Is $-\mathbf{\vec{v}}$ in vector space axioms mean $-1$ multiplied with $\mathbf{\vec{v}}$?
I got this doubt while i am solving the question below.
Let $V$ be the set of positive real numbers satisfying the addition rule $$x\oplus y=xy$$ and scalar multiplication rule $$r \otimes x=rx, r \in \mathbb{R} $$ Is $V$ a vector space? Now I started checking axioms one by one.
$A1.$ $\forall x, y \in \mathbb{R}$ we have $x \oplus y \in \mathbb{R}$, so closure axiom verified.
$A2.$ $x \oplus y=xy=yx=y \oplus x$,so Commutative axiom verified.
$A3.$ We have $x \oplus 1=1 \oplus x=x$, So $1$ is the null vector or additive identity.
$A4.$ To find additive inverse I am very much confused.
We have $-1 \otimes x=-x$
Now $x \oplus -x=-x^2=1$ does not give real solution. So additive inverse does not exist.
Or is it like $x \oplus \frac{1}{x}=1$ which my book has given.
No, $-x$ does not mean, prima facie, $-1 \otimes x$. It refers to the additive inverse of $x$, meaning the (necessarily unique) vector $y$ such that $x \oplus y = y \oplus x = e$, where $e$ is the additive identity, the (necessarily unique) vector that satisfies $x \oplus e = e \oplus x = x$ for all $x$. You can verify that $e = 1$ fits the latter, while $\frac{1}{x}$ fits the former.
As it turns out, the vector space axioms imply that $-1 \otimes x = -x$ in a vector space. So, showing that $-1 \otimes x$ is not an additive inverse to $x$ is indeed a sound strategy for showing a set with attached operations is indeed not a vector space.
More directly you've also shown that $-1 \otimes x$ is not even in the vector space! Your space of positive numbers will not contain $-1 \otimes x$, making the scalar multiplication operation ill-defined.
All that said, I think you should check the question once more. The "standard" scalar multiplication operation in this case is usually $a \otimes x := x^a$, not $ax$.