JG's failed attempt to prove 2+2=4

[Dan Christensen]

Last week, JG set out to prove 2+2=4 as follows:

1. 2+2=1+1+1+1 
2. 4=1+1+1+1 
3. Therefore 2+2=4. 

I tried to patch things up a bit suggesting instead:

Define 2 = 1+1, 3 = 2+1, 4 = 3+1.
Then 2+2 = (1+1) + (1+1) = 1+1+1+1 = (2+1)+1 = 3+1 = 4

This requires associativity of +. It is quite easy to prove the associativity of + starting from Peano's Axioms. After you have constructed the + function, just prove by induction that, for any given x, y in N, we have: For all n in N, x+y+n = x+(y+n).

Of course, JG rejects anything resembling Peano's Axioms, so he cannot do proof by induction in his goofy system and, therefore, cannot establish the associativity of + other than by insults and frantic hand waving, e.g. "as every 4-year-old knows... Only an idiot would question..."

Now, 2+2=4 is not in dispute here. Every 4-year-old actually does know this. What is in dispute here is whether JG's "axioms" are a workable definition of the natural numbers or not. Are they sufficient ON THEIR OWN to prove known properties of the natural numbers? Clearly, they are not. JG cannot even prove that 2+2=4 without citing "what every 4-year-olds" and so on. As such, JG's goofy number system is truly a dead end.


Dan

Download my DC Proof 2.0 software at http://www.dcproof.com
Visit my Math Blog at http://www.dcproof.wordpress.com

[Me]

On Saturday, June 25, 2016 at 4:50:47 PM UTC+2, Dan Christensen wrote:

> Of course, JG rejects anything resembling Peano's Axioms, so he cannot do
> proof by induction in his goofy system and, therefore, cannot establish
> the associativity of + other than by insults and frantic hand waving, e.g.
> "as every 4-year-old knows... Only an idiot would question..."
>
> Now, 2+2=4 is not in dispute here. Every 4-year-old actually does know
> this. What is in dispute here is whether JG's "axioms" are a workable
> definition of the natural numbers or not. Are they sufficient ON THEIR OWN
> to prove known properties of the natural numbers? Clearly, they are not.
> JG cannot even prove that 2+2=4 without citing "what every 4-year-olds"
> and so on. As such, JG's goofy number system is truly a dead end.

Same problem with WM's broken "axiom system" for the natural numbers.

There you cant't even prove/derive the statement

1 e IN .
Not even
IN =/= {} .

[Peter Percival]

Dan Christensen wrote:
> Last week, JG set out to prove 2+2=4 as follows:
>
> 1. 2+2=1+1+1+1
> 2. 4=1+1+1+1
> 3. Therefore 2+2=4.
>
> I tried to patch things up a bit suggesting instead:
>
> Define 2 = 1+1, 3 = 2+1, 4 = 3+1.
> Then 2+2 = (1+1) + (1+1) = 1+1+1+1 = (2+1)+1 = 3+1 = 4
>
> This requires associativity of +.

Requires is too strong.

> It is quite easy to prove the associativity of + starting from Peano's Axioms. After you have constructed the + function, just prove by induction that, for any given x, y in N, we have: For all n in N, x+y+n = x+(y+n).
>
> Of course, JG rejects anything resembling Peano's Axioms, so he cannot do proof by induction in his goofy system and, therefore, cannot establish the associativity of + other than by insults and frantic hand waving, e.g. "as every 4-year-old knows... Only an idiot would question..."
>
> Now, 2+2=4 is not in dispute here. Every 4-year-old actually does know this. What is in dispute here is whether JG's "axioms" are a workable definition of the natural numbers or not. Are they sufficient ON THEIR OWN to prove known properties of the natural numbers? Clearly, they are not. JG cannot even prove that 2+2=4 without citing "what every 4-year-olds" and so on. As such, JG's goofy number system is truly a dead end.
>
>
> Dan
>
> Download my DC Proof 2.0 software at http://www.dcproof.com
> Visit my Math Blog at http://www.dcproof.wordpress.com
>
>

--
A high and rising proportion of children are being born to mothers
least fitted to bring children into the world.... Some are of low
intelligence, most of low educational attainment. They are unlikely
to be able to give children the stable emotional background, the
consistent combination of love and firmness.... They are producing
problem children.... The balance of our human stock, is threatened...
Sir Keith Joseph

[WM]

Am Samstag, 25. Juni 2016 17:02:18 UTC+2 schrieb Me:
> On Saturday, June 25, 2016 at 4:50:47 PM UTC+2, Dan Christensen wrote:

> > Now, 2+2=4 is not in dispute here. Every 4-year-old actually does know
> > this. What is in dispute here is whether JG's "axioms" are a workable
> > definition of the natural numbers or not. Are they sufficient ON THEIR OWN
> > to prove known properties of the natural numbers? Clearly, they are not.

Don't forget that the Peano axioms "on their own" do not define numbers at all except 1 or 0, respectively. His successors could be words of the Bible (and that would be more appropriate).

> > JG cannot even prove that 2+2=4 without citing "what every 4-year-olds"
> > and so on. As such, JG's goofy number system is truly a dead end

if viewed by a fool.

>
> Same problem with WM's broken "axiom system" for the natural numbers.
>
> There you cant't even prove/derive the statement
>
> 1 e IN .

You and he probably cannot. It would require the ability to read text, because the first axiom is 1 e IN.

Regards, WM

[Me]

On Saturday, June 25, 2016 at 5:17:26 PM UTC+2, Peter Percival wrote:
> Dan Christensen wrote:
> > Last week, JG set out to prove 2+2=4 as follows:
> >
> > 1. 2+2=1+1+1+1
> > 2. 4=1+1+1+1
> > 3. Therefore 2+2=4.
> >
> > I tried to patch things up a bit suggesting instead:
> >
> > Define 2 = 1+1, 3 = 2+1, 4 = 3+1.
> > Then 2+2 = (1+1) + (1+1) = 1+1+1+1 = (2+1)+1 = 3+1 = 4
> >
> > This requires associativity of +.
>
> Requires is too strong.

Oh, right, (1 + 1) + (1 + 1) = ((1 + 1) + 1) + 1 would be sufficient.

[Dan Christensen]

On Saturday, June 25, 2016 at 11:17:26 AM UTC-4, Peter Percival wrote:
> Dan Christensen wrote:
> > Last week, JG set out to prove 2+2=4 as follows:
> >
> > 1. 2+2=1+1+1+1
> > 2. 4=1+1+1+1
> > 3. Therefore 2+2=4.
> >
> > I tried to patch things up a bit suggesting instead:
> >
> > Define 2 = 1+1, 3 = 2+1, 4 = 3+1.
> > Then 2+2 = (1+1) + (1+1) = 1+1+1+1 = (2+1)+1 = 3+1 = 4
> >
> > This requires associativity of +.
>
> Requires is too strong.
>

You could prove 2+2=4 by using a definition of + something like:

1. For all x in N: x+1 = S(x)
2. For all x, y in N: x+S(y) = S(x+y)

Using the functionality of S, we can define 2=S(1), 3=S(2) and 4=S(3).

Then 2+2 = 2+S(1) = S(2+1) = S(3) = 4.

But I think this would have confused JG even more since it is derived from Peano's Axioms. While he may foolishly reject them, he cannot reject the associativity of +.

[Peter Percival]

That is what I had in mind.

[Dan Christensen]

On Saturday, June 25, 2016 at 11:25:16 AM UTC-4, WM wrote:
> Am Samstag, 25. Juni 2016 17:02:18 UTC+2 schrieb Me:
> > On Saturday, June 25, 2016 at 4:50:47 PM UTC+2, Dan Christensen wrote:
>
> > > Now, 2+2=4 is not in dispute here. Every 4-year-old actually does know
> > > this. What is in dispute here is whether JG's "axioms" are a workable
> > > definition of the natural numbers or not. Are they sufficient ON THEIR OWN
> > > to prove known properties of the natural numbers? Clearly, they are not.
>
> Don't forget that the Peano axioms "on their own" do not define numbers at all except 1 or 0, respectively. His successors could be words of the Bible (and that would be more appropriate).
>

Peano's Axioms mention only the 5 essential properties the natural numbers. From them, infinitely many others may be derived. That's the whole point of having axioms.

In the context of a formal proof, once you establish Peano's Axioms (by derivation or by assumption), all successors of 1 are equal. So are all successors of them in turn, and so on. This contrary to your frequently repeated lies here, Mucke.

>
> > > JG cannot even prove that 2+2=4 without citing "what every 4-year-olds"
> > > and so on. As such, JG's goofy number system is truly a dead end
>
> if viewed by a fool.

Wrong again, Mucke.

> >
> > Same problem with WM's broken "axiom system" for the natural numbers.
> >
> > There you cant't even prove/derive the statement
> >
> > 1 e IN .
>
> You and he probably cannot. It would require the ability to read text, because the first axiom is 1 e IN.
>

Ah, you have added another axiom! That's 2 now. You just need 3 more.

[Dan Christensen]

I assumed that the order of precedence from left to right, i.e. 1+1+1+1 = ((1+1)+1)+1. Much easier to read without all those brackets.

[Me]

No, it isn't (unfortunately). Your broken "axiom system" actually is (translation by me):

"The set of natural numbers IN = {1, 2, 3, ...} is defined with the following axioms:

(4.1) 1 e M
(4.2) n e M => (n + 1) e M
(4.3) If M satisfies (4.1) and (4.2), then IN c M holds "

You have been told repeatedly, that especially the following axioms are missing:

(1) 1 e IN
(2) n e IN => (n + 1) e IN .

Otherweise you can NOT prove

1 e IN ,

not even

IN =/= {}

from the axioms proper of your system.*)
______________

*) Actually, (4.1) and (4.2) can't be considered _axioms_ of your system (but you are to dumb to get it). Since (4.1) would state that FOR EVERY SET M: 1 e M. Ok, IN THIS CASE we might indeed derive 1 e IN from your "axiom" system. On the other hand, in this case (4.3) would' make any sense, any more. Nuff said.

[Me]

On Saturday, June 25, 2016 at 6:08:23 PM UTC+2, Dan Christensen wrote:
> > > >
> > > > This requires associativity of +.
> > > >

> > > Requires [associativity --me] is too strong.

> > >
> > Oh, right, (1 + 1) + (1 + 1) = ((1 + 1) + 1) + 1 would be sufficient.
> >
> I assumed that the order of precedence from left to right, i.e. 1+1+1+1 =
> ((1+1)+1)+1. Much easier to read without all those brackets.

Right. But I guess Peter wanted to point out that we don't need

x + (y + z) = (x + y) + z ("associativity2)

for any x,y,z e IN _in this case_. But (the special case)

1 + (1 + 1) = (1 + 1) + 1

would actually suffice. Well...

[Me]

On Saturday, June 25, 2016 at 6:15:49 PM UTC+2, Me wrote:
> On Saturday, June 25, 2016 at 5:25:16 PM UTC+2, WM wrote:
> > Am Samstag, 25. Juni 2016 17:02:18 UTC+2 schrieb Me:
> > >
> > > Same problem with WM's broken "axiom system" for the natural numbers.
> > >
> > > There you cant't even prove/derive the statement
> > >
> > > 1 e IN .
> > >
> > You and he probably cannot. It would require the ability to read text,
> > because the first axiom is 1 e IN.
> >
> No, it isn't (unfortunately). Your broken "axiom system" actually is
> (translation by me):
>
> "The set of natural numbers IN = {1, 2, 3, ...} is defined with the
> following axioms:
>
> (4.1) 1 e M
> (4.2) n e M => (n + 1) e M
> (4.3) If M satisfies (4.1) and (4.2), then IN c M holds "
>
> You have been told repeatedly, that especially the following axioms are
> missing:
>
> (1) 1 e IN
> (2) n e IN => (n + 1) e IN .
>

> Otherwise you can NOT prove

>
> 1 e IN ,
>
> not even
>
> IN =/= {}
>
> from the axioms proper of your system.*)
> ______________
>
> *) Actually, (4.1) and (4.2) can't be considered _axioms_ of your system
> (but you are to dumb to get it). Since (4.1) would state that FOR EVERY
> SET M: 1 e M. Ok, IN THIS CASE we might indeed derive 1 e IN from your

> "axiom system". On the other hand, in this case (4.3) wouldn't make sense
> any more.

Hence a MUCH better version of your "axiom system" would be:

(1) 1 e IN
(2) An(n e IN => n + 1 e IN)
(3) AM(1 e M & An(n e M => n + 1 e M) => IN c M) .

Still not good enough, but better than your original approach! (Actually, (3) should be tweaked a little bit in addition.)

At least THIS axiom system would allow for the derivation of the following two theorems:

1 e IN
and
1+1 e IN .

But we STILL could not prove from this axiom system that

1+1 =/= 1 .

To do this an additional axiom would be necessary. For example

(4) An(n e IN => n + 1 =/= 1)

would do.

NOW we could prove

1+1 =/= 1 .

Hence we now would get from our (extended) axiom system:

1 e IN, 1+1 e IN and 1+1 =/= 1 .

With other words

card(IN) >= 2 .

A beginning.

