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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 $(1,1)$?

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    $\begingroup$ What is $V$? Is it $\mathbb{R}^2$? $\endgroup$
    – blub
    Aug 28, 2018 at 19:55
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    $\begingroup$ 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! $\endgroup$ Aug 28, 2018 at 20:02
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    $\begingroup$ 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. $\endgroup$
    – Christoph
    Aug 28, 2018 at 20:42

2 Answers 2

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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 $V$.

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

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  • $\begingroup$ 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! $\endgroup$ Aug 28, 2018 at 20:14
  • $\begingroup$ I've made an edit @EddieDylan $\endgroup$
    – user99914
    Aug 28, 2018 at 20:32

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