z\hat{z}$. Assume $\vec{b}=b_x\hat{x}+b_y\hat{y}+b_z\hat{z}$ is another vector, so, $\vec{a}+\vec{b}$ must be a vector since it is a linear transformation. We define a new vector $\vec{c}=\vec{a}+\vec{b}=(a_x+b_x,a_y+b_y,a_z+b_z)$. Clearly, it is exactly a vector and satisfy any rules a vector must be satisfied, e.g. rotation. So, we say the groups of numbers$(a_x+b_x,a_y+b_y,a_z+b_z)$ constitutes a vector as listed in the order(the order of the three numbers is important). But, not every three numbers form a vector, such as $(a_y+b_y,a_x+b_x,a_z+b_z)$. So What does Feynman mean is the order of the three numbers in the parenthesis is non-communicative, and nothing more. Generally, any group of numbers can form a vector.
