Proof of exponentiation law












1












$begingroup$


I want to prove this : $(ab)^n = a^nb^n$ with a, b and n real numbers.
I know how to prove this when n is an integer but not when n is a real number. I really don't know where to start to prove this. Can you help me ? Thank you.






exponentiation







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    What definition of the exponential function are you using?
    $endgroup$
    – MisterRiemann
    Dec 1 '18 at 23:18










  • $begingroup$
    I don't really understand the question but I think the definition of the exponential I'm using is : $a^b = a*...*a$ where a appears b times.
    $endgroup$
    – Never
    Dec 1 '18 at 23:23










  • $begingroup$
    That definition works if the exponent is an integer. To discuss properties of $a^n$ when $n$ is real one needs to define it first. There are multiple (ultimately equivalent) ways of doing this.
    $endgroup$
    – MisterRiemann
    Dec 1 '18 at 23:25










  • $begingroup$
    Ok,I will look at this. Thank you
    $endgroup$
    – Never
    Dec 1 '18 at 23:27


















1












$begingroup$


I want to prove this : $(ab)^n = a^nb^n$ with a, b and n real numbers.
I know how to prove this when n is an integer but not when n is a real number. I really don't know where to start to prove this. Can you help me ? Thank you.






exponentiation







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    What definition of the exponential function are you using?
    $endgroup$
    – MisterRiemann
    Dec 1 '18 at 23:18










  • $begingroup$
    I don't really understand the question but I think the definition of the exponential I'm using is : $a^b = a*...*a$ where a appears b times.
    $endgroup$
    – Never
    Dec 1 '18 at 23:23










  • $begingroup$
    That definition works if the exponent is an integer. To discuss properties of $a^n$ when $n$ is real one needs to define it first. There are multiple (ultimately equivalent) ways of doing this.
    $endgroup$
    – MisterRiemann
    Dec 1 '18 at 23:25










  • $begingroup$
    Ok,I will look at this. Thank you
    $endgroup$
    – Never
    Dec 1 '18 at 23:27
















1












1








1





$begingroup$


I want to prove this : $(ab)^n = a^nb^n$ with a, b and n real numbers.
I know how to prove this when n is an integer but not when n is a real number. I really don't know where to start to prove this. Can you help me ? Thank you.






exponentiation







share|cite|improve this question









$endgroup$




I want to prove this : $(ab)^n = a^nb^n$ with a, b and n real numbers.
I know how to prove this when n is an integer but not when n is a real number. I really don't know where to start to prove this. Can you help me ? Thank you.






exponentiation




exponentiation






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 1 '18 at 23:17









NeverNever

61




61








  • 1




    $begingroup$
    What definition of the exponential function are you using?
    $endgroup$
    – MisterRiemann
    Dec 1 '18 at 23:18










  • $begingroup$
    I don't really understand the question but I think the definition of the exponential I'm using is : $a^b = a*...*a$ where a appears b times.
    $endgroup$
    – Never
    Dec 1 '18 at 23:23










  • $begingroup$
    That definition works if the exponent is an integer. To discuss properties of $a^n$ when $n$ is real one needs to define it first. There are multiple (ultimately equivalent) ways of doing this.
    $endgroup$
    – MisterRiemann
    Dec 1 '18 at 23:25










  • $begingroup$
    Ok,I will look at this. Thank you
    $endgroup$
    – Never
    Dec 1 '18 at 23:27
















  • 1




    $begingroup$
    What definition of the exponential function are you using?
    $endgroup$
    – MisterRiemann
    Dec 1 '18 at 23:18










  • $begingroup$
    I don't really understand the question but I think the definition of the exponential I'm using is : $a^b = a*...*a$ where a appears b times.
    $endgroup$
    – Never
    Dec 1 '18 at 23:23










  • $begingroup$
    That definition works if the exponent is an integer. To discuss properties of $a^n$ when $n$ is real one needs to define it first. There are multiple (ultimately equivalent) ways of doing this.
    $endgroup$
    – MisterRiemann
    Dec 1 '18 at 23:25










