The space of positive real numbers

Task number: 2504

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: \(a\odot (\mathbf u \oplus \mathbf v) = (\mathbf u \mathbf v)^a=\mathbf u^a \cdot \mathbf v^a = (a \odot \mathbf u) \oplus (a \odot \mathbf v)\).

  • Answer

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

Difficulty level: Easy task (using definitions and simple reasoning)
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