In oder to index all positive fractions

1/1, 1/2, 1/3, 1/4, ...
2/1, 2/2, 2/3, 2/4, ...
3/1, 3/2, 3/3, 3/4, ...
4/1, 4/2, 4/3, 4/4, ...
....

by natural indexes, collect all these indexes 1, 2, 3, ... and put them in the first column, covering the fractions n/1. All other fractions get no indexes. This is expressed by an O's.

For brevity the presence of an index is represented by an X since its particular numerical value is irrelevant. So we get, as the start position:

XOOOOOOOO...
XOOOOOOOO...
XOOOOOOOO...
XOOOOOOOO...
XOOOOOOOO...
XOOOOOOOO...
XOOOOOOOO...
XOOOOOOOO...
XOOOOOOOO...
....

Now every index is moved to the place which Cantor has prescribed: 1/1, 1/2, 2/1, 1/3, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5, 5/1, 1/6, ... . At this place there appears an X and where the index has been taken from there appears an O. After eight steps for instance we see the following picture: 1/1, 1/2, 2/1, 1/3, 3/1, 1/4, 2/3, 3/2 are indexed whereas 4/1, 5/1, 6/1, 7/1, 8/1 have lost their indexes:

XXXXOOOOO...
XOXOOOOOO...
XXOOOOOOO...
OOOOOOOOO...
OOOOOOOOO...
OOOOOOOOO...
OOOOOOOOO...
OOOOOOOOO...
XOOOOOOOO...
....

As the result we find that the number of indexes X does never change, i.e., in every finite step (where it only could change), it does not. The number of O's remains constant and much larger than the number of X's. After every finite step!

If, after all, the X's will cover the whole matrix and all the O's will have disappeared, this can only happen during the so-called Matheologoial Explosion, happening after all finite steps according to the Matheologial Syllogism: The O's will have disappeared never. Never we get into the infinite. ==> The O's will have diappeared in the infinite.

Regards, WM

On 1/15/2022 6:52 AM, WM wrote:
> In oder to index all positive fractions
>
> 1/1, 1/2, 1/3, 1/4, ...
> 2/1, 2/2, 2/3, 2/4, ...
> 3/1, 3/2, 3/3, 3/4, ...
> 4/1, 4/2, 4/3, 4/4, ...
> ...
>
> by natural indexes, collect all these indexes 1, 2, 3, ... and put them in the first column, covering the fractions n/1. All other fractions get no indexes. This is expressed by an O's.

However, Cantor used diagonal threading, which converts the above 2 dimensional matrix, into a 1 dimensional array.

>
> For brevity the presence of an index is represented by an X since its particular numerical value is irrelevant. So we get, as the start position:
>
> XOOOOOOOO...
> XOOOOOOOO...
> XOOOOOOOO...
> XOOOOOOOO...
> XOOOOOOOO...
> XOOOOOOOO...
> XOOOOOOOO...
> XOOOOOOOO...
> XOOOOOOOO...
> ...
>
> Now every index is moved to the place which Cantor has prescribed: 1/1, 1/2, 2/1, 1/3, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5, 5/1, 1/6, ... . At this place there appears an X and where the *index has been taken* from there appears an O. After eight steps for instance we see the following picture: 1/1, 1/2, 2/1, 1/3, 3/1, 1/4, 2/3, 3/2 are indexed whereas 4/1, 5/1, 6/1, 7/1, 8/1 *have lost their indexes*:

your mistakes are highlighted.

>
> XXXXOOOOO...
> XOXOOOOOO...
> XXOOOOOOO...
> OOOOOOOOO...
> OOOOOOOOO...
> OOOOOOOOO...
> OOOOOOOOO...
> OOOOOOOOO...
> XOOOOOOOO...
> ...
>
> As the result we find that the number of indexes X does never change, i.e., in every finite step (where it only could change), it does not. The number of O's remains constant and much larger than the number of X's. After every finite step!

look, google core memory

It was early memory for computers using tiny ferrite rings, structured in a large matrix, with the reset line threaded diagonally through the matrix
just like Cantors Enumeration. no bits were "lost", or "taken from"

it is radiation hardened memory, and was used in outer space up till about 1980

>
> If, after all, the X's will cover the whole matrix and all the O's will have disappeared, this can only happen during the so-called Matheologoial Explosion, happening after all finite steps according to the Matheologial Syllogism: The O's will have disappeared never. Never we get into the infinite. ==> The O's will have diappeared in the infinite.

no, you *disappeared them* in your construct.

