Are "dark" numbers the same as non-standard numbers?

[David Petry]

I don't understand WM's explanation of his intuition behind "dark" integers, but maybe they're the same as "non-standard" integers.

Maybe this is his intuition:

Consider the sequence of sets S_n = {n, n+1, ... 2n-1}

If that sequence has a limit, then that limit must contain no standard integers, but must contain infinitely many integers that are bigger than any standard integer, and these "bigger" integers can be called "dark" integers, or perhaps "non-standard" integers.

I see no reason to believe that such an intuition couldn't be formalized in a consistent way.

[WM]

david...@gmail.com schrieb am Mittwoch, 13. April 2022 um 21:12:45 UTC+2:
> I don't understand WM's explanation of his intuition behind "dark" integers, but maybe they're the same as "non-standard" integers.

That is deplorable. Consider the endsegments E(n) = {n, n+1, n+2, ...}. They are all infinite but have an empty intersection. That is a contradiction, because they are inclusion-monotonic: All are contained in their predecessors and therefore have an infinite intersection with them

∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵ₀

An empty intersection cannot result from standard infinite endsegments

|∩{E(k) : k ∈ ℕ_def}| = ℵ₀

On the other hand, no natural numbers can be found in the intersection. They are dark.

>
> Maybe this is his intuition:
>
> Consider the sequence of sets S_n = {n, n+1, ... 2n-1}
>
> If that sequence has a limit, then that limit must contain no standard integers, but must contain infinitely many integers that are bigger than any standard integer, and these "bigger" integers can be called "dark" integers, or perhaps "non-standard" integers.

Standard integers n are such which can be defined (a decimal representation can be found and the finite initial segments {1, 2, 3, ..., n} can be found). The sequence of sets

S_n = {n, n+1, ... 2n-1}

or the sequence of sets
E(n) = {n, n+1, n+2, ...}
must contain infinitely many integers that are bigger than any standard integer.

Almost all integers are non-standard. I call them dark. The standard integers are a tiny, potentially infinite subset.

Regards, WM

[sergio]

On 4/13/2022 2:57 PM, WM wrote:
> david...@gmail.com schrieb am Mittwoch, 13. April 2022 um 21:12:45 UTC+2:
>> I don't understand WM's explanation of his intuition behind "dark" integers, but maybe they're the same as "non-standard" integers.
>
> That is deplorable. Consider the endsegments E(n) = {n, n+1, n+2, ...}. They are all infinite but have an empty intersection. That is a contradiction,

Wrong. the intersection of all endsegments is empty, as k is not in E(k+1), when will you learn simple math ?


> because they are bla bla: All are contained inside their mommy predecessors and therefore have an indefinite interdiction with them
>
> ∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} = { }

>
> An empty intersection cannot result from standard infinite endsegments

the intersection is not an endsegment, confused one.

>
>
> On the other hand, no un-natural numbers can be found in the interdictyon. They are tan, almost dark.

there are no dark numbers.

>>
>> Maybe this is his intuition:
>>
>> Consider the sequence of sets S_n = {n, n+1, ... 2n-1}

>>
>> If that sequence has a limit, then that limit must contain no standard integers, but must contain infinitely many integers that are bigger than any standard integer, and these "bigger" integers can be called "dark" integers, or perhaps "non-standard" integers.
>
> Standard integers n are such which can be defined (a decimal representation can be found and the finite initial segments {1, 2, 3, ..., n} can be found).

Wrong. integers do not have decimal representation, dummy. Google Counting Numbers

> The sequence of sets
> S_n = {n, n+1, ... 2n-1}

that is one set, where are the others ??

> or the sequence of sets
> E(n) = {n, n+1, n+2, ...}

that is one set called an endsegment, where are the others ? did you forget them ?

> must contain infinitely many integers that are bigger than any standard integer.

elements or sets ? are you confused again ?

>
> Almost all integers are non-standard.

Wrong. Your math is non-standard, you are a crank, troll.


> I call them dark.

because you are still in your closet.



