9 mathematicians are at an international conference. Of any

>>7672488

Why

>>7672473
Yes.

>>7672473
>No mathematician speaks more than 3 languages.

BULLSHIT, to get a PhD you need to be a polyglot

123
1xx
x2x
xx3

456
4xx
x5x
xx6

If #9 is 789 that contradicts the premise (123 456 789, no common ground).
So #9 must have at least one number among 123456, and so there is one language with three people.

If it's not obvious why I arranged the numbers like this, try taking one of them (WLOG, 1xx) away. You have to add either 1xx back or add 789 or add another number that's listed twice. In any of the three events we're back where we started.

So yes.

>>7672473
If no mathematician speaks more than two languages than the most mathematicians I can get there to be is six.

ACE
ADG
BCF
BDH
EF
GH

>>7672509
This assumes the existence of two people who speak 6 different languages between them (123, 456)
Moreover, it assumes that all of these 6 languages are also spoken by someone else in the room

Not saying you're wrong, but your proof doesn't work without these unjustified assumptions (yet)

>>7672509
>>7672522
>high schoolers trying to do a proof

Assume that the statement is not necessarily true.
Then every language spoken by at most 2 people and every mathematician speaks at most 3 languages,
which implies that every mathematician shares a language with no more than 3 other mathematicians.
Now pick one arbitrary mathematician M1 (8 remain)
Eliminate the (at most) three mathematicians that share a language with M1 (>=5 remain)
Pick another arbitrary mathematician M2 from the remaining ones (>=4 remain)
Eliminate those that share a language with M2 (>=1 remain)
By picking a third one we have constructed a set of Ms with no shared language.
Therefore our assumption is false and the answer to OPs question is 'yes'.

>>7672650
I didn't say I was doing a proof, faggot.

How does this account for guarantee that at least two in any three share a language?

>>7672629
Do I really have to spell out that if the fifth person shares any language between the first person then the there are already three languages shared? Or that if any of the two three four people share more than one language then the options limit even faster without opening up new ones? It's not that complicated.

>Moreover, it assumes that all of these 6 languages are also spoken by someone else in the room
No it doesn't. It is necessarily true, I demonstrate that by assuming the last person doesn't and we break the premise.

I skipped some trivial steps, but the proof is sound. You can whine how "it isn't first order logic", but tell it to Archimedes, because I'm not doing it the Euclid way. Too tedious and insulting to the reader.

Assume that there is no language that is spoken by at least 3 mathematicians.


Without loss of generality there exist only 27 different languages [math]L_1, L_2, \ldots ,L_{27}[/math].
Consider the variables [math]x_{i,j}[/math] which is a equal to [math]1[/math] if mathematician [math]i[/math] can speak the language [math]L_j[/math] and equal to [math]0[/math] otherwise.

The constraints give a set of inequalities:

[math] \sum_j (x_{i_1,j} x_{i_2,j} + x_{i_1,j} x_{i_3,j} + x_{i_2,j} x_{i_3,j}) \geq 1[/math] for all tupels [math](i_1, i_2, i_3) \in \{1,2,3,4,5,6,7,8,9 \}^3[/math].
[math]\sum_j x_{i,j} \leq 3[/math] for [math]i \in \{1,2,3,4,5,6,7,8,9 \}[/math]
[math] \sum_i x_{i,j} \leq 2[/math] for all [math]j \in \{L_1, L_2, \ldots, L_{27}\}[/math]
[math]x_{i,j} \in \{0,1\}[/math] for all [math]i \in \{1,2,3,4,5,6,7,8,9 \}[/math] and [math]j \in \{L_1, L_2, \ldots, L_{27}\}[/math].


Just plug this system of constraints in an ILP solver to see that no solution exists.

>>7672730
Meant to quote >>7672650