The nature of the number $a$ where $log_p(q)=a$
$begingroup$
Let's look at $p,q in mathbb P, pne q$. $log_p(q)=a$ means $p^a=q$.
$a$ is irrational, because $a=frac{r}{s}, r,sin mathbb N Rightarrow p^{frac{r}{s}}=q Rightarrow p^r=q^s$ which violates the fundamental theorem of arithmetic.
Recently, in another context, the Gelfond Schneider Theorem was brought to my attention, and at first it seemed to clash with the notion that $a$ in the above example is irrational. I think I have found the resolution to my confusion.
$p$ being an integer is plainly an algebraic number. The Gelfond Schneider Theorem proves that for algebraic numbers $m,n; mne 0,1$; $m^n$ is transcendental.
Plainly $q$ is not transcendental. Thus, although $a$ is irrational, it cannot be an irrational algebraic number. The forced conclusion is that $a$ is transcendental, and so must be a great majority of logarithms. Moreover, the implication is that algebraic numbers raised to a transcendental power need not be transcendental, but can (in at least some cases) be integers or rational numbers. Gelfond Schneider does not necessarily hold when an exponent $n$ is transcendental.
My simple question is: Have I understood matters correctly?
number-theory soft-question prime-numbers logarithms
edited Dec 21 '18 at 19:35
Shaun
9,789113684
asked Dec 21 '18 at 18:43
Keith BackmanKeith Backman
1,5041812
$endgroup$
add a comment |
$begingroup$
Let's look at $p,q in mathbb P, pne q$. $log_p(q)=a$ means $p^a=q$.
$a$ is irrational, because $a=frac{r}{s}, r,sin mathbb N Rightarrow p^{frac{r}{s}}=q Rightarrow p^r=q^s$ which violates the fundamental theorem of arithmetic.
Recently, in another context, the Gelfond Schneider Theorem was brought to my attention, and at first it seemed to clash with the notion that $a$ in the above example is irrational. I think I have found the resolution to my confusion.
$p$ being an integer is plainly an algebraic number. The Gelfond Schneider Theorem proves that for algebraic numbers $m,n; mne 0,1$; $m^n$ is transcendental.
Plainly $q$ is not transcendental. Thus, although $a$ is irrational, it cannot be an irrational algebraic number. The forced conclusion is that $a$ is transcendental, and so must be a great majority of logarithms. Moreover, the implication is that algebraic numbers raised to a transcendental power need not be transcendental, but can (in at least some cases) be integers or rational numbers. Gelfond Schneider does not necessarily hold when an exponent $n$ is transcendental.
My simple question is: Have I understood matters correctly?
number-theory soft-question prime-numbers logarithms
edited Dec 21 '18 at 19:35
Shaun
9,789113684
asked Dec 21 '18 at 18:43
Keith BackmanKeith Backman
1,5041812
$endgroup$
4
$begingroup$
Yeah your understanding is good, except your last sentence. If you're looking at $a^b$, where $b$ is transcendental, the issue isn't that "Gelfond Schneider does not necessarily hold". The conditions of the theorem haven't been met, since a transcendental number is not algebraic by definition!
$endgroup$
– Joe
Dec 21 '18 at 18:52
2
$begingroup$
@Joe Poor wording on my part. You're right that the issue isn't 'not holding,' but being irrelevant, which is what I was trying to get at.
$endgroup$
– Keith Backman
Dec 21 '18 at 18:59
1
$begingroup$
Your statement of the Gelfond-Schneider Theorem is incorrect. If $m,n$ are algebraic with $0ne mne 1$ and $nnot in Bbb Q$ then $m^n$ is transcendental. So if $p,q$ are unequal primes and $p^a=q$ then $ a not in Bbb Q,$ so by the G-S Theorem, if $a$ were algebraic then $p^a=q$ would be transcendental. So $a$ is not algebraic.
$endgroup$
– DanielWainfleet
Dec 22 '18 at 4:49
add a comment |
$begingroup$
Let's look at $p,q in mathbb P, pne q$. $log_p(q)=a$ means $p^a=q$.
$a$ is irrational, because $a=frac{r}{s}, r,sin mathbb N Rightarrow p^{frac{r}{s}}=q Rightarrow p^r=q^s$ which violates the fundamental theorem of arithmetic.
