Mixing
Prove that $b^{r+s} = b^{r}b^{s}$ if r and s are rational. Specific detail with regards to proof.
$begingroup$
Prove that $b^{r+s} = b^{r}b^{s}$ if r and s are rational. Also $b > 1, b in mathbb{R}$
My question has to do with a specific aspect of this proof. I'm taking the question from Principles of Mathematical Analysis by Rudin.
So say we have defined $r = frac{p}{q}$ and $s = frac{m}{n}$, where $p, q, m, n in mathbb{Z}$
To prove this statement I will end up showing that both sides are equivalent to a particular expression. That expression being: $$b^{np}b^{mq}$$
In proving from the side where $(b^{r+s})^{nq}$, I arrive at the following expression: $$b^{pn + mq}$$
In all of the solutions that I have seen, they make the jump from $$b^{pn + mq} = b^{pn} b^{mq}$$
Of course this is how the properties of exponents behave, but it is this exact property that I am trying to prove, so how is making that jump to the final step even possible?
real-analysis proof-explanation
asked Jan 24 at 19:52
dc3rddc3rd
1,50711138
$endgroup$
add a comment |
$begingroup$
Prove that $b^{r+s} = b^{r}b^{s}$ if r and s are rational. Also $b > 1, b in mathbb{R}$
My question has to do with a specific aspect of this proof. I'm taking the question from Principles of Mathematical Analysis by Rudin.
So say we have defined $r = frac{p}{q}$ and $s = frac{m}{n}$, where $p, q, m, n in mathbb{Z}$
To prove this statement I will end up showing that both sides are equivalent to a particular expression. That expression being: $$b^{np}b^{mq}$$
In proving from the side where $(b^{r+s})^{nq}$, I arrive at the following expression: $$b^{pn + mq}$$
In all of the solutions that I have seen, they make the jump from $$b^{pn + mq} = b^{pn} b^{mq}$$
Of course this is how the properties of exponents behave, but it is this exact property that I am trying to prove, so how is making that jump to the final step even possible?
real-analysis proof-explanation
asked Jan 24 at 19:52
dc3rddc3rd
1,50711138
$endgroup$
$begingroup$
Have your proven that $(b^a)^c=b^{ac}$ for rational exponents? Because you're using it, and you shouldn't if you haven't proven it. @DietrichBurde I feel that a natural middle ground would be positive real.
$endgroup$
– Arthur
Jan 24 at 19:57
$begingroup$
@DietrichBurde I edited my question. $b>1, bin mathbb{R}$
$endgroup$
– dc3rd
Jan 24 at 20:04
add a comment |
$begingroup$
Prove that $b^{r+s} = b^{r}b^{s}$ if r and s are rational. Also $b > 1, b in mathbb{R}$
My question has to do with a specific aspect of this proof. I'm taking the question from Principles of Mathematical Analysis by Rudin.
So say we have defined $r = frac{p}{q}$ and $s = frac{m}{n}$, where $p, q, m, n in mathbb{Z}$
To prove this statement I will end up showing that both sides are equivalent to a particular expression. That expression being: $$b^{np}b^{mq}$$
In proving from the side where $(b^{r+s})^{nq}$, I arrive at the following expression: $$b^{pn + mq}$$
In all of the solutions that I have seen, they make the jump from $$b^{pn + mq} = b^{pn} b^{mq}$$
Of course this is how the properties of exponents behave, but it is this exact property that I am trying to prove, so how is making that jump to the final step even possible?
real-analysis proof-explanation
asked Jan 24 at 19:52
dc3rddc3rd
1,50711138
$endgroup$
Prove that $b^{r+s} = b^{r}b^{s}$ if r and s are rational. Also $b > 1, b in mathbb{R}$
My question has to do with a specific aspect of this proof. I'm taking the question from Principles of Mathematical Analysis by Rudin.
So say we have defined $r = frac{p}{q}$ and $s = frac{m}{n}$, where $p, q, m, n in mathbb{Z}$
To prove this statement I will end up showing that both sides are equivalent to a particular expression. That expression being: $$b^{np}b^{mq}$$
In proving from the side where $(b^{r+s})^{nq}$, I arrive at the following expression: $$b^{pn + mq}$$
In all of the solutions that I have seen, they make the jump from $$b^{pn + mq} = b^{pn} b^{mq}$$
Of course this is how the properties of exponents behave, but it is this exact property that I am trying to prove, so how is making that jump to the final step even possible?
real-analysis proof-explanation
real-analysis proof-explanation
asked Jan 24 at 19:52
dc3rddc3rd
1,50711138
asked Jan 24 at 19:52
dc3rddc3rd
1,50711138
edited Jan 24 at 20:04
dc3rd
asked Jan 24 at 19:52
dc3rddc3rd
1,50711138
asked Jan 24 at 19:52
dc3rddc3rd
1,50711138
asked Jan 24 at 19:52
dc3rddc3rd
1,50711138
1,50711138
$begingroup$
Have your proven that $(b^a)^c=b^{ac}$ for rational exponents? Because you're using it, and you shouldn't if you haven't proven it. @DietrichBurde I feel that a natural middle ground would be positive real.