[Virgil]

In article <8a6d9f99-82c2-4a96...@googlegroups.com>,

WM <wolfgang.m...@hs-augsburg.de> wrote:

> Am Samstag, 25. Juni 2016 17:02:18 UTC+2 schrieb Me:
> > On Saturday, June 25, 2016 at 4:50:47 PM UTC+2, Dan Christensen wrote:
>
> > > Now, 2+2=4 is not in dispute here. Every 4-year-old actually does know
> > > this. What is in dispute here is whether JG's "axioms" are a workable
> > > definition of the natural numbers or not. Are they sufficient ON THEIR
> > > OWN
> > > to prove known properties of the natural numbers? Clearly, they are not.
>
> Don't forget that the Peano axioms "on their own" do not define numbers at
> all except 1 or 0, respectively. His successors could be words of the Bible
> (and that would be more appropriate).

Actually, Peano successorship requires an UNENDING sequence which the
finite number of words of the bible cannot provide,
So WM is
WRONG !
AGAIN ! !
AS USUAL ! ! !


In Proper Mathematics, Cantor is provably right,
claiming Card(Q) = Card(N) and Card(N) < Card(R)
So in Mathematics Card(Q) < Card(R), with fewer rationals than reals.

In his WMytheology, WM claims Cantor is wrong,
there WM claims Card(Q) > Card(N) and Card(N) = Card(R)
So in WM's witless worthless wacky world of WMytheology,
Card(Q) > Card(R), with more rationals than reals.

But there are not any rationals that are not also reals in Mathematics,
only in WM's witless worthless wacky world of WMytheology!

So WM is
LYING !
AGAIN ! !
AS USUAL ! ! !
--
Virgil
"Mit der Dummheit kampfen Gotter selbst vergebens." (Schiller)

[Jim Burns]

On 6/25/2016 11:44 AM, Peter Percival wrote:
> Me wrote:
>> On Saturday, June 25, 2016 at 5:17:26 PM UTC+2,
>> Peter Percival wrote:
>>> Dan Christensen wrote:

>>>> Last week, JG set out to prove 2+2=4 as follows:
>>>>
>>>> 1. 2+2=1+1+1+1
>>>> 2. 4=1+1+1+1
>>>> 3. Therefore 2+2=4.
>>>>
>>>> I tried to patch things up a bit suggesting instead:
>>>>
>>>> Define 2 = 1+1, 3 = 2+1, 4 = 3+1.
>>>> Then 2+2 = (1+1) + (1+1) = 1+1+1+1 = (2+1)+1 = 3+1 = 4
>>>>
>>>> This requires associativity of +.
>>>
>>> Requires is too strong.
>>
>> Oh, right, (1 + 1) + (1 + 1) = ((1 + 1) + 1) + 1
> would be sufficient.
>
> That is what I had in mind.

If we change our focus for a moment from what WM has said
to what would be reasonable and useful to say (I know --
a very different thing), I think it is fair to say
something close to associativity of + is required.

In order to prove 2 + 2 = 4 , we can assume the usual
definitions of 2,3,4 and one instance of associativity
2 = 1 + 1
3 = 2 + 1
4 = 3 + 1
2 + (1 + 1) = (2 + 1) + 1

But it's unreasonable and useless to have axioms that _only_
prove 2 + 2 = 4 .

Suppose we take those four assumptions as our axiom scheme
and then extend it.

We extend our definitions to the rest of the numbers
(but not to 0, like a good medieval Christian)
2 = 1 + 1
3 = 2 + 1
4 = 3 + 1
...

and we make a general rule -- association-like --
all x,y, x + (y + 1) = (x + y) + 1

This general rule looks like half of the typical definition
of + from S , where we define S(x) = (x + 1) ,
the other half being
all x, x + 0 = x
which we can't say, since we have no 0s .

I suspect that this is _enough_ to derive all the usual specific
results of addition, 2 + 2 = 4, 678 + 592 = 1270 , etc,
which would be enough for many merchants and bankers of that
time, I think.

(Is it _not enough_ of we leave that rule out?
I'm not sure how to ask that question formally.)

The longer I contemplate what we need to assume in order to
have numbers, the more impressed I become with their
ineradicable nature, like metaphysical crab grass.

But very beautiful and elegant metaphysical crab grass,
of course.

[Me]

On Saturday, June 25, 2016 at 7:44:49 PM UTC+2, Jim Burns wrote:

> We extend our definitions to the rest of the numbers
> (but not to 0, like a good medieval Christian)
> 2 = 1 + 1
> 3 = 2 + 1
> 4 = 3 + 1
> ...
>
> and we make a general rule -- association-like --
>
> all x,y, x + (y + 1) = (x + y) + 1
>
> This general rule looks like half of the typical definition
> of + from S , where we define S(x) = (x + 1) ,

Right, and hence you DON'T have to formulate it as a rule, when you've already introduced "+" the usual way, such that the two recursive equations

An e IN: n + 1 = S(n)
An,m e IN: n + S(m) = S(n + m)

are satified. Your "rule" immediately follows from these two equations.

(The first of these "defining" equations usually is adopted when starting the natural numbers with 1 instead of 0. Ax. 1 e IN.)

[WM]

Am Samstag, 25. Juni 2016 17:58:17 UTC+2 schrieb Dan Christensen:


> Peano's Axioms mention only the 5 essential properties the natural numbers.

No, they omit the most important one.

> From them, infinitely many others may be derived. That's the whole point of having axioms.

But the most important one cannot be derived from them.

> In the context of a formal proof, once you establish Peano's Axioms (by derivation or by assumption), all successors of 1 are equal.

That is not mentioned in the axioms. On the contrary.

> > > Same problem with WM's broken "axiom system" for the natural numbers.
> > >
> > > There you cant't even prove/derive the statement
> > >
> > > 1 e IN .
> >
> > You and he probably cannot. It would require the ability to read text, because the first axiom is 1 e IN.
> >
>
> Ah, you have added another axiom!

No, it has been there all the time. You forgot to read it.

That's 2 now. You just need 3 more.

When using the successor notion, certainly. But not in the context of addition of 1. Certainly you would also employ a stoker in a space shuttle.

Regards, WM

[Me]

On Saturday, June 25, 2016 at 8:28:02 PM UTC+2, WM wrote:
> Am Samstag, 25. Juni 2016 17:58:17 UTC+2 schrieb Dan Christensen:
> > > >

> > > > You cant't even prove/derive the statement

> > > >
> > > > 1 e IN .
> > > >
> > > You and he probably cannot. It would require the ability to read text,
> > > because the first axiom is 1 e IN.
> > >
> > Ah, you have added another axiom!
> >
> No, it has been there all the time. You forgot to read it.

Huh?!

Your broken "axiom system" actually is (translation by me):

"The set of natural numbers IN = {1, 2, 3, ...} is defined with the following axioms:

(4.1) 1 e M
(4.2) n e M => (n + 1) e M
(4.3) If M satisfies (4.1) and (4.2), then IN c M holds "

So where can we read "1 e IN"? I'm sorry, but I'm not able to spot it in any of your "axioms".

Actually, you have been told repeatedly, that especially the following axioms are MISSING in your system:

(1) 1 e IN
(2) n e IN => (n + 1) e IN .

Otherweise you can NOT prove

1 e IN ,

not even

IN =/= {}

from the axioms proper of your system.*)
______________

*) Actually, (4.1) and (4.2) can't be considered _axioms_ of your system (but you are to dumb to get it). Since (4.1) would state that FOR EVERY SET M: 1 e M. Ok, IN THIS CASE we might indeed derive 1 e IN from your "axiom" system. On the other hand, in this case (4.3) wouldn't make sense any more.

[WM]

Am Samstag, 25. Juni 2016 18:15:49 UTC+2 schrieb Me:


> "The set of natural numbers IN = {1, 2, 3, ...} is defined with the following axioms:
>
> (4.1) 1 e M
> (4.2) n e M => (n + 1) e M
> (4.3) If M satisfies (4.1) and (4.2), then IN c M holds "
>

> I am too dumb to get it.

Therefore I will explain it again:

Consider the sets denoted by M with the properties 1 in M and with n also n+1 in M. |N is the intersection of these sets.

Regards, WM

[WM]

Am Samstag, 25. Juni 2016 18:51:50 UTC+2 schrieb Me:


> 1 e IN
> and
> 1+1 e IN .
>
> But we STILL could not prove from this axiom system that
>
> 1+1 =/= 1 .

That is wrong. Here we have not the clumsy and failed notation of successors but addition of 1. That implies 1 + 1 =/= 1. Compare Lorenzen)

Regards, WM

[Me]

On Saturday, June 25, 2016 at 8:36:21 PM UTC+2, WM wrote:
> Am Samstag, 25. Juni 2016 18:15:49 UTC+2 schrieb Me:
> >
> > "The set of natural numbers IN = {1, 2, 3, ...} is defined with the
> > following axioms:
> >
> > (4.1) 1 e M
> > (4.2) n e M => (n + 1) e M
> > (4.3) If M satisfies (4.1) and (4.2), then IN c M holds "
> >
> > I am too dumb to get it.

No, I did't write that, asshole!

But YOU wrote:

"the first axiom is 1 e IN."

Liar!

[Me]

On Saturday, June 25, 2016 at 8:40:21 PM UTC+2, WM wrote:
> Am Samstag, 25. Juni 2016 18:51:50 UTC+2 schrieb Me:

> Here we have not the clumsy and failed notation of successors but addition
> of 1. That implies 1 + 1 =/= 1.

Well, if this is the case, where's your proof?

Do you know the meaning of the notion /proof/ in mathematics, Mückenheim?

Hint:
https://en.wikipedia.org/wiki/Mathematical_proof

[Peter Percival]

(1+1)+(1+1) requires (Ux,y,z)(x+(y+z)=(x+y)+z) is *false*. Had Dan
claimed that (Ux,y,z)(x+(y+z)=(x+y)+z) was sufficient for (1+1)+(1+1) I
wouldn't have complained.

To see the falsity of the first claim consider that it is equivalent to

if (1+1)+(1+1)=1+1+1+1 then (Ux,y,z)(x+(y+z)=(x+y)+z).

The second claim is equivalent to

if (Ux,y,z)(x+(y+z)=(x+y)+z) then (1+1)+(1+1)=1+1+1+1

which is true with the left association assumption, i.e., the assumption
that 1+1+1+1=((1+1)+1)+1.

[WM]

Am Samstag, 25. Juni 2016 21:27:05 UTC+2 schrieb Me:
> On Saturday, June 25, 2016 at 8:40:21 PM UTC+2, WM wrote:
> > Am Samstag, 25. Juni 2016 18:51:50 UTC+2 schrieb Me:
>
> > Here we have not the clumsy and failed notation of successors but addition
> > of 1. That implies 1 + 1 =/= 1.
>
> Well, if this is the case, where's your proof?

You show an incredible degree of stupidity.
There is no proof necessary or possible because the construction by addition of the unit cannot result in a smaller set of units.

>
> Do you know the meaning of the notion /proof/ in mathematics, Mückenheim?

Do you wish to criticize Lorenzen or Zermelo or von Neumann or all mathematicians with exception of Christensen?

Regards, WM

[WM]

Am Samstag, 25. Juni 2016 20:56:03 UTC+2 schrieb Me:
> On Saturday, June 25, 2016 at 8:36:21 PM UTC+2, WM wrote:
> > Am Samstag, 25. Juni 2016 18:15:49 UTC+2 schrieb Me:
> > >
> > > "The set of natural numbers IN = {1, 2, 3, ...} is defined with the
> > > following axioms:
> > >
> > > (4.1) 1 e M
> > > (4.2) n e M => (n + 1) e M
> > > (4.3) If M satisfies (4.1) and (4.2), then IN c M holds "
> > >
> > > I am too dumb to get it.
>

> Sorry, I did't write that I am an asshole!

Of course there is no reason why you shouldn't call yourself what you like. As long as you don't insult others.

>
> But YOU wrote:
>
> "the first axiom is 1 e IN."

Yes, of course. It is so because |N is the intersection of all sets which contain 1.

Sorry that I am thinking a bit too fast for you.

Regards, WM

[Peter Percival]

WM wrote:
> Am Samstag, 25. Juni 2016 20:56:03 UTC+2 schrieb Me:
>> On Saturday, June 25, 2016 at 8:36:21 PM UTC+2, WM wrote:
>>> Am Samstag, 25. Juni 2016 18:15:49 UTC+2 schrieb Me:
>>>>
>>>> "The set of natural numbers IN = {1, 2, 3, ...} is defined with the
>>>> following axioms:
>>>>
>>>> (4.1) 1 e M
>>>> (4.2) n e M => (n + 1) e M
>>>> (4.3) If M satisfies (4.1) and (4.2), then IN c M holds "
>>>>
>>>> I am too dumb to get it.
>>
>> Sorry, I did't write that I am an asshole!
>
> Of course there is no reason why you shouldn't call yourself what you like. As long as you don't insult others.
>>
>> But YOU wrote:
>>
>> "the first axiom is 1 e IN."
>
> Yes, of course. It is so because |N is the intersection of all sets which contain 1.

Do you mean that? If it did follow that 1 e |N because |N is the
intersection of all sets which contain 1, then 1 e |N would not be an
axiom since it is proved from something else. Or, to be more precise,
it would be a redundant axiom.

Also, the intersection of all sets which contain 1, is {1}.

