Proving the existence of a real exponential function

[Portuga]

Homework Statement

Proof that $a^r < \gamma < a^s$ for all rational $r$ and $s$, $r<x<s$, and $x \in \mathbb{R}$ known.

Relevant Equations

Assume that $f(x)=a^x$ is defined only for $x \in \mathbb{Q}$. Assume that $f(x)$ is strictly growing.
Lemma 1.
If $a>1$ is a know real, $\forall \varepsilon > 0$, $\exists n \in \mathbb{N}$ such that $$a^{\frac{1}{n}} - 1 < \varepsilon$$.
Lemma 2.
If $a>1$ and $x$ are a known reals, $\forall \varepsilon > 0$, $\exists r, s \in \mathbb{Q}$, $r<x<s$ such that $$a^s - a^r < \varepsilon$$.

Ok, first I tried to show that $A = \left \{a^{r}|r\in\mathbb{Q},r<x \right \}$ does not have a maximum value.
Assume $\left\{ a^{r}\right\}$ has a maximum, $a^{r_m}$. By this hypothesis, $r_{m}<x$ and $r_{m}>r\forall r\neq r_{m}\in\mathbb{Q}$. Consider now that $q\in\mathbb{Q}|q>0$ such that $$ q<x-r_{m}.$$ Of course, $$a^{q}<a^{x},$$because $q<x$.
As $q>0$, $r_{m}+q>r_{m}$, a contradiction. So, $A$ does not have a maximum value.
If $x$ is a rational, $a^x$ is the supreme for $A$ and the infimus for $B=\left\{ a^{s}|s\in\mathbb{Q},s>x\right\}$ and the proof is finished.
If not, by Lemma 2, there are $r_1$ and $s_1$, $r_1 \in A$, $s_1 \in B$ so that $$a^{s_{1}}-a^{r_{1}}<\epsilon.$$
Now, if so, then $a^{r}=a^{s}$ and the suprema and infimus of $A$ and $B$ are equal.

 

[epenguin]

Is the statem you are trying to prove true for all rational number? E.g. for 0 < a < 1
and others

 

[Portuga]

Yes, for rationals, $a \in \mathbb{R}$, $a>1$.

 

[Portuga]

Now I see that I have made a real mess.
Let me try to make things more clear.
Homework Statement: consider $a>1$ a known real number. Prove that, for all real $x$, there is only one real $\gamma$ such that $$a^r<\gamma<a^s$$ for any rationals $r$ and $s$, with $r<x<s$.
Relevant Equations: $\max A$ is the maximum element of $A$. The upper quote of a set $A$ is a value greater than any of its elements. The supremum of a set is its smallest upper quote.

My attempt of a solution.
The set $A=\left\{ a^{r}|r\in\mathbb{Q},r<x\right\}$ is not empty and superiorly limited by any $a^s$, $s$ rational and $s>x$; so this set has a supremum, but not a maximum.
To show that it has not a maximum, suppose that $r_m$ is it. Thus, $r_{m}<x$ and $r_{m}>r\forall r\neq r_{m}\in A$. But it is also possible to generate a rational $q\in \mathbb{Q}|q>0$ such that $$q<x-r_{m},$$ using $$q=r_{m}+\frac{1}{a},$$ $a$ being the least integer for which $r_{m}+\frac{1}{a} < x.$ It is clear that $q+r_m<x$, so it is in $A$ and $q+r_m>r_m$ (because $\frac{1}{a}>0)$ contradicting the initial hypothesis that $r_m$ is a maximum.
Using analogous reasoning, it is proved that $B=\left\{ a^{s}|r\in\mathbb{Q},s>x\right\}$ has a infimum but not a minimum.
This should make evident that $$a^r < \gamma < a^s.$$
Is this a correct reasoning?

 

[Delta2]

Not sure if your proof that A doesn't have a maximum correct, however I think what is definitely needed is to prove that $$supA=infB(=\gamma)$$

EDIT: Sorry I only read your second post, now that I read your first post, you try there to prove that $supA=infB$ but I don't understand that proof. I have slept only 4 hours last night, I 'll come back to this when I have slept more.

 

[Portuga]

Not sure if your proof that A doesn't have a maximum correct, however I think what is definitely needed is to prove that $$supA=infB(=\gamma)$$

EDIT: Sorry I only read your second post, now that I read your first post, you try there to prove that $supA=infB$ but I don't understand that proof. I have slept only 4 hours last night, I 'll come back to this when I have slept more.

Ok, thank you very much.