Re: Simple enough for every reader?

On 30.05.2025 02:05, Ben Bacarisse wrote:

WM <wolfgang.mueckenheim@tha.de> writes:

Ah, so you can't do what you suggested and talk to the student, after
the exam, to try to persuade them that you are right!  That did sound a
bit crazy.

Do you never have oral exams in England?

 

In the exam there are questions like these:
>
- Beschreiben Sie, was man unter der Abzählbarkeit aller positiven Brüche
  versteht und erörtern Sie ein Gegenargument.
- Beschreiben Sie, was man unter der Überabzählbarkeit der reellen Zahlen
  versteht, und erörtern Sie ein Gegenargument.
- Beschreiben Sie das Spiel "Wir erobern den Binären Baum" und die damit
   verknüpfte Aussage.
- Sehen Sie eine Parallele zwischen MacDuck und der Nummerierung aller
   Brüche? Wenn ja, welche?
- Was halten Sie vom wissenschaftlichen Wert der Mengenlehre unter
   Berücksichtigung von Banach-Tarski-Paradoxon und Verteilung der Brüche in
  (0, 1) und (1, oo).

  From this, it is not obvious that you want students to say anything I'd
consider to be wrong in an exam.  So maybe someone could indeed get full
marks without having to deny mathematics.

Correct mathematics should not be denied.

Are there any claims in your
lectures that someone at the university down the road would object to?

Of course.

 

Try to answer. Then I will give you marks.

 I'll try a couple of questions...
 

- Beschreiben Sie, was man unter der Abzählbarkeit aller positiven Brüche
  versteht und erörtern Sie ein Gegenargument.

 The positive fractions are said to be countable because the function
    b(0) = 1
   b(n+1) = s(b(n))
   where s(q) = 1 / (2*floor(q) - q + 1)
 is a bijection between the natural numbers and the positive fractions
according to the definition in Prof. Mückenheim's textbook.
 I am not aware of a valid counter argument since this is simply a
definition of what the term "countable" means.

It has been shown to the student by many arguments that the bijection fails. The simplest is McDuck or this:
All positive fractions
     1/1, 1/2, 1/3, 1/4, ...
     2/1, 2/2, 2/3, 2/4, ...
     3/1, 3/2, 3/3, 3/4, ...
     4/1, 4/2, 4/3, 4/4, ...
     ...
can be indexed by the Cantor function k = (m + n - 1)(m + n - 2)/2 + m which attaches the index k to the fraction m/n in Cantor's sequence
1/1, 1/2, 2/1, 1/3, 2/2, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5, 2/4, 3/3, 4/2, 5/1, 1/6, 2/5, 3/4, ... .
Its terms can be represented by matrices. When we attach all indeXes k = 1, 2, 3, ..., for clarity represented by X, to the integer fractions m/1 and indicate missing indexes by hOles O, then we get the matrix M(0) as starting position:
XOOO...    XXOO...    XXOO...    XXXO...
XOOO...    OOOO...    XOOO...    XOOO...
XOOO...    XOOO...    OOOO...    OOOO...
XOOO...    XOOO...    XOOO...    OOOO...
and so on, as you know it already.
The shortest is this:
All Cantor's natural numbers can be manipulated collectively, for instance subtracted: ℕ \ {1, 2, 3, ...} = { }. Here all have disappeared.
Could all Cantor's natural numbers be distinguished, then this subtraction could also happen but, caused by the well-order, a last one would disappear. Contradiction.
Without giving one of these arguments (if desired also showing that and why it fails) you would get only half of the full score.

- Beschreiben Sie, was man unter der Überabzählbarkeit der reellen Zahlen
  versteht, und erörtern Sie ein Gegenargument.

 The real numbers are said to be uncountable because no bijective
function exists between N and R.
 I am not aware of any valid counterargument because this theorem is
well-established.

In my lesson you could have learned many counter arguments. The simplest is that countability already fails: Conquer the Binary Tree.

The way you word the questions does seem to allow for correct answers.
What does your mark scheme say for these questions?

You would get half of the possible points, when you could not describe any counter arguments. Criticizing them would be welcome.

Would you accept
any answers that I would consider to be wrong?

Yes. But you could explain why you think it is wrong. You would already have discussed this during the lesson. I could have convinced you.

 

Do you not have to write marking schemes for your exams?  And if in
fact you do, what do yours say about alternative answers?

 

If a student had ever disproved my proofs he would have got additional points. But up to now that has never happened. Try it. You would be the first: Could all Cantor's natural numbers be distinguished, then this subtraction could also happen but, caused by the well-order, a last one would disappear. Strive for the additional points!
Regards, WM