> Sorry that I am thinking a bit too fast for you.
>
> Regards, WM
>

[WM]

Am Samstag, 25. Juni 2016 22:41:49 UTC+2 schrieb Peter Percival:
> WM wrote:
> > Am Samstag, 25. Juni 2016 20:56:03 UTC+2 schrieb Me:
> >> On Saturday, June 25, 2016 at 8:36:21 PM UTC+2, WM wrote:
> >>> Am Samstag, 25. Juni 2016 18:15:49 UTC+2 schrieb Me:
> >>>>
> >>>> "The set of natural numbers IN = {1, 2, 3, ...} is defined with the
> >>>> following axioms:
> >>>>
> >>>> (4.1) 1 e M
> >>>> (4.2) n e M => (n + 1) e M
> >>>> (4.3) If M satisfies (4.1) and (4.2), then IN c M holds "
> >>>>
> >>>> I am too dumb to get it.
> >>
> >> Sorry, I did't write that I am an asshole!
> >
> > Of course there is no reason why you shouldn't call yourself what you like. As long as you don't insult others.
> >>
> >> But YOU wrote:
> >>
> >> "the first axiom is 1 e IN."
> >
> > Yes, of course. It is so because |N is the intersection of all sets which contain 1.
>
> Do you mean that? If it did follow that 1 e |N because |N is the
> intersection of all sets which contain 1, then 1 e |N would not be an
> axiom since it is proved from something else. Or, to be more precise,
> it would be a redundant axiom.

That does not matter. There are many redundant axioms even in ZFC. See
https://www.hs-augsburg.de/~mueckenh/Transfinity/Transfinity/pdf
p. 39 ff.

>
> Also, the intersection of all sets which contain 1, is {1}.

|N is the intersection of all sets that contain 1 (according to 4.1) and have the inductive property (4.2).

Regards, WM

[Me]

On Saturday, June 25, 2016 at 10:18:55 PM UTC+2, WM wrote:

> > But YOU wrote:
> >
> > "the first axiom is 1 e IN."
> >
> Yes, of course. It is so because |N is the intersection of all sets which
> contain 1.

Huh?!

Man, I'm sorry to say that, but I think that you are completely screwed up.

[Peter Percival]

Ok, that's fine.

[Me]

On Saturday, June 25, 2016 at 10:41:49 PM UTC+2, Peter Percival wrote:
> WM wrote:
> > Am Samstag, 25. Juni 2016 20:56:03 UTC+2 schrieb Me:
> >> On Saturday, June 25, 2016 at 8:36:21 PM UTC+2, WM wrote:
> >>> Am Samstag, 25. Juni 2016 18:15:49 UTC+2 schrieb Me:
> >>>>
> >>>> "The set of natural numbers IN = {1, 2, 3, ...} is defined with the
> >>>> following axioms:
> >>>>
> >>>> (4.1) 1 e M
> >>>> (4.2) n e M => (n + 1) e M
> >>>> (4.3) If M satisfies (4.1) and (4.2), then IN c M holds "
> >>>>

> >> But YOU wrote:
> >>
> >> "the first axiom is 1 e IN."
> >
> > Yes, of course. It is so because |N is the intersection of all sets which contain 1.
>
> Do you mean that? If it did follow that 1 e |N because |N is the
> intersection of all sets which contain 1, then 1 e |N would not be an
> axiom since it is proved from something else. Or, to be more precise,
> it would be a redundant axiom.
>
> Also, the intersection of all sets which contain 1, is {1}.

Maybe a recent reply to one of WM's posts may help:

================

Obviously you tried to "reproduce" the introduction (definition) of the natural numbers _as a certain subset of of the real numbers_ (after THEY have been introduced, say, axiomatically).

To accomplish this we refer to the notion of an /inductive set/.

(Def.:) A subset M of IR is called /inductive/ if the following conditions are satisfied:

1. 1 e M
2. x e M => x + 1 e M

Looks familiar, doesn't it?

(Def.:) /IN/ is the intersection of all inductive sets.

Source:
https://de.wikipedia.org/wiki/Nat%C3%BCrliche_Zahl#Die_natürlichen_Zahlen_als_Teilmenge_der_reellen_Zahlen

Now your "axiom" (4.3) is NOT just a paraphrase of the definition mentioned above. It is a STATEMENT concerning IN (not a proper definition).

You completely screwed up EVERYTHING!

First of all, this approach presupposes "the real number system". Hence Addition (+) is already introduced and hence "available" for defining the natural numbers.

Secondly, that's the REASON why such an approach does not have to mention the two "Peano axioms"

An e IN: n + 1 =/= 1 ,
An,m e IN: n + 1 = m + 1 => n = m .

I guess that's the reason why they do not appear in your "system". (Though you obviously didn't realize the dependency of this approach on (IR, +).)

Finally, after defining /IN/ the way mentioned above it's easy enough to prove that IN is the smallest inductive set. (That's where your statement (4.3) comes into play. Though in the context of this approach it's not an axiom, but a theorem.)

My GUESS is that for your "axiom system" you borrowed material from the book "Analysis I" by Martin Barner and Friedrich Flohr, pp. 22-23 - things that you quite obviously didn't understand.

================

[Me]

On Saturday, June 25, 2016 at 10:52:54 PM UTC+2, WM wrote:

> |N is the intersection of all sets that contain 1 (according to 4.1) and
> have the inductive property (4.2).

===========

> IN is the intersection of all sets having these properties.

THIS obviously is a (so called) proper definition for /IN/. Funnily enough this definition is _not stated_ in the context of your "system".

In short, your "axiom system" is total crap.

Obviously you tried to "reproduce" the introduction (definition) of the natural numbers as a certain subset of of the real numbers (after THEY have been introduced axiomatically).

To accomplish this we refer to the notion of an /inductive set/.

A subset M of IR is called /inductive/ if the following conditions are satisfied:

1. 1 e M
2. x e M => x + 1 e M

Looks familiar, doesn't it?

(Def.:) Then IN is the intersection of all inductive sets.

Source:
https://de.wikipedia.org/wiki/Natürliche_Zahl#Die_nat.C3.BCrlichen_Zahlen_als_Teilmenge_der_reellen_Zahlen

Now your "axiom" (4.3) is (again) NOT just a paraphrase of the definition mentioned above. It is a STATEMENT concerning IN (not a proper definition).

You completely screwed up EVERYTHING!

First of all, this approach presupposes "the real number system". Hence Addition (+) is already introduced and hence "available" for defining the natural numbers.

Secondly, that's the REASON why such an approach does not have to mention the two "Peano axioms"

An e IN: n + 1 =/= 1 ,
An,m e IN: n + 1 = m + 1 => n = m .

I guess that's the reason why they do not appear in your "system". (Though you obviously didn't realize the dependency of this approach on (IR, +))

Finally, after defining /IN/ the way mentioned above it's easy enough to prove that IN is the smallest inductive set. (That's where your statement (4.3) comes into play. Though in the context of this approach it's not an axiom, but a theorem.)

How about giving up mathematics an becoming a gardener?

(My GUESS is that for your "axiom system" you borrowed material from Analysis I by Martin Barner and Friedrich Flohr, pp. 22-23 - things that you quite obviously didn't understand.)

[Me]

On Saturday, June 25, 2016 at 11:35:12 PM UTC+2, Peter Percival wrote:
> WM wrote:
> > Am Samstag, 25. Juni 2016 22:41:49 UTC+2 schrieb Peter Percival:
> > >
> > > Also, the intersection of all sets which contain 1, is {1}.
> > >

> > IN is the intersection of all sets that contain 1 (according to 4.1)

> > and have the inductive property (4.2).
> >
> Ok, that's fine.
>

No, that's NOT fine, since this definition is _not stated_ in the context of his "system".

And even if it WERE, there would be no JUSTIFICATION for that. (Which axioms ensure that there is at least ONE such set?)

[Me]

On Saturday, June 25, 2016 at 8:36:21 PM UTC+2, WM wrote:

*lol* This approach only works when it's already clear (i.e. can be proved) that there ARE such sets. Otherwise /IN/ is not even defined. (Since the notion of "the intersection of these sets" is meaningless in this case.)

Hint: In the context of set theory the axiom of infinity ensures the existence of so called /inductive sets/ (adopting the definitions 0 := {} and s(x) := x u {x}, due to von Neumann).

You are really such a fool, man.

[Jim Burns]

On 6/25/2016 2:17 PM, Me wrote:
> On Saturday, June 25, 2016 at 7:44:49 PM UTC+2,
> Jim Burns wrote:

>> We extend our definitions to the rest of the numbers
>> (but not to 0, like a good medieval Christian)
>> 2 = 1 + 1
>> 3 = 2 + 1
>> 4 = 3 + 1
>> ...
>>
>> and we make a general rule -- association-like --
>>
>> all x,y, x + (y + 1) = (x + y) + 1
>>
>> This general rule looks like half of the typical definition
>> of + from S , where we define S(x) = (x + 1) ,
>
> Right, and hence you DON'T have to formulate it as a rule,
> when you've already introduced "+" the usual way,
> such that the two recursive equations
>
> An e IN: n + 1 = S(n)
> An,m e IN: n + S(m) = S(n + m)
>
> are satified. Your "rule" immediately follows from these
> two equations.

I agree that my rule above follows from these equations.
In fact, the first equation also follows from the second with
all n, n + 0 = n and S(0) = 1

I am a great fan of the usual recursive definition of +
It seems to have reached the Euclidean ideal of obviousness
and also fully describe all the needed properties of +
in order for it not to be anything else.

My only slight improvement on its obviousness (IMO) is
to spell the second
all x,y,z, x + y = z <-> x + S(y) = S(z)

What I was trying to do here (in what is likely to be an
amateurish fashion) is imagine what rules an early adopter
of these "numbers" ( a _very_ early adopter -- possibly
pre-historic?) would use and what they could derive from them.
This imagining of mine seems not quite such a special
pleading as WM's "axioms" (which apparently only have the
virtue of allowing WM to pretend he knows what he is
talking about).

It's only speculation, but what I speculate is that by the
time we have learned what addition _is_ facts such as
2 + 2 = 4 are already set in stone.

> (The first of these "defining" equations usually is adopted
> when starting the natural numbers with 1 instead of 0.
> Ax. 1 e IN.)

Sorry, I don't get this.

[Me]

On Sunday, June 26, 2016 at 3:01:59 AM UTC+2, Jim Burns wrote:

> > (The first of these "defining" equations usually is adopted

> > when starting the natural numbers with 1 instead of 0. ...)

> >
> Sorry, I don't get this.

When the first "Peano axiom" of our axiom system reads

1 e IN

we usually adopt the "defining" equations

An e IN: n + 1 = S(n)
An,m e IN: n + S(m) = S(n + m)

for the recursive definition of +.

When on the other hand the first "Peano axiom" of our axiom system reads

0 e IN

we usually adopt the "defining" equations

An e IN: n + 0 = n

An,m e IN: n + S(m) = S(n + m)

for the recursive definition of +.

[Jim Burns]

On 6/25/2016 8:54 PM, Me wrote:
> On Saturday, June 25, 2016 at 8:36:21 PM UTC+2, WM wrote:

>> Consider the sets denoted by M with the properties 1 in M and
>> with n also n+1 in M. |N is the intersection of these sets.
>
> *lol* This approach only works when it's already clear
> (i.e. can be proved) that there ARE such sets. Otherwise /IN/
> is not even defined. (Since the notion of "the intersection of
> these sets" is meaningless in this case.)

Meaningless is one way to look at it, a way that perhaps is
kinder to WM than he deserves.

Another way to look at the intersection of no sets
(which WM may be using as his definition of IN ) is as
the set of all elements contained in every set in {} .
Since there are no sets in {}, this is vacuously true of
every element, which would make IN the universal class.

My purely personal preference is that WM would take this
interpretation, in the hopes that he would then try to defend
his choice. Pure comedy gold!

[Jim Burns]

Ah! I get it now. Thank you for your patience.

I've never had to define + without 0 , and it didn't
occur to me that that is what you were doing (though
you'd said you were doing that, I think)

It makes sense now that, just as 0 is "the thing that
when added to anything gives you the thing back",
1 is "the thing that when added to anything gives
its successor". So these are both suitable definitions.

I'm a bit surprised that either (with the second equation)
is all we need to properly define + (ignoring 0).
But maybe I shouldn't be.

[Me]

On Sunday, June 26, 2016 at 3:29:42 AM UTC+2, Jim Burns wrote:

> I've never had to define + without 0, and it didn't

> occur to me that that is what you were doing (though

> you'd said you were doing that, I think).

I see.

Indeed this alternative set of "defining" equation is not even mentioned here:
https://en.wikipedia.org/wiki/Peano_axioms#Addition

> It makes sense now that, just as 0 is "the thing that
> when added to anything gives you the thing back",
> 1 is "the thing that when added to anything gives
> its successor".

Precisely.

> So these are both suitable definitions.

Guess so.

Actually, I like the elegant definition of < (or <=) the just defined operation (+) allows for:

a < b :<-> Ec(c e IN & a + c = b) (when Ax. 1: 1 e IN)
(then a <= b :<-> a < b v a = b)
or
a <= b :<-> Ec(c e IN & a + c = b) (when Ax. 1: 0 e IN)
(then a < b :<-> a <= b & a =/= b)

We then immediately can prove the nice result

An e IN ~Em e IN: n < m < S(n) .

With other words, for any number n, there is no number "between" n and its successor S(n), as expected (but not _immediately_ clear from the Peano axioms).

[Peter Percival]

You should try selling that to Dan.

> You are really such a fool, man.
>

[Virgil]

In article <8c1a19b5-e1b6-4fbd...@googlegroups.com>,

WM <wolfgang.m...@hs-augsburg.de> wrote:

> Am Samstag, 25. Juni 2016 18:51:50 UTC+2 schrieb Me:
>
>
> > 1 e IN
> > and
> > 1+1 e IN .
> >
> > But we STILL could not prove from this axiom system that
> >
> > 1+1 =/= 1 .
>
> That is wrong. Here we have not the clumsy and failed notation of successors

It only fails in WM's witless worthless wacky world of WMytheology,
where everything fails. In mathematics, Peano succeeds brilliantly!

