Spivak, Calculus: how to show that $a^x<a^y$ for $a>1$ and $a^x>a^y$ for $a<1$, using the "usual algebraic definition" of $a^x$ for rational $x$.

Question

The following is a problem from Ch. 22 of Spivak's Calculus

29. Let $a>0$, and for rational $x$ let $f(x)=a^x$, as defined in the usual elementary algebraic way. This problem shows directly that $f$ can be extended to a continuous function $\bar{f}$ on the whole line. Problem 28 provided the necessary machinery.

(a) For rational $x<y$, show that $a^x<a^y$ for $a>1$ and $a^x>a^y$ for $a<1$.

What is the "usual elementary algebraic way" that $a^x$ is defined for rational $x$?

Here is the solution manual solution to item (a)

Since $a^y=a^{y-x}\cdot a^x$ we just need that for rational $z$ we have $a^z>1$ for $a>1$ and $a^z<1$ for $a<1$, and this follows immediately from the elementary definitions.

I don't quite comprehend this solution.

There is a result in Chapter 7 proving that every positive number has an $n^{th}$ root, $n\in\mathbb{N}$, and if $n$ is odd then every number has an $n^{th}$ root.

In Chapter 18, we do have a more lengthy discussion about the function $a^x$. Perhaps I am answering my own question here, but it seems the definition of $f(x)=a^x$ for rational $x$ is a function such that

$$a^1=a$$

$$\forall x,y \in \mathbb{Q} \implies f(x+y)=f(x)f(y)=a^{x+y}$$

For this to be true, there are some implicit definitions

• $n\in\mathbb{N}\implies a^n=a\cdot a \cdot a\cdot ...\cdot a$ (n times)
• $a^{-n}=\frac{1}{a^n}$
• $a^{1/n}=\sqrt[n]{a}$
• $a^{m/n}=a^{m/n}=\sqrt[n]{a^m}$

Note that these definitions take care of the cases of a rational $x$ that is either $0$, $-n$, $1/n$, or $m/n$.

Answer 1

I think the definition of $a^x$ for a rational number $x=\frac{q}{p}$, where $q$ is an integer and $p$ is a positive integer in the usual elementary algebraic way is $a^x:=(a^{\frac{1}{p}})^q.$

---

Let $z$ be a positive rational number.
We can write $z=\frac{q}{p}$, where $p$ and $q$ are positive integers.

Let $a>1$.
If $a^{\frac{1}{p}}\leq1,$ then $a=(a^{\frac{1}{p}})^p\leq 1^p=1$.
This is a contradiction.
So, $a^{\frac{1}{p}}>1.$
So, $a^z=(a^{\frac{1}{p}})^q>1^q=1.$

Let $0<a<1$.
If $a^{\frac{1}{p}}\geq 1,$ then $a=(a^{\frac{1}{p}})^p\geq 1^p=1$.
This is a contradiction.
So, $a^{\frac{1}{p}}<1.$
So, $a^z=(a^{\frac{1}{p}})^q<1^q=1.$

---

Let $x,y$ be rational numbers such that $x<y$.
Then, we can write $y=(y-x)+x$.
Since $x<y$, $0<y-x$.
Let $z:=y-x$.
Then, $a^y=a^{(y-x)+x}=a^{y-x}a^{x}=a^za^x.$

If $a>1$, then $a^z>1$ since $z$ is a positive rational number.
So, $a^y=a^za^x>a^x.$

If $0<a<1$, then $a^z<1$ since $z$ is a positive rational number.
So, $a^y=a^za^x<a^x.$