I've been having a discussion with a friend in private email

about the business of numbers with fractional exponents

not following the Fundamental Theorem of Arithmetic. He

was generous enough to send me a very long and detailed

explanation, and Paul has gone over this with me more concisely

in the past, but in spite of all of the explanation, I still

don't get it.

I understand that many different combinations of prime-factors

and fractional exponents can be found which *approach any

floating-point value arbitrarily closely*, but EXACT values

are STILL INCOMMENSURABLE!!

Why must numbers of the form a^(b/c) be understood in

terms of their floating-point decimal value? If we stick

to the a^(b/c) form we get exact values and can manipulate

them the same way as our regular rational numbers of the

form x^y.

If b and c in a^(b/c) are always integers, the simple

calculations needed for tuning math always gives results

which are also integers, so there's never any error at all.

I've been trying to understand this for three years...

someone please help.

-monz

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--- In tuning-math@y..., "monz" <joemonz@y...> wrote:

>

> I've been having a discussion with a friend in private email

> about the business of numbers with fractional exponents

> not following the Fundamental Theorem of Arithmetic. He

> was generous enough to send me a very long and detailed

> explanation, and Paul has gone over this with me more concisely

> in the past, but in spite of all of the explanation, I still

> don't get it.

>

> I understand that many different combinations of prime-factors

> and fractional exponents can be found which *approach any

> floating-point value arbitrarily closely*, but EXACT values

> are STILL INCOMMENSURABLE!!

>

> Why must numbers of the form a^(b/c) be understood in

> terms of their floating-point decimal value? If we stick

> to the a^(b/c) form we get exact values and can manipulate

> them the same way as our regular rational numbers of the

> form x^y.

>

> If b and c in a^(b/c) are always integers, the simple

> calculations needed for tuning math always gives results

> which are also integers, so there's never any error at all.

JG: Monz, can you give an example of a prime number

taken to a rational power (where the power's numerator and

denominator are integer) which (directly, as numerically

evaluated)equals an integer? I couldn't. Of course, the

rational valued exponent must not be equal (itself) to

an integer value...

>

> I've been trying to understand this for three years...

> someone please help.

JG: You're welcome to post my entire email from yesterday.

J Gill

>

>

>

> -monz

>

>

>

>

>

> _________________________________________________________

> Do You Yahoo!?

> Get your free @yahoo.com address at http://mail.yahoo.com

> From: unidala <JGill99@imajis.com>

> To: <tuning-math@yahoogroups.com>

> Sent: Saturday, December 29, 2001 10:23 AM

> Subject: [tuning-math] Re: non-uniqueness of a^(b/c) type numbers

>

> > [me, monz]

> > If b and c in a^(b/c) are always integers, the simple

> > calculations needed for tuning math always gives results

> > which are also integers, so there's never any error at all.

>

> JG: Monz, can you give an example of a prime number

> taken to a rational power (where the power's numerator and

> denominator are integer) which (directly, as numerically

> evaluated)equals an integer? I couldn't. Of course, the

> rational valued exponent must not be equal (itself) to

> an integer value...

No, I don't think I can give an example of that either.

But does it matter?

I'm simply having a hard time understanding how

a^(b/c) can possibly equal d^(e/f) EXACTLY.

As far as I can see, there are no six integers

that will satisfy that equation.

-monz

_________________________________________________________

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monz wrote:

> I'm simply having a hard time understanding how

> a^(b/c) can possibly equal d^(e/f) EXACTLY.

>

> As far as I can see, there are no six integers

> that will satisfy that equation.

Um, how about 2^(2/3) = 2^(4/6) = 4^(1/3)

Graham

> From: <graham@microtonal.co.uk>

> To: <tuning-math@yahoogroups.com>

> Sent: Saturday, December 29, 2001 12:50 PM

> Subject: [tuning-math] Re: non-uniqueness of a^(b/c) type numbers

>

>

> monz wrote:

>

> > I'm simply having a hard time understanding how

> > a^(b/c) can possibly equal d^(e/f) EXACTLY.

> >

> > As far as I can see, there are no six integers

> > that will satisfy that equation.

>

> Um, how about 2^(2/3) = 2^(4/6) = 4^(1/3)

Oops! So did I really mean to write:

"no six *coprime* integers that will satisfy that equation"?

Help... I'm sinking...

-monz

_________________________________________________________

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--- In tuning-math@y..., "monz" <joemonz@y...> wrote:

> I understand that many different combinations of prime-factors

> and fractional exponents can be found which *approach any

> floating-point value arbitrarily closely*, but EXACT values

> are STILL INCOMMENSURABLE!!