> Regards, WM

[FredJeffries]

See the late Ed Nelson's 'Internal Set Theory'

https://web.math.princeton.edu/~nelson/papers/ist.pdf
https://web.math.princeton.edu/~nelson/books/1.pdf
https://en.wikipedia.org/wiki/Internal_set_theory
https://en.wikipedia.org/wiki/Edward_Nelson

[zelos...@gmail.com]

it means nothing because he is an idiot

[zelos...@gmail.com]

onsdag 13 april 2022 kl. 21:57:40 UTC+2 skrev WM:
> david...@gmail.com schrieb am Mittwoch, 13. April 2022 um 21:12:45 UTC+2:
> > I don't understand WM's explanation of his intuition behind "dark" integers, but maybe they're the same as "non-standard" integers.
> That is deplorable. Consider the endsegments E(n) = {n, n+1, n+2, ...}. They are all infinite but have an empty intersection. That is a contradiction,

No it isn't you fucking retard!

>because they are inclusion-monotonic: All are contained in their predecessors and therefore have an infinite intersection with them

But those are only intersections of FINITE number of sets, that has no bearing on the infinite number of sets intersection!

>
> ∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵ₀

THAT IS FINITE NUMBER OF SETS!

>
> An empty intersection cannot result from standard infinite endsegments
>
> |∩{E(k) : k ∈ ℕ_def}| = ℵ₀

WRONG! Your N_def is equal to N

[Jim Burns]

On 4/13/2022 3:12 PM, David Petry wrote:

> I don't understand WM's explanation of his intuition
> behind "dark" integers, but maybe they're the same as
> "non-standard" integers.

I think that the best description of WM's dark numbers
that I've been able to devise is that they're
general-purpose counter-examples.
Everlasting proof-stoppers.

If we matheologians claim that natural numbers
gyre and gimble in the wabe, then WM claims that
_dark numbers_ do not gyre and gimble in the wabe.

This produces the appearance of WM being able to discuss
gyring and gimbling in the wabe, without requiring him
to find out what it means to gyre and gimble in the wabe.

Whatever the defects this strategy may have, I admit
that it is very efficient in its use of WM's time.

I too have, in the past, thought that dark numbers
could be non-standard numbers. That never quite worked
out, somehow. It seems to me now that non-standard
numbers are contrary to the purpose of dark numbers,
because one can prove claims about them, infinitely-
-many though they are. It seems to me that the purpose
of dark numbers to to keep things from being proved.

> Maybe this is his intuition:
>
> Consider the sequence of sets S_n = {n, n+1, ... 2n-1}
>
> If that sequence has a limit, then that limit must
> contain no standard integers,

If the limit is taken for each natural, each n
leaves the sequence and does not return.
The limit would be the empty set.
And, yes, it does not contain any standard integers.
But this is utter matheology.

> but must contain infinitely many integers that are
> bigger than any standard integer, and these "bigger"
> integers can be called "dark" integers, or perhaps
> "non-standard" integers.
>
> I see no reason to believe that such an intuition
> couldn't be formalized in a consistent way.

I expect that part of WM's intuitions can be formalized
in a consistent way. Robinson arithmetic springs to mind.

All of WM's intuitions?
That seems very unlikely to me.
WM explains inconsistencies in his posts
as proof of the incorrectness of everyone else.
I think that claiming that everyone else is wrong
is something he would be very slow to give up.

I don't know how this could be tested, but I suspect
that, if somehow all WM's intuitions were found
to be consistent, he would turn to new intuitions.
Being an iconoclast seems to be very important to him.

[David Petry]

On Wednesday, April 13, 2022 at 12:57:40 PM UTC-7, WM wrote:
> david...@gmail.com schrieb am Mittwoch, 13. April 2022 um 21:12:45 UTC+2:
> > I don't understand WM's explanation of his intuition behind "dark" integers, but maybe they're the same as "non-standard" integers.

> That is deplorable.

Could I ask you to be more precise? Precisely in what way is what I wrote "deplorable" ?

