Re: 2N=E

On 10/22/2024 2:24 PM, WM wrote:

On 22.10.2024 19:24, Jim Burns wrote:

On 10/22/2024 10:11 AM, WM wrote:

On 22.10.2024 11:43, joes wrote:

The larger numbers have already
all been "created".

>
Not before multiplying them.

>
"Creation" of these numbers
is confusingly imagined to be
creation of these numbers.

>
These numbers did not belong to the set
when it was multiplied.

Any time and any place the set is a
well.ordered.successored.non.0.predecessored
set, each in the set has its double in the set.
The all.the.doubles _claim_
follows the
well.ordered.successored.non.0.predecessored
claim not.first.falsely,
so it can't be false.

These numbers are described by axioms.

>
By axiom also
all numbers of the set are present and
available to be multiplied.

And their products
( i×(j+1)=(i×j)+i, i+(j+1)=(i+j)+1, i≠j⇒i+1≠j+1 )
are in the
well.ordered.successored.non.0.predecessored
set.

The numbers and their doubles
always were/will.be being.described.by.those.axioms
always were/will.be existing.

>
They are existing when multiplied.

They are in the
well.ordered.successored.non.0.predecessored
set.

We cannot perform supertasks like
counting infinitely.many.

>
But we can use all existing numbers,
for instance for mappings.

We accept subsets of all triples of set.elements.
Do you accept them, too?
There is a set '×' of all ⟨i,j,k⟩ such that
i,j,k are in the
well.ordered.successored.non.0.predecessored
set  and  i×j=k