# 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?

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 $(1,1)$?

• What is $V$? Is it $\mathbb{R}^2$?
– blub
Aug 28, 2018 at 19:55
• To define a vector space, you need a set, a field, an internal law on the set (the addition) and a scalar multiplication. You haven’t told us what are the set, the field and the scalar multiplication. It would be good to provide us with those information! Aug 28, 2018 at 20:02
• After verifying you do indeed get a real vector space, note that $\mathbb R^2\to V$ given by $(x,y) \mapsto (\exp x, \exp y)$ is an isomorphism of real vector spaces, where $\mathbb R^2$ here carries the usual structure. Aug 28, 2018 at 20:42

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 $V$.

• I have added the question word for word from my textbook. Thank you! Aug 28, 2018 at 20:16

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 $V$.

• Oh! That is the relationship between additive inverse and identity. I need a better textbook that has better explanation than the one I have on hand then. Thank you! Aug 28, 2018 at 20:14
• I've made an edit @EddieDylan
– user99914
Aug 28, 2018 at 20:32