>Consider the endsegments E(n) = {n, n+1, n+2, ...}. They are all infinite but have an empty intersection. That is a contradiction, because they are inclusion-monotonic: All are contained in their predecessors and therefore have an infinite intersection with them
>
> ∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵ₀
>
> An empty intersection cannot result from standard infinite endsegments
>
> |∩{E(k) : k ∈ ℕ_def}| = ℵ₀
>
> On the other hand, no natural numbers can be found in the intersection. They are dark.
> >
> > Maybe this is his intuition:
> >
> > Consider the sequence of sets S_n = {n, n+1, ... 2n-1}
> >
> > If that sequence has a limit, then that limit must contain no standard integers, but must contain infinitely many integers that are bigger than any standard integer, and these "bigger" integers can be called "dark" integers, or perhaps "non-standard" integers.
> Standard integers n are such which can be defined (a decimal representation can be found and the finite initial segments {1, 2, 3, ..., n} can be found). The sequence of sets
> S_n = {n, n+1, ... 2n-1}
> or the sequence of sets
> E(n) = {n, n+1, n+2, ...}
> must contain infinitely many integers that are bigger than any standard integer.
>
> Almost all integers are non-standard. I call them dark. The standard integers are a tiny, potentially infinite subset.

I just don't know what your point is. Are you saying that my attempt to understand your argument is flawed?

[David Petry]

On Wednesday, April 13, 2022 at 9:31:33 PM UTC-7, Jim Burns wrote:
> On 4/13/2022 3:12 PM, David Petry wrote:
>
> > I don't understand WM's explanation of his intuition
> > behind "dark" integers, but maybe they're the same as
> > "non-standard" integers.
> I think that the best description of WM's dark numbers
> that I've been able to devise is that they're
> general-purpose counter-examples.
> Everlasting proof-stoppers.
>
> If we matheologians claim that natural numbers
> gyre and gimble in the wabe, then WM claims that
> _dark numbers_ do not gyre and gimble in the wabe.
>
> This produces the appearance of WM being able to discuss
> gyring and gimbling in the wabe, without requiring him
> to find out what it means to gyre and gimble in the wabe.

Finally, at long last, somebody clears things up for me. Thank you, thank you, thank you.

>
> Whatever the defects this strategy may have, I admit
> that it is very efficient in its use of WM's time.
>
> I too have, in the past, thought that dark numbers
> could be non-standard numbers. That never quite worked
> out, somehow. It seems to me now that non-standard
> numbers are contrary to the purpose of dark numbers,
> because one can prove claims about them, infinitely-
> -many though they are. It seems to me that the purpose
> of dark numbers to to keep things from being proved.
> > Maybe this is his intuition:
> >
> > Consider the sequence of sets S_n = {n, n+1, ... 2n-1}
> >
> > If that sequence has a limit, then that limit must
> > contain no standard integers,
> If the limit is taken for each natural, each n
> leaves the sequence and does not return.
> The limit would be the empty set.
> And, yes, it does not contain any standard integers.
> But this is utter matheology.

But now you're confusing me a little bit.

> > but must contain infinitely many integers that are
> > bigger than any standard integer, and these "bigger"
> > integers can be called "dark" integers, or perhaps
> > "non-standard" integers.
> >
> > I see no reason to believe that such an intuition
> > couldn't be formalized in a consistent way.
> I expect that part of WM's intuitions can be formalized
> in a consistent way. Robinson arithmetic springs to mind.
>
> All of WM's intuitions?
> That seems very unlikely to me.
> WM explains inconsistencies in his posts
> as proof of the incorrectness of everyone else.
> I think that claiming that everyone else is wrong
> is something he would be very slow to give up.
>
> I don't know how this could be tested, but I suspect
> that, if somehow all WM's intuitions were found
> to be consistent, he would turn to new intuitions.
> Being an iconoclast seems to be very important to him.

I'd really like to be able to take WM seriously, but he does make it difficult.

[zelos...@gmail.com]

why would you want to take a complete crank seriosuly?

[FromTheRafters]

David Petry wrote :

My interpretation of it is that he considers the "dark" numbers to be
the difference set between the potentially infinite and the actual
infinite. He may as well try flying a kite in the aether wind.

[Gus Gassmann]

On Thursday, 14 April 2022 at 01:34:41 UTC-3, david...@gmail.com wrote:
> On Wednesday, April 13, 2022 at 12:57:40 PM UTC-7, WM wrote:
> > david...@gmail.com schrieb am Mittwoch, 13. April 2022 um 21:12:45 UTC+2:
> > > I don't understand WM's explanation of his intuition behind "dark" integers, but maybe they're the same as "non-standard" integers.
>
> > That is deplorable.
> Could I ask you to be more precise? Precisely in what way is what I wrote "deplorable" ?