In Mathematics, Cantor is provably right,
saying Card(Q) = Card(N) and Card(N) < Card(R)
So Card(Q) < Card(R), with fewer rationals than reals

In WMaths/WMytheology, WM wrongly claims Cantor is wrong,
WM claims Card(Q) > Card(N) and Card(N) = Card(R)
So WM must falsely be claiming Card(Q) > Card(R),
and thus wrongly claiming more rationals than reals.

So WM is
LYING !
AGAIN ! !
AS USUAL ! ! !
--
Virgil
"Mit der Dummheit kampfen Gotter selbst vergebens." (Schiller)

[Virgil]

In article <f862117c-5e8a-4d86...@googlegroups.com>,

WM <wolfgang.m...@hs-augsburg.de> wrote:

> Am Samstag, 25. Juni 2016 21:27:05 UTC+2 schrieb Me:
> > On Saturday, June 25, 2016 at 8:40:21 PM UTC+2, WM wrote:
> > > Am Samstag, 25. Juni 2016 18:51:50 UTC+2 schrieb Me:
> >
> > > Here we have not the clumsy and failed notation of successors but
> > > addition
> > > of 1. That implies 1 + 1 =/= 1.
> >
> > Well, if this is the case, where's your proof?
>
> You show an incredible degree of stupidity.

Not anywhere nearly as much as WM shows!

> There is no proof necessary or possible

If true, proof would be both necessary and possible!

> >
> > Do you know the meaning of the notion /proof/ in mathematics, Mückenheim?
>
> Do you wish to criticize

WM wants to but doesn't know how to do it properly,
so he always messes it up!!

In Mathematics, Cantor is provably right,

Card(Q) = Card(N) and Card(N) < Card(R)
So Card(Q) < Card(R), with fewer rationals than reals

In WMaths/WMytheology, WM claims Cantor is wrong,

WM claims Card(Q) > Card(N) and Card(N) = Card(R)

So Card(Q) > Card(R), with more rationals than reals.

But there aren't more rational than reals in proper mathematics,
only in WM's witless worthless wacky world of WMytheology.

So WM is
WRONG !

[Virgil]

In article <691037ff-710d-4595...@googlegroups.com>,

WM <wolfgang.m...@hs-augsburg.de> wrote:

> Am Samstag, 25. Juni 2016 18:15:49 UTC+2 schrieb Me:
>
>
> > "The set of natural numbers IN = {1, 2, 3, ...} is defined with the
> > following axioms:
> >
> > (4.1) 1 e M
> > (4.2) n e M => (n + 1) e M

Not without predefining n + 1, which you have not done.

Peano does it better!

In Mathematics, Cantor is provably right,
Card(Q) = Card(N) and Card(N) < Card(R)
So Card(Q) < Card(R), with fewer rationals than reals

In WMaths/WMytheology, WM claims Cantor is wrong,
WM claims Card(Q) > Card(N) and Card(N) = Card(R)
So Card(Q) > Card(R), with more rationals than reals.

SO WM claims what he cannot supply, more rationals than reals.

So WM is
LYING !

[Virgil]

In article <66df0e7e-06f7-4100...@googlegroups.com>,

WM <wolfgang.m...@hs-augsburg.de> wrote:

> Am Samstag, 25. Juni 2016 20:56:03 UTC+2 schrieb Me:
> > On Saturday, June 25, 2016 at 8:36:21 PM UTC+2, WM wrote:
> > > Am Samstag, 25. Juni 2016 18:15:49 UTC+2 schrieb Me:
> > > >
> > > > "The set of natural numbers IN = {1, 2, 3, ...} is defined with the
> > > > following axioms:
> > > >
> > > > (4.1) 1 e M
> > > > (4.2) n e M => (n + 1) e M
> > > > (4.3) If M satisfies (4.1) and (4.2), then IN c M holds "
> > > >
> > > > I am too dumb to get it.
> >
> > Sorry, I did't write that I am an asshole!
>
> Of course there is no reason why you shouldn't call yourself what you like.
> As long as you don't insult others.
> >
> > But YOU wrote:
> >
> > "the first axiom is 1 e IN."
>
> Yes, of course.

But where did you get your "1" or your "|N"?

> It is so because |N is the intersection of all sets which
> contain 1.

Maybe INSIDE of WM's witless worthless wacky world of WMytheology.

But everywhere OUTSIDE of WM's witless worthless wacky world of
WMytheology, {1} is the intersection of all sets which contain 1.

At least among those less idiotically arrogant than WM.

>
> Sorry that I am thinking a bit too fast for you.

And, as usual, much too fast to get anything right!

In WMaths/WMytheology, WM claims Cantor is wrong,
WM claims Card(Q) > Card(N) and Card(N) = Card(R)

So WM claims more rationals than reals
even though every rational is a real!
>
> Regards, WM

[Virgil]

In article <238b824a-1802-4035...@googlegroups.com>,

Then in WM's witless worthless wacky world of WMytheology, but not in
any proper form of mathematics, |N is a subset of {1}.

So WM is
LYING !
AGAIN ! !
AS USUAL ! ! !

[Virgil]

In article <c96056f8-f95f-43f0...@googlegroups.com>,
WM <wolfgang.m...@hs-augsburg.de> wrote:

> Am Samstag, 25. Juni 2016 17:58:17 UTC+2 schrieb Dan Christensen:
>
>
> > Peano's Axioms mention only the 5 essential properties the natural numbers.
>
> No, they omit the most important one.
>
> > From them, infinitely many others may be derived. That's the whole point of
> > having axioms.
>
> But the most important one cannot be derived from them.
>
> > In the context of a formal proof, once you establish Peano's Axioms (by
> > derivation or by assumption), all successors of 1 are equal.
>
> That is not mentioned in the axioms. On the contrary.
>
> > > > Same problem with WM's broken "axiom system" for the natural numbers.
> > > >
> > > > There you cant't even prove/derive the statement
> > > >
> > > > 1 e IN .
> > >
> > > You and he probably cannot. It would require the ability to read text,
> > > because the first axiom is 1 e IN.
> > >
> >
> > Ah, you have added another axiom!
>
> No, it has been there all the time. You forgot to read it.

Which one of the following is it, WM?


PEANO AXIOMS:
There is a set S and an object o, and a function F such that
1. o is a member of S
2. if x is in S then F(x) is in S
3. For every x in S, o =/= F(x)
4. For every x and y in S, if F(x) = F(y) then x = y
5. If a set T is such that o is a member of T and whenever x is a
member of T then also F(x) is a member of T then S is a subset of T.


> When using the successor notion, certainly. But not in the context of
> addition of 1.

The Peano axioms above do not mention either addition or 1.


> Certainly you would also employ a stoker in a space shuttle.

In a space shuttle designed by WM, certainly!

In a space shuttle designed by someone competent, never!



In Proper Mathematics, Cantor is provably right,
claiming Card(Q) = Card(N) and Card(N) < Card(R)
So in Mathematics Card(Q) < Card(R), with fewer rationals than reals.

In his WMytheology, WM claims Cantor is wrong,
there WM claims Card(Q) > Card(N) and Card(N) = Card(R)
So in WM's witless worthless wacky world of WMytheology, he claims

Card(Q) > Card(R), with more rationals than reals.

But there are no rationals that are not also reals in the world of
mathemaiccs free of WM's witless worthless wacky world of WMytheology1

[Dan Christensen]

On Saturday, June 25, 2016 at 4:15:06 PM UTC-4, WM wrote:
> Am Samstag, 25. Juni 2016 21:27:05 UTC+2 schrieb Me:
> > On Saturday, June 25, 2016 at 8:40:21 PM UTC+2, WM wrote:
> > > Am Samstag, 25. Juni 2016 18:51:50 UTC+2 schrieb Me:
> >
> > > Here we have not the clumsy and failed notation of successors but addition
> > > of 1. That implies 1 + 1 =/= 1.
> >
> > Well, if this is the case, where's your proof?
>
> You show an incredible degree of stupidity.
> There is no proof necessary or possible because the construction by addition of the unit cannot result in a smaller set of units.

You cannot prove the existence of even one number in your goofy system unless you add another axiom, Mucke. And here you are claim that there are at least two distinct numbers: 1 and 1+1??? It doesn't work that way.


Dan
Download my DC Proof 2.0 software at http://www.dcproof.com
Visit my Math Blog at http://www.dcproof.wordpress.com

[John Gabriel]

On Saturday, 25 June 2016 16:50:47 UTC+2, Dan Christensen wrote:
> Last week, JG set out to prove 2+2=4 as follows:
>
> 1. 2+2=1+1+1+1 
> 2. 4=1+1+1+1 
> 3. Therefore 2+2=4. 
>
> I tried to patch things up a bit suggesting instead:
>
> Define 2 = 1+1, 3 = 2+1, 4 = 3+1.
> Then 2+2 = (1+1) + (1+1) = 1+1+1+1 = (2+1)+1 = 3+1 = 4

I am laughing my arse off at the local idiot DC.

One doesn't have to define 2 = 1 + 1 you moron. It was defined that way over 2000 years ago. Pay attention idiot!

2 is the measure of a magnitude composed of 1 unit and another unit.
3 is the measure of a magnitude composed of 2 units and another unit.

Nothing is gained by Peano's crapaxioms. Associativity has NOTHING to do with any of this you brainwashed idiot!!!!! NOTHING is gained by trying to define numbers recursively as in Peano's dismal failure. Peano was an absolute MORON. That makes you ten times the imbecile he was.

Numbers come before the operations on them. Association is a concept that arose AFTER the natural numbers were established. It was done so geometrically using areas. The same with all the other NON-REMARKABLE properties such as commutation and distribution.

As Dullrich used to say, a definition DOES NOT require proof. I always agreed with him on that. However, I disagreed that a definition is WELL-FORMED, unless it can be reified flawlessly, that means no contradictions or paradoxes. You can build a lot of theory on crap but eventually it all comes falling down as it rots.

WM has seen the light. On the other hand, you and your fellow morons will NEVER see the light. You will perish in your ignorance. Nor are you helping those by teaching them the useless rot you never understood and still fail to understand.

Peano was a fool:

https://www.youtube.com/watch?v=lAD-pfSNdSs

>
> This requires associativity of +. It is quite easy to prove the associativity of + starting from Peano's Axioms. After you have constructed the + function, just prove by induction that, for any given x, y in N, we have: For all n in N, x+y+n = x+(y+n).
>
> Of course, JG rejects anything resembling Peano's Axioms, so he cannot do proof by induction in his goofy system and, therefore, cannot establish the associativity of + other than by insults and frantic hand waving, e.g. "as every 4-year-old knows... Only an idiot would question..."
>
> Now, 2+2=4 is not in dispute here. Every 4-year-old actually does know this. What is in dispute here is whether JG's "axioms" are a workable definition of the natural numbers or not. Are they sufficient ON THEIR OWN to prove known properties of the natural numbers? Clearly, they are not. JG cannot even prove that 2+2=4 without citing "what every 4-year-olds" and so on. As such, JG's goofy number system is truly a dead end.

[WM]

Am Samstag, 25. Juni 2016 23:54:55 UTC+2 schrieb Me:
> On Saturday, June 25, 2016 at 11:35:12 PM UTC+2, Peter Percival wrote:
> > WM wrote:
> > > Am Samstag, 25. Juni 2016 22:41:49 UTC+2 schrieb Peter Percival:
> > > >
> > > > Also, the intersection of all sets which contain 1, is {1}.
> > > >
> > > IN is the intersection of all sets that contain 1 (according to 4.1)
> > > and have the inductive property (4.2).
> > >
> > Ok, that's fine.
> >
> No, that's NOT fine, since this definition is _not stated_ in the context of his "system".

I have no system. But the definition is stated in my book W. Mückenheim: "Mathematik für die ersten Semester", 4th ed., De Gruyter, Berlin 2015. ISBN 978-3-11-037733-0.

>
> And even if it WERE, there would be no JUSTIFICATION for that. (Which axioms ensure that there is at least ONE such set?)

Only a fool could write or read "1 in M" and be in doubt whether there was a set containing 1.

Regards, WM

[WM]

Am Sonntag, 26. Juni 2016 03:11:49 UTC+2 schrieb Me:
> On Sunday, June 26, 2016 at 3:01:59 AM UTC+2, Jim Burns wrote:
>
> > > (The first of these "defining" equations usually is adopted
> > > when starting the natural numbers with 1 instead of 0. ...)
> > >
> > Sorry, I don't get this.
>
> When the first "Peano axiom" of our axiom system reads
>
> 1 e IN
>
> we usually adopt the "defining" equations
>
> An e IN: n + 1 = S(n)
> An,m e IN: n + S(m) = S(n + m)
>
> for the recursive definition of +.

That is not contained in the Peano axioms.

>
> When on the other hand the first "Peano axiom" of our axiom system reads
>
> 0 e IN
>
> we usually adopt the "defining" equations
>
> An e IN: n + 0 = n
> An,m e IN: n + S(m) = S(n + m)
>
> for the recursive definition of +.

That is not contained in the Peano axioms either.

According to the Peano axioms you could adopt n + 1 = n as well as n + 0 = S(n). You see that your belief in axioms is blatent nonsense. You don't use them in defining the important facts. You are self-deceiving. But many "logicians" seem to like that.

