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mejercit@hotmail.com
science forum beginner
Joined: 21 Mar 2006
Posts: 9
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 Posted: Mon Apr 17, 2006 9:33 pm Post subject:
Question about the logarithm of 3 to the base 2
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It is obvious that the logarithm of 3 to the base 2 is not rational.
Is there any proof on whether or not this number is transcendental?
Michael |
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Proginoskes
science forum Guru
Joined: 29 Apr 2005
Posts: 2593
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| Posted: Mon Apr 17, 2006 10:19 pm Post subject:
Re: Question about the logarithm of 3 to the base 2
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Michael Ejercito wrote:
| Quote: | It is obvious that the logarithm of 3 to the base 2 is not rational.
Is there any proof on whether or not this number is transcendental?
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Yes. Lindemann's Theorem:
THEOREM. Suppose a(1), a(2), ..., a(k) are distinct algebraic numbers,
and b(1), ..., b(k) are rational numbers. Then, if
b(1) e^(a(1)) + b(2) e^(a(2)) + ... + b(k) e^(a(k)) = 0,
then all the b(i)'s are 0.
--- Christopher Heckman |
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Rob Johnson
science forum Guru
Joined: 26 May 2005
Posts: 318
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| Posted: Mon Apr 17, 2006 10:42 pm Post subject:
Re: Question about the logarithm of 3 to the base 2
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In article <1145309637.036927.135040@t31g2000cwb.googlegroups.com>,
"Michael Ejercito" <mejercit@hotmail.com> wrote:
| Quote: | It is obvious that the logarithm of 3 to the base 2 is not rational.
Is there any proof on whether or not this number is transcendental?
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It is pretty simple to show that log_2(3) is irrational. Suppose
log_2(3) = p/q, then 2^p = 3^q. By the Fundamental Theorem of
Arithmetic, p = q = 0.
Rob Johnson <rob@trash.whim.org>
take out the trash before replying
to view any ASCII art, display article in a monospaced font |
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The World Wide Wade
science forum Guru
Joined: 24 Mar 2005
Posts: 790
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| Posted: Mon Apr 17, 2006 10:42 pm Post subject:
Re: Question about the logarithm of 3 to the base 2
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In article
<1145309637.036927.135040@t31g2000cwb.googlegroups.com>,
"Michael Ejercito" <mejercit@hotmail.com> wrote:
| Quote: | It is obvious that the logarithm of 3 to the base 2 is not rational.
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If it were rational, then there would be positive integers m and
n such that 2^(m/n) = 3, or 2^m = 3^n -><-. |
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cablecar@gmail.com
science forum beginner
Joined: 17 Apr 2006
Posts: 1
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| Posted: Mon Apr 17, 2006 11:05 pm Post subject:
Re: Question about the logarithm of 3 to the base 2
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if log_2(3) is rational then:
log_2(3) = a/b a,b in Z and b is not 0 and a,b are relatively prime
antilog it to show
3 = 2^(a/b)
rearrange it so it looks like
ln3/ln2 = a/b
0 = aln2 - bln3
0 = ln( 2^a / 3^b )
so 2^a / 3^b = 1
note that all powers of 2 are even (2^a = 0 (mod 2) all a in Z+)
and all powers of 3 are odd (3^b = 1 (mod 2) all b in Z+)
so 0 / 1 = 1 (mod 2)
which is false so the original assumption is false.
Lambert'll probably give a better proof. |
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fermat
science forum beginner
Joined: 17 Apr 2006
Posts: 2
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| Posted: Mon Apr 17, 2006 11:29 pm Post subject:
Re: Question about the logarithm of 3 to the base 2
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1. ln(x) is transcendental if x is positive, rational and ≠1, i.e. ln(2) and ln(3) are transcendental.
2. Any non-constant algebraic function of a single variable yields a transcendental value when applied to a transcendental argument. So for example, from knowing that π is transcendental, we can immediately deduce that 5π, (π-3)/√2, (√π-√3)8 and (π5+7)1/7 are transcendental as well.
