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I was working on a problem displaying the expansion of the definition of exponents, and naturally the final question was to prove the exponent laws when the exponents are real numbers.

For $a>1$, define $a^x=\sup\{a^r:r\in\mathbb Q, r\le x\}$ for any $x\in\mathbb R$. Prove the exponent laws hold for real exponents and $a>1$.

Now I have proved that $a^{x+y}=a^xa^y$ for any real number $x,y$. The next thing I was to prove was $(a^x)^y=a^{xy}$. So I first fixed $x$ and let a rational number $r$ less or equal to $y$, and show $(a^x)^r=a^{rx}$. This is my attempt:

First suppose that $(a^x)^r<\sup\{a^{rs}:s\in\mathbb Q,s\le x\}$. Then there exists a rational number $t\le x$ such that $(a^x)^r<(a^t)^r\le\sup\{a^{rs}:s\in\mathbb Q,s\le x\}$, and hence $a^x<a^t$, so we reach a contradiction.

Now I have to show a contradiction when $(a^x)^r>\sup\{a^{rs}:s\in\mathbb Q,s\le x\}$, and I'm completely stuck. Can anyone help me out with this? Any help would be appreciated.

WishofStar
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1 Answers1

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$ \def\qq{\mathbb{Q}} $ ... there exists a rational number $t≤x$ such that $(a^x)^r < (a^t)^r ≤ \sup\{a^{rs}:s∈\qq,s≤x\}$ ...

This is correct. After this you then concluded $a^x < a^t$, but I am not sure whether you actually understood how to get it, because you cannot simply "take $r$-th root" unless you have already proven the needed inequalities involving rational powers of reals. Some work is needed here.

Moreover, you should not immediately seek a proof by contradiction when doing real analysis. Instead, focus on the underlying structure. Here, we can in fact directly prove the desired result. Here is a sketch (I'll leave the proof of each substep to you):

$(a^x)^y = \sup \{ ( \sup\{ a^r : r∈\qq_{<x} \} )^s : s∈\qq_{<y} \}$

$ = \sup \{ \sup\{ (a^r)^s : r∈\qq_{<x} \} : s∈\qq_{<y} \}$

$ = \sup \{ (a^r)^s : r∈\qq_{<x} ∧ s∈\qq_{<y} \}$

$ = \cdots$

It should be easy to finish now, using the definition of multiplication of reals and the properties of exponentiation for rationals.

user21820
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    I should point out that using the supremum to define exponentiation for reals is real painful and slightly unintuitive, and should be discarded in favour of using Cauchy sequences, if you want to truly understand real numbers. – user21820 Jun 20 '20 at 15:55
  • Yeah, the thing you mentioned in your first line was a typo. I'll edit it right away. Also, thanks for the advice! The book I'm using hasn't introduced Cauchy sequences at the point, so I think it's legit to only use supremums. However, it must be a good exercise anyway. – WishofStar Jun 20 '20 at 15:59
  • @WishofStar: Ok, I have edited my answer in response to your edited question. I understand that you are following a particular approach using supremums and hence Cauchy sequences are not currently relevant, which is why I only gave a side remark about that in a comment. =) – user21820 Jun 20 '20 at 16:48
  • +1 for the answer as well as your 1st comment. – Paramanand Singh Jun 21 '20 at 02:55