# Axioms Vector Space

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

• Take the logarithm of each component of an element of $V$ to see what's going on. Apr 7, 2021 at 16:16
• 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. Apr 7, 2021 at 16:19
• 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 $(x^{-1},y^{-1})\oplus (x,y) = (1,1)$ Apr 7, 2021 at 16:21
• Thank you, and everything is working correctly this was the last axiom I needed to satisfy. Apr 7, 2021 at 16:22
• 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. Apr 7, 2021 at 16:23

Yes, in this context $$0$$ is $$(1,1)$$. And $$-(x,y)=\left(\frac1x,\frac1y\right)$$. Can you confirm it?
• @Koro What can I say? I type fast. $\ddot\smile$ Apr 7, 2021 at 16:15