If $a,b,x\in\mathbb{R}^+$ and $a

Question

I am reading in a real analysis text about the definition of exponentiation when the exponent is real.

I have at my disposal all the properties of natural/integer/rational exponentiation and the definition of real exponentiation, which is as follows.

If $a\in\mathbb{R}^+$ and $x\in\mathbb{R}$ we define $$a^x:=\begin{cases}\sup\{a^r:r\in\mathbb{Q}, r<x\}=\inf\{a^s:s\in\mathbb{Q},s>x\} & \text{ if }a\geq 1\\ \inf\{a^r:r\in\mathbb{Q}, r<x\}=\sup\{a^s:s\in\mathbb{Q},s>x\} & \text{ if }0<a\leq 1\end{cases}.$$

Now I am trying to prove that the usual properties of exponentiation hold; in particular, I am currently trying to prove that if $a,b,x\in\mathbb{R}^+$ and $a<b$ then $a^x<b^x$.

I have been able to prove that if $0<a<b$ and $x\in\mathbb{R}^+$ then $a^x\leq b^x$ as follows.

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$\fbox{Suppose $1\leq a<b.$}$ Then if we take $\varepsilon>0$ by definition we know that there exist $r,r'\in\mathbb{Q}$ such that $r<x, r'<x$, $a^x-\varepsilon<a^r\leq a^x$ and $b^x-\varepsilon<b^{r'}\leq b^x$ so if we set $\rho=\max\{r,r'\}$ we have $a^x-\varepsilon<a^\rho\leq a^x$ and $b^x-\varepsilon<b^\rho\leq b^x$ hence $0<b^\rho-a^\rho<b^x-a^x+\varepsilon$, where the first inequality follows from the properties of rational exponentiation. In particular $0<b^x-a^x+\varepsilon$. This last inequality implies, due to the arbitrariness of $\varepsilon>0$, that $0\leq b^x-a^x$ i.e. $a^x\leq b^x$.

$\fbox{Suppose now $a<b\leq 1.$}$ Then if we take $\varepsilon>0$ by definition we know that there exist $s,s'\in\mathbb{Q}$ such that $x<s$, $x<s'$, $a^s-\varepsilon<a^s\leq a^x$ and $b^{s'}-\varepsilon<b^{s'}\leq b^x$ so if we set $\rho=\max\{s,s'\}$ we have $a^s-\varepsilon<a^\rho\leq a^x$ and $b^{s'}-\varepsilon<b^\rho\leq b^x$ hence $0<b^\rho-a^\rho<b^x-a^x+\varepsilon$, where the first inequality follows from the properties of rational exponentiation. In particular $0<b^x-a^x+\varepsilon$. This last inequality implies, due to the arbitrariness of $\varepsilon>0$, that $0<b^x-a^x$ i.e. $a^x\leq b^x$.

$\fbox{Suppose lastly that $a<1<b.$}$ Then if we take $\varepsilon>0$ by definition we know that there exist $s,s'\in\mathbb{Q}$ such that $x<s$, $x<s'$, $a^x-\frac{\varepsilon}{2}<a^s\leq a^x$ and $b^x\leq b^{s'}< b^x+\frac{\varepsilon}{2}$ hence $0<b^{s'}-b^s<b^x-a^x+\varepsilon$ and due to the arbitrariness of $\varepsilon>0$ it follows that $0\leq b^x-a^x$ i.e. $a^x\leq b^x.$

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What I haven't been able to prove is the strict inequality $a^x<b^x$ so I would be grateful if someone would give me an hint about how to prove this fact. Thanks.

Answer 1

Take $x>0$.

Take $0<a<b$. Pick $r_0\in(0,x)\cap\Bbb Q$. I know that for rational $r>r_0$: $$b^r=a^r\cdot(b/a)^r\ge a^r\cdot(b/a)^{r_0}$$Since you've hopefully established the increasing property of exponentiation for rational exponents. Define $c:=(b/a)^{r_0}$; we have $c>1$.

The main idea here is that we don't just have the flimsy inequality $b^r>a^r$, but we have it with a constant difference in size that we can use for any rational. If we don't control the size of $b^r-a^r$ somehow, it is going to be very hard to deduce the inequality. The choice of $r_0$ is unimportant.

Suppose now $1\le a$. We know $a^x=\sup_{r_0<r<x\\r\in\Bbb Q}a^r$ and likewise for $b^x$. Suppose $b^x<c\cdot a^x$. Then, since multiplication by $c$ is preserves suprema, by definition of supremum there is a rational $r_0<r<x$ with $c\cdot a^r>b^x\ge b^r$. However, we know this is impossible. It follows $b^x\ge c\cdot a^x>a^x$, since $a^x>0$ is easy to prove. Therefore we have a strict inequality.

Suppose $a<b<1$. We know $a^x=\inf_{r_0<r<x\\r\in\Bbb Q}a^r$ and similarly for $b^x$; if $b^x<c\cdot a^x$, then because multiplication by $c$ preserves infima, by definition of infimum there would exist a rational $r_0<r<x$ with $b^r<c\cdot a^x\le c\cdot a^r$. However, we again know this is impossible, so $b^x\ge c\cdot a^x$ and $b^x>a^x$ follows.

Exercise: be inspired to figure out the case $a<1\le b$ using these ideas. We need not fiddle with epsilons.