How much more precise do you want it? WM thinks his gobbledygook about dark numbers must be perfectly and instantly understandable to anyone who comes in contact with it. (It is not; it is contradictory garbage.) That you are evidently skeptical makes it (and you) deplorable --- in his eyes. More is not necessary to know.

[WM]

zelos...@gmail.com schrieb am Donnerstag, 14. April 2022 um 06:28:17 UTC+2:
> onsdag 13 april 2022 kl. 21:57:40 UTC+2 skrev WM:
> > david...@gmail.com schrieb am Mittwoch, 13. April 2022 um 21:12:45 UTC+2:
> > > I don't understand WM's explanation of his intuition behind "dark" integers, but maybe they're the same as "non-standard" integers.
> > That is deplorable. Consider the endsegments E(n) = {n, n+1, n+2, ...}. They are all infinite but have an empty intersection. That is a contradiction,
> No it isn't

In mathematics an inclusion monotonic decreasing sequence of sets like the endsegments has an empty intersection only when an empty set is reached. Before that there would exist a non-empty minimum.

> >because they are inclusion-monotonic: All are contained in their predecessors and therefore have an infinite intersection with them
> But those are only intersections of FINITE number of sets, that has no bearing on the infinite number of sets intersection!

Every definable endsegment E(n) belongs to a finite number of sets, namely to the n first endsegments.

> >
> > ∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵ₀
> THAT IS FINITE NUMBER OF SETS!

Every definable endsegment E(n) belongs to a finite number of sets, namely to the n first endsegments. Define an endsegment that does not!

> >
> > An empty intersection cannot result from standard infinite endsegments
> >
> > |∩{E(k) : k ∈ ℕ_def}| = ℵ₀
> WRONG! Your N_def is equal to N

It is equal to the potentially infinite collection of definable natural numbers. If those were all natural numbers, then there would no actual infinity (and no endsegment) exist.

Regards, WM

[WM]

Jim Burns schrieb am Donnerstag, 14. April 2022 um 06:31:33 UTC+2:
> On 4/13/2022 3:12 PM, David Petry wrote:

> > If that sequence has a limit, then that limit must
> > contain no standard integers,
> If the limit is taken for each natural, each n
> leaves the sequence and does not return.
> The limit would be the empty set.

If you go through all natural numbers and drop them into a garbage can, then you end up with the empty set? Or with ω? Or is it impossible to go through all because there is no all?

The latter would be fact in potentialinfinity.

> And, yes, it does not contain any standard integers.
> But this is utter matheology.

Either matheology is wrong, because we have only potential infinity. Then there will be no limit at all. On the other hand, even in actual infinity, we will never reach ω or an empty set with *defined* natural numbers. Therefore, if there are all natural numbers, then there must be some which we can not drop into the garbage can. ZThey must even be the actuallyinfinite part of ℕ, because FISONs are not actually infinite.

> I don't know how this could be tested, but I suspect
> that, if somehow all WM's intuitions were found
> to be consistent, he would turn to new intuitions.
> Being an iconoclast seems to be very important to him.

Chuckle. I would be very glad if countermathematical ideas, which are provably wrong, as an empty intersection of infinite inclusion monotonic sets, would be eradicated. Can you really support the idea that the intersection could be empty because infinitely many sets are involved? Would infinitely many sets {1} give an empty intersection? But infinite sets should give an empty intersection???

Regards. WM

[WM]

FromTheRafters schrieb am Donnerstag, 14. April 2022 um 10:48:51 UTC+2:

> My interpretation of it is that he considers the "dark" numbers to be
> the difference set between the potentially infinite and the actual
> infinite.

You have hit the nail on the head.

Regards, WM

[WM]

david...@gmail.com schrieb am Donnerstag, 14. April 2022 um 06:34:41 UTC+2:
> On Wednesday, April 13, 2022 at 12:57:40 PM UTC-7, WM wrote:
> > david...@gmail.com schrieb am Mittwoch, 13. April 2022 um 21:12:45 UTC+2:
> > > I don't understand WM's explanation of his intuition behind "dark" integers, but maybe they're the same as "non-standard" integers.
>
> > That is deplorable.
> Could I ask you to be more precise? Precisely in what way is what I wrote "deplorable" ?

