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

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.

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?