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Let $V$ be the set of all ordered pairs of positive real numbers;

Given addition of elements in our set defined as $$(x_1,y_1)\oplus (x_2,y_2) = (x_1x_2,~y_1y_2)$$

and given scalar multiplication defined as $$k\odot (x,y)=(x^k,y^k)$$

Axioms for vector addition

A4. An element $0$ in $V$ exists such that $v+0 = v = 0+v$ for every $v$ in $V$.

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

I've identified the element "$0$" as being $(1,1)$, but I can't seem to verify A5

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  • $\begingroup$ Take the logarithm of each component of an element of $V$ to see what's going on. $\endgroup$ Apr 7, 2021 at 16:16
  • $\begingroup$ As an aside... assuming everything is working correctly and your operations and set will actually satisfy the axioms... you will find that $-v$, the additive inverse of $v$ in your proposed vector space, will be equal to $(-1)\odot v$ where $-1$ is the additive inverse of unity in the scalar field. Similarly, you will find that $0$, the vector which is the additive identity in your proposed vector space, will be equal to $0\odot v$ where this $0$ was the additive identity in the scalar field. $\endgroup$
    – JMoravitz
    Apr 7, 2021 at 16:19
  • $\begingroup$ Note here that $0\odot (x,y) = (x^0,y^0)=(1,1)$ and that $(-1)\odot (x,y) = (x^{-1},y^{-1})$ and you do indeed have $(x^{-1},y^{-1})\oplus (x,y) = (1,1)$ $\endgroup$
    – JMoravitz
    Apr 7, 2021 at 16:21
  • $\begingroup$ Thank you, and everything is working correctly this was the last axiom I needed to satisfy. $\endgroup$ Apr 7, 2021 at 16:22
  • $\begingroup$ Note also here the importance of our set only containing pairs of positive reals... since otherwise we wouldn't have "additive inverses" of pairs containing zero since $0^{-1}=\frac{1}{0}$ is undefined. If the set in question were the set of all pairs of real numbers, this would have caused it to fail to be a vector space. $\endgroup$
    – JMoravitz
    Apr 7, 2021 at 16:23

1 Answer 1

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Yes, in this context $0$ is $(1,1)$. And $-(x,y)=\left(\frac1x,\frac1y\right)$. Can you confirm it?

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  • $\begingroup$ I was just in the middle of posting this :) $\endgroup$
    – Koro
    Apr 7, 2021 at 16:14
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    $\begingroup$ @Koro What can I say? I type fast. $\ddot\smile$ $\endgroup$ Apr 7, 2021 at 16:15

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