I found it deploarable that you don't understand, as you said. But as it turned out, you have understood. Your example resembles my example of endsegments.

> I just don't know what your point is.

*Infinite* inclusion monotonic sets cannot have an empty intersection. The idea that the intersection of *infinite* endsegments could be empty because infinitely many sets are involved is wrong. Infinitely many sets {1} would not give an empty intersection either.

> Are you saying that my attempt to understand your argument is flawed?

No. It is a good example.

Regards, WM

[Jim Burns]

On 4/14/2022 12:41 AM, David Petry wrote:
> On Wednesday, April 13, 2022 at 9:31:33 PM UTC-7,
> Jim Burns wrote:
>> On 4/13/2022 3:12 PM, David Petry wrote:

>>> Maybe this is his intuition:
>>>
>>> Consider the sequence of sets S_n = {n, n+1, ... 2n-1}
>>>
>>> If that sequence has a limit, then that limit must
>>> contain no standard integers,
>>
>> If the limit is taken for each natural, each n
>> leaves the sequence and does not return.
>> The limit would be the empty set.
>> And, yes, it does not contain any standard integers.
>> But this is utter matheology.
>
> But now you're confusing me a little bit.

Sorry, my bad.

"Matheology" is WM's standard insult to
regular mathematics.
I mean to say the empty limit would not be part of
a formalization acceptable to WM.

[sergio]

On 4/14/2022 5:22 AM, WM wrote:
> zelos...@gmail.com schrieb am Donnerstag, 14. April 2022 um 06:28:17 UTC+2:
>> onsdag 13 april 2022 kl. 21:57:40 UTC+2 skrev WM:
>>> david...@gmail.com schrieb am Mittwoch, 13. April 2022 um 21:12:45 UTC+2:
>>>> I don't understand WM's explanation of his intuition behind "dark" integers, but maybe they're the same as "non-standard" integers.
>>> That is deplorable. Consider the endsegments E(n) = {n, n+1, n+2, ...}. They are all infinite but have an empty intersection. That is a contradiction,
>> No it isn't
>
> In mathematics an inclusion monotonic decreasing sequence of sets like the endsegments has an empty intersection only when an empty set is reached. Before that there would exist a non-empty minimum.

however the intersection of all endsegments is empty,
trivial proof, assume any number k is in an endsegment, E(k), then it is not in the E(k+1) endsegment.

>
>>> because they are inclusion-monotonic: All are contained in their predecessors and therefore have an infinite intersection with them
>> But those are only intersections of FINITE number of sets, that has no bearing on the infinite number of sets intersection!
>
> Every definable endsegment E(n) belongs to a finite number of sets, namely to the n first endsegments.

"definable" is a troll defined term

>>>
>>> ∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵ₀

Wrong. assume any number k is in an endsegment, E(k), then it is not in the E(k+1) endsegment.

>> THAT IS FINITE NUMBER OF SETS!
>
> Every definable endsegment E(n) belongs to a finite number of sets, namely to the n first endsegments. Define an endsegment that does not!

assume any number k is in an endsegment, E(k), then it is not in the E(k+1) endsegment.

>>>
>>> An empty intersection cannot result from standard infinite endsegments
>>>
>>> |∩{E(k) : k ∈ ℕ_def}| = ℵ₀
>> WRONG! Your N_def is equal to N
>

> assume any number k is in an endsegment, E(k), then it is not in the E(k+1) endsegment.
>
> Regards, WM

[sergio]

On 4/14/2022 5:57 AM, WM wrote:

> david...@gmail.com schrieb am Donnerstag, 14. April 2022 um 06:34:41 UTC+2:
>> On Wednesday, April 13, 2022 at 12:57:40 PM UTC-7, WM wrote:
>>> david...@gmail.com schrieb am Mittwoch, 13. April 2022 um 21:12:45 UTC+2:
>>>> I don't understand WM's explanation of his intuition behind "dark" integers, but maybe they're the same as "non-standard" integers.
>>
>>> That is deplorable.
>> Could I ask you to be more precise? Precisely in what way is what I wrote "deplorable" ?
>

> I found it deploarable that you don't understand, as you said. But as it turned out, you have understood. Your example resembles my example of endsegments.

his failed example.