$endgroup$
– Arthur
Jan 24 at 19:57
$begingroup$
@DietrichBurde I edited my question. $b>1, bin mathbb{R}$
$endgroup$
– dc3rd
Jan 24 at 20:04
add a comment |
$begingroup$
Have your proven that $(b^a)^c=b^{ac}$ for rational exponents? Because you're using it, and you shouldn't if you haven't proven it. @DietrichBurde I feel that a natural middle ground would be positive real.
$endgroup$
– Arthur
Jan 24 at 19:57
$begingroup$
@DietrichBurde I edited my question. $b>1, bin mathbb{R}$
$endgroup$
– dc3rd
Jan 24 at 20:04
$begingroup$
Have your proven that $(b^a)^c=b^{ac}$ for rational exponents? Because you're using it, and you shouldn't if you haven't proven it. @DietrichBurde I feel that a natural middle ground would be positive real.
$endgroup$
– Arthur
Jan 24 at 19:57
$begingroup$
Have your proven that $(b^a)^c=b^{ac}$ for rational exponents? Because you're using it, and you shouldn't if you haven't proven it. @DietrichBurde I feel that a natural middle ground would be positive real.
$endgroup$
– Arthur
Jan 24 at 19:57
$begingroup$
@DietrichBurde I edited my question. $b>1, bin mathbb{R}$
$endgroup$
– dc3rd
Jan 24 at 20:04
$begingroup$
@DietrichBurde I edited my question. $b>1, bin mathbb{R}$
$endgroup$
– dc3rd
Jan 24 at 20:04
add a comment |
1 Answer
1
active
oldest
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$begingroup$
It is not the exact property you are trying to prove. In the latter the exponents are integers; and it can be proven easily by induction (or almost by definition) if you prefer, as it is just counting $b$'s. That is, $b^{m+n}$ means to multiply $m+n$ instances of $b$; while $b^mb^n$ means to multiply $m$ instances of $b$ with $n$ instances of $b$ for a total of $m+n$ $b$'s.
answered Jan 24 at 20:02
Martin ArgeramiMartin Argerami
128k1184184
$endgroup$
$begingroup$
I don't completely follow what you are alluding to, $r$ and $s$ are stated to be rationals and not integers. So how would I use the integer properties to arrive at the solution?
$endgroup$
– dc3rd
Jan 27 at 23:04
$begingroup$
By doing exactly as you wrote in your question, you reduce the problem to the integer case.
$endgroup$
– Martin Argerami
Jan 28 at 0:52
$begingroup$
Hmmm...now that you mention that and looking at the solution again that is very interesting. I never looked at it with that perspective. Thank you for the help.
$endgroup$
– dc3rd
Jan 28 at 2:11
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
It is not the exact property you are trying to prove. In the latter the exponents are integers; and it can be proven easily by induction (or almost by definition) if you prefer, as it is just counting $b$'s. That is, $b^{m+n}$ means to multiply $m+n$ instances of $b$; while $b^mb^n$ means to multiply $m$ instances of $b$ with $n$ instances of $b$ for a total of $m+n$ $b$'s.
answered Jan 24 at 20:02
Martin ArgeramiMartin Argerami
128k1184184
$endgroup$
$begingroup$
I don't completely follow what you are alluding to, $r$ and $s$ are stated to be rationals and not integers. So how would I use the integer properties to arrive at the solution?
$endgroup$
– dc3rd
Jan 27 at 23:04
$begingroup$
By doing exactly as you wrote in your question, you reduce the problem to the integer case.
$endgroup$
– Martin Argerami
Jan 28 at 0:52
$begingroup$
Hmmm...now that you mention that and looking at the solution again that is very interesting. I never looked at it with that perspective. Thank you for the help.
$endgroup$
– dc3rd
Jan 28 at 2:11
add a comment |
$begingroup$
It is not the exact property you are trying to prove. In the latter the exponents are integers; and it can be proven easily by induction (or almost by definition) if you prefer, as it is just counting $b$'s. That is, $b^{m+n}$ means to multiply $m+n$ instances of $b$; while $b^mb^n$ means to multiply $m$ instances of $b$ with $n$ instances of $b$ for a total of $m+n$ $b$'s.
answered Jan 24 at 20:02
Martin ArgeramiMartin Argerami
128k1184184
$endgroup$
$begingroup$
I don't completely follow what you are alluding to, $r$ and $s$ are stated to be rationals and not integers. So how would I use the integer properties to arrive at the solution?
