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

On 11/15/2024 5:10 AM, WM wrote:

On 14.11.2024 19:31, Jim Burns wrote:

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

then
corresponding sides are in the same ratio

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

>
Your writing is unreadable

A geometric representation of
square.root, multiplication, and division exist.
One representation uses similar triangles.
Also, a geometric representation of
addition, subtraction, and order exist.
Cantor's bijection ⟨i,j⟩ ↦ k ↦ ⟨i,j⟩
⎛ k = (i+j-1)⋅(i+j-2)/2+i
⎜ i = k-⌈(2⋅k+¼)¹ᐟ²-1/2⌉⋅⌈(2⋅k+¼)¹ᐟ²-3/2⌉/2
⎝ j = ⌈(2⋅k+¼)¹ᐟ²+1/2⌉⋅⌈(2⋅k+¼)¹ᐟ²-1/2⌉/2-1-k
is composed of
square.root, multiplication, division, addition,
subtraction, and ⌈ceiling⌉ (order),
for all of which geometric representations exist.

but that does not matter because
of course only a disproof is possible,
since there are no bijections.

After all bijections are excluded,
of course there are no bijections.
On the other hand,
⎛ k = (i+j-1)⋅(i+j-2)/2+i
⎜ i = k-⌈(2⋅k+¼)¹ᐟ²-1/2⌉⋅⌈(2⋅k+¼)¹ᐟ²-3/2⌉/2
⎝ j = ⌈(2⋅k+¼)¹ᐟ²+1/2⌉⋅⌈(2⋅k+¼)¹ᐟ²-1/2⌉/2-1-k
exists.

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.

⟨k,1⟩ ↦ ⟨i,j⟩ ↤ ⟨k,1⟩
⎛ i = k-⌈(2⋅k+¼)¹ᐟ²-1/2⌉⋅⌈(2⋅k+¼)¹ᐟ²-3/2⌉/2
⎜ j = ⌈(2⋅k+¼)¹ᐟ²+1/2⌉⋅⌈(2⋅k+¼)¹ᐟ²-1/2⌉/2-1-k
⎝ k = (i+j-1)⋅(i+j-2)/2+i
Each ⟨k,1⟩ sends X to ⟨i,j⟩
Each ⟨i,j⟩ receives X from ⟨k,1⟩
According to geometry.
Which I predict makes geometry wrong[WM], too.