WM ALWAYS posts BAD MATH. you will find error is EVERY one of his posts.

>
>> I just don't know what your point is.
>

> *Infinite* inclusion monotonic sets cannot have an empty intersection.

assume any number k is in an endsegment, E(k), then it is not in the E(k+1) endsegment.

> The idea that the intersection *infinite endsegments could be empty because infinitely many sets are involved is wrong.

assume any number k is in an endsegment, E(k), then it is not in the E(k+1) endsegment.

> Infinitely many sets {1} would not give an empty intersection either.

you have no idea what you are spewing about.

>
>> Are you saying that my attempt to understand your argument is flawed?
>

> No. It is a good example.

[sergio]

On 4/14/2022 6:04 AM, WM wrote:
> david...@gmail.com schrieb am Donnerstag, 14. April 2022 um 06:34:41 UTC+2:
>> On Wednesday, April 13, 2022 at 12:57:40 PM UTC-7, WM wrote:
>>> david...@gmail.com schrieb am Mittwoch, 13. April 2022 um 21:12:45 UTC+2:
>>>> I don't understand WM's explanation of his intuition behind "dark" integers, but maybe they're the same as "non-standard" integers.
>>
>>> That is deplorable.
>> Could I ask you to be more precise? Precisely in what way is what I wrote "deplorable" ?
>
> I found it deploarable that you don't understand, as you said. But as it turned out, you have understood. Your example resembles my example of endsegments.
>
>> I just don't know what your point is.
>

assume any number k is in an endsegment, E(k), then it is not in the E(k+1) endsegment.
>

[sergio]

On 4/14/2022 5:45 AM, WM wrote:
> Jim Burns schrieb am Donnerstag, 14. April 2022 um 06:31:33 UTC+2:
>> On 4/13/2022 3:12 PM, David Petry wrote:
>
>>> If that sequence has a limit, then that limit must
>>> contain no standard integers,
>> If the limit is taken for each natural, each n
>> leaves the sequence and does not return.
>> The limit would be the empty set.
>
> If you go through all natural numbers and drop them into a garbage can,

so you have nothing, and N is in a garbage can, now what ?

> then you end up with the empty set? Or with ω? Or is it impossible to go through all because there is no all?

are you looking at your hands, or in the garbage can ?

>
> The latter would be fact in potentialinfinity.

fake word.

>
>> And, yes, it does not contain any standard integers.
>> But this is utter matheology.
>

<snip delusion>

>
>> I don't know how this could be tested, but I suspect
>> that, if somehow all WM's intuitions were found
>> to be consistent, he would turn to new intuitions.
>> Being an iconoclast seems to be very important to him.
>
> Chuckle. I would be very glad if countermathematical ideas, which are provably wrong, as an empty intersection of infinite inclusion monotonic sets, would be eradicated. Can you really support the idea that the intersection could be empty because infinitely many sets are involved? Would infinitely many sets {1} give an empty intersection? But infinite sets should give an empty intersection???

Eat this, you Chumley;

assume any number k is in an endsegment, E(k), then it is not in the E(k+1) endsegment.

>

> Regards. WM

[sergio]

you snipped out the important part;

[sergio]

WM is a crank, every single post of his has intentional errors, he posts no Math content.
You would like him if your into pseudo math or pretend math.

[FromTheRafters]

After serious thinking sergio wrote :

> On 4/14/2022 5:45 AM, WM wrote:
>> Jim Burns schrieb am Donnerstag, 14. April 2022 um 06:31:33 UTC+2:
>>> On 4/13/2022 3:12 PM, David Petry wrote:
>>
>>>> If that sequence has a limit, then that limit must
>>>> contain no standard integers,
>>> If the limit is taken for each natural, each n
>>> leaves the sequence and does not return.
>>> The limit would be the empty set.
>>
>> If you go through all natural numbers and drop them into a garbage can,
>
> so you have nothing, and N is in a garbage can, now what ?

Composting, making composted numbers. Go green!