>
> Regards, WM

For the sake of argument we will say that the fractions cannot be indexed. This does not change the fact that N_P is a Peano set and thus we can use induction to show the facts that no element of N_P is dark, every element of N_P can be considered, and no element of N_P is larger than every element of N_P. The fact that we can use induction the show that every element of N_P is followed by an infinite number of elements of N_P, each of which is not dark, can be considered and is not larger than every element of N_P, changes nothing.

--
William Hughes

William schrieb am Samstag, 15. Januar 2022 um 18:15:42 UTC+1:
> For the sake of argument we will say that the fractions cannot be indexed.

I agree.

> This does not change the fact that N_P is a Peano set and thus we can use induction to show the facts that no element of N_P is dark, every element of N_P can be considered, and no element of N_P is larger than every element of N_P.

I agree.

> The fact that we can use induction the show that every element of N_P is followed by an infinite number of elements of N_P, each of which is not dark,

This infinite number however cannot be a completed infinity. In a linear order completion implies last.

Complete, infinite, order cannot exist together.

Regards, WM

On Saturday, January 15, 2022 at 2:23:42 PM UTC-4, WM wrote:
> William schrieb am Samstag, 15. Januar 2022 um 18:15:42 UTC+1:
> > ... no element of N_P is larger than every element of N_P.
> I agree.

So, as no element of N_P is larger than every element of N_P, there is no element of N_P between N_P and omega.

--

William Hughes

On 1/15/2022 12:23 PM, WM wrote:
> William schrieb am Samstag, 15. Januar 2022 um 18:15:42 UTC+1:
>> For the sake of argument we will say that the fractions cannot be indexed.
>
> I agree.
>
>> This does not change the fact that N_P is a Peano set and thus we can use induction to show the facts that no element of N_P is dark, every element of N_P can be considered, and no element of N_P is larger than every element of N_P.
>
> I agree.
>
>> The fact that we can use induction the show that every element of N_P is followed by an infinite number of elements of N_P, each of which is not dark,
>
> This infinite number however cannot be a completed infinity. In a linear order completion implies last.

no.

count the naturals 1,2,3,4,... it is in linear order and there is no last

>
> Complete, infinite, order cannot exist together.

you forget you can prove completeness by showing no element is missing...

>
> Regards, WM

sergio schrieb am Samstag, 15. Januar 2022 um 19:45:00 UTC+1:
> On 1/15/2022 12:23 PM, WM wrote:
> > William schrieb am Samstag, 15. Januar 2022 um 18:15:42 UTC+1:
> >> For the sake of argument we will say that the fractions cannot be indexed.
> >
> > I agree.
> >
> >> This does not change the fact that N_P is a Peano set and thus we can use induction to show the facts that no element of N_P is dark, every element of N_P can be considered, and no element of N_P is larger than every element of N_P.
> >
> > I agree.
> >
> >> The fact that we can use induction the show that every element of N_P is followed by an infinite number of elements of N_P, each of which is not dark,
> >
> > This infinite number however cannot be a completed infinity. In a linear order completion implies last.
> no.
>
> count the naturals 1,2,3,4,... it is in linear order and there is no last

You cannot know till to meet it.
> >
> > Complete, infinite, order cannot exist together.
> you forget you can prove completeness by showing no element is missing...

I showed that no O is missing in

> XXXXOOOOO...
> XOXOOOOOO...
> XXOOOOOOO...
> OOOOOOOOO...
> OOOOOOOOO...
> OOOOOOOOO...
> OOOOOOOOO...
> OOOOOOOOO...
> XOOOOOOOO...
> ...

even after all possible transpositions.

Regards, WM

William schrieb am Samstag, 15. Januar 2022 um 19:36:31 UTC+1:
> On Saturday, January 15, 2022 at 2:23:42 PM UTC-4, WM wrote:
> > William schrieb am Samstag, 15. Januar 2022 um 18:15:42 UTC+1:
> > > ... no element of N_P is larger than every element of N_P.
> > I agree.
>
> So, as no element of N_P is larger than every element of N_P, there is no element of N_P between N_P and omega.