  • $begingroup$
    Ok,I will look at this. Thank you
    $endgroup$
    – Never
    Dec 1 '18 at 23:27










1




1




$begingroup$
What definition of the exponential function are you using?
$endgroup$
– MisterRiemann
Dec 1 '18 at 23:18




$begingroup$
What definition of the exponential function are you using?
$endgroup$
– MisterRiemann
Dec 1 '18 at 23:18












$begingroup$
I don't really understand the question but I think the definition of the exponential I'm using is : $a^b = a*...*a$ where a appears b times.
$endgroup$
– Never
Dec 1 '18 at 23:23




$begingroup$
I don't really understand the question but I think the definition of the exponential I'm using is : $a^b = a*...*a$ where a appears b times.
$endgroup$
– Never
Dec 1 '18 at 23:23












$begingroup$
That definition works if the exponent is an integer. To discuss properties of $a^n$ when $n$ is real one needs to define it first. There are multiple (ultimately equivalent) ways of doing this.
$endgroup$
– MisterRiemann
Dec 1 '18 at 23:25




$begingroup$
That definition works if the exponent is an integer. To discuss properties of $a^n$ when $n$ is real one needs to define it first. There are multiple (ultimately equivalent) ways of doing this.
$endgroup$
– MisterRiemann
Dec 1 '18 at 23:25












$begingroup$
Ok,I will look at this. Thank you
$endgroup$
– Never
Dec 1 '18 at 23:27






$begingroup$
Ok,I will look at this. Thank you
$endgroup$
– Never
Dec 1 '18 at 23:27












1 Answer
1






active

oldest

votes


















1












$begingroup$

If you're using the usual interpretation of the exponent of real numbers, you can use the fact that $(ab)^n=e^{nlog(ab)}=e^{n(log a+log b)}=e^{nlog a+nlog b}=e^{nlog a}cdot e^{nlog b}=a^ncdot b^n$



Edit:
The fact that $e^{x+y}=e^{x}e^{y}$ comes from the way we define the exponent.
If it is as the solution to the differential equation $f(z)=f'(z); f(0)=1$ this is pretty easy, as then $$D(e^{c-z}e^z)=e^{c-z}e^z-e^{c-z}e^z=0$$ which gives us that $e^{c-z}e^z$ is constant and the result follows for $x,y$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I didn't know this interpretation of the exponent, I will look at this. Thank you.
    $endgroup$
    – Never
    Dec 1 '18 at 23:25










  • $begingroup$
    This solution assumes that one already knows that $log(ab) = log a + log b$ as well as $e^{x+y}=e^xe^y$, which may or may not be the case, depending on the definition that one starts with.
    $endgroup$
    – MisterRiemann
    Dec 1 '18 at 23:26










  • $begingroup$
    Ok. Thank you .
    $endgroup$
    – Never
    Dec 1 '18 at 23:29










  • $begingroup$
    What does $D(e^(c−z)e^z)$ mean ?
    $endgroup$
    – Never
    Dec 1 '18 at 23:34










  • $begingroup$
    It's just another way to write $frac{d(e^{c-z}e^{z})}{dz}$, the derivative of $e^{c-z}e^{z}$ with respect to z
    $endgroup$
    – NL1992
    Dec 1 '18 at 23:37













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1 Answer
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1 Answer
1






active

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active

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active

oldest

votes









1












$begingroup$

If you're using the usual interpretation of the exponent of real numbers, you can use the fact that $(ab)^n=e^{nlog(ab)}=e^{n(log a+log b)}=e^{nlog a+nlog b}=e^{nlog a}cdot e^{nlog b}=a^ncdot b^n$



Edit:
The fact that $e^{x+y}=e^{x}e^{y}$ comes from the way we define the exponent.
If it is as the solution to the differential equation $f(z)=f'(z); f(0)=1$ this is pretty easy, as then $$D(e^{c-z}e^z)=e^{c-z}e^z-e^{c-z}e^z=0$$ which gives us that $e^{c-z}e^z$ is constant and the result follows for $x,y$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I didn't know this interpretation of the exponent, I will look at this. Thank you.
    $endgroup$
    – Never
    Dec 1 '18 at 23:25