Recently, in another context, the Gelfond Schneider Theorem was brought to my attention, and at first it seemed to clash with the notion that $a$ in the above example is irrational. I think I have found the resolution to my confusion.
$p$ being an integer is plainly an algebraic number. The Gelfond Schneider Theorem proves that for algebraic numbers $m,n; mne 0,1$; $m^n$ is transcendental.
Plainly $q$ is not transcendental. Thus, although $a$ is irrational, it cannot be an irrational algebraic number. The forced conclusion is that $a$ is transcendental, and so must be a great majority of logarithms. Moreover, the implication is that algebraic numbers raised to a transcendental power need not be transcendental, but can (in at least some cases) be integers or rational numbers. Gelfond Schneider does not necessarily hold when an exponent $n$ is transcendental.
My simple question is: Have I understood matters correctly?
number-theory soft-question prime-numbers logarithms
edited Dec 21 '18 at 19:35
Shaun
9,789113684
asked Dec 21 '18 at 18:43
Keith BackmanKeith Backman
1,5041812
$endgroup$
Let's look at $p,q in mathbb P, pne q$. $log_p(q)=a$ means $p^a=q$.
$a$ is irrational, because $a=frac{r}{s}, r,sin mathbb N Rightarrow p^{frac{r}{s}}=q Rightarrow p^r=q^s$ which violates the fundamental theorem of arithmetic.
Recently, in another context, the Gelfond Schneider Theorem was brought to my attention, and at first it seemed to clash with the notion that $a$ in the above example is irrational. I think I have found the resolution to my confusion.
$p$ being an integer is plainly an algebraic number. The Gelfond Schneider Theorem proves that for algebraic numbers $m,n; mne 0,1$; $m^n$ is transcendental.
Plainly $q$ is not transcendental. Thus, although $a$ is irrational, it cannot be an irrational algebraic number. The forced conclusion is that $a$ is transcendental, and so must be a great majority of logarithms. Moreover, the implication is that algebraic numbers raised to a transcendental power need not be transcendental, but can (in at least some cases) be integers or rational numbers. Gelfond Schneider does not necessarily hold when an exponent $n$ is transcendental.
My simple question is: Have I understood matters correctly?
number-theory soft-question prime-numbers logarithms
number-theory soft-question prime-numbers logarithms
edited Dec 21 '18 at 19:35
Shaun
9,789113684
asked Dec 21 '18 at 18:43
Keith BackmanKeith Backman
1,5041812
edited Dec 21 '18 at 19:35
Shaun
9,789113684
asked Dec 21 '18 at 18:43
Keith BackmanKeith Backman
1,5041812
edited Dec 21 '18 at 19:35
Shaun
9,789113684
edited Dec 21 '18 at 19:35
Shaun
9,789113684
edited Dec 21 '18 at 19:35
Shaun
9,789113684
9,789113684
asked Dec 21 '18 at 18:43
Keith BackmanKeith Backman
1,5041812
asked Dec 21 '18 at 18:43
Keith BackmanKeith Backman
1,5041812
asked Dec 21 '18 at 18:43
Keith BackmanKeith Backman
1,5041812
1,5041812
4
$begingroup$
Yeah your understanding is good, except your last sentence. If you're looking at $a^b$, where $b$ is transcendental, the issue isn't that "Gelfond Schneider does not necessarily hold". The conditions of the theorem haven't been met, since a transcendental number is not algebraic by definition!
$endgroup$
– Joe
Dec 21 '18 at 18:52
2
$begingroup$
@Joe Poor wording on my part. You're right that the issue isn't 'not holding,' but being irrelevant, which is what I was trying to get at.
$endgroup$
– Keith Backman
Dec 21 '18 at 18:59
1
$begingroup$
Your statement of the Gelfond-Schneider Theorem is incorrect. If $m,n$ are algebraic with $0ne mne 1$ and $nnot in Bbb Q$ then $m^n$ is transcendental. So if $p,q$ are unequal primes and $p^a=q$ then $ a not in Bbb Q,$ so by the G-S Theorem, if $a$ were algebraic then $p^a=q$ would be transcendental. So $a$ is not algebraic.