I'm not sure what you mean by this, in any sense which would not also be true of ordinary rational numbers.

> Why must numbers of the form a^(b/c) be understood in

> terms of their floating-point decimal value? If we stick

> to the a^(b/c) form we get exact values and can manipulate

> them the same way as our regular rational numbers of the

> form x^y.

Since a>0, there is no problem interpreting these exponents. If you want to use prime factorization, you get a vector space over the rational numbers from the exponents, which in some ways is easier to treat than the abelian group you get from confining yourself to the rationals, with integer exponents.

--- In tuning-math@y..., "unidala" <JGill99@i...> wrote:

> JG: Monz, can you give an example of a prime number

> taken to a rational power (where the power's numerator and

> denominator are integer) which (directly, as numerically

> evaluated)equals an integer? I couldn't. Of course, the

> rational valued exponent must not be equal (itself) to

> an integer value...

If x = p^(a/b), where a/b is in its lowest terms, then

x^b = p^a, and x^b - p^a = 0 is an irreducible polynomial of degree b. Any solution will therefore be an algebraic integer (since p is an integer) but not an ordinary "rational integer".

In other words, it can't happen. :)

--- In tuning-math@y..., "monz" <joemonz@y...> wrote:

> Oops! So did I really mean to write:

> "no six *coprime* integers that will satisfy that equation"?

The first leads to the polynomial x^c - a^b = 0, and the second to

x^f - d^e = 0. Since gcd(a,d)=1 they have no common factor.

--- In tuning-math@y..., "monz" <joemonz@y...> wrote:

>

> I've been having a discussion with a friend in private email

> about the business of numbers with fractional exponents

> not following the Fundamental Theorem of Arithmetic. He

> was generous enough to send me a very long and detailed

> explanation, and Paul has gone over this with me more concisely

> in the past, but in spite of all of the explanation, I still

> don't get it.

>

> I understand that many different combinations of prime-factors

> and fractional exponents can be found which *approach any

> floating-point value arbitrarily closely*, but EXACT values

> are STILL INCOMMENSURABLE!!

>

> Why must numbers of the form a^(b/c) be understood in

> terms of their floating-point decimal value? If we stick

> to the a^(b/c) form we get exact values and can manipulate

> them the same way as our regular rational numbers of the

> form x^y.

>

> If b and c in a^(b/c) are always integers, the simple

> calculations needed for tuning math always gives results

> which are also integers, so there's never any error at all.

>

> I've been trying to understand this for three years...

> someone please help.

I'm not sure what you're trying to understand here, so please clarify.

--- In tuning-math@y..., "monz" <joemonz@y...> wrote:

> No, I don't think I can give an example of that either.

> But does it matter?

>

> I'm simply having a hard time understanding how

> a^(b/c) can possibly equal d^(e/f) EXACTLY.

>

> As far as I can see, there are no six integers

> that will satisfy that equation.

You mean, if a and d are prime? Looks about right. Now let's look at

a more relevant example, how about 7/26-comma meantone. How did you

express the fifth and the major third of this tuning again?

> From: paulerlich <paul@stretch-music.com>

> To: <tuning-math@yahoogroups.com>

> Sent: Saturday, December 29, 2001 4:00 PM

> Subject: [tuning-math] Re: non-uniqueness of a^(b/c) type numbers

>

>

> --- In tuning-math@y..., "monz" <joemonz@y...> wrote:

>

> > No, I don't think I can give an example of that either.

> > But does it matter?

> >

> > I'm simply having a hard time understanding how

> > a^(b/c) can possibly equal d^(e/f) EXACTLY.

> >

> > As far as I can see, there are no six integers

> > that will satisfy that equation.

>

> You mean, if a and d are prime? Looks about right. Now let's look at

> a more relevant example, how about 7/26-comma meantone. How did you

> express the fifth and the major third of this tuning again?

