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I've formally defined the power whose base is a positive number and its exponent is a real non-zero number like this:

$$x^{\alpha}=\sup\left\{x^p\in\mathbb{R}\mid p\in\mathbb{Q},0<p\leq\alpha\right\},\text{ if $\alpha>0$.}$$ $$x^{\alpha}=\inf\left\{x^p\in\mathbb{R}\mid p\in\mathbb{Q},\alpha\leq p<0\right\},\text{ if $\alpha<0$.}$$

I want to prove the properties which are true for rational numbers ($x>0,\alpha,\beta\neq 0$):

  1. $x^\alpha\cdot x^\beta=x^{\alpha+\beta}$.
  2. $(x^\alpha)^\beta=x^{\alpha\cdot\beta}$.
  3. $(x\cdot y)^\alpha=x^\alpha\cdot y^\alpha$.

Any ideas? My problem is that there're 2 supremum and I don't have any idea to handle that. I hope, I didn't commit any mistake (sorry for my English!).

vitamin d
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Raúl Filigrana Villalba
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  • Please show what you have done so far. – Kavi Rama Murthy Aug 03 '21 at 09:40
  • If $A,B$ are nonempty sets of nonnegative real numbers then $(\sup A)(\sup B) = \sup (AB)$ where $AB = { ab \mid a\in A,\ b\in B }$. – md2perpe Aug 03 '21 at 09:45
  • I haven't do anything. I know that property and that I must apply in the first property when $\alpha,\beta>0$. But with the rest is different. I got stuck at the beginning. For example, for the second it would be (\alpha,\beta>0):

    $$(x^\alpha)^\beta=\sup\left{(x^\alpha)^p\middle|0<p\leq \beta\right}=$$ $$\sup\left{(\sup\left{x^p\middle|0<p\leq\alpha\right})^q\middle|0<q\leq \beta\right}.$$

    And now, what? How can I go from that to:

    $$\left{x^p\middle|0<p\leq\alpha\cdot\beta\right}.$$

     – Raúl Filigrana Villalba Aug 03 '21 at 09:53
  • BTW, your English is good. Tú inglés es bueno. – md2perpe Aug 03 '21 at 09:55
  • For nested suprema: $$\sup { \sup { f(x,y) \mid x\in X } \mid y\in Y } = \sup { f(x,y) \mid x\in X,\ y\in Y }$$ – md2perpe Aug 03 '21 at 11:18

1 Answers1

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If $A,B$ are nonempty sets of nonnegative real numbers then $(\sup A)(\sup B) = \sup (AB)$ and $(\inf A)(\inf B)=\inf (AB),$ where $AB = \{ ab \mid a\in A,\ b\in B \}$.

So for example, if $\alpha,\beta>0,$ then $$\begin{align} x^\alpha \cdot x^\beta &= \sup\{x^q \mid q\in\mathbb{Q},\ 0<q\leq\alpha\} \cdot \sup\{x^r \mid r\in\mathbb{Q},\ 0<r\leq\beta\} \\ &= \sup\{x^q\cdot x^r \mid q,r\in\mathbb{Q},\ 0<q\leq\alpha,\ 0<r\leq\beta\}) \\ &= \sup\{x^{q+r} \mid q,r\in\mathbb{Q},\ 0<q+r\leq\alpha+\beta\}) \\ &= \sup\{x^{s} \mid s\in\mathbb{Q},\ 0<s\leq\alpha+\beta\}) \\ &= x^{\alpha+\beta} . \end{align}$$

md2perpe
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  • Yes, when $\alpha$ and $\beta$ are both negatives, it is obvious too. My problem is when $\alpha\cdot\beta<0$. Maybe if, for example, $\beta<0<\alpha$, I should prove that:

    $$x^\alpha=\inf\left{x^p\middle|\alpha\leq p\right}.$$

     – Raúl Filigrana Villalba Aug 03 '21 at 10:15
  • I think the hard property would be the second one. I don't know what to do with two supremum. By the way, thank you very much. – Raúl Filigrana Villalba Aug 03 '21 at 10:16
  • @RaúlFiligranaVillalba. You mean when you have nested suprema like $\sup { \sup { (x^q)^r \mid q\in Q } \mid r\in R }$ (for some sets $Q,R$)? Then you can just write them as one supremum: $\sup { (x^q)^r \mid q\in Q,\ r\in R }$. – md2perpe Aug 03 '21 at 10:48
  • And why can you do that? – Raúl Filigrana Villalba Aug 04 '21 at 10:53
  • @RaúlFiligranaVillalba. Do you accept $\max(a_1,\ldots,a_m, b_1,\ldots, b_n) = \max(\max(a1_1,\ldots,a_m), \max(b_1,\ldots,b_n))$? If so, can you also accept that $\sup(\sup(A), \sup(B), \sup(C)) = \sup(A \cup B \cup C)$? – md2perpe Aug 04 '21 at 12:05