Mathematics does not require "Axioms". The job of a pure mathematician is not to build some elaborate castle in the sky, and to proclaim that it stands up on the strength of some arbitrarily chosen assumptions. The job is to investigate the mathematical reality of the world in which we live. For this, no assumptions are necessary. ... The difficulty with the current reliance on "Axioms" arises from a grammatical confusion ... People use the term "Axiom" when often they really mean definition. Thus the "axioms" of group theory are in fact just definitions. We say exactly what we mean by a group, that's all.
[N.J. Wildberger: "Set Theory: Should You Believe?"]

Regards, WM

[WM]

Am Samstag, 25. Juni 2016 22:59:24 UTC+2 schrieb Me:
> On Saturday, June 25, 2016 at 10:18:55 PM UTC+2, WM wrote:
>
> > > But YOU wrote:
> > >
> > > "the first axiom is 1 e IN."
> > >
> > Yes, of course. It is so because |N is the intersection of all sets which
> > contain 1.
>
> Huh?!
>

> Man, I'm sorry to say that, but I think that I am completely screwed up.

Don't be sad. Not everyone is endowed with perception of logic.

Regards, WM

[WM]

Am Samstag, 25. Juni 2016 23:38:50 UTC+2 schrieb Me:


>
> Obviously you tried to "reproduce" the introduction (definition) of the natural numbers _as a certain subset of of the real numbers_ (after THEY have been introduced, say, axiomatically).

Not at all. The assition of 1 does not even require fractions.
But don't be sad. Not everyone is endowed with perception of logic.

> First of all, this approach presupposes "the real number system".

Did the ancient Egypts know real numbers? Dis they know how to count?
But don't be sad. Not everyone is endowed with perception of logic.

Regards, WM

[WM]

Am Sonntag, 26. Juni 2016 08:57:37 UTC+2 schrieb John Gabriel:


> Numbers come before the operations on them.

Correct!

> Association is a concept that arose AFTER the natural numbers were established.

Correct!

> It was done so geometrically using areas. The same with all the other NON-REMARKABLE properties such as commutation and distribution.

>

> On the other hand, you and your fellow morons will NEVER see the light. You will perish in your ignorance. Nor are you helping those by teaching them the useless rot you never understood and still fail to understand.

Occasionally logicians inquire as to whether the current "Axioms" need to be changed further, or augmented. The more fundamental question – whether mathematics requires any Axioms – is not up for discussion. That would be like trying to get the high priests on the island of Okineyab to consider not whether the Divine Ompah's Holy Phoenix has twelve or thirteen colours in her tail (a fascinating question on which entire tomes have been written), but rather whether the Divine Ompah exists at all. Ask that question, and icy stares are what you have to expect, then it's off to the dungeons, mate, for a bit of retraining. Mathematics does not require "Axioms". The job of a pure mathematician is not to build some elaborate castle in the sky, and to proclaim that it stands up on the strength of some arbitrarily chosen assumptions. The job is to investigate the mathematical reality of the world in which we live. For this, no assumptions are necessary. Careful observation is necessary, clear definitions are necessary, and correct use of language and logic are necessary. But at no point does one need to start invoking the existence of objects or procedures that we cannot see, specify, or implement. The difficulty with the current reliance on "Axioms" arises from a grammatical confusion [...] People use the term "Axiom" when often they really mean definition. Thus the "axioms" of group theory are in fact just definitions. We say exactly what we mean by a group, that's all. [N.J. Wildberger: "Set Theory: Should You Believe?"]

Regards, WM

[WM]

Am Sonntag, 26. Juni 2016 06:31:13 UTC+2 schrieb Dan Christensen:
> On Saturday, June 25, 2016 at 4:15:06 PM UTC-4, WM wrote:
> > Am Samstag, 25. Juni 2016 21:27:05 UTC+2 schrieb Me:
> > > On Saturday, June 25, 2016 at 8:40:21 PM UTC+2, WM wrote:
> > > > Am Samstag, 25. Juni 2016 18:51:50 UTC+2 schrieb Me:
> > >
> > > > Here we have not the clumsy and failed notation of successors but addition
> > > > of 1. That implies 1 + 1 =/= 1.
> > >
> > > Well, if this is the case, where's your proof?
> >
> > You show an incredible degree of stupidity.
> > There is no proof necessary or possible because the construction by addition of the unit cannot result in a smaller set of units.
>
> You cannot prove the existence of even one number

Wrong. 1 in M shows that 1 exists.

> And here you are claim that there are at least two distinct numbers: 1 and 1+1??? It doesn't work that way.

It does punctiliously work that way. Perhaps you would like to belong to the guild of logicians?

Most (but not all) of the difficulties of Set Theory arise from the insistence that there exist "infinite sets", and that it is the job of mathematics to study them and use them. In perpetuating these notions, modern mathematics takes on many of the aspects of a religion. It has its essential creed - namely Set Theory, and its unquestioned assumptions, namely that mathematics is based on "Axioms", in particular the Zermelo-Fraenkel "Axioms of Set Theory". It has its anointed priesthood, the logicians, who specialize in studying the foundations of mathematics, a supposedly deep and difficult subject that requires years of devotion to master. Other mathematicians learn to invoke the official mantras when questioned by outsiders, but have only a hazy view about how the elementary aspects of the subject hang together logically. Training of the young is like that in secret societies - immersion in the cult involves intensive undergraduate memorization of the standard thoughts before they are properly understood, so that comprehension often follows belief instead of the other (more healthy) way around. ... The large international conferences let the fellowship gather together and congratulate themselves on the uniformity and sanity of their world view, though to the rare outsider that sneaks into such events the proceedings no doubt seem characterized by jargon, mutual incomprehensibility and irrelevance to the outside world. [N.J. Wildberger: "Set Theory: Should You Believe?"]

But I am in doubt whether the guild would accept you. They'd know at least that the Peano axioms do not define the natural numbers without some handwaving about n + 0 =/= n + 1. Your only advantage is that your ideas and "proofs" of 1 + 1 = 2 are of absolute irrelevance to the outside world.

Regards, WM

[WM]

Am Sonntag, 26. Juni 2016 03:20:04 UTC+2 schrieb Jim Burns:


> Meaningless is one way to look at it, a way that perhaps is
> kinder to WM than he deserves.
>
> Another way to look at the intersection of no sets
> (which WM may be using as his definition of IN )

Even when denying the existence of any sets and set theory, the axiom "1 in M" states the existence of 1. With set theory the axiom states the existence of a set M = {1}.

You should try to write more intelligent texts in future.

Regards, WM

[WM]

"pi is the thing that when added to anything gives its successor" would be equally possible. You see that your support of DC was not justified.

Regards, WM

[Dan Christensen]

On Sunday, June 26, 2016 at 8:24:45 AM UTC-4, WM wrote:
> Am Sonntag, 26. Juni 2016 06:31:13 UTC+2 schrieb Dan Christensen:
> > On Saturday, June 25, 2016 at 4:15:06 PM UTC-4, WM wrote:
> > > Am Samstag, 25. Juni 2016 21:27:05 UTC+2 schrieb Me:
> > > > On Saturday, June 25, 2016 at 8:40:21 PM UTC+2, WM wrote:
> > > > > Am Samstag, 25. Juni 2016 18:51:50 UTC+2 schrieb Me:
> > > >
> > > > > Here we have not the clumsy and failed notation of successors but addition
> > > > > of 1. That implies 1 + 1 =/= 1.
> > > >
> > > > Well, if this is the case, where's your proof?
> > >
> > > You show an incredible degree of stupidity.
> > > There is no proof necessary or possible because the construction by addition of the unit cannot result in a smaller set of units.
> >
> > You cannot prove the existence of even one number
>
> Wrong. 1 in M shows that 1 exists.
>

You never actually say that 1 in N. Why can you not introduce that as an axiom? While you are at, why not add that every element of has a unique successor using whatever notation you prefer (S(), +, ++ or +1). And that 1 has not predecessor. And that the successor relation is injective? It would solve all your problems here, Mucke.

> > And here you are claim that there are at least two distinct numbers: 1 and 1+1??? It doesn't work that way.
>
> It does punctiliously work that way. Perhaps you would like to belong to the guild of logicians?
>

No thanks. I'm doing just fine on my own.

> Most (but not all) of the difficulties of Set Theory arise from the insistence that there exist "infinite sets", and that it is the job of mathematics to study them and use them.

[snip]

In mathematics, we study all kinds of sets, finite and infinite, countable and uncountable. And it is import to be able distinguish between them. So, enough of your crank conspiracy theories, Mucke.

[Me]

On Sunday, June 26, 2016 at 2:24:45 PM UTC+2, WM wrote:
> Am Sonntag, 26. Juni 2016 06:31:13 UTC+2 schrieb Dan Christensen:
> >

> > You cannot prove the existence of even one number.

> >
> Wrong. 1 in M shows that 1 exists.

OMG!

Look, Mückenheim. If (4.1) in your "axiom system" really WERE an axiom, then it would state that EVER SET contains 1 as an element.

The reason for this is the following: in mathematics free variables in statements are considert to be implicitely bound by universal quantifiers. (Otherwise the "statements" wouldn't be "declarative sentences" or "Aussagen" etc.)

Hence if we state the second Peano axiom the following way:

x e IN --> S(x) e IN ,

the actual axiom (when written down in full) would be

Ax(x e IN --> S(x) e IN) .

In the same way your "axiom" (4.1) if considered as an axiom would actually read (when written down in full):

(4.1) AM(1 e M)

And then, right, we might indeed derive

1 e IN

from it (by specialization). On the other hand, we also could derive

1 e {} .

Contradiction! (Since there is no element in {}.)

So we have shown that it is UNREASONABLE to consider (4.1) an axiom.

THE SAME is true for your "axiom" (4.2).

Actually (4.1) and (4.2) are only "outsourced" parts of your _single_ AXIOM (4.3).

Hence your whole system would better read:

(A3) For any set M, if the M satisfies the follwing two conditions
(a) 1 e M, (b) n e M => n + 1 e M, for all n e IN
then IN c M.

Now even YOU should be able to see that you CAN'T derive 1 e IN from it (after all, this axiom would hold for IN = {} too).

So a rather straightforward approach to remedy this rather unlucky state of affairs is to add the follwing axiom CONCERNING IN:

(A1) 1 e IN .

Now, we actually GET 1 e IN from our axioms (since we explicitely stated it as an axiom).

In addition

(A2) n e IN => n + 1 e IN

would be rather hepful. (Otherwise we would not be able to prove 1+1 e IN.)

So "finally" we would get the following system:

(A1) 1 e IN
(A2) n e IN => n + 1 e IN
(A3) For any set M, if the M satisfies the follwing two conditions
(a) 1 e M, (b) n e M => n + 1 e M, for all n e IN
then IN c M

which obviously resembles your own approach somehow (but is free of the quirks mentioned above).

[Dan Christensen]

On Sunday, June 26, 2016 at 2:57:37 AM UTC-4, John Gabriel wrote:
> On Saturday, 25 June 2016 16:50:47 UTC+2, Dan Christensen wrote:
> > Last week, JG set out to prove 2+2=4 as follows:
> >
> > 1. 2+2=1+1+1+1 
> > 2. 4=1+1+1+1 
> > 3. Therefore 2+2=4. 
> >
> > I tried to patch things up a bit suggesting instead:
> >
> > Define 2 = 1+1, 3 = 2+1, 4 = 3+1.
> > Then 2+2 = (1+1) + (1+1) = 1+1+1+1 = (2+1)+1 = 3+1 = 4
>
> I am laughing my arse off at the local idiot DC.
>
> One doesn't have to define 2 = 1 + 1 you moron. It was defined that way over 2000 years ago. Pay attention idiot!
>

JG cleary has no clue about what axioms are how they are used.

> 2 is the measure of a magnitude composed of 1 unit and another unit.
> 3 is the measure of a magnitude composed of 2 units and another unit.
>

An just how can you apply this nonsense in proving anything about the natural numbers, e.g. that 2+2=4? You can't, of course. That such a simple result eludes you should tell that your "system" as it stands is useless, Troll Boy.

> Nothing is gained by Peano's crapaxioms. Associativity has NOTHING to do with any of this you brainwashed idiot!!!!!

You can't even do basic arithmetic without the associativity of addition and multiplication, Troll Boy. But you would know this if you had ever finished high school.

> NOTHING is gained by trying to define numbers recursively as in Peano's dismal failure.

From Peano's Axioms, we can formally derive all of modern number theory. As we see here, from your goofy number system, we cannot even cannot even derive 2+2=4 as a first step!

********************************

John Gabriel is a buffoon. He knows NOTHING about mathematics, but seems to delight in confusing and misleading students here, "messing with their minds" as he probably thinks of it. What readers should know about Psycho Troll John Gabriel, in his own words:

JG's God Complex:

“I am the Creator of this galaxy.”
-- March 19, 2015

“I am the greatest mathematician ever.”
-- June 21, 2016

“I am the last word on everything.”
-- May 6, 2015

“Whatever I imagine is real because whatever I imagine is well defined.”
-- March 26, 2015

“Unless I think it's logic, it's not... There are no rules in mathematics... As I have repeatedly stated, if there were to be rules, I'd be making the rules.”
-- March 17, 2015

"There are no axioms required when concepts are well defined. My mathematics is well defined."
--June 21, 2016

JG's Final Solution:

“Hitler was a genius and a very talented artist... As from a moral point of view, again his actions can't be judged, because his morals are different.” (Like JG's morals?)
-- March 18, 2015

“All those who don't accept New Calculus, you better say goodbye to your kids. Because John Gabriel is coming.”
-- July 9, 2014

Some months later, JG claimed this posting of July 9, was the work of an impostor. Do we believe him? I'm not inclined to since only four days later, on July 13, he also posted and subsequently confirmed:

"I will point out a few facts about Hitler that most of you arrogant idiots didn't know or refused to acknowledge because your Jewish overlords do not allow you... Unfortunately, Hitler's henchmen got the wrong Jews...