3. Therefore log2(3)=ln(3)/ln(2) is also transcendental. |
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Rob Johnson
science forum Guru
Joined: 26 May 2005
Posts: 318
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| Posted: Tue Apr 18, 2006 12:19 am Post subject:
Re: Question about the logarithm of 3 to the base 2
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In article <1145309637.036927.135040@t31g2000cwb.googlegroups.com>,
"Michael Ejercito" <mejercit@hotmail.com> wrote:
| Quote: | It is obvious that the logarithm of 3 to the base 2 is not rational.
Is there any proof on whether or not this number is transcendental?
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In article <22426652.1145316591301.JavaMail.jakarta@nitrogen.mathforum.org>,
fermat <phecht_@_dc.uba.ar> wrote:
| Quote: | 1. ln(x) is transcendental if x is positive, rational and != 1, i.e. ln(2) and ln(3) are transcendental.
2. Any non-constant algebraic function of a single variable yields a transcendental value when applied to a transcendental argument. So for example, from knowing that pi is transcendental, we can immediately deduce that 5pi, (pi-3)/sqrt(2), (sqrt(pi)-sqrt(3))8 and (pi5+7)1/7 are transcendental as well.
3. Therefore log2(3)=ln(3)/ln(2) is also transcendental.
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Please quote relevant parts of the article to which you are replying.
If 3 is supposed to follow from 1 and 2, please explain how.
Rob Johnson <rob@trash.whim.org>
take out the trash before replying
to view any ASCII art, display article in a monospaced font |
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Gerry Myerson
science forum Guru
Joined: 28 Apr 2005
Posts: 871
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| Posted: Tue Apr 18, 2006 12:21 am Post subject:
Re: Question about the logarithm of 3 to the base 2
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In article
<22426652.1145316591301.JavaMail.jakarta@nitrogen.mathforum.org>,
fermat <phecht_@_dc.uba.ar> wrote:
| Quote: | 1. ln(x) is transcendental if x is positive, rational and ‚ 1, i.e. ln(2) and
ln(3) are transcendental.
2. Any non-constant algebraic function of a single variable yields a
transcendental value when applied to a transcendental argument. So for
example, from knowing that ¼ is transcendental, we can immediately deduce
that 5¼, (¼-3)/ˆ2, (ˆ¼-ˆ3)8 and (¼5+7)1/7 are transcendental as well.
3. Therefore log2(3)=ln(3)/ln(2) is also transcendental.
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Everything above is correct, except for the "therefore".
--
Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) |
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The World Wide Wade
science forum Guru
Joined: 24 Mar 2005
Posts: 790
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| Posted: Tue Apr 18, 2006 12:28 am Post subject:
Re: Question about the logarithm of 3 to the base 2
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In article
<waderameyxiii-D285C0.15422117042006@comcast.dca.giganews.com>,
The World Wide Wade <waderameyxiii@comcast.remove13.net> wrote:
| Quote: | In article
1145309637.036927.135040@t31g2000cwb.googlegroups.com>,
"Michael Ejercito" <mejercit@hotmail.com> wrote:
It is obvious that the logarithm of 3 to the base 2 is not rational.
If it were rational, then there would be positive integers m and
n such that 2^(m/n) = 3, or 2^m = 3^n -><-.
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My mistake, I read "It is" as "Is it". |
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mensanator@aol.compost
science forum Guru
Joined: 24 Mar 2005
Posts: 826
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| Posted: Tue Apr 18, 2006 12:42 am Post subject:
Re: Question about the logarithm of 3 to the base 2
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Gerry Myerson wrote:
| Quote: | In article
22426652.1145316591301.JavaMail.jakarta@nitrogen.mathforum.org>,
fermat <phecht_@_dc.uba.ar> wrote:
1. ln(x) is transcendental if x is positive, rational and , 1, i.e. ln(2) and
ln(3) are transcendental.