But omega is the limit, the border before which all must happen. omega would be the last ordinal unless a smaller one next to omega could be found (if all was in linear order). You cannot deny it before you have reached it.

Regards, WM

The explosion started at math zero...
Mathematical God creates gravity...

On 1/15/2022 1:20 PM, WM wrote:
> sergio schrieb am Samstag, 15. Januar 2022 um 19:45:00 UTC+1:
>> On 1/15/2022 12:23 PM, WM wrote:
>>> William schrieb am Samstag, 15. Januar 2022 um 18:15:42 UTC+1:
>>>> For the sake of argument we will say that the fractions cannot be indexed.
>>>
>>> I agree.
>>>
>>>> This does not change the fact that N_P is a Peano set and thus we can use induction to show the facts that no element of N_P is dark, every element of N_P can be considered, and no element of N_P is larger than every element of N_P.
>>>
>>> I agree.
>>>
>>>> The fact that we can use induction the show that every element of N_P is followed by an infinite number of elements of N_P, each of which is not dark,
>>>
>>> This infinite number however cannot be a completed infinity. In a linear order completion implies last.
>> no.
>>
>> count the naturals 1,2,3,4,... it is in linear order and there is no last
>
> You cannot know till to meet it.

If you meet it, it is finite.

The set of Natural number is infinite, there is no last.

>>>
>>> Complete, infinite, order cannot exist together.
>> you forget you can prove completeness by showing no element is missing...
>
> I showed that no O is missing in
>
>> XXXXOOOOO...
>> XOXOOOOOO...
>> XXOOOOOOO...
>> OOOOOOOOO...
>> OOOOOOOOO...
>> OOOOOOOOO...
>> OOOOOOOOO...
>> OOOOOOOOO...
>> XOOOOOOOO...
>> ...
>
> even after all possible transpositions.
>
> Regards, WM

you can show completion by proving there are no missing numbers,

take the rationals, n/k where n and k are natural numbers.
take the set composed of "for any n, and any k, the rational n/k exists"

all rationals are included, none are missing, it is complete.

"Complete, infinite, order cannot exist together." is bogus

all natural numbers are in N, there are none missing, it is complete, they are ordered, and infinite...

so...

On 1/15/2022 2:25 PM, WM wrote:

> But omega is the limit, the border before which all must happen.

If there exists a totally-ordered collection {0,...,k}
for which each cut C is a step
( max(C) = i min({0,...,k}\C = j )
and each step is a count
( j = i+1 )
and which begins at 0 and ends at k,
then k is before omega.

That's what omega is.

> omega would be the last ordinal unless a smaller one
> next to omega could be found (if all was in linear order).

| Suppose psi was a smaller one next to omega.
| A totally-ordered collection {0,...,psi} exists
| for which each cut is a step and each step is a count
| and which begins at 0 and ends at psi.
| | psi+1 exists and
| a totally-ordered collection {0,...,psi+1} exists
| for which each cut is a step and each step is a count
| and which begins at 0 and ends at psi+1.
| | Because "smaller one next to omega" psi+1 = omega
| | omega exists and
| a totally-ordered collection {0,...,omega} exists
| for which each cut is a step and each step is a count
| and which begins at 0 and ends at omega.
| | omega+1 exists and
| a totally-ordered collection {0,...,omega+1} exists
| for which each cut is a step and each step is a count
| and which begins at 0 and ends at omega+1.
| | Therefore omega+1 is before omega,
| by definition of omega.
| However, omega+1 is not in {0,...,omega}
| Contradiction.

Therefore,
there is no smaller one next to omega.

> You cannot deny it before you have reached it.

We can deny that omega is something other than
what we have defined omega to be.
We don't need to reach omega to do that.

From that, by reliable statement-steps,
we can learn things about omega, without ever reaching it.
For example: there is no smaller one next to omega.

