Vector Space with unusual addition? I'm studying before my class starts in a few weeks and I encountered this question in one of the practice problems:
The addition it has given me is defined as,  

$(a,b)+(c,d)= (ac,bd)$

It's asking me if this is a vector of space and I am stuck after proving this, 

There is an element $0$ in $V$ so that $v + 0 = v$ for all $v$ in $V$.

I did this -> $(a,b)+(1,1) = (1a,1b) = (a,b)$


Stuck right here,

For each $v$ in $V$ there is an element $-v$ in $V$ so that $v+(-v) = 0$.

$(a,b)+(0,0) = (0a,0b) = (0,0)$


Is $(0,0)$ $a$ $-v$ when there's no such thing as '$-0$'?
Do I stop proving right at the step? 
So this is not a vector of space?
Thank you for your time.
Edit: Thank you everyone! The question is stated exactly like so,

Show that the set of ordered pairs of positive real numbers is a vector space under the addition and scalar multiplication.
  $$(a,b)+(c,d) = (ac,bd),$$
  $$c(a,b) = (a^c, b^c).$$

So the additive inverse is an element that, when added to $(a,b)$, will give me the additive identity, which in this case is $(1,1)$?

As you have the neutral element $o=(1,1)$ you need to make sure your inverses are relative to that. Assuming $V=\{(a,b): a,b\in\mathbb{R}, a,b>0\}$ or something of that kind you could use $(a,b)+(\frac1a,\frac1b)=(1,1)$.
What you still need is to tell us how your base field acts on $V$. 
