I'm looking to learn Real Analysis on my own. Am reading Elements of Real Analysis by Bartle. I came across this project which defines the powers of real numbers i.e. exponentiation.
Firstly I am unsure about my solutions and they are too long to be posted here. Furthermore, the author restricts the development to bases that are strictly greater than $1$. I wanted to bypass this section for now and return later, but the theorem:
$$\forall \ \ a, b \gt 0: a^x \gt b^x \iff a \gt b \;\; \forall \ x \gt 0 $$
is something that is used extensively and I feel like I'm cheating when I take it for granted without seeing a proof.
Can anyone direct me to a reference which defines $a^x$ for real $x$ so that I can move forward. I am only just beginning to finish off the Topology chapters so my knowledge on sequences isn't all that great either.
Any advice is appreciated. Thanks in advance.
You should proceed with the definition $a^{x} = \lim_{n \to \infty}a^{x_{n}}$ where $x_{n}$ is a sequence of rationals tending to $x$ and $a > 0$. This route is bit difficult compared to the standard route of using logarithms via integral. But providing a rigorous justification of all the usual properties of exponents using the above definition turns out to be a good exercise in real analysis.
The first thing one needs to do is to show that the limit $\lim_{n \to \infty}a^{x_{n}}$ exists and the definition is unambiguous i.e. if $x_{n}, y_{n}$ are sequences of rationals both tending to $x$ then $\lim_{n \to \infty}a^{x_{n}} = \lim_{n \to \infty}a^{y_{n}}$. The algebraic properties like $a^{x + y} = a^{x}a^{y}$ don't pose a major challenge.
About your inequalities let us say we have $a > b > 0$ and $x_{n}$ is a sequence of rationals tending to $x$. We need to show $a^{x} > b^{x}$. Clearly by the rules of rational exponents we have $a^{x_{n}} > b^{x_{n}}$ but taking limits as $n \to \infty$ weakens the inequality to $\geq $. So what we need are the following inequalities $$ra^{r - 1}(a - b) > a^{r} - b^{r} > rb^{r - 1}(a - b)$$ and $$sa^{s - 1}(a - b) < a^{s} - b^{s} < sb^{s - 1}(a - b)$$ where $r, s$ are rationals with $0 < s < 1 < r$. These are established in this answer.
Let $x > 1$ then we can suppose after a certain value of $n$, $x_{n} > 1$. Clearly then we have $$a^{x_{n}} - b^{x_{n}} > x_{n}b^{x_{n} - 1}(a - b)$$ Taking limits as $n \to \infty$ we get $$a^{x} - b^{x} \geq xb^{x - 1}(a - b) > 0$$ so that $a^{x} > b^{x}$. Similarly we can use other inequality (dealing with $s$) to handle the case when $0 < x < 1$. For negative values of $x$ the inequality is reversed i.e. $a^{x} < b^{x}$. This can be easily done via the rule that $a^{-x} = 1/a^{x}$.