On Saturday, January 15, 2022 at 3:25:41 PM UTC-4, WM wrote:
> William schrieb am Samstag, 15. Januar 2022 um 19:36:31 UTC+1:

> > So, as no element of N_P is larger than every element of N_P, there is no element of N_P between N_P and omega.
> But omega

We do not need omega. We have

(i) for every n element of N_P, there are are an infinite number of elements (not the same elements for every n) of N_P that are
larger than n,

(ii) There is no element of N_P that is larger than every element of N_P

[(i) you have repeated many times. (ii) is a statement that you agree with]
--

William Hughes

On 1/15/2022 1:25 PM, WM wrote:
> William schrieb am Samstag, 15. Januar 2022 um 19:36:31 UTC+1:
>> On Saturday, January 15, 2022 at 2:23:42 PM UTC-4, WM wrote:
>>> William schrieb am Samstag, 15. Januar 2022 um 18:15:42 UTC+1:
>>>> ... no element of N_P is larger than every element of N_P.
>>> I agree.
>>
>> So, as no element of N_P is larger than every element of N_P, there is no element of N_P between N_P and omega.
>
> But omega is the limit, the border before which all must happen. omega would be the last ordinal unless a smaller one next to omega could be found (if all was in linear order). You cannot deny it before you have reached it.
>
> Regards, WM

and you will never reach omega.

assume your at the largest number k

there is always a k + 1

conflict. so k is not the largest number.

Perhaps "You cannot deny it before you have reached it", but I deny that you will ever reach it as I say there is always a k+1, and you never will get
to omega. Simple

On Saturday, 15 January 2022 at 15:25:41 UTC-4, WM wrote:
> William schrieb am Samstag, 15. Januar 2022 um 19:36:31 UTC+1:
> > On Saturday, January 15, 2022 at 2:23:42 PM UTC-4, WM wrote:
> > > William schrieb am Samstag, 15. Januar 2022 um 18:15:42 UTC+1:
> > > > ... no element of N_P is larger than every element of N_P.
> > > I agree.
> >
> > So, as no element of N_P is larger than every element of N_P, there is no element of N_P between N_P and omega.
> But omega is the limit, the border before which all must happen. omega would be the last ordinal unless a smaller one next to omega could be found (if all was in linear order). You cannot deny it before you have reached it.

What a crock. WM does not know what a linear order is. Never has, never will. A linear order does not imply the existence of a smallest element, nor a largest, nor an element "next to" omega. The only fact is that every natural number n is smaller than omega, with an infinite number of other natural numbers k satisfying n < k < omega.

William schrieb am Samstag, 15. Januar 2022 um 21:48:05 UTC+1:
> On Saturday, January 15, 2022 at 3:25:41 PM UTC-4, WM wrote:
> > William schrieb am Samstag, 15. Januar 2022 um 19:36:31 UTC+1:
>
> > > So, as no element of N_P is larger than every element of N_P, there is no element of N_P between N_P and omega.
> > But omega
> We do not need omega.

Then we have no finished infinity and no infinite set.

We have
>
> (i) for every n element of N_P, there are are an infinite number of elements (not the same elements for every n) of N_P that are
> larger than n,
>
> (ii) There is no element of N_P that is larger than every element of N_P

So it is. For every element of N_P that you choose there is a larger element of N_P. But that does not mean completed infinity.
>
> [(i) you have repeated many times. (ii) is a statement that you agree with]

Of course. I call it potential infinity.

Regards, WM

sergio schrieb am Samstag, 15. Januar 2022 um 20:45:17 UTC+1:
> On 1/15/2022 1:20 PM, WM wrote:
>
> >> you forget you can prove completeness by showing no element is missing...
> >
> > I showed that no O is missing in
> >
> >> XXXXOOOOO...
> >> XOXOOOOOO...
> >> XXOOOOOOO...
> >> OOOOOOOOO...
> >> OOOOOOOOO...
> >> OOOOOOOOO...
> >> OOOOOOOOO...
> >> OOOOOOOOO...
> >> XOOOOOOOO...
> >> ...
> >
> > even after all possible transpositions.

> you can show completion by proving there are no missing numbers,

I have proved that after all transpositions, there are infinitely many fractions without index.
>
> all rationals are included, none are missing, it is complete.

How do you bring that into line with the fact that there are infinitely many fractions without index?

Regards, WM

Jim Burns schrieb am Samstag, 15. Januar 2022 um 21:15:35 UTC+1:
> On 1/15/2022 2:25 PM, WM wrote:

> > omega would be the last ordinal unless a smaller one
> > next to omega could be found (if all was in linear order).
> | Suppose psi was a smaller one next to omega.
> | A totally-ordered collection {0,...,psi} exists

No.

