Vector Space with unusual addition?

Question

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 $(1,1)$?

Answer 1

First of all $(0,0)$ is not the "zero vector $\vec 0$", from what you did $\vec 0 = (1,1)$. So finding

$$ v+ (0,0) = (0,0)$$

does not mean that $(0,0)$ is the negative of $v$. You are looking for $(c,d)$ so that

$$ (a, b) + (c, d) = \vec 0 = (1,1).$$

Edit (as per the new edit of the OP): The set $V$ is the set of ordered pair of positive real number. So $(0,0)$ is just not an element of $V$.

Answer 2

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 $V$.