Krdytkyu

Prove that if $a > b > 0, p > 0$, then $a^p > b^p.$

As I was reading baby Rudin, this fact was a step that Rudin skipped (Theorem 3.20a), but it is not obvious to me how to prove this.

Thanks in advance.

EDIT (relevant definitions and results from exercise 6, chapter 1):

EDIT 2 - Presume $b > 1$.

Let $r = m/n, n>0$, where $m$ and $n$ are integers. Then $b^r = (b^m)^{1/n}$.
It is proved that $b^{r+s} = b^rb^s$. If $x$ is real and if we let $$B(x) = { b^r | r in mathbb{Q}, rleq x }$$
then $b^r = sup B(r)$ and we define $b^x = sup B(x) $ for any real $x$.
Also it is proved that $b^{r+s} = b^xb^y$, for real $x$ and $y$.

Hopefully that helps, my bad for not including it the first time around.

Let $a>b>0$. We go to prove that $a^{p}>b^{p}$ for any $pinmathbb{Q}cap(0,infty)$.

Firstly, we prove that $a^{n}>b^{n}$ for any $ninmathbb{N}$. This
can be proved easily by induction. For, the formula is obviously true
for $n=1$. Suppose that the formula is true for $n=k$, i.e., $a^{k}>b^{k}$,
then $a^{k+1}=acdot a^{k}>acdot b^{k}>bcdot b^{k}=b^{k+1}$. By
mathematical induction, the formula is true for all $ninmathbb{N}$.

Next, we show that $a^{frac{1}{n}}>b^{frac{1}{n}}$ for any $ninmathbb{N}$
(Here, we assume that for any $x>0$, $ninmathbb{N}$, there exists
$y>0$ such that $y^{n}=x$. Note that $y$ can be shown unique and
is denoted by $x^{frac{1}{n}}.$). Prove by contradiction, suppose
the contrary that there exists $ninmathbb{N}$ such that $a^{frac{1}{n}}leq b^{frac{1}{n}}$.
If $a^{frac{1}{n}}=b^{frac{1}{n}},$ we have $a=left(a^{frac{1}{n}}right)^{n}=left(b^{frac{1}{n}}right)^{n}=b$,
which is a contradiction. If $a^{frac{1}{n}}<b^{frac{1}{n}}$, then
by the previous result, $left(a^{frac{1}{n}}right)^{n}<left(b^{frac{1}{n}}right)^{n}$.
That is, $a<b$, which is also a contradiction.

Finally, let $pinmathbb{Q}cap(0,infty)$. Choose $m,ninmathbb{N}$
such that $p=frac{m}{n}$. Then by the first part, $a^{m}>b^{m}$.
By the second part, $left(a^{m}right)^{frac{1}{n}}>left(b^{m}right)^{frac{1}{n}}$.
Hence $a^{p}>b^{p}$.

We know that $f^prime(x)=px^{p-1}$. And we know that both $dfrac{1}{x}>0$ and $x^p>0$ so it follows that $x^{p-1}>0$. So $f^prime(x)=px^{p-1}>0$.

So $f$ is increasing on the interval $(0,infty)$.

So a power function $f(x)=x^p$ is an increasing function when $p>0$.

So by the definition of an increasing function on $(0,infty)$ $a>b$ if and only if $a^p>b^p$.

You are using Rudin's "Principals of Mathematical Induction" and you are doing Chapter 1, Excercise 6. Which relies very heavily on the Theorem 1.21 and the proof thereof that;

For any $b > 1$ and $n in mathbb N$ there is a unique positive $c$ so that $c^n =b$. We call such a $c:= b^{frac 1n}$.

The proof makes use of the least upper bound property and the archimedian principal and the fact that for all $c < b$ then $c^n < b*c^{n-1} < b^2*c^{n-2} < ....< b^n$. We then consider $C= sup {c|c^n < b}$ and... the proof writes itself.

But HERE's the thing. In the process of doing this we have established that for all $c < b$ than so that $d= c^n < b$ that $c < b^{frac 1n} = sup {c|c^n < b}$. Thus for $d < b$ we have $c = d^{frac 1n} < b^{frac 1n}$.

And if that WASN't immediately clear, it'd have to be by contradiction:

If $d^{frac 1n} ge b^{frac 1n}$ then $d=(d^{frac 1n})^n ge (b^{frac 1n})^n = b$ which is a contradiction.

So if we show that it is consistent to define for $p =frac mn$ that $b^p = (b^{frac 1n})^m$ we would have $0 < a < b iff 0 < a^{frac 1n} < b^{frac 1n} iff a^{frac mn} < b^{frac mn}$.

And it'd only take a line to extend that result to $a^x = sup {a^q|q< x; qin mathbb Q}< sup{a^q|q< x; qin mathbb Q} = b^x$.

Which is why Rudin "took it for granted".

====in recap ==

For natural numbers it's clear by induction.

If $a^n > b^n > 0$ then $a^{n+1} =a^n*a > a^n*b > b^n*b = b^{n+1}$.

For $p = frac 1n; nin mathbb N$ it's clear by contradiction.

If $a^{frac 1n}le b^{frac 1n}$ we'd have $a = (a^{frac 1n})^n le (b^{frac 1n})^n = b$.

So for rational $p = frac nm; n,min mathbb n$ then $a^p = (a^n)^{frac 1m} > (b^n)^{frac 1m} = b^p$.

ANd for any real $x>0$ we have $a^x = sup{a^q|q < x} > sup {b^q|q < x}$ [admittedly that step would need a sentence or two but it'd be straight forward] $= b^x$.

If $pin mathbb{N}$ induction mathematical.