$endgroup$
– dc3rd
Jan 27 at 23:04
$begingroup$
By doing exactly as you wrote in your question, you reduce the problem to the integer case.
$endgroup$
– Martin Argerami
Jan 28 at 0:52
$begingroup$
Hmmm...now that you mention that and looking at the solution again that is very interesting. I never looked at it with that perspective. Thank you for the help.
$endgroup$
– dc3rd
Jan 28 at 2:11
add a comment |
$begingroup$
It is not the exact property you are trying to prove. In the latter the exponents are integers; and it can be proven easily by induction (or almost by definition) if you prefer, as it is just counting $b$'s. That is, $b^{m+n}$ means to multiply $m+n$ instances of $b$; while $b^mb^n$ means to multiply $m$ instances of $b$ with $n$ instances of $b$ for a total of $m+n$ $b$'s.
answered Jan 24 at 20:02
Martin ArgeramiMartin Argerami
128k1184184
$endgroup$
It is not the exact property you are trying to prove. In the latter the exponents are integers; and it can be proven easily by induction (or almost by definition) if you prefer, as it is just counting $b$'s. That is, $b^{m+n}$ means to multiply $m+n$ instances of $b$; while $b^mb^n$ means to multiply $m$ instances of $b$ with $n$ instances of $b$ for a total of $m+n$ $b$'s.
answered Jan 24 at 20:02
Martin ArgeramiMartin Argerami
128k1184184
answered Jan 24 at 20:02
Martin ArgeramiMartin Argerami
128k1184184
answered Jan 24 at 20:02
Martin ArgeramiMartin Argerami
128k1184184
answered Jan 24 at 20:02
Martin ArgeramiMartin Argerami
128k1184184
128k1184184
$begingroup$
I don't completely follow what you are alluding to, $r$ and $s$ are stated to be rationals and not integers. So how would I use the integer properties to arrive at the solution?
$endgroup$
– dc3rd
Jan 27 at 23:04
$begingroup$
By doing exactly as you wrote in your question, you reduce the problem to the integer case.
$endgroup$
– Martin Argerami
Jan 28 at 0:52
$begingroup$
Hmmm...now that you mention that and looking at the solution again that is very interesting. I never looked at it with that perspective. Thank you for the help.
$endgroup$
– dc3rd
Jan 28 at 2:11
add a comment |
$begingroup$
I don't completely follow what you are alluding to, $r$ and $s$ are stated to be rationals and not integers. So how would I use the integer properties to arrive at the solution?
$endgroup$
– dc3rd
Jan 27 at 23:04
$begingroup$
By doing exactly as you wrote in your question, you reduce the problem to the integer case.
$endgroup$
– Martin Argerami
Jan 28 at 0:52
$begingroup$
Hmmm...now that you mention that and looking at the solution again that is very interesting. I never looked at it with that perspective. Thank you for the help.
$endgroup$
– dc3rd
Jan 28 at 2:11
$begingroup$
I don't completely follow what you are alluding to, $r$ and $s$ are stated to be rationals and not integers. So how would I use the integer properties to arrive at the solution?
$endgroup$
– dc3rd
Jan 27 at 23:04
$begingroup$
I don't completely follow what you are alluding to, $r$ and $s$ are stated to be rationals and not integers. So how would I use the integer properties to arrive at the solution?
$endgroup$
– dc3rd
Jan 27 at 23:04
$begingroup$
By doing exactly as you wrote in your question, you reduce the problem to the integer case.
$endgroup$
– Martin Argerami
Jan 28 at 0:52
$begingroup$
By doing exactly as you wrote in your question, you reduce the problem to the integer case.
$endgroup$
– Martin Argerami
Jan 28 at 0:52
$begingroup$
Hmmm...now that you mention that and looking at the solution again that is very interesting. I never looked at it with that perspective. Thank you for the help.
$endgroup$
– dc3rd
Jan 28 at 2:11
$begingroup$
Hmmm...now that you mention that and looking at the solution again that is very interesting. I never looked at it with that perspective. Thank you for the help.
$endgroup$
– dc3rd
Jan 28 at 2:11
add a comment |
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$begingroup$
Have your proven that $(b^a)^c=b^{ac}$ for rational exponents? Because you're using it, and you shouldn't if you haven't proven it. @DietrichBurde I feel that a natural middle ground would be positive real.
$endgroup$
– Arthur
Jan 24 at 19:57
$begingroup$
@DietrichBurde I edited my question. $b>1, bin mathbb{R}$
$endgroup$
– dc3rd
Jan 24 at 20:04