## The space of positive real numbers

Let $$\mathbb R^+$$ be all the positive real numbers and define operation $$\oplus$$ on $$\mathbb R^+$$ and $$\odot: \mathbb Q\times \mathbb R^+\to \mathbb R^+$$ as follows: $$\mathbf u\oplus \mathbf v =\mathbf u\mathbf v, \ a\odot \mathbf u = \mathbf u^a$$.

Is it true, that $$(\mathbb R^+,\oplus,\odot)$$ is a vector space over the rational numbers $$\mathbb Q$$?

• #### Resolution

Addition of vectors, that is $$(\mathbb R^+,\oplus)=(\mathbb R^+,\cdot)$$ forms an Abelian group. Note that its neutral element, or the zero vector, is the positive real number one, i.e. $$\mathbf 0=1$$ because $$\mathbf u\oplus \mathbf 0=\mathbf u \oplus 1 =\mathbf u\cdot 1=\mathbf u$$.

Similarly, we choose the opposite vector $$\ominus\mathbf u=\frac1{\mathbf u}\in \mathbb R^+$$ and we have $$\mathbf u\oplus (\ominus\mathbf u)=\mathbf u\cdot \frac1{\mathbf u}=1=\mathbf 0$$.

Axiom of multiplication by one: $$1\odot \mathbf u = \mathbf u^1=\mathbf u$$.

The axiom of "associativity" of a scalar multiple: $$(ab)\odot \mathbf u = \mathbf u^{ab}=(\mathbf u^b)^a = a \odot ( b \odot \mathbf u)$$.

The axiom of "distributivity" from the right: $$(a+b)\odot \mathbf u = \mathbf u^{a+b}=\mathbf u^a \cdot \mathbf u^b = (a \odot \mathbf u) \oplus (b \odot \mathbf u)$$.

Axiom of distributivity from the left: $$a\odot (\mathbf u \oplus \mathbf v) = (\mathbf u \mathbf v)^a=\mathbf u^a \cdot \mathbf v^a = (a \odot \mathbf u) \oplus (a \odot \mathbf v)$$.

Yes, $$(\mathbb R^+,\oplus,\odot)$$ forms a vector space over $$\mathbb Q$$.