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

On 11/15/2024 09:55 AM, Jim Burns wrote:

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:

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

>
Consider geometry.

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For two triangles △A′B′C′ and △A″B″C″
if
△A′B′C′ and △A″B″C″ are similar triangles

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then
corresponding sides are in the same ratio

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

>
Your writing is unreadable

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A geometric representation of
square.root, multiplication, and division exist.
One representation uses similar triangles.
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Also, a geometric representation of
addition, subtraction, and order exist.
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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.
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but that does not matter because
of course only a disproof is possible,
since there are no bijections.

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After all bijections are excluded,
of course there are no bijections.
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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.
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Setting aside for a moment
what you _think_ Cantor's bijection is,
what part of _that_
is impossible to represent geometrically?

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

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⟨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⟩
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According to geometry.
Which I predict makes geometry wrong[WM], too.
>
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Non-standard models of integers exist.
Russell's retro-thesis "ordinary infinity"
is sort of a lie - if there's infinity
it's extra-ordinary - somebody like Dana Scott
had introduced "circle" and "box" modalities,
because he was into modal logic and relevance logic,
and "each" is not always "all".
Of course it's well-known that any mere stipulation
is formally refutable, rather trivially. That
doesn't excuse absence of reason by any means.
"Statistics" does not "predict", though
"guesses" I suppose may be said -
retro-troll.