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I'm given the following:

Let V be the set of ordered pairs $(x,y)$ of positive real numbers with addition and scalar multiplication defined by

$$(x_1, y_1) + (x_2,y_2) =(x_1x_2,y_1y_2) $$

and

$$a(x,y) = (x^a,y^a)$$

My question is with the 5th addition axiom given by

A5: For each element v in V, an element -v in V exists such that $-v+v=0 $ and $v + (-v)=0$

I don't really know how to go about verifying this axiom and judging by the questions wording "Show that V is a vector space by verifying all the axioms" I can't get outta this by saying that V isn't a vector space.

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First of all, we need to identify what is $0$, it should refers to the addition identity.

Notice that we have $(x,y)+(1,1)=(x,y)$ and $(1,1)+(x,y)=(x,y)$, hence the addition identity is $(1,1)$.

Remark: note that the addition operation is commutative.

Now let $v=(x,y)$ and we should try to find $-v=(p,q)$.

Hence we need $(x,y)+(p,q)=(1,1)$

That is $(xp,yq)=(1,1)$, try to solve for $p$ and $q$ in terms of $x$ and $y$.

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The '0' of a vector space is the identity operator of the addition operator. It has nothing to do with the number 0. For verifying the inverse axiom, you have to first be clear on what 0 the axiom means. Look at the 0 of the axiom based on definition of the addition operator. And then you can find out what -v is for v

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In this vector space, $0=(1,1)$. And if $v=(x,y)$, then $-v=\left(\frac1x,\frac1y\right)$, since$$\left(\frac1x,\frac1y\right)+(x,y)=(1,1)=0.$$

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  • $\begingroup$ Ah I see, I realized (0,0) become (1,1) but completely forgot about it the second I got to this axiom. Makes so much more sense now, Thank you $\endgroup$ – AFC Jan 13 '18 at 19:33
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There is no contradiction because, due to the addition law, the neutral element instead of being (0,0), is (1,1).

Thus, for the existence of an opposite, you have to check not $-v+v=(0,0)$ but $-v+v=(1,1)$ which is accomplished, if $v:=(a,b)$ by defining $-v:=(1/a,1/b)$.

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