2. Any non-constant algebraic function of a single variable yields a
transcendental value when applied to a transcendental argument. So for
example, from knowing that ¼ is transcendental, we can immediately deduce
that 5¼, (¼-3)/ˆ2, (ˆ¼-ˆ3)8 and (¼5+7)1/7 are transcendental as well.
3. Therefore log2(3)=ln(3)/ln(2) is also transcendental.
Everything above is correct, except for the "therefore".
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Is it because there are two transcendental arguments, not
"a transcendental argument"?
| Quote: |
--
Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) |
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David McAnally
science forum Guru Wannabe
Joined: 08 May 2005
Posts: 136
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| Posted: Tue Apr 18, 2006 12:49 am Post subject:
Re: Question about the logarithm of 3 to the base 2
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"Proginoskes" <CCHeckman@gmail.com> writes:
| Quote: | Michael Ejercito wrote:
It is obvious that the logarithm of 3 to the base 2 is not rational.
Is there any proof on whether or not this number is transcendental?
Yes. Lindemann's Theorem:
THEOREM. Suppose a(1), a(2), ..., a(k) are distinct algebraic numbers,
and b(1), ..., b(k) are rational numbers. Then, if
b(1) e^(a(1)) + b(2) e^(a(2)) + ... + b(k) e^(a(k)) = 0,
then all the b(i)'s are 0.
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The statement of Lindemann's Theorem above is wrong. The actual statement
is stronger, because you only require b(1), ..., b(k) to be algebraic
numbers.
How do you get to the transcendence of log_2(3) from Lindemann's Theorem?
You can prove the transcendence of log_2(3) from the Gelfond-Schneider
Theorem: if a(1), a(2), ..., a(k) are algebraic numbers, and the values of
log(a(1)), log(a(2)), ..., log(a(k)) are linearly independent over the
rational numbers, then 1, log(a(1)), ..., log(a(k)) are linearly
independent over the algebraic numbers.
----- |
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David McAnally
science forum Guru Wannabe
Joined: 08 May 2005
Posts: 136
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| Posted: Tue Apr 18, 2006 12:55 am Post subject:
Re: Question about the logarithm of 3 to the base 2
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fermat <phecht_@_dc.uba.ar> writes:
| Quote: | 1. ln(x) is transcendental if x is positive, rational and not equal to
1, i.e. ln(2) and ln(3) are transcendental.
2. Any non-constant algebraic function of a single variable yields a
transcendental value when applied to a transcendental argument.
3. Therefore log2(3)=ln(3)/ln(2) is also transcendental.
|
Statement 3 does not follow from statements 1 and 2. For example,
1. ln(x) is transcendental if x is positive, rational and not equal to
1, i.e. ln(2) and ln(4) are transcendental.
2. Any non-constant algebraic function of a single variable yields a
transcendental value when applied to a transcendental argument.
3. log2(4)=ln(4)/ln(2) = 2 is algebraic.
Your problem was that you were taking an algebraic function of TWO
transcendental arguments. Therefore your observation 2 was no longer
applicable.
The answer to the original question is to use the Gelfond-Schneider
Theorem.
----- |
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Chip Eastham
science forum Guru
Joined: 01 May 2005
Posts: 412
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| Posted: Tue Apr 18, 2006 2:25 am Post subject:
Re: Question about the logarithm of 3 to the base 2
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mensanator@aol.compost wrote:
| Quote: | Gerry Myerson wrote:
In article
22426652.1145316591301.JavaMail.jakarta@nitrogen.mathforum.org>,
fermat <phecht_@_dc.uba.ar> wrote:
1. ln(x) is transcendental if x is positive, rational and , 1, i.e. ln(2) and
ln(3) are transcendental.
2. Any non-constant algebraic function of a single variable yields a
transcendental value when applied to a transcendental argument. So for
example, from knowing that ¼ is transcendental, we can immediately deduce
that 5¼, (¼-3)/ˆ2, (ˆ¼-ˆ3)8 and (¼5+7)1/7 are transcendental as well.
3. Therefore log2(3)=ln(3)/ln(2) is also transcendental.
Everything above is correct, except for the "therefore".
