Re: [tuning] Commas and the ABC conjecture

----- Original Message -----
From: <genewardsmith@juno.com>
To: <tuning@yahoogroups.com>
Sent: Friday, September 28, 2001 8:27 PM
Subject: [tuning] Commas and the ABC conjecture

> I had thought of examining the Farey sequence for useful intervals,
> but it occurred to me that the same thing may be accomplised more
> easily simply by looking at ratios belonging to a p-limit. If b is a
> fixed small integer, it follows from a very deep and recent
> conjecture in elementary number theory called the ABC conjecture that
> in the p-limit, there will be only a finite number of ratios (a+b)/a.

Is there any mathematical proof for this?

> From this we get the following list:
>
> 3-limit
>
> 2, 3/2, 4/3, 9/8
>
> 5-limit
>
> 5/4, 6/5, 10/9, 16/15, 25/24, 81/80
>
> 7-limit
>
> 7/6, 8/7, 15/14, 21/20, 28/27, 36/35, 49/48, 50/49, 64/63, 126/125,
> 225/224, 2401/2400, 4375/4374
>
> 11-limit
>
> 11/10, 12/11, 22/21, 33/32, 45/44, 55/54, 56/55, 99/98, 100/99,
> 121/120, 176/175, 243/242, 385/384, 441/440, 540/539, 3025/3024,
> 9801/9800
>
> There is no guarantee that these lists are complete, or even finite,

I can give a proof that the first list is complete:

Let's consider the 3-limit interval as

a=2^c
b=3^d

where c & d are integers
The interval is

b/a where a=b-1

or a/b where b=a-1

a=b-1 ==>
(1) 2^c = ( 3^d ) -1

if d is odd let's put d1= d-1 then 2^c = ( 3^d1 ) * 3 -1 ==> 2^c -2 = ( 3^d1 ) * 3 -3

where d1 then is even.

==> (2 ^ ( c-1 ) -1 ) * 2 = (( 3 ^ d1 ) -1 ) * 3

this means ((3 ^ d1) -1 ) = [( 3^ (d1/2) ) -1 ] * [ ( 3 ^ ( d1/2) ) +1 ] is even (2).
and unless c =1 we have [ ( 3^ ( d1/2 ) ) -1 ] * [ ( 3 ^ ( d1/2 ) ) +1 ] /2 is odd value

Let's put :
f = [ ( 3 ^ ( d1/2 ) ) -1 ]
g= [ ( 3 ^ ( d1/2 ) ) +1 ]

g * f / 2 is odd value means g or f is odd , but here we have f = g +2 this means f & g are both odd which means f * g is odd which cannot be ( from ( 2) )

c=1 means ( 3^d ) -1 =2 means (3^d) = 3 means d =1. the result 3/2

Now if d is even, from (1):

2^c = [ ( 3^(d/2) ) -1 ] * [ ( 3^(d/2) ) +1 ]
=e =f
e = 2^ee

f= 2^ff

2^ee = 2^ff + 2

2^(ee-1) = 2^(ff-1) + 1

this cannot be unless 2^(ee-1) =1 ==> ee-1 = 0 ==> ee =1 ==> e=2, f=4 ==> c=3 the result is 9/8
or ( 2^(ff-1) ) =1 ==> ff-1=0 ==> ff=1 ==> f=2, e=0 ==> 2^c= 0 connot be

for b=a-1... I'm sorry it is exhausting, do it yourself!

Rami V.

--- In tuning@y..., "Rami Vitale" <alfred1@s...> wrote:

> > it follows from a very deep and recent
> > conjecture in elementary number theory called the ABC conjecture
that
> > in the p-limit, there will be only a finite number of ratios
(a+b)/a.

> Is there any mathematical proof for this?

There isn't for the ABC conjecture, which is why it is a conjecture.
The question of how strong the above statement is, and if it is
provable without ABC, is an interesting one to a number theorist if
not, perhaps, to the people on this list. This was posted here by
accident, and I cancelled it, but once again something I cancelled
was followed up anyway!

I think it might be possible to prove it with Baker's theorem; I'll
think about that.

> I can give a proof that the first list is complete:

It is indeed, and let me see if I can complete and clarify your proof.

If 2^n = 3^d - 1, and d = 2m is even, then 2^n = (3^m + 1)(3^m - 1),
so both 3^m + 1 and 3^m - 1 are powers of 2; this can only happen
when m=1, so we get the interval 9/8. If d is odd, then
(3^d - 1)/2 = 1 + 3 + ... + 3^(d-1) is an odd sum of odd numbers, and
hence is odd; the only odd power of 2 is 2^0 = 1, and we have found
the interval 3/2.

If 3^n = 2^d - 1 and d is odd, then since 2 is congruent to -1 mod 3,
2^d is congruent to -1 and 2^d-1 is congruent to 1; the only power of
3 congruent to 1 is 3^0 = 1, and we have found the interval 2/1.
Finally, if d = 2m is even then 3^n = (2^m + 1)(2^m - 1); both
factors are powers of 3 and this can only happen if m=1, and we have
found the interval 4/3; this completes the list.

I intended to post this on tuning-math, where another discussion of
this posting is taking place.

--- In tuning@y..., genewardsmith@j... wrote:

> I think it might be possible to prove it with Baker's theorem; I'll
> think about that.

I thought about it and Baker's theorem does the trick--details on the
math group. The thing to remember is that in any p-limit, there are
only a finite number of ratios p/q where |p - q| < D, for some fixed
bound D.