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












2












$begingroup$



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.










share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    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






  • 1




    $begingroup$
    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 ${a^x mid xmbox{ is rational}}$.
    $endgroup$
    – Danny Pak-Keung Chan
    Dec 24 '18 at 20:51
















2












$begingroup$



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.










share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    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






  • 1




    $begingroup$
    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 ${a^x mid xmbox{ is rational}}$.
    $endgroup$
    – Danny Pak-Keung Chan
    Dec 24 '18 at 20:51














2












2








2





$begingroup$



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.










share|cite|improve this question











$endgroup$





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.







real-analysis






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 24 '18 at 22:22







Steven Wagter

















asked Dec 24 '18 at 20:20









Steven WagterSteven Wagter

1789




1789








  • 2




    $begingroup$
    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






  • 1




    $begingroup$
    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 ${a^x mid xmbox{ is rational}}$.
    $endgroup$
    – Danny Pak-Keung Chan
    Dec 24 '18 at 20:51














  • 2




    $begingroup$
    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






  • 1




    $begingroup$
    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 ${a^x mid xmbox{ is rational}}$.
    $endgroup$
    – Danny Pak-Keung Chan
    Dec 24 '18 at 20:51








2




2




$begingroup$
Firstly, you need to ask how is $a^p$ defined.
$endgroup$
– Danny Pak-Keung Chan
Dec 24 '18 at 20:25




$begingroup$
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




$begingroup$
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




1




1




$begingroup$
Then Steven, please include exercise 6, chapter 1, in an edit to your question post.
$endgroup$
– Namaste
Dec 24 '18 at 20:34




$begingroup$
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




$begingroup$
@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 ${a^x mid xmbox{ is rational}}$.
$endgroup$
– Danny Pak-Keung Chan
Dec 24 '18 at 20:51




$begingroup$
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 ${a^x mid xmbox{ is rational}}$.
$endgroup$
– Danny Pak-Keung Chan
Dec 24 '18 at 20:51










5 Answers
5






active

oldest

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2












$begingroup$

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 $a^{p}>b^{p}$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    For the case that $pinmathbb{R}cap(0,infty)$, 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



















1












$begingroup$

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 $a^p>b^p$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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 $a>b implies a^pgt b^p$ 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



















1












$begingroup$

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] $= b^x$.






share|cite|improve this answer











$endgroup$





















    0












    $begingroup$

    If $pin mathbb{N}$ induction mathematical.






    share|cite|improve this answer











    $endgroup$









    • 1




      $begingroup$
      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



















    0












    $begingroup$

    Here is another potential route through this.



    Since $agt bgt 0$ we have $frac abgt 1$ and we might be in a position to say that $frac ab=1+r$ with $rgt 0$ and $left(frac abright)^p=(1+r)^pgt 1$.



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






    share|cite|improve this answer









    $endgroup$














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      5 Answers
      5






      active

      oldest

      votes








      5 Answers
      5






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2












      $begingroup$

      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 $a^{p}>b^{p}$.






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        For the case that $pinmathbb{R}cap(0,infty)$, 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
















      2












      $begingroup$

      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 $a^{p}>b^{p}$.






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        For the case that $pinmathbb{R}cap(0,infty)$, 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














      2












      2








      2





      $begingroup$

      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 $a^{p}>b^{p}$.






      share|cite|improve this answer









      $endgroup$



      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 $a^{p}>b^{p}$.







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered Dec 24 '18 at 20:38









      Danny Pak-Keung ChanDanny Pak-Keung Chan

      2,58438




      2,58438












      • $begingroup$
        For the case that $pinmathbb{R}cap(0,infty)$, 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


















      • $begingroup$
        For the case that $pinmathbb{R}cap(0,infty)$, 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
















      $begingroup$
      For the case that $pinmathbb{R}cap(0,infty)$, 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






      $begingroup$
      For the case that $pinmathbb{R}cap(0,infty)$, 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




      $begingroup$
      The asker included the definition from Rudin, in a comment above.
      $endgroup$
      – Namaste
      Dec 24 '18 at 20:42











      1












      $begingroup$

      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 $a^p>b^p$.






