Re: Incompleteness of Cantor's enumeration of the rational numbers

On 14.11.2024 19:31, Jim Burns wrote:

On 11/14/2024 5:20 AM, WM wrote:

On 14.11.2024 00:16, Jim Burns wrote:

 

Therefore
a geometric representation let alone proof of
most of Cantor's bijections is impossible.

 Consider geometry.
 For two triangles △A′B′C′ and △A″B″C″
if
△A′B′C′ and △A″B″C″ are similar triangles
⎛ μ∠C′A′B′ = μ∠C″A″B″
⎜ μ∠A′B′C′ = μ∠A″B″C″
⎜ μ∠B′C′A′ = μ∠B″C″A″
⎝ △A′B′C′ ∼ △A″B″C″
then
corresponding sides are in the same ratio
( μA︫︭′︭B︫′/μA︫︭″︭B︫″ = μB︫︭′︭C︫′/μB︫︭″︭C︫″ = μA︫︭′︭C︫′/μA︫︭″︭C︫″
 For similar △A′B′C′ ∼ △A″B″C″
if μA︫︭′︭B︫′ = 1
  and μB︫︭″︭C︫″ = x
then 1/μA︫︭″︭B︫″ = μB︫︭′︭C︫′/x
  and μA︫︭″︭B︫″ = μB︫︭′︭C︫′ = x¹ᐟ²
 For similar △A′B′C′ ∼ △A″B″C″
if μA︫︭′︭B︫′ = 1
  and μA︫︭″︭B︫″ = x
  and μB︫︭′︭C︫′ = y
then 1/x = y/μB︫︭″︭C︫″
  and μB︫︭″︭C︫″ = x⋅y
 For similar △A′B′C′ ∼ △A″B″C″
if μA︫︭′︭B︫′ = 1
  and μA︫︭″︭B︫″ = x
  and μB︫︭″︭C︫″ = y
then 1/x = μB︫︭′︭C︫′/y
  and μB︫︭′︭C︫′ = y/x
 The ceiling ⌈x⌉ of x is the first integer ≥ x
 

Therefore
a geometric representation let alone proof of
most of Cantor's bijections is impossible.

Your writing is unreadable but that does not matter because of course only a disproof is possible, since there are no bijections. One simple disproof is based on these intervals I(n) = [n - 1/10, n + 1/10].

 Cantor's bijection is
k ↦ i/j
i︭+︭j := ⌈(2⋅k+¼)¹ᐟ²+½⌉
i := k-(i︭+︭j-1)⋅(i︭+︭j-2)/2
j := i︭+︭j-i
(i+j-1)⋅(i+j-2)/2+i = k
 Setting aside for a moment
what you _think_ Cantor's bijection is,
what part of _that_
is impossible to represent geometrically?

It is impossible to cover the matrix
XOOO...
XOOO...
XOOO...
XOOO...
...
by shuffling, shifting, reordering the X, because they are not distinguishable.
Regards, WM