0
$\begingroup$

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 $\mathbb{R}+$?

$\endgroup$
0
$\begingroup$

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 $\frac1u$.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.