Prove that if $a > b > 0, p > 0$, then $a^p > b^p$
$begingroup$
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.
real-analysis
$endgroup$
|
show 9 more comments
$begingroup$
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.
real-analysis
$endgroup$
2
$begingroup$
Firstly, you need to ask how is $a^p$ defined.
$endgroup$
– Danny Pak-Keung Chan
Dec 24 '18 at 20:25
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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
|
show 9 more comments
$begingroup$
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.
real-analysis
$endgroup$
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
real-analysis
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
|
show 9 more comments
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
|
show 9 more comments
5 Answers
5
active
oldest
votes
$begingroup$
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 $a^{p}>b^{p}$.
$endgroup$
$begingroup$
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
add a comment |
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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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I know that, but isn't that a restatement of the result I am trying to prove?
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– Steven Wagter
Dec 24 '18 at 20:29
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I will add more to my answer.
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– John Wayland Bales
Dec 24 '18 at 20:36
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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.
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– Namaste
Dec 24 '18 at 20:37
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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
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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
add a comment |
$begingroup$
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] $= b^x$.
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add a comment |
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If $pin mathbb{N}$ induction mathematical.
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1
$begingroup$
I am seeking a proof where p can be any real number.
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– Steven Wagter
Dec 24 '18 at 20:28
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If $a^n leq b^n Rightarrow ln a leq ln b$ Is a contradiction, since $a>b Rightarrow ln a> ln b$
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– 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
add a comment |
$begingroup$
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.
$endgroup$
add a comment |
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5 Answers
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5 Answers
5
active
oldest
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active
oldest
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active
oldest
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$begingroup$
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 $a^{p}>b^{p}$.
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $a^{p}>b^{p}$.
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $a^{p}>b^{p}$.
$endgroup$
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 $a^{p}>b^{p}$.
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
add a comment |
$begingroup$
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
add a comment |
$begingroup$
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 $a^p>b^p$.
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $a^p>b^p$.
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $a^p>b^p$.
$endgroup$
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 $a^p>b^p$.
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
add a comment |
$begingroup$
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
add a comment |
$begingroup$
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] $= b^x$.
$endgroup$
add a comment |
$begingroup$
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] $= b^x$.
$endgroup$
add a comment |
$begingroup$
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] $= b^x$.
$endgroup$
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] $= b^x$.
edited Dec 24 '18 at 22:36
answered Dec 24 '18 at 20:59
fleabloodfleablood
1
1
add a comment |
add a comment |
$begingroup$
If $pin mathbb{N}$ induction mathematical.
$endgroup$
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
add a comment |
$begingroup$
If $pin mathbb{N}$ induction mathematical.
$endgroup$
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
add a comment |
$begingroup$
If $pin mathbb{N}$ induction mathematical.
$endgroup$
If $pin mathbb{N}$ induction mathematical.
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
add a comment |
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
add a comment |
$begingroup$
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.
$endgroup$
add a comment |
$begingroup$
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.
$endgroup$
add a comment |
$begingroup$
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.
$endgroup$
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.
answered Dec 24 '18 at 20:47
Mark BennetMark Bennet
82.1k984182
82.1k984182
add a comment |
add a comment |
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2
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Firstly, you need to ask how is $a^p$ defined.
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– Danny Pak-Keung Chan
Dec 24 '18 at 20:25
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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.
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– parsiad
Dec 24 '18 at 20:28
1
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Then Steven, please include exercise 6, chapter 1, in an edit to your question post.
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– Namaste
Dec 24 '18 at 20:34
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@amWhy you're right, should have included this. Editing the question now.
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– Steven Wagter
Dec 24 '18 at 20:39
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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 ${a^x mid xmbox{ is rational}}$.
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– Danny Pak-Keung Chan
Dec 24 '18 at 20:51