Prove $x < y \implies a^x < a^y$

Question

I want to prove the following statement $$\forall x,y \in \mathbb R ,\ a >1 :\\ \ \ \ \ x < y \iff a^x < a^y$$

I can use all the exponentiation laws for rational numbers and I would like to prove the statement by using rational sequences $q_n \rightarrow x$ and $r_n \rightarrow y$ which converge to $x$ and $y$ for a rising $n$.

I tried proving using the contrapositive version of the statement for my left - to - right proof, but I do not think that it is correct. My proof was the following:

$$a^x \ge a^y \implies x\geq y$$

If $a^x \ge a^y$ then, if you choose an $n$ large enough, then any $q_n$ is larger than $r_n$. This implies also that $\lim_{n\to\infty}q_n \geq \lim_{n\to\infty} r_n $and thus concludes the statement.

Any hints?

Answer 1

Hint:

It trivial for $\mathbf N$. Deduce it first for $\mathbf Z$, then $\mathbf Q$.

Next, show that if $x,y$ are two real numbers such that $x<y$, there exists sequences $(x_n)$ and $(y_n)$ which converge to $x$ and $y$ respectively, and such that $x_n<y_n$ for all $n$.

Then use that passing to the limit preserves non-strict inequalities.

Last step, to show the limit inequality is strict, prove there exists $u,v\in \mathbf Q\cap (x,y)$ such that $ x_n\le u<v\le y_n$ if $n$ is large enough.

Answer 2

Two basic ideas:

With logarithms: $\;\log_ax\;$ is monotone ascending since $\;a>1\;$ , and thus

$$a^x<a^y\iff x=\log_aa^x<\log_aa^y=y$$

Without logarithms:

$$a^y>a^x\stackrel{\div \,a^x}\iff a^{y-x}>1\iff y-x>0\;\ldots$$

Answer 3

Suppose $${a^x>a^y}$$ $$a^{y/x}<1$$ $$a^{y/x}<a$$ $${y/x}<1$$

We get $y<x$, a contradiction. Hence, $a^x<a^y$