Linear Algebra: Determine Zero Vector 
Let $u = (u_1, u_2)$ and $v = (v_1, v_2)$ be any two positive vectors. Define $u + v = (u_1v_1, u_2/v_2)$ and $ku = (u_1^k, ku_2)$.
(a) Find the zero vector.
(b) Find $-u$.
(c) Does $(-1)u = -u$, why?

I was solving this problem. By the operations defined, I got two different zero vectors, one is $(1, 1)$, and another is $(1, u_2^2)$. So, should I say that the zero vector does not exist and hence the following questions could not be solved?
 A: I don't really understand the revised scalar multiplication operation. And the "addition operation" isn't commutative:
$$\mathbf{u} + \mathbf{v} = (u_1v_1,u_2/v_2) \neq \mathbf{v} + \mathbf{u} = (v_1u_1,v_2/u_2)$$
This doesn't sound like a consistent definition.
Additional details below.

The zero vector $\mathbf{0}$ is a unique member of the vector space $V$ such that:

*

*Additive identity: $\forall \mathbf{a}\in V\;\;\;\mathbf{0}+\mathbf{a}= \mathbf{a}$

*Scalar multiplication by zero: $\forall \mathbf{a}\in V\;\;0\mathbf{a}= \mathbf{0}$
So we need to find a vector $\mathbf{0}=(a_1,a_2) \in V$ that gives us:
(1) $(u_1a_1,u_2/a_2) = (u_1,u_2) = (a_1u_1,a_2/u_2)$ and
(2) $(u_1^0,0)=(a_1,a_2)$
$(2)$ uniquely identifies $\mathbf{0}=(1,0)$.
However, $(1,0)$ cannot serve as an additive inverse:
$$\mathbf{0} + \mathbf{u} = (u_1,0)\neq \mathbf{u} \neq \mathbf{u} + \mathbf{0} = (1,\infty)$$
If we fix scalar multiplication to be $k\mathbf{u} = (u_1^k,u_2^k)$ then scalar multiplication will be consistent with the additive identity.
Was there a typo? Also, if $u$ is supposed to be a positive vector
