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Jul
7
comment IMC 2008 first problem first day. Finding continuous functions so $x-y\in \mathbb Q \implies f(x)-f(y)\in \mathbb Q$
ok, thanks, I am going to compete in the IMC so I don't know if I can just say that $g(x)$ is clearly continuous, I think so, but I don't want to lose points.
Jul
7
comment IMC 2008 first problem first day. Finding continuous functions so $x-y\in \mathbb Q \implies f(x)-f(y)\in \mathbb Q$
it doesn't work if you make a hint box very big.
Jul
7
revised IMC 2008 first problem first day. Finding continuous functions so $x-y\in \mathbb Q \implies f(x)-f(y)\in \mathbb Q$
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Jul
7
asked IMC 2008 first problem first day. Finding continuous functions so $x-y\in \mathbb Q \implies f(x)-f(y)\in \mathbb Q$
Jul
7
comment 2014 IMC first problem first day (eigenvalues of a product of symmetric matrices).
I don't understand why you say we must diagonalize $A^2$ and not $A$, anyways here is a link to the official solution: imc-math.org.uk/imc2013/IMC2013-day1-solutions.pdf
Jul
7
comment 2014 IMC first problem first day (eigenvalues of a product of symmetric matrices).
My bad, the problem asks to prove $|\lambda|>1$
Jul
7
revised 2014 IMC first problem first day (eigenvalues of a product of symmetric matrices).
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Jul
7
comment Graph theory, $n$ people sitting around table.
Yes, very nice.
Jul
7
comment Graph theory, $n$ people sitting around table.
So for $n$ an odd prime the answer for $1$ is $(n-1)/2$ and the answer for $2$ is yes.
Jul
7
comment Graph theory, $n$ people sitting around table.
Partial solution: If $n$ is prime everybody gets to sit next to everyone else, consider a circle with $n$ dots placed around regularly, if you fix a length size then the edges of that length form a hamiltonian cycle.
Jul
7
comment Graph theory, $n$ people sitting around table.
Essentially the problem asks you to find disjoint hamiltonian cycles in $K_n$
Jul
7
comment Graph theory, $n$ people sitting around table.
What high school do you attend? seems hard for precalculus.
Jul
7
comment Pythagorean theorem expressed without roots in an old Tamilian (Indian) statement
Oh, I hadn't thought of minimizing the integral, that is very interesting. Of course there is the problem that when $k$ goes to infinity the proposed approximation gets worse and worse, approaching $\frac{7}{8}$'ths of what it should be.
Jul
7
revised Prove that the graph dual to Eulerian planar graph is bipartite.
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Jul
7
answered If $G$ is abelian, it has a subgroup in every order of $|G|'s$ divisors?
Jul
7
comment If $G$ is abelian, it has a subgroup in every order of $|G|'s$ divisors?
Groups with this property are called CLT groups (converse langrange theorem). The set of CLT groups is properly contained in the set of soluble groups. The set of supersoluble groups is properly contained in the set of CLT groups. In other words, CLT implies soluble and supersoluble implies CLT. Since abelian implies supersoluble we have that abelian implies CLT.
Jul
7
revised Pythagorean theorem expressed without roots in an old Tamilian (Indian) statement
added 1367 characters in body
Jul
7
awarded  Enlightened
Jul
7
awarded  Nice Answer
Jul
7
revised Pythagorean theorem expressed without roots in an old Tamilian (Indian) statement
added 118 characters in body