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08:38
@amWhy @XanderHenderson: This is really not math. This user has been posting a ton of "is this true?" questions most of which are like this. Given the absolutely zero context, I would guess that this user just factorized the difference between the two prime factors given, and out popped 59...
This is worse.
@EnzoCreti: How many times are you planning to change your question? Everytime user Henry comes us with a counter-example, you change your condtions. — Tito Piezas III Feb 22 at 11:22
 
2 hours later…
DRF
DRF
11:12
@user21820 #14 would be a nice problem if you allowed infinite number of students and problems but asked whether there must be at least one problem solved by infinitely many students.:)
@DRF What does "more than half" mean in that case?
DRF
DRF
@user21820 right that's why you have to change that bit. Either to finitely many or just one. You can't use half reasonably. You could ask if there must be infinitely many problems solved by infinitely many students but that's too trivial to show. limiting to 1 works better.
3 messages moved from CRUDE
@DRF Could you state in full your version of the problem?
DRF
DRF
@user21820 Assuming we have a countably infinite class of students and a test with countably infinitely many questions and given that every student solved infinitely many questions, must it follow that at least one problem was solved by infinitely many students?
DRF
DRF
11:24
@user21820 Yeap it's even possible that each problem was only solved by a single student. Assuming AC.
No need AC. Just create a test with problems labelled by a pair of natural numbers.
DRF
DRF
@user21820 Hmm I think you're assuming AC at some point. I wonder where. Countably many countable sets can be uncountable without AC. So I suppose yeah if you label them to begin with you're good. But I wonder if there must be a labeling.
Your question is "is it possible", and the answer is "no, ∃ ..." and we don't need AC to prove that.
DRF
DRF
@user21820 Ahh no you're right. We can split $\mathbb{N}$ into countably many disjoint countable subsets. You can just use some arithmetic I think.
The correct generalization would be: A class has students S and a test has questions Q, and for each x∈S let f(x)⊆Q be the questions that x solved. If for every x∈S there is no injection from f(x) into Q∖f(x), must there exist y∈Q such that there is no injection from { x : x∈S ∧ y∈f(x) } into { x : x∈S ∧ y∉f(x) }?
DRF
DRF
12:22
@user21820 That is very nice.
 
11 hours later…

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