  • $begingroup$
    This solution assumes that one already knows that $log(ab) = log a + log b$ as well as $e^{x+y}=e^xe^y$, which may or may not be the case, depending on the definition that one starts with.
    $endgroup$
    – MisterRiemann
    Dec 1 '18 at 23:26










  • $begingroup$
    Ok. Thank you .
    $endgroup$
    – Never
    Dec 1 '18 at 23:29










  • $begingroup$
    What does $D(e^(c−z)e^z)$ mean ?
    $endgroup$
    – Never
    Dec 1 '18 at 23:34










  • $begingroup$
    It's just another way to write $frac{d(e^{c-z}e^{z})}{dz}$, the derivative of $e^{c-z}e^{z}$ with respect to z
    $endgroup$
    – NL1992
    Dec 1 '18 at 23:37


















1












$begingroup$

If you're using the usual interpretation of the exponent of real numbers, you can use the fact that $(ab)^n=e^{nlog(ab)}=e^{n(log a+log b)}=e^{nlog a+nlog b}=e^{nlog a}cdot e^{nlog b}=a^ncdot b^n$



Edit:
The fact that $e^{x+y}=e^{x}e^{y}$ comes from the way we define the exponent.
If it is as the solution to the differential equation $f(z)=f'(z); f(0)=1$ this is pretty easy, as then $$D(e^{c-z}e^z)=e^{c-z}e^z-e^{c-z}e^z=0$$ which gives us that $e^{c-z}e^z$ is constant and the result follows for $x,y$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I didn't know this interpretation of the exponent, I will look at this. Thank you.
    $endgroup$
    – Never
    Dec 1 '18 at 23:25










  • $begingroup$
    This solution assumes that one already knows that $log(ab) = log a + log b$ as well as $e^{x+y}=e^xe^y$, which may or may not be the case, depending on the definition that one starts with.
    $endgroup$
    – MisterRiemann
    Dec 1 '18 at 23:26










  • $begingroup$
    Ok. Thank you .
    $endgroup$
    – Never
    Dec 1 '18 at 23:29










  • $begingroup$
    What does $D(e^(c−z)e^z)$ mean ?
    $endgroup$
    – Never
    Dec 1 '18 at 23:34










  • $begingroup$
    It's just another way to write $frac{d(e^{c-z}e^{z})}{dz}$, the derivative of $e^{c-z}e^{z}$ with respect to z
    $endgroup$
    – NL1992
    Dec 1 '18 at 23:37
















1












1








1





$begingroup$

If you're using the usual interpretation of the exponent of real numbers, you can use the fact that $(ab)^n=e^{nlog(ab)}=e^{n(log a+log b)}=e^{nlog a+nlog b}=e^{nlog a}cdot e^{nlog b}=a^ncdot b^n$



Edit:
The fact that $e^{x+y}=e^{x}e^{y}$ comes from the way we define the exponent.
If it is as the solution to the differential equation $f(z)=f'(z); f(0)=1$ this is pretty easy, as then $$D(e^{c-z}e^z)=e^{c-z}e^z-e^{c-z}e^z=0$$ which gives us that $e^{c-z}e^z$ is constant and the result follows for $x,y$.






share|cite|improve this answer











$endgroup$



If you're using the usual interpretation of the exponent of real numbers, you can use the fact that $(ab)^n=e^{nlog(ab)}=e^{n(log a+log b)}=e^{nlog a+nlog b}=e^{nlog a}cdot e^{nlog b}=a^ncdot b^n$