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        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 $a>b implies a^pgt b^p$ 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
















      1












      $begingroup$

      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 $a^p>b^p$.






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        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 $a>b implies a^pgt b^p$ 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














      1












      1








      1





      $begingroup$

      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 $a^p>b^p$.






      share|cite|improve this answer











      $endgroup$



      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 $a^p>b^p$.







      share|cite|improve this answer














      share|cite|improve this answer



      share|cite|improve this answer








      edited Dec 24 '18 at 20:41

























      answered Dec 24 '18 at 20:27









      John Wayland BalesJohn Wayland Bales

      15.4k21238




      15.4k21238












      • $begingroup$
        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 $a>b implies a^pgt b^p$ 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


















      • $begingroup$
        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 $a>b implies a^pgt b^p$ 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
















      $begingroup$
      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 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




      $begingroup$
      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




      $begingroup$
      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 $a>b implies a^pgt b^p$ for $a>b>0, p>0$.
      $endgroup$
      – Namaste
      Dec 24 '18 at 20:39






      $begingroup$
      However, you don't need to prove all of "if and only if"; you need only prove $a>b implies a^pgt b^p$ 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




      $begingroup$
      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











      1












      $begingroup$

      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] $= b^x$.






      share|cite|improve this answer











      $endgroup$


















        1












        $begingroup$

        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] $= b^x$.






        share|cite|improve this answer











        $endgroup$
















          1












          1








          1





          $begingroup$

          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] $= b^x$.






          share|cite|improve this answer











          $endgroup$



          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] $= b^x$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 24 '18 at 22:36

























          answered Dec 24 '18 at 20:59









          fleabloodfleablood

          1




          1























              0












              $begingroup$

              If $pin mathbb{N}$ induction mathematical.






              share|cite|improve this answer











              $endgroup$









              • 1




                $begingroup$
                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
















              0












              $begingroup$

              If $pin mathbb{N}$ induction mathematical.






              share|cite|improve this answer











              $endgroup$









              • 1




                $begingroup$
                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














              0












              0








              0





              $begingroup$

              If $pin mathbb{N}$ induction mathematical.






              share|cite|improve this answer











              $endgroup$



              If $pin mathbb{N}$ induction mathematical.







              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              edited Dec 24 '18 at 20:29

























              answered Dec 24 '18 at 20:28









              Julio Trujillo GonzalezJulio Trujillo Gonzalez

              856




              856








              • 1




                $begingroup$
                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














              • 1




                $begingroup$
                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








              1




              1




              $begingroup$
              I am seeking a proof where p can be any real number.
              $endgroup$
              – Steven Wagter
              Dec 24 '18 at 20:28




              $begingroup$
              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




              $begingroup$
              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




              $begingroup$
              natural logarithm is a function monotonically increasing
              $endgroup$
              – Julio Trujillo Gonzalez
              Dec 24 '18 at 20:48











              0












              $begingroup$

              Here is another potential route through this.



              Since $agt bgt 0$ we have $frac abgt 1$ and we might be in a position to say that $frac ab=1+r$ with $rgt 0$ and $left(frac abright)^p=(1+r)^pgt 1$.



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






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                Here is another potential route through this.



                Since $agt bgt 0$ we have $frac abgt 1$ and we might be in a position to say that $frac ab=1+r$ with $rgt 0$ and $left(frac abright)^p=(1+r)^pgt 1$.



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






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  Here is another potential route through this.



                  Since $agt bgt 0$ we have $frac abgt 1$ and we might be in a position to say that $frac ab=1+r$ with $rgt 0$ and $left(frac abright)^p=(1+r)^pgt 1$.



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






                  share|cite|improve this answer









                  $endgroup$



                  Here is another potential route through this.



                  Since $agt bgt 0$ we have $frac abgt 1$ and we might be in a position to say that $frac ab=1+r$ with $rgt 0$ and $left(frac abright)^p=(1+r)^pgt 1$.



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







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 24 '18 at 20:47









                  Mark BennetMark Bennet

                  82.1k984182




                  82.1k984182






























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