Is it because there are two transcendental arguments, not
"a transcendental argument"?
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Yes, for example ln(2) and ln(4) transcendental tell us what
about ln(4)/ln(2)?
Sorry to beat the horse. I assure you it is not dead, only
sleeping.
-- c |
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Proginoskes
science forum Guru
Joined: 29 Apr 2005
Posts: 2593
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| Posted: Tue Apr 18, 2006 6:32 am Post subject:
Re: Question about the logarithm of 3 to the base 2
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David McAnally wrote:
| Quote: | "Proginoskes" <CCHeckman@gmail.com> writes:
Michael Ejercito wrote:
It is obvious that the logarithm of 3 to the base 2 is not rational.
Is there any proof on whether or not this number is transcendental?
Yes. Lindemann's Theorem:
THEOREM. Suppose a(1), a(2), ..., a(k) are distinct algebraic numbers,
and b(1), ..., b(k) are rational numbers. Then, if
b(1) e^(a(1)) + b(2) e^(a(2)) + ... + b(k) e^(a(k)) = 0,
then all the b(i)'s are 0.
The statement of Lindemann's Theorem above is wrong. The actual statement
is stronger, because you only require b(1), ..., b(k) to be algebraic
numbers.
How do you get to the transcendence of log_2(3) from Lindemann's Theorem?
You can prove the transcendence of log_2(3) from the Gelfond-Schneider
Theorem: if a(1), a(2), ..., a(k) are algebraic numbers, and the values of
log(a(1)), log(a(2)), ..., log(a(k)) are linearly independent over the
rational numbers, then 1, log(a(1)), ..., log(a(k)) are linearly
independent over the algebraic numbers.
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That's right ... I got the wrong theorem again. (Maybe one of these
days I'll remember it correctly. See
http://groups.google.com/group/sci.math/msg/1245933ff1bb80e9?dmode=source
for another time I got it wrong.)
--- Christopher Heckman |
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mejercit@hotmail.com
science forum beginner
Joined: 21 Mar 2006
Posts: 9
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| Posted: Sun May 07, 2006 5:55 pm Post subject:
Re: Question about the logarithm of 3 to the base 2
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David McAnally wrote:
| Quote: | fermat <phecht_@_dc.uba.ar> writes:
1. ln(x) is transcendental if x is positive, rational and not equal to
1, i.e. ln(2) and ln(3) are transcendental.
2. Any non-constant algebraic function of a single variable yields a
transcendental value when applied to a transcendental argument.
3. Therefore log2(3)=ln(3)/ln(2) is also transcendental.
Statement 3 does not follow from statements 1 and 2. For example,
1. ln(x) is transcendental if x is positive, rational and not equal to
1, i.e. ln(2) and ln(4) are transcendental.
2. Any non-constant algebraic function of a single variable yields a
transcendental value when applied to a transcendental argument.
3. log2(4)=ln(4)/ln(2) = 2 is algebraic.
Your problem was that you were taking an algebraic function of TWO
transcendental arguments. Therefore your observation 2 was no longer
applicable.
The answer to the original question is to use the Gelfond-Schneider
Theorem.
-----
the Gelfdonf Schneider-Theorem shows that a^b is transcendental if a |
is an algebraic number and b is algebraic and irrational number.
The logarithm of 3 to the base 2 is irrational because if it were
rational, there would be two integers m and n with no common factors
such that 2^(m/n)=3 which means 2^m=3^m, which is impossible for any
two integers. So log 2(3) is irrational.
If log 2(3) is algebraic, then the Gelfond-Schneider theorem implies
that 2^(log2(3)) is transcendental, since 2 is algebraic (it is a
solution of x^2-4=0) and log2(3) is ASSUMED to be algebraic.
2^(log2(3))=3, which is a solution to x^3-27=0. Therefore, log2(3)
can not be algebraic and must be transcendental.
I wonder about the natural logarithm of 3 (ln(3)). The base of the
natural logarithm is not algebraic.
Michael |
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