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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 $$B(x) = \{ b^r | r \in \mathbb{Q}, r\leq x \}$$ 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.

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    $\begingroup$ Firstly, you need to ask how is $a^p$ defined. $\endgroup$ – Danny Pak-Keung Chan Dec 24 '18 at 20:25
  • $\begingroup$ If you know the derivative of $x \mapsto x^p$, then it's very easy. But as Danny points out, this involves knowing the definition of this function. $\endgroup$ – parsiad Dec 24 '18 at 20:28
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    $\begingroup$ Then Steven, please include exercise 6, chapter 1, in an edit to your question post. $\endgroup$ – Namaste Dec 24 '18 at 20:34
  • $\begingroup$ @amWhy you're right, should have included this. Editing the question now. $\endgroup$ – Steven Wagter Dec 24 '18 at 20:39
  • $\begingroup$ I do not have the book "Rudin". For the sake of completeness, it is better for you to type the definition. Moreover, $a^p$, being a real number, is obviously NOT the set $\{a^x \mid x\mbox{ is rational}\}$. $\endgroup$ – Danny Pak-Keung Chan Dec 24 '18 at 20:51
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Let $a>b>0$. We go to prove that $a^{p}>b^{p}$ for any $p\in\mathbb{Q}\cap(0,\infty)$.

Firstly, we prove that $a^{n}>b^{n}$ for any $n\in\mathbb{N}$. 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 $a^{k+1}=a\cdot a^{k}>a\cdot b^{k}>b\cdot b^{k}=b^{k+1}$. By mathematical induction, the formula is true for all $n\in\mathbb{N}$.

Next, we show that $a^{\frac{1}{n}}>b^{\frac{1}{n}}$ for any $n\in\mathbb{N}$ (Here, we assume that for any $x>0$, $n\in\mathbb{N}$, 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 $n\in\mathbb{N}$ 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 $p\in\mathbb{Q}\cap(0,\infty)$. Choose $m,n\in\mathbb{N}$ 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 $a^{p}>b^{p}$.

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  • $\begingroup$ For the case that $p\in\mathbb{R}\cap(0,\infty)$, we need the precise definition of $a^p$. (Notice that there are many different but equivalent ways to define $a^p$). As I do not have the book nor the precise definition of $a^p$, so I skip the proof for the most general case. $\endgroup$ – Danny Pak-Keung Chan Dec 24 '18 at 20:40
  • $\begingroup$ The asker included the definition from Rudin, in a comment above. $\endgroup$ – Namaste Dec 24 '18 at 20:42
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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 $a^p>b^p$.

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  • $\begingroup$ I know that, but isn't that a restatement of the result I am trying to prove? $\endgroup$ – Steven Wagter Dec 24 '18 at 20:29
  • $\begingroup$ I will add more to my answer. $\endgroup$ – John Wayland Bales Dec 24 '18 at 20:36
  • $\begingroup$ Does the proposition say anything about $f(x) = x^p$ being an increasing function? That's an observation you need to argue that $a> b \iff a^p >b^p$. For a non-increasing function, we can't conclude that. You need to use information about the function you defined in comments. $\endgroup$ – Namaste Dec 24 '18 at 20:37
  • $\begingroup$ However, you don't need to prove all of "if and only if"; you need only prove $a>b \implies a^p\gt b^p$ for $a>b>0, p>0$. $\endgroup$ – Namaste Dec 24 '18 at 20:39
  • $\begingroup$ I'm sorry, John Wayland Bales, my second to last comment above was addressing the asker, not you or your answer. $\endgroup$ – Namaste Dec 24 '18 at 20:45
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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 $a^x = \sup \{a^q|q< x; q\in \mathbb Q\}< \sup\{a^q|q< x; q\in \mathbb Q\} = b^x$.

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 $p = \frac 1n; n\in \mathbb N$ 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 $p = \frac nm; n,m\in \mathbb n$ 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] $= b^x$.

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If $p\in \mathbb{N}$ induction mathematical.

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    $\begingroup$ I am seeking a proof where p can be any real number. $\endgroup$ – Steven Wagter Dec 24 '18 at 20:28
  • $\begingroup$ If $a^n \leq b^n \Rightarrow \ln a \leq \ln b$ Is a contradiction, since $a>b \Rightarrow \ln a> \ln b$ $\endgroup$ – Julio Trujillo Gonzalez Dec 24 '18 at 20:41
  • $\begingroup$ natural logarithm is a function monotonically increasing $\endgroup$ – Julio Trujillo Gonzalez Dec 24 '18 at 20:48
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Here is another potential route through this.

Since $a\gt b\gt 0$ we have $\frac ab\gt 1$ and we might be in a position to say that $\frac ab=1+r$ with $r\gt 0$ and $\left(\frac ab\right)^p=(1+r)^p\gt 1$.

For example we can show that $(1+r)^n\gt 1^m$ for integer $n, m$ so we can do this for $p$ a positive rational.

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