Let $\mathbb{R}+$ be the set of all positive real numbers. Define the operations of addition and scalar multiplication as follows:

$u + v = u.v$ $\forall u,v \in \mathbb{R}+$

$au = u^a$ $\forall u \in \mathbb{R}+$ and real scalar $a$.

Prove that $\mathbb{R}+$ is a real vector space.

I am able to verify all the axioms for it to be vector space except inverse element axiom. Is question correct? Should it be defined over $\mathbb{R}$ instead of $\mathbb{R}+$?


The inverse of $u\in\mathbb{R}^+$ will be $\frac1u$ in that vector space. Note that the zero element of that vector space is $1$, since $(\forall u\in\mathbb{R}^+):1.u=u.1=u$. So, since $u.\frac1u=\frac1u.u=1$, the inverse of $u$ is $\frac1u$.

  • $\begingroup$ Thank you for the comment. I am confused about additive inverse which is defined as an element when added to "u" results into zero. $\endgroup$ – Pankaj Jun 19 '18 at 9:53
  • $\begingroup$ @Pankaj You seem to forget that the zero vector here is the number $1$. $\endgroup$ – José Carlos Santos Jun 19 '18 at 9:54
  • $\begingroup$ I got that. I am actually confused about additive inverse. What would be the additive inverse for this vector space. (additive inverse-- an element when added to the 'x' results into zero) $\endgroup$ – Pankaj Jun 19 '18 at 9:56
  • $\begingroup$ @Pankaj As I wrote in my answer, the additive inverse of $u$ is $\frac1u$. $\endgroup$ – José Carlos Santos Jun 19 '18 at 9:56
  • $\begingroup$ ok I got it. Thank you. $\endgroup$ – Pankaj Jun 19 '18 at 9:57

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