Edit:
The fact that $e^{x+y}=e^{x}e^{y}$ comes from the way we define the exponent.
If it is as the solution to the differential equation $f(z)=f'(z); f(0)=1$ this is pretty easy, as then $$D(e^{c-z}e^z)=e^{c-z}e^z-e^{c-z}e^z=0$$ which gives us that $e^{c-z}e^z$ is constant and the result follows for $x,y$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 1 '18 at 23:32

























answered Dec 1 '18 at 23:22









NL1992NL1992

8311




8311












  • $begingroup$
    I didn't know this interpretation of the exponent, I will look at this. Thank you.
    $endgroup$
    – Never
    Dec 1 '18 at 23:25










  • $begingroup$
    This solution assumes that one already knows that $log(ab) = log a + log b$ as well as $e^{x+y}=e^xe^y$, which may or may not be the case, depending on the definition that one starts with.
    $endgroup$
    – MisterRiemann
    Dec 1 '18 at 23:26










  • $begingroup$
    Ok. Thank you .
    $endgroup$
    – Never
    Dec 1 '18 at 23:29










  • $begingroup$
    What does $D(e^(c−z)e^z)$ mean ?
    $endgroup$
    – Never
    Dec 1 '18 at 23:34










  • $begingroup$
    It's just another way to write $frac{d(e^{c-z}e^{z})}{dz}$, the derivative of $e^{c-z}e^{z}$ with respect to z
    $endgroup$
    – NL1992
    Dec 1 '18 at 23:37




















  • $begingroup$
    I didn't know this interpretation of the exponent, I will look at this. Thank you.
    $endgroup$
    – Never
    Dec 1 '18 at 23:25










  • $begingroup$
    This solution assumes that one already knows that $log(ab) = log a + log b$ as well as $e^{x+y}=e^xe^y$, which may or may not be the case, depending on the definition that one starts with.
    $endgroup$
    – MisterRiemann
    Dec 1 '18 at 23:26










  • $begingroup$
    Ok. Thank you .
    $endgroup$
    – Never
    Dec 1 '18 at 23:29










  • $begingroup$
    What does $D(e^(c−z)e^z)$ mean ?
    $endgroup$
    – Never
    Dec 1 '18 at 23:34










  • $begingroup$
    It's just another way to write $frac{d(e^{c-z}e^{z})}{dz}$, the derivative of $e^{c-z}e^{z}$ with respect to z
    $endgroup$
    – NL1992
    Dec 1 '18 at 23:37


















$begingroup$
I didn't know this interpretation of the exponent, I will look at this. Thank you.
$endgroup$
– Never
Dec 1 '18 at 23:25




$begingroup$
I didn't know this interpretation of the exponent, I will look at this. Thank you.
$endgroup$
– Never
Dec 1 '18 at 23:25












$begingroup$
This solution assumes that one already knows that $log(ab) = log a + log b$ as well as $e^{x+y}=e^xe^y$, which may or may not be the case, depending on the definition that one starts with.
$endgroup$
– MisterRiemann
Dec 1 '18 at 23:26




$begingroup$
This solution assumes that one already knows that $log(ab) = log a + log b$ as well as $e^{x+y}=e^xe^y$, which may or may not be the case, depending on the definition that one starts with.
$endgroup$
– MisterRiemann
Dec 1 '18 at 23:26












$begingroup$
Ok. Thank you .
$endgroup$
– Never
Dec 1 '18 at 23:29




$begingroup$
Ok. Thank you .
$endgroup$
– Never
Dec 1 '18 at 23:29












$begingroup$
What does $D(e^(c−z)e^z)$ mean ?
$endgroup$
– Never
Dec 1 '18 at 23:34




$begingroup$
What does $D(e^(c−z)e^z)$ mean ?
$endgroup$
– Never
Dec 1 '18 at 23:34












$begingroup$
It's just another way to write $frac{d(e^{c-z}e^{z})}{dz}$, the derivative of $e^{c-z}e^{z}$ with respect to z
$endgroup$
– NL1992
Dec 1 '18 at 23:37






$begingroup$
It's just another way to write $frac{d(e^{c-z}e^{z})}{dz}$, the derivative of $e^{c-z}e^{z}$ with respect to z
$endgroup$
– NL1992
Dec 1 '18 at 23:37




















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