"Alex J Van Wey" <kalishiva24@hotmail.com> wrote:
> Will someone explain these symbols for me?
>I've been "lurking" on this list for about 9 months, and I don't
>remember ever seeing an explanation for what looks vaguely like a
>molecular model or archetecture.
See http://www.cix.co.uk/~gbreed/lattice.htm
 Dave Keenan
http://dkeenan.com
Dear all,
I read the posts about lattices. I usually don't write lattices with 3:2 lying along one axis, instead I choose to write (apologies if mangled in advance)
etc etc etc
36/259/5  9/8 etc
  
6/5  3/2  15/8 etc
  
1/1  5/4  25/16 etc
  
5/3 25/24125/96 etc
A sevenlimit structure would stick out of the page with 7/6 lines lying over the 1/1 and 12/7 below. Higher dimensions need a little imagination.
My article has this with letter names, and also can be used for ET/EDO or whatever you want to call ntones per 2/1 scale. I wonder if
n /
\/ 2, with some abbreviation for the root could do? nR2, like 12R2 or 22R2. Then you could notate 12R3/2 for dividing the 3/2 into 12 equal parts. Then meantone becomes 2R5/4, sort of...
As for proving 3^m != 2^n:
Suppose they are equal.
3^m = 2^n
log both sides
mlog3 = nlog2
n/m = log3/log2
log3/log2 is irrational, so no integer ratio can ever equal that fraction, so no integer values for n and m exist that satisfy
3^m = 2^n
But y'all knew that anyway...
Mark
hi Mark,
(if using the Yahoo web interface, you have to use
"Message Index" and then "Expand Messages" to view this properly)
> From: <mark.gould@argonet.co.uk>
> To: <tuning@yahoogroups.com>
> Sent: Tuesday, February 19, 2002 12:23 AM
> Subject: [tuning] Lattices
>
>
> Dear all,
>
> I read the posts about lattices. I usually don't write
> lattices with 3:2 lying along one axis, instead I choose
> to write (apologies if mangled in advance)
>
> etc etc etc
> 36/259/5  9/8 etc
>   
> 6/5  3/2  15/8 etc
>   
> 1/1  5/4  25/16 etc
>   
> 5/3 25/24125/96 etc
that's essentially the same as the "triangular" lattices we
draw here ... you're just not drawing lines to connect the
3/2s, but the 3/2s on your lattice d o form a linear axis.
i use either rectangular or triangular lattices. my lattice
formula, which creates the graphics on my webpages, creates a
form of rectangular lattice. but usually when i post something
here i adopt the triangular format which is kind of a standard
for email.
primefactoring your lattice, the exponents of [3 5] gives you:
[ 2 2][ 2 1][ 2 0]
[ 1 1][ 1 0][ 1 1]
[ 0 0][ 0 1][ 0 2]
[1 1][1 2][1 3]
the simplest email version of my rectangular lattice,
as well as the format i use in my Excel spreadsheets, as at
http://www.ixpres.com/interval/dict/eqtemp.htm
follows the [3 5] pattern:
[1 1][ 0 1][ 1 1]
[1 0][ 0 0][ 1 0]
[1 1][ 0 1][ 1 1] etc.
or in simplest terms:
[ 0 +1]
[1 0][ 0 0][+1 0]
[ 0 1]
exponents of 3 increase/decrease along the horizontal axis,
and exponents of 5 increase/decrease along the vertical.
so in my rectangular format the notes in your lattice would go:
[1 3]
[1 2][ 0 2]
[1 1][ 0 1][ 1 1]
[ 0 0][ 1 0][ 2 0]
[ 1 1][ 2 1]
[ 2 2]
the ratio lattice of that would be:
125/96