“In the early 20th century, there was a eugenics program in the United States. Too bad it was halted... It would be a very good idea to round up all the academic idiots, gas them and incinerate the useless lot. Only those that pass John Gabriel's exam should be allowed to live.”
-- July 13, 2014

(Links to an archive of original postings available on request. Serious enquiries only.)


JG's Just Plain Stupid:

“100 years have shown that nothing Einstein predicted is correct.” (“Jewish” science, right, JG?)
-- March 23, 2016

“To claim that all the natural numbers are in the set N, one must be able to list them all from beginning to the end.”
-- December 2, 2015

“1/0 is not undefined.”
-- May 19, 2015

“1/3 does NOT mean 1 divided by 3 and never has meant that”
-- February 8, 2015

“The square root of 2 and pi are NOT numbers.”
-- May 28, 2015

“By definition, a line is the distance between two points.”
-- April 13, 2015

“So, 'is a member of' = 'is a subset of.'”
-- May 16, 2015

“There is no such thing as a continuous real number line.”
-- March 24, 2015

“Indeed, there is no such thing as an instantaneous speed -- certainly not with respect to the calculus.” (Note: Instantaneous speed is indicated by the speedometer in a car. Another international Jewish conspiracy, JG?)
-- March 17, 2015

“Proofs had nothing to do with calculus.”
-- May 30, 2015

"You don't need associativity or commutativity or any other crap."
--June 21, 2106

JG doesn't like proofs. In his wacky system, he cannot even prove that 2+2=4. It seems unlikely to me that he would have anything worthwhile to say about mathematics. On the contrary, it seems he is deliberately trying to mislead and confuse students here.

A special word of caution to students: Do not attempt to use JG's “system” in any course work in any high school, college or university on the planet. You will fail miserably. His system is certainly no “shortcut” to success in mathematics. It is truly a dead-end.

[Dan Christensen]

On Sunday, June 26, 2016 at 8:24:02 AM UTC-4, WM wrote:

>
> Occasionally logicians inquire as to whether the current "Axioms" need to be changed further, or augmented. The more fundamental question – whether mathematics requires any Axioms – is not up for discussion.

“Axioms are rubbish!”
--WM, sci.math, 2014/11/19


More absurd quotes from Wolfgang Muckenheim (WM):

“In my system, two different numbers can have the same value.”
-- sci.math, 2014/10/16

“1+2 and 2+1 are different numbers.”
-- sci.math, 2014/10/20

“1/9 has no decimal representation.”
-- sci.math, 2015/09/22

"0.999... is not 1."
-- sci.logic 2015/11/25

“No set is countable, not even |N.”
-- sci.logic, 2015/08/05

“Countable is an inconsistent notion.”
-- sci.math, 2015/12/05

“A [natural] number with aleph_0 digits is not less than aleph_0.”
-- sci.math, 2015/08/12

“The notion of aleph_0 is not meaningful.”
-- sci.math, 2015/08/28


Slipping ever more deeply into madness...

“There is no actually infinite set |N.”
-- sci.math, 2015/10/26

“|N is not covered by the set of natural numbers.”
-- sci.math, 2015/10/26

“The set of all rationals can be shown not to exist.”
--sci.math, 2015/11/28

“Everything is in the list of everything and therefore everything belongs to a not uncountable set.”
-- sci.math, 2015/11/30

"'Not equal' and 'equal can mean the same.”
-- sci.math, 2016/06/09

A special word of caution to students: Do not attempt to use WM's “system” (MuckeMath) in any course work in any high school, college or university on the planet. You will fail miserably. MuckeMath is certainly no shortcut to success in mathematics.

Using WM's “axioms” for the natural numbers, he cannot prove that 1=/=2 or the existence of even a single number. It is truly a dead-end.

[Me]

On Sunday, June 26, 2016 at 2:52:34 PM UTC+2, Dan Christensen wrote:
> On Sunday, June 26, 2016 at 8:24:45 AM UTC-4, WM wrote:
> >

> > [...]

> >
> You never actually say that 1 in N. Why can you not introduce that as an

> axiom? While you are at [it], why not add that every element of N has a

> unique successor using whatever notation you prefer (S(), +, ++ or +1).

Well, I guess this idiot (erroneously) believes that these two axioms are already "included" (don't ask!) in his two "axioms" (or whatever):

(4.1) 1 e M
(4.2) n e M => n+1 e M

*sigh*

> And that 1 has not predecessor. And that the successor relation is
> injective? It would solve all your problems here, Mucke.

Nope. These two axioms would "force" IN to be infinite. I guess that's the REAL REASON why he is not willing to adopt these two axioms.

Of course, in this case, IN (in the context of his system) might just be the set {1}.

[Dan Christensen]

On Saturday, June 25, 2016 at 2:36:21 PM UTC-4, WM wrote:
> Am Samstag, 25. Juni 2016 18:15:49 UTC+2 schrieb Me:
>
>
> > "The set of natural numbers IN = {1, 2, 3, ...} is defined with the following axioms:
> >

> > (4.1) 1 e M

> > (4.2) n e M => (n + 1) e M
> > (4.3) If M satisfies (4.1) and (4.2), then IN c M holds "
> >
>
> > I am too dumb to get it.
>

> Therefore I will explain it again:
>
> Consider the sets denoted by M with the properties 1 in M and with n also n+1 in M. |N is the intersection of these sets.
>

Wrong again, Mucke. Go back to your original source. I'm quite sure that you will find the above axiom was described there as the principle of mathematical induction. It is only one part of what defines the natural numbers.

[WM]

Am Sonntag, 26. Juni 2016 14:52:34 UTC+2 schrieb Dan Christensen:
> On Sunday, June 26, 2016 at 8:24:45 AM UTC-4, WM wrote:
> > Am Sonntag, 26. Juni 2016 06:31:13 UTC+2 schrieb Dan Christensen:
> > > On Saturday, June 25, 2016 at 4:15:06 PM UTC-4, WM wrote:
> > > > Am Samstag, 25. Juni 2016 21:27:05 UTC+2 schrieb Me:
> > > > > On Saturday, June 25, 2016 at 8:40:21 PM UTC+2, WM wrote:
> > > > > > Am Samstag, 25. Juni 2016 18:51:50 UTC+2 schrieb Me:
> > > > >
> > > > > > Here we have not the clumsy and failed notation of successors but addition
> > > > > > of 1. That implies 1 + 1 =/= 1.
> > > > >
> > > > > Well, if this is the case, where's your proof?
> > > >
> > > > You show an incredible degree of stupidity.
> > > > There is no proof necessary or possible because the construction by addition of the unit cannot result in a smaller set of units.
> > >
> > > You cannot prove the existence of even one number
> >
> > Wrong. 1 in M shows that 1 exists.
> >
>
> You never actually say that 1 in N. Why can you not introduce that as an axiom?

1 in M proves that 1 exists. So, when applying set theory, which is fine in the finite domain, {1} = M exists. Can't you see this? Therefore the intersection |N contains 1.

> While you are at, why not add that every element of has a unique successor

That is not necessary when havin a set of n elements and adding 1. Cantor did it already:
1 = E0. (1)
Man vereinige nun mit E0 ein anderes Ding e1, die Vereinigungsmenge heiße E1, so daß

E1 = (E0, e1) = (e0, e1). (2)

290 Die Kardinalzahl von E1 heißt "Zwei" und wird mit 2 bezeichnet:

2 = E1. (3)

Durch Hinzufügung neuer Elemente erhalten wir die Reihe der Mengen

E2 = (E1, e2), E3 = (E2, e3), . . . ,

No use for further axioms.

> using whatever notation you prefer (S(), +, ++ or +1). And that 1 has not predecessor.

Not necessary. Do youreally wish to criticize Cantor, Zermelo, v. Neumann, and myself? All of us refrain from defining uniquness because it is guaranteed automatically.

> And that the successor relation is injective? It would solve all your problems here

I have no problems.

Regards, WM

[WM]

Am Samstag, 25. Juni 2016 19:44:49 UTC+2 schrieb Jim Burns:
> On 6/25/2016 11:44 AM, Peter Percival wrote:
> > Me wrote:
> >> On Saturday, June 25, 2016 at 5:17:26 PM UTC+2,

> >> Peter Percival wrote:
> >>> Dan Christensen wrote:
>
> >>>> Last week, JG set out to prove 2+2=4 as follows:
> >>>>
> >>>> 1. 2+2=1+1+1+1
> >>>> 2. 4=1+1+1+1
> >>>> 3. Therefore 2+2=4.
> >>>>
> >>>> I tried to patch things up a bit suggesting instead:
> >>>>
> >>>> Define 2 = 1+1, 3 = 2+1, 4 = 3+1.
> >>>> Then 2+2 = (1+1) + (1+1) = 1+1+1+1 = (2+1)+1 = 3+1 = 4
> >>>>

> >>>> This requires associativity of +.
> >>>

> >>> Requires is too strong.
> >>
> >> Oh, right, (1 + 1) + (1 + 1) = ((1 + 1) + 1) + 1
> > would be sufficient.
> >
> > That is what I had in mind.
>
> If we change our focus for a moment from what WM has said
> to what would be reasonable and useful to say (I know --
> a very different thing)

You believe to know that you know?
What about subsets of countable sets? Are they always countable?

What about the Binary Tree emptied of path and nodes in countably many steps. Does it contain further paths (which then cannot be defined by nodes).

Regards, WM

[Dan Christensen]

On Sunday, June 26, 2016 at 10:38:49 AM UTC-4, WM wrote:
> Am Sonntag, 26. Juni 2016 14:52:34 UTC+2 schrieb Dan Christensen:
> > On Sunday, June 26, 2016 at 8:24:45 AM UTC-4, WM wrote:
> > > Am Sonntag, 26. Juni 2016 06:31:13 UTC+2 schrieb Dan Christensen:
> > > > On Saturday, June 25, 2016 at 4:15:06 PM UTC-4, WM wrote:
> > > > > Am Samstag, 25. Juni 2016 21:27:05 UTC+2 schrieb Me:
> > > > > > On Saturday, June 25, 2016 at 8:40:21 PM UTC+2, WM wrote:
> > > > > > > Am Samstag, 25. Juni 2016 18:51:50 UTC+2 schrieb Me:
> > > > > >
> > > > > > > Here we have not the clumsy and failed notation of successors but addition
> > > > > > > of 1. That implies 1 + 1 =/= 1.
> > > > > >
> > > > > > Well, if this is the case, where's your proof?
> > > > >
> > > > > You show an incredible degree of stupidity.
> > > > > There is no proof necessary or possible because the construction by addition of the unit cannot result in a smaller set of units.
> > > >
> > > > You cannot prove the existence of even one number
> > >
> > > Wrong. 1 in M shows that 1 exists.
> > >
> >
> > You never actually say that 1 in N. Why can you not introduce that as an axiom?
>
> 1 in M proves that 1 exists.

It doesn't prove that 1 is an element of N. That requires a separate axiom.

Again, you have mistakenly taken a statement of the principle of induction for the definition of the set of natural numbers. It is only ONE PART.

[Me]

On Sunday, June 26, 2016 at 4:38:49 PM UTC+2, WM wrote:

> 1 in M proves that 1 exists.

*lol* If /M/ is a constant, this SAYS that 1 is an element in the set M. And with existental introduction we can even derive that there is a set such that it contains the element 1: Ex(1 e x).

If on the other /M/ is a variable "1 in M" is not even a statement. Hence we can "prove" nothing from it.

Since in your "system" /1/ is an /undefined notion/ (i. e. a primitive term), you can immediately state

1 = 1

and from that derive,

Ex(x = 1)

"1 exists".

No additional (non-logcal) axioms needed!

[Dan Christensen]

On Sunday, June 26, 2016 at 10:52:32 AM UTC-4, WM wrote:
> Am Samstag, 25. Juni 2016 19:44:49 UTC+2 schrieb Jim Burns:

> > If we change our focus for a moment from what WM has said
> > to what would be reasonable and useful to say (I know --
> > a very different thing)
>
> You believe to know that you know?
> What about subsets of countable sets? Are they always countable?
>

Using only the notation of set theory, please construct any countable set A and an uncountable subset B of A. I am quite certain, you will find it impossible. No diagrams, please. Just set theory. (Don't hold your breath, folks. Hee, hee!)

[Virgil]

In article <c963efad-41a5-499c...@googlegroups.com>,

WM <wolfgang.m...@hs-augsburg.de> wrote:

> You believe to know that you know?
> What about subsets of countable sets? Are they always countable?

Those free from the corruption of WMytheology know whether subsets of
countable sets are always countable. WM does not!

>
> What about the Binary Tree emptied of path and nodes in countably many steps.

Outside of WM's witless worthless wacky world of WMytheology one can
empty a Complete Infinite Binary Tree of paths by removing only the root
node of the tree, which is a necessary member of each path.

