Does the following set satisfy the conditions for a vector space?

Question

Question: Let V be the set of all positive real numbers. Determine whether V is a vector space with the operations below.

$x + y = xy$

$cx=x^c$

I can prove 8 theorems but I am getting a different answer from my book on axiom of additive identity and inverse.

My approach: Let x belong to V

Additive Identity: x+0=x0=0 (The book says zero vector is 1.Why?)

Additive inverse x+(-x)=$x(-x)=-x^2$ (The book says additive inverse is 1/x Why?)

Answer 1

Zero vector is defined by the element $e \in V$ such that $x+e = x$ for all $x.$ Thus, is $x+e = xe = x$, then $e = 1$. Your additive inverse $-x$ is then defined as $x + (-x) = e$. So $x + (-x) = x*(-x) = 1 \implies -x = \frac1x$. Note that $-x$ does not mean "negative" $x$ like we are used to, but is just notation for the additive inverse which you find by solving $x + (-x) = e$.

Answer 2

Try the following options: The zero vector is the number 1, and the additive inverse of $x$ is $1/x$.