In each vector [a b c] below,

a = exponent of prime-factor 2

b = exponent of prime-factor 3

c = exponent of prime-factor 5

7/26-comma meantone

-------------------

"5th" =

[-1*(26/26) 1*(26/26) 0*(26/26)] = 3/2 = Pythagorean "perfect 5th"

- [-4*(7/26) 4*(7/26) -1*(7/26) ] = (81/80)^(7/26) = 7/26-comma

-----------------------------------

[ 2/26 -2/26 7/26] = (3/2)/ ((81/80)^(7/26))

= [ 1/13 -1/13 7/26] = 7/26-comma meantone "5th"

"major 3rd" =

[ 1/13 -1/13 7/26] = 7/26-comma meantone "5th"

* [ 4/1 4/1 4/1 ] = 4 times (= 4 generators)

-----------------------

[ 4/13 -4/13 28/26] = ~5/1 = 7/26-meantone "5th harmonic"

= [ 4/13 -4/13 14/13] (= reduced)

= [ 4/13 -4/13 14/13] = 7/26-meantone "5th harmonic"

- [ 2*(13/13) 0 0 ] = subtract 2 "8ves"

------------------------

[ -22/13 -4/13 14/13] = 7/26-comma meantone "major 3rd"

And how about throwing in the 55-tone 1/6-comma meantone

as well, since I'm actually exploring that a lot lately?

1/6-comma meantone

-------------------

"5th" =

[-1*(6/6) 1*(6/6) 0*(6/6)] = 3/2 = Pythagorean "perfect 5th"

- [-4*(1/6) 4*(1/6) -1*(1/6)] = (81/80)^(7/26) = 1/6-comma

-------------------------------

[ -2/6 2/6 1/6 ] = (3/2)/ ((81/80)^(1/6))

= [ -1/3 1/3 1/6 ] = 1/6-comma meantone "5th"

"major 3rd" =

[-1/3 1/3 1/6] = 1/6-comma meantone "5th"

* [ 4/1 4/1 4/1] = 4 times (= 4 generators)

-----------------------

[-4/3 4/3 4/6] = ~5/1 = 1/6-meantone 5th harmonic

= [-4/3 4/3 2/3] (= reduced)

= [ -4/3 4/3 2/3] = 1/6-meantone 5th harmonic

- [2*(3/3) 0 0 ] = subtract 2 "8ves"

------------------------

[ -10/3 4/3 2/3] = 1/6-comma meantone "major 3rd

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--- In tuning-math@y..., "monz" <joemonz@y...> wrote:

> > From: paulerlich <paul@s...>

> > To: <tuning-math@y...>

> > Sent: Saturday, December 29, 2001 4:00 PM

> > Subject: [tuning-math] Re: non-uniqueness of a^(b/c) type numbers

> >

> >

> > --- In tuning-math@y..., "monz" <joemonz@y...> wrote:

> >

> > > No, I don't think I can give an example of that either.

> > > But does it matter?

> > >

> > > I'm simply having a hard time understanding how

> > > a^(b/c) can possibly equal d^(e/f) EXACTLY.

> > >

> > > As far as I can see, there are no six integers

> > > that will satisfy that equation.

> >

> > You mean, if a and d are prime? Looks about right. Now let's look

at

> > a more relevant example, how about 7/26-comma meantone. How did

you

> > express the fifth and the major third of this tuning again?

>

>

> In each vector [a b c] below,

>

> a = exponent of prime-factor 2

> b = exponent of prime-factor 3

> c = exponent of prime-factor 5

>

>

>

> 7/26-comma meantone

> -------------------

>

> "5th" =

>

> [-1*(26/26) 1*(26/26) 0*(26/26)] = 3/2 =

Pythagorean "perfect 5th"

> - [-4*(7/26) 4*(7/26) -1*(7/26) ] = (81/80)^(7/26) = 7/26-

comma

> -----------------------------------

> [ 2/26 -2/26 7/26] = (3/2)/ ((81/80)^(7/26))

> = [ 1/13 -1/13 7/26] = 7/26-comma meantone "5th"

>

>

>

>

> "major 3rd" =

>

>

> [ 1/13 -1/13 7/26] = 7/26-comma meantone "5th"

> * [ 4/1 4/1 4/1 ] = 4 times (= 4 generators)

> -----------------------

> [ 4/13 -4/13 28/26] = ~5/1 = 7/26-meantone "5th harmonic"

> = [ 4/13 -4/13 14/13] (= reduced)

>

>

> = [ 4/13 -4/13 14/13] = 7/26-meantone "5th harmonic"

> - [ 2*(13/13) 0 0 ] = subtract 2 "8ves"

> ------------------------

> [ -22/13 -4/13 14/13] = 7/26-comma meantone "major 3rd"

Well, now I'm confused, because I thought that when you were

latticing 7/26-comma meantone using this method (and ignoring

octaves), you were somehow getting a two-dimensional arrangement of

points. Is that not correct?