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

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.

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.

• Firstly, you need to ask how is $a^p$ defined. – Danny Pak-Keung Chan Dec 24 '18 at 20:25
• 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. – parsiad Dec 24 '18 at 20:28
• Then Steven, please include exercise 6, chapter 1, in an edit to your question post. – Namaste Dec 24 '18 at 20:34
• @amWhy you're right, should have included this. Editing the question now. – Steven Wagter Dec 24 '18 at 20:39
• 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 $\{a^x \mid x\mbox{ is rational}\}$. – Danny Pak-Keung Chan Dec 24 '18 at 20:51

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 $$a^{\frac{1}{n}}, then by the previous result, $$\left(a^{\frac{1}{n}}\right)^{n}<\left(b^{\frac{1}{n}}\right)^{n}$$. That is, $$a, 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 $$a^{p}>b^{p}$$.

• 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. – Danny Pak-Keung Chan Dec 24 '18 at 20:40
• The asker included the definition from Rudin, in a comment above. – Namaste Dec 24 '18 at 20:42

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 $$a^p>b^p$$.

• I know that, but isn't that a restatement of the result I am trying to prove? – Steven Wagter Dec 24 '18 at 20:29
• I will add more to my answer. – John Wayland Bales Dec 24 '18 at 20:36
• 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. – Namaste Dec 24 '18 at 20:37
• 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 $a>b>0, p>0$. – Namaste Dec 24 '18 at 20:39
• I'm sorry, John Wayland Bales, my second to last comment above was addressing the asker, not you or your answer. – Namaste Dec 24 '18 at 20:45

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] $$= b^x$$.

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

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

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.