In Proper Mathematics, Cantor is provably right,
claiming Card(Q) = Card(N) and Card(N) < Card(R)
So in Mathematics Card(Q) < Card(R), with fewer rationals than reals.

In his WMytheology, WM claims Cantor is wrong,
there WM claims Card(Q) > Card(N) and Card(N) = Card(R)
So in WM's witless worthless wacky world of WMytheology,

Card(Q) > Card(R), with more rationals than reals.

Except there are no rationals which are not reals in mathematics.

[Virgil]

In article <fa5e773b-c918-4cef...@googlegroups.com>,

Not when, as is the case, pi already has a definition incompatible with
that one!

So WM is
WRONG !

AGAIN ! !
AS USUAL ! ! !

[Virgil]

In article <582ab881-e6ea-45a8...@googlegroups.com>,

WM <wolfgang.m...@hs-augsburg.de> wrote:

> Am Sonntag, 26. Juni 2016 03:20:04 UTC+2 schrieb Jim Burns:
>
>
> > Meaningless is one way to look at it, a way that perhaps is
> > kinder to WM than he deserves.
> >
> > Another way to look at the intersection of no sets
> > (which WM may be using as his definition of IN )
>
> Even when denying the existence of any sets and set theory, the axiom "1 in
> M" states the existence of 1.

But no one other than he who rejects all axioms will ever claims any
such axiom, and his rejection kills it.


WM claims Card(Q) > Card(|N) = Card(R) which requires the existence of
rational numbers which are not reals numbers, but such numbers do not
exist even in WM's witless worthless wacky world of WMytheology.

[Virgil]

In article <6d7bea6d-d171-4eca...@googlegroups.com>,

WM <wolfgang.m...@hs-augsburg.de> wrote:

> But I am in doubt whether the guild would accept you.

The guild of mathematicians clearly rejects WM and his witless

worthless wacky world of WMytheology.

> They'd know at least
> that the Peano axioms do not define the natural numbers without some
> handwaving about n + 0 =/= n + 1.

The Peano axioms are totally silent, and wave no hands at all, about 0.


PEANO AXIOMS:
There is a set S and an object u, and a function F such that
1. u is a member of S

2. if x is in S then F(x) is in S

3. For every x in S, u =/= F(x)

4. For every x and y in S, if F(x) = F(y) then x = y

5. If a set T is such that u is a member of T and whenever x is a

member of T then also F(x) is a member of T then S is a subset of T.

> Your only advantage is that your ideas and
> "proofs" of 1 + 1 = 2 are of absolute irrelevance to the outside world.

So in WM's witless worthless wacky world of WMytheology outside of
proper mathematics, is WM claiming that 1 + 1 =/= 2?

In Proper Mathematics, Cantor is provably right,
claiming Card(Q) = Card(N) and Card(N) < Card(R)
So in Mathematics Card(Q) < Card(R), with fewer rationals than reals.

In WM's witless worthless wacky world of WMytheology,
WM claims Cantor is wrong!
There WM claims Card(Q) > Card(N) and Card(N) = Card(R)

So in WM's witless worthless wacky world of WMytheology,
Card(Q) > Card(R), with more rationals than reals.

But WM cannot name even one such unreal rational.
So WM is
LYING !

AGAIN ! !
AS USUAL ! ! !

[WM]

Am Sonntag, 26. Juni 2016 17:32:17 UTC+2 schrieb Dan Christensen:


> > What about subsets of countable sets? Are they always countable?
> >
>
> Using only the notation of set theory, please construct any countable set A and an uncountable subset B of A.

The other way round. Try to describe in set theory the fact that all texts in all languages are countable, but the subset of all real numbers is not.

If that is impossible set theory is useless even in moonshine examples.

Regards, WM

[WM]

Am Sonntag, 26. Juni 2016 17:31:31 UTC+2 schrieb Me:
> On Sunday, June 26, 2016 at 4:38:49 PM UTC+2, WM wrote:
>
> > 1 in M proves that 1 exists.
>
> *lol* If /M/ is a constant,

...

> If on the other /M/ is a variable

...

1 is given as existing. The character of M is irrelevant.

>

> "1 exists".
>
> No additional (non-logcal) axioms needed!

Of course! Nothing else is needed. Why do you make a fuss about that fact?

Regards, WM

[Virgil]

In article <53c0e39a-3446-4dbb...@googlegroups.com>,

WM <wolfgang.m...@hs-augsburg.de> wrote:

> Am Sonntag, 26. Juni 2016 14:52:34 UTC+2 schrieb Dan Christensen:
> > On Sunday, June 26, 2016 at 8:24:45 AM UTC-4, WM wrote:
> > > Am Sonntag, 26. Juni 2016 06:31:13 UTC+2 schrieb Dan Christensen:
> > > > On Saturday, June 25, 2016 at 4:15:06 PM UTC-4, WM wrote:
> > > > > Am Samstag, 25. Juni 2016 21:27:05 UTC+2 schrieb Me:
> > > > > > On Saturday, June 25, 2016 at 8:40:21 PM UTC+2, WM wrote:
> > > > > > > Am Samstag, 25. Juni 2016 18:51:50 UTC+2 schrieb Me:
> > > > > >
> > > > > > > Here we have not the clumsy and failed notation of successors but
> > > > > > > addition
> > > > > > > of 1. That implies 1 + 1 =/= 1.
> > > > > >
> > > > > > Well, if this is the case, where's your proof?
> > > > >
> > > > > You show an incredible degree of stupidity.
> > > > > There is no proof necessary or possible because the construction by
> > > > > addition of the unit cannot result in a smaller set of units.
> > > >
> > > > You cannot prove the existence of even one number
> > >
> > > Wrong. 1 in M shows that 1 exists.

Not until you have proved that M exists, which you have not done!

> >
> > You never actually say that 1 in N. Why can you not introduce that as an
> > axiom?
>
> 1 in M proves that 1 exists.

Not until you have proved that your M exists, which you have not done!

> So, when applying set theory, which is fine in
> the finite domain, {1} = M exists. Can't you see this?

We cannot afford take WM's word for anything, since he has so often
been caught lying.

>
> > While you are at, why not add that every element of has a unique successor
>
> That is not necessary when havin a set of n elements and adding 1.
>

> Not necessary. Do youreally wish to criticize Cantor, Zermelo, v. Neumann,
> and myself?

Only yourself, as you are the one of those who does not belong with the
others.

> All of us refrain from defining uniquness because it is
> guaranteed automatically.

By what?

>
> > And that the successor relation is injective? It would solve all your
> > problems here
>
> I have no problems.

WM is
LYING !
AGAIN ! !
AS USUAL ! ! !

That WM does not acknowledge the many problems in WM's witless worthless
wacky world of WMytheology does not make them invisible to others.

WM is so insanely anti-Cantor that WM claims nonsense rather than accept
anything by Cantor:

In Mathematics, Cantor says

Card(Q) = Card(N) and Card(N) < Card(R)

So he says Card(Q) < Card(R), with fewer rationals than reals

In WMaths/WMytheology, WM claims Cantor is always wrong,
Thus WM claims Card(Q) > Card(N) and Card(N) = Card(R)
So Card(Q) > Card(R), with more rationals than reals.

But in any mathematics free from WM's witless worthless wacky world of
WMytheology, there aren't any rationals that are not also realS.

[Virgil]

In article <700cd7a5-6c49-49ae...@googlegroups.com>,

WM <wolfgang.m...@hs-augsburg.de> wrote:

> Am Sonntag, 26. Juni 2016 08:57:37 UTC+2 schrieb John Gabriel:

The Sum total of WM's and JG's knowledge of mathematics is negative.

WM is so insanely anti-Cantor that WM claims nonsense rather than accept
anything by Cantor:

In Mathematics, Cantor says
Card(Q) = Card(N) and Card(N) < Card(R)
So he says Card(Q) < Card(R), with fewer rationals than reals

In WMaths/WMytheology, WM claims Cantor is always wrong,
Thus WM claims Card(Q) > Card(N) and Card(N) = Card(R)
So Card(Q) > Card(R), with more rationals than reals.

But in any mathematics free from WM's witless worthless wacky world of

WMytheology, there aren't any rationals that are not reals

[Virgil]

In article <6db19894-d359-4a54...@googlegroups.com>,

WM <wolfgang.m...@hs-augsburg.de> wrote:

> Am Samstag, 25. Juni 2016 23:38:50 UTC+2 schrieb Me:
>
>
> >
> > Obviously you tried to "reproduce" the introduction (definition) of the
> > natural numbers _as a certain subset of of the real numbers_ (after THEY
> > have been introduced, say, axiomatically).
>
> Not at all. The assition of 1 does not even require fractions.

The "assition of 1" does require something, possibly a spell-check.

In Proper Mathematics, Cantor is provably right,

claiming Card(Q) = Card(N) and Card(N) < Card(R)
So in Mathematics Card(Q) < Card(R), with fewer rationals than reals.

In his WMytheology, WM claims Cantor is wrong,
there WM claims Card(Q) > Card(N) and Card(N) = Card(R)
So in WM's witless worthless wacky world of WMytheology,

Card(Q) > Card(R), with more rationals than reals.

Even though outside of WM's witless worthless wacky world of WMytheology
there are no rationals that are not also reals!

So WM's witless worthless wacky world of WMytheology is WRONG!

[Virgil]

In article <64a50f0b-d03f-4f02...@googlegroups.com>,
WM <wolfgang.m...@hs-augsburg.de> wrote:

> Am Samstag, 25. Juni 2016 22:59:24 UTC+2 schrieb Me:
> > On Saturday, June 25, 2016 at 10:18:55 PM UTC+2, WM wrote:
> >
> > > > But YOU wrote:
> > > >
> > > > "the first axiom is 1 e IN."
> > > >
> > > Yes, of course. It is so because |N is the intersection of all sets which
> > > contain 1.

Then WM is also claiming that |N = {1}!

Which may work in WM's witless worthless wacky world of WMytheology,
but does NOT work in proper mathematics.

[Dan Christensen]

On Sunday, June 26, 2016 at 12:25:49 PM UTC-4, WM wrote:
> Am Sonntag, 26. Juni 2016 17:32:17 UTC+2 schrieb Dan Christensen:
>
>
> > > What about subsets of countable sets? Are they always countable?
> > >
> >
> > Using only the notation of set theory, please construct any countable set A and an uncountable subset B of A.
>
> The other way round. Try to describe in set theory the fact that all texts in all languages are countable, but the subset of all real numbers is not.
>

So, you cannot actually prove that there exists a countable set and an uncountable subset of that set. Thought so.

[Virgil]

In article <ffd3e099-6833-40c6...@googlegroups.com>,
WM <wolfgang.m...@hs-augsburg.de> wrote:

> Am Sonntag, 26. Juni 2016 03:11:49 UTC+2 schrieb Me:
> > On Sunday, June 26, 2016 at 3:01:59 AM UTC+2, Jim Burns wrote:
> >
> > > > (The first of these "defining" equations usually is adopted
> > > > when starting the natural numbers with 1 instead of 0. ...)
> > > >
> > > Sorry, I don't get this.
> >
> > When the first "Peano axiom" of our axiom system reads
> >
> > 1 e IN
> >
> > we usually adopt the "defining" equations
> >
> > An e IN: n + 1 = S(n)
> > An,m e IN: n + S(m) = S(n + m)
> >
> > for the recursive definition of +.
>

> That is not contained in the Peano axioms.

> >
> > When on the other hand the first "Peano axiom" of our axiom system reads
> >
> > 0 e IN
> >
> > we usually adopt the "defining" equations
> >
> > An e IN: n + 0 = n
> > An,m e IN: n + S(m) = S(n + m)
> >
> > for the recursive definition of +.
>

> That is not contained in the Peano axioms either.
>
> According to the Peano axioms you could adopt n + 1 = n
> as well as n + 0 = S(n).

Maybe in WM's corrupt version of them but not as they actually are.

n + 0 = n and/or n + 1 = S(n)
but not n + 1 = n or n + 0 = S(n).

PEANO AXIOMS:
There is a set S and an object o, and a function F such that
1. o is a member of S

2. if x is in S then F(x) is in S

3. For every x in S, o =/= F(x)

4. For every x and y in S, if F(x) = F(y) then x = y

5. If a set T is such that o is a member of T and whenever x is a

member of T then also F(x) is a member of T then S is a subset of T.

> You see that your belief in axioms is blatent nonsense.

NOt anywhere nearly as either blatant or nonsensical as any fool part of
WM's witless worthless wacky world of WMytheology!




> Mathematics does not require "Axioms".

WM's witless worthless wacky world of WMytheology may not benefit by
them being witless, worthless and wacky anyway but mathematics does.





> The job of a pure mathematician

Since WM is far more physicist than mathematician,
he can speak only for physicists and not for mathematicians!

Cantor was a mathematician and could speak for mathematicians.

In Mathematics, Cantor is provably right,

Card(Q) = Card(N) and Card(N) < Card(R)

So Card(Q) < Card(R), with fewer rationals than reals

In WMaths/WMytheology, WM claims Cantor is wrong,

WM claims Card(Q) > Card(N) and Card(N) = Card(R)

So Card(Q) > Card(R), with more rationals than reals.

But in proper mathematics there are NOT any rationals which are not
reals.