> | for which each cut is a step and each step is a count
> | and which begins at 0 and ends at psi.
> |
> | psi+1 exists and
> | a totally-ordered collection {0,...,psi+1} exists
> | for which each cut is a step and each step is a count
> | and which begins at 0 and ends at psi+1.
> |
> | Because "smaller one next to omega" psi+1 = omega

Therefore psi is not accessible.
> |
> Therefore,
> there is no smaller one next to omega.

Then it is not next to omega. Where is the first smaller one?
What is between it an omega?

Regards, WM

On 1/15/2022 4:38 PM, WM wrote:
> Jim Burns schrieb am Samstag, 15. Januar 2022 um 21:15:35 UTC+1:
>> On 1/15/2022 2:25 PM, WM wrote:
>
>>> omega would be the last ordinal unless a smaller one
>>> next to omega could be found (if all was in linear order).
>> | Suppose psi was a smaller one next to omega.
>> | A totally-ordered collection {0,...,psi} exists
>
> No.
>
>> | for which each cut is a step and each step is a count
>> | and which begins at 0 and ends at psi.
>> |
>> | psi+1 exists and
>> | a totally-ordered collection {0,...,psi+1} exists
>> | for which each cut is a step and each step is a count
>> | and which begins at 0 and ends at psi+1.
>> |
>> | Because "smaller one next to omega" psi+1 = omega
>
> Therefore psi is not accessible.
>> |
>> Therefore,
>> there is no smaller one next to omega.
>
> Then it is not next to omega. Where is the first smaller one?
> What is between it an omega?
>
> Regards, WM
>
>

he already told you psi+1, or try psi, or psi-1

these are very very simple proofs there is no number "next to" omega

and yet again you fail, which convinces everyone that you never took algebra I.

Can you convince people you are right ? no, if they had Algebra I, then no.

On 1/15/2022 4:30 PM, WM wrote:
> William schrieb am Samstag, 15. Januar 2022 um 21:48:05 UTC+1:
>> On Saturday, January 15, 2022 at 3:25:41 PM UTC-4, WM wrote:
>>> William schrieb am Samstag, 15. Januar 2022 um 19:36:31 UTC+1:
>>
>>>> So, as no element of N_P is larger than every element of N_P, there is no element of N_P between N_P and omega.
>>> But omega
>> We do not need omega.
>
> Then we have no finished infinity and no infinite set.

so you stopped, stopped at k again. because you can only deal with the finite.

>
>
> We have
>>
>> (i) for every n element of N_P, there are are an infinite number of elements (not the same elements for every n) of N_P that are
>> larger than n,
>>
>> (ii) There is no element of N_P that is larger than every element of N_P
>
> So it is. For every element of N_P that you choose there is a larger element of N_P. But that does not mean completed infinity.

sure it does, as you can show all numbers are in the set, it is complete.

>>
>> [(i) you have repeated many times. (ii) is a statement that you agree with]
>
> Of course. I call it potential infinity.

which is finite, because you stopped at k.

>
> Regards, WM

On 1/15/2022 4:34 PM, WM wrote:
> sergio schrieb am Samstag, 15. Januar 2022 um 20:45:17 UTC+1:
>> On 1/15/2022 1:20 PM, WM wrote:
>>
>>>> you forget you can prove completeness by showing no element is missing...
>>>
>>> I showed that no O is missing in
>>>
>>>> XXXXOOOOO...
>>>> XOXOOOOOO...
>>>> XXOOOOOOO...
>>>> OOOOOOOOO...
>>>> OOOOOOOOO...
>>>> OOOOOOOOO...
>>>> OOOOOOOOO...
>>>> OOOOOOOOO...
>>>> XOOOOOOOO...
>>>> ...
>>>
>>> even after all possible transpositions.
>
>> you can show completion by proving there are no missing numbers,
>
> I have proved that after all transpositions, there are infinitely many fractions without index.

it is your bad math, you deleted them.

>>
>> all rationals are included, none are missing, it is complete.
>
> How do you bring that into line with the fact that there are infinitely many fractions without index?

you deleted them, your construct is just bad math.

"At this place there appears an X and where the *index has been taken from* there appears an O. After eight steps for instance we see the following
picture: 1/1, 1/2, 2/1, 1/3, 3/1, 1/4, 2/3, 3/2 are indexed whereas 4/1, 5/1, 6/1, 7/1, 8/1 *have lost their indexes*"

Highlighted is where you deleted numbers.

your missing Os are deception.

