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.
Best Answer
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.