So in proper mathematics, WM is wrong and Cantor is right

[Virgil]

In article <f00d5330-2899-4496...@googlegroups.com>,
WM <wolfgang.m...@hs-augsburg.de> wrote:

> Am Samstag, 25. Juni 2016 23:54:55 UTC+2 schrieb Me:
> > On Saturday, June 25, 2016 at 11:35:12 PM UTC+2, Peter Percival wrote:
> > > WM wrote:
> > > > Am Samstag, 25. Juni 2016 22:41:49 UTC+2 schrieb Peter Percival:
> > > > >
> > > > > Also, the intersection of all sets which contain 1, is {1}.
> > > > >
> > > > IN is the intersection of all sets that contain 1 (according to 4.1)
> > > > and have the inductive property (4.2).
> > > >
> > > Ok, that's fine.
> > >
> > No, that's NOT fine, since this definition is _not stated_ in the context
> > of his "system".
>
> I have no system.

You have your witless worthless wacky world of WMytheology,
which is systematically wrong!

[Virgil]

In article <a870088c-43e3-43d1...@googlegroups.com>,

WM <wolfgang.m...@hs-augsburg.de> wrote:

> Am Sonntag, 26. Juni 2016 17:31:31 UTC+2 schrieb Me:
> > On Sunday, June 26, 2016 at 4:38:49 PM UTC+2, WM wrote:
> >
> > > 1 in M proves that 1 exists.
> >
> > *lol* If /M/ is a constant,
> ...
> > If on the other /M/ is a variable
> ...
>
> 1 is given as existing. The character of M is irrelevant.

If the character of M is such th it does not cannot have members,
that would be relevant!

So WM is
LYING !
AGAIN ! !
AS USUAL ! ! !

[Virgil]

In article <60395259-7647-47f8...@googlegroups.com>,

WM <wolfgang.m...@hs-augsburg.de> wrote:

> Am Sonntag, 26. Juni 2016 17:32:17 UTC+2 schrieb Dan Christensen:
>
>
> > > What about subsets of countable sets? Are they always countable?
> > >
> >
> > Using only the notation of set theory, please construct any countable set A
> > and an uncountable subset B of A.
>
> The other way round. Try to describe in set theory the fact that all texts in
> all languages are countable, but the subset of all real numbers is not.

But real numbers are not languages, and for most real numbers there are
no names in any language. So no problem!

Other than WM's trying to make one where none exists!

[WM]

Am Sonntag, 26. Juni 2016 19:09:27 UTC+2 schrieb Dan Christensen:
> On Sunday, June 26, 2016 at 12:25:49 PM UTC-4, WM wrote:
> > Am Sonntag, 26. Juni 2016 17:32:17 UTC+2 schrieb Dan Christensen:
> >
> >
> > > > What about subsets of countable sets? Are they always countable?
> > > >
> > >
> > > Using only the notation of set theory, please construct any countable set A and an uncountable subset B of A.
> >
> > The other way round. Try to describe in set theory the fact that all texts in all languages are countable, but the subset of all real numbers is not.
> >
>
> So, you cannot actually prove that there exists a countable set

You don't understand that the set of all texts is countable and that the real nubers are its subset? Is that wishful blindness or true stupidity? If you cannot think by yourself and with your DC proof, try to understand more intelligent mathematicians. They will tell you.

The possible combinations of finitely many letters form a countable set, and since every determined real number must be definable by a finite number of words, there can exist only countably many real numbers - in contradiction to Cantor's classical theorem and its proof. [H. Weyl: "Das Kontinuum", Veit, Leipzig (1918) p. 18]

If we define the real numbers in a strictly formal system, where only finite derivations and fixed symbols are permitted, then these real numbers can certainly be enumerated because the formulas and derivations on the basis of their constructive definition are countable. [Kurt Schütte: "Beweistheorie", Springer (1960)]

One emphasizes that an infinite sequence or a decimal fraction can be given only by an arithmetical law, and one regards the continuum as a set of elements defined by such laws. [...]
Nonetheless, if we pursue the thought that each real number is defined by an arithmetical law the idea of the totality of real numbers is no longer indispensable, and the axiom of choice is not at all evident. [Paul Bernays: "On Platonism in Mathematics", (1934) p. 5ff]

It is obvious that the set of all real numbers having a finite description is countable (because the set of all words over a finite alphabet is countable, and the set of all finitely describable numbers is one of its subsets).
[Wolfgang Thumser, "Das Kalenderblatt 100117", de.sci.mathematik (23 Jan 2010)]


Regards, WM

[WM]

Am Sonntag, 26. Juni 2016 19:23:38 UTC+2 schrieb Virgil:
> In article <60395259-7647-47f8...@googlegroups.com>,
> WM <wolfgang.m...@hs-augsburg.de> wrote:
>
> > Am Sonntag, 26. Juni 2016 17:32:17 UTC+2 schrieb Dan Christensen:
> >
> >
> > > > What about subsets of countable sets? Are they always countable?
> > > >
> > >
> > > Using only the notation of set theory, please construct any countable set A
> > > and an uncountable subset B of A.
> >
> > The other way round. Try to describe in set theory the fact that all texts in
> > all languages are countable, but the subset of all real numbers is not.
>
> But real numbers are not languages, and for most real numbers there are
> no names in any language. So no problem!

How can you well-order undefinable quantities? Note that according to Zermelo's proof all elements can be distinguished. He says: "If there existed a first element m' such that M' differed from M''". How can a first element with a certain property exist if all elements are undefined and undefinable?

Regards, WM

[Dan Christensen]

On Sunday, June 26, 2016 at 2:47:05 PM UTC-4, WM wrote:
> Am Sonntag, 26. Juni 2016 19:09:27 UTC+2 schrieb Dan Christensen:
> > On Sunday, June 26, 2016 at 12:25:49 PM UTC-4, WM wrote:
> > > Am Sonntag, 26. Juni 2016 17:32:17 UTC+2 schrieb Dan Christensen:
> > >
> > >
> > > > > What about subsets of countable sets? Are they always countable?
> > > > >
> > > >
> > > > Using only the notation of set theory, please construct any countable set A and an uncountable subset B of A.
> > >
> > > The other way round. Try to describe in set theory the fact that all texts in all languages are countable, but the subset of all real numbers is not.
> > >
> >
> > So, you cannot actually prove that there exists a countable set
>
> You don't understand that the set of all texts is countable and that the real nubers are its subset?

You don't understand that the real numbers exist independently of any set of strings of characters. But thanks of confirming once again that you cannot actually prove that there exists a countable set and an uncountable subset of that set.

[Dan Christensen]

On Sunday, June 26, 2016 at 2:52:13 PM UTC-4, WM wrote:
> Am Sonntag, 26. Juni 2016 19:23:38 UTC+2 schrieb Virgil:
> > In article <60395259-7647-47f8...@googlegroups.com>,
> > WM <wolfgang.m...@hs-augsburg.de> wrote:
> >
> > > Am Sonntag, 26. Juni 2016 17:32:17 UTC+2 schrieb Dan Christensen:
> > >
> > >
> > > > > What about subsets of countable sets? Are they always countable?
> > > > >
> > > >
> > > > Using only the notation of set theory, please construct any countable set A
> > > > and an uncountable subset B of A.
> > >
> > > The other way round. Try to describe in set theory the fact that all texts in
> > > all languages are countable, but the subset of all real numbers is not.
> >
> > But real numbers are not languages, and for most real numbers there are
> > no names in any language. So no problem!
>
> How can you well-order undefinable quantities?

The notion of "undefinable quantities" has no place in mathematics these days. It is in the realm of philosophical navel gazing, not mathematics.

[WM]

Am Sonntag, 26. Juni 2016 21:03:05 UTC+2 schrieb Dan Christensen:


> The notion of "undefinable quantities" has no place in mathematics these days. It is in the realm of philosophical navel gazing, not mathematics.

Very good standpoint! But the set of definable quantities is not uncountable. Therefore uncountable sets ... Can you complete this sentence?

Regards, WM

[WM]

Am Sonntag, 26. Juni 2016 20:55:39 UTC+2 schrieb Dan Christensen:


> You don't understand that the real numbers exist independently of any set of strings of characters.

No. I don't. And I can't understand that anybody believes so, in particular because they cannot be distinguished and therefore do not obey the axiom of extensionality.

Regards, WM

[WM]

Am Sonntag, 26. Juni 2016 21:03:05 UTC+2 schrieb Dan Christensen:

> the real numbers exist independently of any set of strings of characters.

> The notion of "undefinable quantities" has no place in mathematics these days

Regards, WM

[Me]

On Sunday, June 26, 2016 at 9:15:59 PM UTC+2, WM wrote:

> because they cannot be distinguished and therefore do not obey the axiom of extensionality.

Twas brillig, and the slithy toves
Did gyre and gimble in the wabe;
All mimsy were the borogoves,
And the mome raths outgrabe.

[Dan Christensen]

Utter nonsense! Real numbers can be seen as sets of rational numbers (Dedekind cuts). Two reals numbers are distinct if and only if there exists an rational number in one, but not the other (by the axiom of extensionality).

[Me]

On Sunday, June 26, 2016 at 8:55:39 PM UTC+2, Dan Christensen wrote:

> You don't understand that the real numbers exist independently of any
> set of strings of characters. But thanks of confirming once again that you
> cannot actually prove that there exists a countable set and an uncountable
> subset of that set.

On the other hand, could he prove "2 + 2 = 4" already, using his bogus "axiom system" for "the natural numbers"?

It seems to me that he's now trying to avoid this topic.

[Dan Christensen]

Correct. Again, it is in the realm of philosophical navel gazing, not mathematics. It is a totally useless concept. No one has ever even bothered to formalize it. As I have said, if it can't be formalized, it isn't mathematics.

[Dan Christensen]

Yup. You can't prove 2+2=4 by the induction axiom alone. Though he seems oblivious, this is precisely what his "axioms" amounts to -- actually only one axiom. Contrary to Mucke's delusions, it is NOT, by itself, a definition of the natural numbers.

[WM]

Am Sonntag, 26. Juni 2016 21:26:30 UTC+2 schrieb Dan Christensen:
> On Sunday, June 26, 2016 at 3:15:59 PM UTC-4, WM wrote:
> > Am Sonntag, 26. Juni 2016 20:55:39 UTC+2 schrieb Dan Christensen:
> >
> >
> > > You don't understand that the real numbers exist independently of any set of strings of characters.
> >
> > No. I don't. And I can't understand that anybody believes so, in particular because they cannot be distinguished and therefore do not obey the axiom of extensionality.
> >
>
> Utter nonsense! Real numbers can be seen as sets of rational numbers (Dedekind cuts). Two reals numbers are distinct if and only if there exists an rational number in one, but not the other (by the axiom of extensionality).

How can you know that, if you can't define most real numbers?
(By the way, in every Dedekind cut all rational numbers exist.)

Regards, WM

[WM]

Am Sonntag, 26. Juni 2016 21:33:55 UTC+2 schrieb Me:
> On Sunday, June 26, 2016 at 8:55:39 PM UTC+2, Dan Christensen wrote:
>
> > You don't understand that the real numbers exist independently of any
> > set of strings of characters. But thanks of confirming once again that you
> > cannot actually prove that there exists a countable set and an uncountable
> > subset of that set.
>
> On the other hand, could he prove "2 + 2 = 4" already,

To prove 2 + 2 = 4 in my approach based on counting +1 (or to require such a proof) would only be a proof of missingt brain.

> It seems to me that he's now trying to avoid this topic.

More interesting is the question how the Dedekind cuts of undefinable real numbers can be compared.

Regards, WM

[WM]

Am Sonntag, 26. Juni 2016 22:03:33 UTC+2 schrieb Dan Christensen:
> On Sunday, June 26, 2016 at 3:33:55 PM UTC-4, Me wrote:
> > On Sunday, June 26, 2016 at 8:55:39 PM UTC+2, Dan Christensen wrote:
> >
> > > You don't understand that the real numbers exist independently of any
> > > set of strings of characters. But thanks of confirming once again that you
> > > cannot actually prove that there exists a countable set and an uncountable
> > > subset of that set.
> >
> > On the other hand, could he prove "2 + 2 = 4" already, using his bogus "axiom system" for "the natural numbers"?
> >
> > It seems to me that he's now trying to avoid this topic.
>
> Yup. You can't prove 2+2=4 by the induction axiom alone. Though he seems oblivious, this is precisely what his "axioms" amounts to -- actually only one axiom.

Like Cantor's, Zermelo's, and von Neumann's approach as well as Lorenzen's and Brouwer's. We all simply fail in the eyes of the great logician! We should be ashamed alltogether. But I am the only one of us who is alive. Should I carry all that shame alone?

Please help!

Regards, WM

[WM]

Am Sonntag, 26. Juni 2016 21:34:03 UTC+2 schrieb Dan Christensen:
> On Sunday, June 26, 2016 at 3:18:51 PM UTC-4, WM wrote:
> > Am Sonntag, 26. Juni 2016 21:03:05 UTC+2 schrieb Dan Christensen:
> >
> >
> >
> > > the real numbers exist independently of any set of strings of characters.
> >
> > > The notion of "undefinable quantities" has no place in mathematics these days
> >
>
> Correct.

But if there is no string of characters, then they cannot be defined by any string of characters. Then they are undefinable quantities, aren't they?

Regards, WM

[Me]

On Sunday, June 26, 2016 at 10:56:11 PM UTC+2, WM wrote:
> Am Sonntag, 26. Juni 2016 22:03:33 UTC+2 schrieb Dan Christensen:
>
> Like Cantor's, Zermelo's, and von Neumann's approach as well as Lorenzen's

> and Brouwer's. We all ...

I see, brothers in arms!