>
> Regards, WM

On 1/15/2022 5:38 PM, WM wrote:
> Jim Burns schrieb
> am Samstag, 15. Januar 2022 um 21:15:35 UTC+1:
>> On 1/15/2022 2:25 PM, WM wrote:

>>> omega would be the last ordinal unless a smaller one
>>> next to omega could be found (if all was in linear order).
>>
>> | Suppose psi was a smaller one next to omega.
>> | A totally-ordered collection {0,...,psi} exists
>
> No.

{0,...,psi} exists because,
for every k smaller than omega, {0,...,k} exists.

omega is the first infinite ordinal.
Anything before omega is a finite ordinal.
Any finite ordinal is before omega.

Being a finite ordinal k is equivalent to
{0,...,k} existing, starting at 0, ending at k,
with a counting-order
(each cut is a step, each step is a count).

"No" means that what you intend by your claim is not
what is broadly meant by those words in your claim.
Since we are not mind-readers we must depend upon
what is broadly meant. So, "no" means that you
have picked yourself up by the scruff of your neck
and thrown yourself in the dumpster.
Carry on. Don't let me interrupt your self-dumping.

>> | for which each cut is a step and each step is a count
>> | and which begins at 0 and ends at psi.
>> |
>> | psi+1 exists and
>> | a totally-ordered collection {0,...,psi+1} exists
>> | for which each cut is a step and each step is a count
>> | and which begins at 0 and ends at psi+1.
>> |
>> | Because "smaller one next to omega" psi+1 = omega
>
> Therefore psi is not accessible.

If psi < omega then {0,...,psi} exists.
When you aren't rigging the game, that means that
psi is accessible.

>> |
>> Therefore,
>> there is no smaller one next to omega.
>
> Then it is not next to omega.
> Where is the first smaller one?

How could the first smaller than omega not be
next to omega? Do you know what "next" means?

> What is between it an omega?

There is no first smaller.
For each smaller there are infinitely-many between
-- which is why there is no first smaller.

On Saturday, January 15, 2022 at 6:30:12 PM UTC-4, WM wrote:
> William schrieb am Samstag, 15. Januar 2022 um 21:48:05 UTC+1:

> We have
> >
> > (i) for every n element of N_P, there are are an infinite number of elements (not the same elements for every n) of N_P that are
> > larger than n,
> >
> > (ii) There is no element of N_P that is larger than every element of N_P

> So it is.

N_P does not change. A potentially infinite object changes. N_P is not potentially infinite.

--
William Hughes

William schrieb am Sonntag, 16. Januar 2022 um 02:50:30 UTC+1:

> N_P does not change. A potentially infinite object changes. N_P is not potentially infinite.

Construct the largest number you can. It is the the largest one, but only as long as you do not construct a larger one. And after that one, you can construct an even larger one. That is potential infinity.

Regards, WM

Jim Burns schrieb am Sonntag, 16. Januar 2022 um 02:16:18 UTC+1:
> On 1/15/2022 5:38 PM, WM wrote:
> > Jim Burns schrieb
> > am Samstag, 15. Januar 2022 um 21:15:35 UTC+1:

> There is no first smaller.
> For each smaller there are infinitely-many between
> -- which is why there is no first smaller.

But not all can be chosen. Each chosen one is the last one only as long as you do not construct a larger one.
That is potential infinity.

Regards, WM

On Sunday, January 16, 2022 at 9:46:10 AM UTC+1, WM wrote:
> Jim Burns schrieb am Sonntag, 16. Januar 2022 um 02:16:18 UTC+1:
> > On 1/15/2022 5:38 PM, WM wrote:
> > > Jim Burns schrieb
> > > am Samstag, 15. Januar 2022 um 21:15:35 UTC+1:
> > There is no first smaller.
> > For each smaller there are infinitely-many between
> > -- which is why there is no first smaller.
> But not all can be chosen. Each chosen one is the last one only as long as you do not construct a larger one.
> That is potential infinity.
>
> Regards, WM

The man is completely tireless, repeating the same nonsense now for more than 20 years.
\pi(x,y) = \frac{1}{2} (x + y) (x + y + 1) + y is bijective.

Now shut up until you have understood this fact.