$endgroup$
– DanielWainfleet
Dec 22 '18 at 4:49
add a comment |
4
$begingroup$
Yeah your understanding is good, except your last sentence. If you're looking at $a^b$, where $b$ is transcendental, the issue isn't that "Gelfond Schneider does not necessarily hold". The conditions of the theorem haven't been met, since a transcendental number is not algebraic by definition!
$endgroup$
– Joe
Dec 21 '18 at 18:52
2
$begingroup$
@Joe Poor wording on my part. You're right that the issue isn't 'not holding,' but being irrelevant, which is what I was trying to get at.
$endgroup$
– Keith Backman
Dec 21 '18 at 18:59
1
$begingroup$
Your statement of the Gelfond-Schneider Theorem is incorrect. If $m,n$ are algebraic with $0ne mne 1$ and $nnot in Bbb Q$ then $m^n$ is transcendental. So if $p,q$ are unequal primes and $p^a=q$ then $ a not in Bbb Q,$ so by the G-S Theorem, if $a$ were algebraic then $p^a=q$ would be transcendental. So $a$ is not algebraic.
$endgroup$
– DanielWainfleet
Dec 22 '18 at 4:49
4
4
$begingroup$
Yeah your understanding is good, except your last sentence. If you're looking at $a^b$, where $b$ is transcendental, the issue isn't that "Gelfond Schneider does not necessarily hold". The conditions of the theorem haven't been met, since a transcendental number is not algebraic by definition!
$endgroup$
– Joe
Dec 21 '18 at 18:52
$begingroup$
Yeah your understanding is good, except your last sentence. If you're looking at $a^b$, where $b$ is transcendental, the issue isn't that "Gelfond Schneider does not necessarily hold". The conditions of the theorem haven't been met, since a transcendental number is not algebraic by definition!
$endgroup$
– Joe
Dec 21 '18 at 18:52
2
2
$begingroup$
@Joe Poor wording on my part. You're right that the issue isn't 'not holding,' but being irrelevant, which is what I was trying to get at.
$endgroup$
– Keith Backman
Dec 21 '18 at 18:59
$begingroup$
@Joe Poor wording on my part. You're right that the issue isn't 'not holding,' but being irrelevant, which is what I was trying to get at.
$endgroup$
– Keith Backman
Dec 21 '18 at 18:59
1
1
$begingroup$
Your statement of the Gelfond-Schneider Theorem is incorrect. If $m,n$ are algebraic with $0ne mne 1$ and $nnot in Bbb Q$ then $m^n$ is transcendental. So if $p,q$ are unequal primes and $p^a=q$ then $ a not in Bbb Q,$ so by the G-S Theorem, if $a$ were algebraic then $p^a=q$ would be transcendental. So $a$ is not algebraic.
$endgroup$
– DanielWainfleet
Dec 22 '18 at 4:49
$begingroup$
Your statement of the Gelfond-Schneider Theorem is incorrect. If $m,n$ are algebraic with $0ne mne 1$ and $nnot in Bbb Q$ then $m^n$ is transcendental. So if $p,q$ are unequal primes and $p^a=q$ then $ a not in Bbb Q,$ so by the G-S Theorem, if $a$ were algebraic then $p^a=q$ would be transcendental. So $a$ is not algebraic.
$endgroup$
– DanielWainfleet
Dec 22 '18 at 4:49
add a comment |
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$begingroup$
Yeah your understanding is good, except your last sentence. If you're looking at $a^b$, where $b$ is transcendental, the issue isn't that "Gelfond Schneider does not necessarily hold". The conditions of the theorem haven't been met, since a transcendental number is not algebraic by definition!
$endgroup$
– Joe
Dec 21 '18 at 18:52
2
$begingroup$
@Joe Poor wording on my part. You're right that the issue isn't 'not holding,' but being irrelevant, which is what I was trying to get at.
$endgroup$
– Keith Backman
Dec 21 '18 at 18:59
1
$begingroup$
Your statement of the Gelfond-Schneider Theorem is incorrect. If $m,n$ are algebraic with $0ne mne 1$ and $nnot in Bbb Q$ then $m^n$ is transcendental. So if $p,q$ are unequal primes and $p^a=q$ then $ a not in Bbb Q,$ so by the G-S Theorem, if $a$ were algebraic then $p^a=q$ would be transcendental. So $a$ is not algebraic.
$endgroup$
– DanielWainfleet
Dec 22 '18 at 4:49