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,
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 $(1,1)$?