25/24  25/16
 
5/3  5/4  15/8
 
1/1  3/2  9/8
 
6/5  9/5

36/25
the triangular format simply skews that a bit
so that the basic 5limit concordances, 6/5 and 5/4,
are like mirrorimages of each other with reference
to the 1/1; here's the [3 5] pattern:
[1 1]  [ 0 1]  [ 1 1]
/ \ / \ /
[1 0]  [ 0 0]  [ 1 0]
/ \ / \ /
[1 1]  [ 0 1]  [ 1 1] etc.
or in simplest terms:
[ 0 +1]
/
[1 0]  [ 0 0]  [+1 0]
/
[ 0 1]
so the typical "triangular" version of your lattice is:
[1 3]
/ \
[1 2]  [ 0 2]
/ \ / \
[1 1]  [ 0 1]  [ 1 1]
\ / \ / \
[ 0 0]  [ 1 0]  [ 2 0]
\ / \ /
[ 1 1]  [ 2 1]
\ /
[ 2 2]
and in ratios (we usually use : instead of /
to avoid confusion with the connecting lines)
125:96
/ \
/ \
25:2425:16
/ \ / \
/ \ / \
5:35:415:8
\ / \ / \
\ / \ / \
1:13:29:8
\ / \ /
\ / \ /
6:59:5
\ /
\ /
36:25
>
> A sevenlimit structure would stick out of the page with 7/6 lines lying
over the 1/1 and 12/7 below. Higher dimensions need a little imagination.
>
> My article has this with letter names, and also can be used for ET/EDO or
whatever you want to call ntones per 2/1 scale. I wonder if
>
> n /
> \/ 2, with some abbreviation for the root could do? nR2, like 12R2 or
22R2. Then you could notate 12R3/2 for dividing the 3/2 into 12 equal parts.
Then meantone becomes 2R5/4, sort of...
but our standard way of notating this stuff mathematically
is essentially equivalent to what you're proposing:
12R3/2 = (3/2)^(1/12)
2R5/4 = (5/4)^(1/2) etc.
monz
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HI Mark,
An easier way to show that 2^n can't equal 3^n is
to proceed by showing that any power of 3 is always odd, so can't
be divided by 2.
It's easier in the sense that you can prove it entirely
from first principles, while your proof needs extra
steps to show that the ratio of the two logs is indeed
irrational
What one is really doing here is using moduli (clock
arithmetic)
Recall  one writes 7 = 2 (mod 5) to mean that
7 has the remainder 2 on dividing by 5.
Start with 2, and the remainder on dividing by 2 is 0.
Multiply by 2, and mod 2, that's the same as multiplying
0 by 2, so result is again 0. So any power is the same
as 0 mod 2.
2^m = 0 (mod 2)
Now start with 3. That's the same as 1 mod 2. Multiply 1 by 3
and you get 3 again,which is 1 mod 2 again. So no matter how
many times you multiply by 3, the result is always the same
as 1 mod 2, so can never equal 0 mod 2, so can never be
even.
3^n = 1 (mod 2)
Same will work for any powers. E.g. 5 and 7. Work mod 5, then
5 = 0 mod 5, while 7 = 2 mod 5.
Now, 2*7 = 14 = 4 mod 5.
4*7 = 28 = 3 mod 5
3*7 = 21 = 1 mod 5
1*7 = 7 = 2 mod 5 again.
Note that we've gone through all the remainders that are possible
on dividing by 5 except for 0.
That's because of a group theory theorem, called Fermat's little
theorem.
http://www.cuttheknot.com/blue/Fermat.html
http://www.utm.edu/research/primes/notes/proofs/FermatsLittleTheorem.html
It's a neat theorem :).
By that theorem we have to go through all the possible remainders
other than 0, and can never reach 0 unless we start from it.
So if two numbers p and q are prime, p is non zero mod q,
so any power of p will also be non zero mod q, and so
can never equal a power of q, as those will all be zero
mod q.
In the case of our example, 7 is non zero mod 5, so any
power of 7 will also be non zero mod 5, and so can never
be a power of 5 as those will all be zero mod 5.
7^n !=0 (mod 5)
5^m = 0 (mod 5)
I think this is the standard number theoretic way of proving
the result, if I remember rightly, though it is a good many
years ago that I did it.
Robert
 In tuning@y..., "monz" <joemonz@y...> wrote:
> > n /
> > \/ 2, with some abbreviation for the root could do? nR2, like
12R2 or
> 22R2. Then you could notate 12R3/2 for dividing the 3/2 into 12
equal parts.
> Then meantone becomes 2R5/4, sort of...
>
>
> but our standard way of notating this stuff mathematically
> is essentially equivalent to what you're proposing:
>
> 12R3/2 = (3/2)^(1/12)
> 2R5/4 = (5/4)^(1/2) etc.
not really monz. we've used the notation 12ED(3/2), for example, for
mark's 12R3/2. it really means (3/2)^(n/12) for all integers n, not
just (3/2)^(1/12) as you indicated.
 In tuning@y..., mark.gould@a... wrote:
> Then meantone becomes 2R5/4, sort of...
4R5 (or 4ED5), with octave equivalents . . .
 In tuning@y..., "Robert Walker" <robertwalker@n...> wrote:
> I think this is the standard number theoretic way of proving
> the result, if I remember rightly, though it is a good many
> years ago that I did it.
Right. It's usually considered preferable to use a lowpowered theorem to get your result if one is available, though sometimes we are showing off the power of a highpowered theorem, or it gives us a very short proof, so we use it instead. In this case we could use the highoctane facts about the number theoretic properties of the logarithm, but elementary number theory gives the same result just as readily.
i meant to add that it should be easy to see that
these are all simply transformations of each other
in particular, Mark, the triangular lattice is
very similar to yours.
> From: monz <joemonz@yahoo.com>
> To: <tuning@yahoogroups.com>
> Sent: Tuesday, February 19, 2002 1:12 PM
> Subject: Re: [tuning] Lattices
>
>
> > From: <mark.gould@argonet.co.uk>
> > To: <tuning@yahoogroups.com>
> > Sent: Tuesday, February 19, 2002 12:23 AM
> > Subject: [tuning] Lattices
> >
> >
> > etc etc etc
> > 36/259/5  9/8 etc
> >   
> > 6/5  3/2  15/8 etc
> >   
> > 1/1  5/4  25/16 etc
> >   
> > 5/3 25/24125/96 etc
>
> ...
>
> 125/96
> 
> 25/24  25/16
>  
> 5/3  5/4  15/8
>  
> 1/1  3/2  9/8
>  
> 6/5  9/5
> 
> 36/25
> ...
>
> 125:96
> / \
> / \
> 25:2425:16
> / \ / \
> / \ / \
> 5:35:415:8
> \ / \ / \
> \ / \ / \
> 1:13:29:8
> \ / \ /
> \ / \ /
> 6:59:5
> \ /
> \ /
> 36:25
>
monz
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