Recent content by Periwinkle

There is no transitive operation of ##G## on a set with ##8## elements ##G## is a transformation group, so a permutation of the eight-element set corresponds to each element of the group by a homeomorphism. Each permutation can be produced as a product of distinct cyclic permutations in one...

Aristotle distinguished two forms of existence in Metaphysics: potential existence and actual existence. The housebuilder is potentially a builder even if he is not building. Because he has the ability to build. When the builder makes a house, then he naturally has an actual existence as a...

In mathematical circles, if there is an error in a mathematical proof, they correctly specify where the error is located. It was so memorable to me that even the great Newton taught in one of his books, solving the word problems for the students, or rather for their teachers. For example, I...

I. My reasoning above, which I wrote yesterday before and hastened, is based on that for each ##X## person, we count the number of pairs of individuals for which this ##X## is involved for each pair of ##X, Y## persons, count the number of other persons involved in this pair. That is, the...

Mark participants in the conference with ##A_1, A_2, \dots A_n##. Person ##A_i## knows ##k_i## other persons, does not knows ##l_i##. Then it is not involved in ## \binom {k_i} 2 ## pairs because knows both of them, in case of ##\binom {l_i} 2 ## it is not involved, because does not know any...

I also thought of prime numbers for the first time, but here we are talking about sums. Then I thought of binary writing of the numbers. It was confusing that 32 and 64 were not needed, only 16.

From 1 to 31, all natural numbers can be written in one unique way in the following form $$ a\cdot 1 + b\cdot 2 + c\cdot 2^2 +d\cdot 2^3 + e \cdot 2^4 = a\cdot 1 +b\cdot 2 +c\cdot 4 +d\cdot 8+ e\cdot 16$$ wherein a, b, c, d, e are each 0 or 1.In the first vessel, we leave one weight, in the...

Thank you very much for your comment. I realized that my proof was completely wrong. Interestingly, proving such an easy-to-understand theorem - at least for me is undoubtedly so difficult.

Originally, I interpreted the question as follows there are at least ##\lfloor \frac{n}{2} \rfloor - 1## persons {each of which knows both the aforementioned two persons} or else {each of which knows no one from them} but I couldn't solve it. However, if I interpret it as follows: there are...

According to Weierstrass's maximum value theorem, if a function is continuous on ##[a,b]##, then it is bounded on this interval and takes its maximum and minimum. If function ##f(x)## is not constant, then its maximum or minimum is different from ##f(a) = f(b)=0##. Let this be the maximum that...

$$S = \sum_{n \in \mathbb{N}}a_n \\ W = \sum_{n \in \mathbb{N}_0}2^na_{2^n} $$ If ##~W## converges then ##S## also converges $$ a_1 \leq a_1 \\ a_2 + a_3 \leq 2a_2 \\ a_4 + a_5+a_6 + a_7 \leq 4a_4 \\ \dots \\a_{2^n} + a_{2^n+1}+ \dots + a_{2^{n+1}-1} \leq 2^n a_{2^n} \\ \dots$$ If ##~S##...

Horace, Ode I.11 Tu ne quaesieris - scire nefas! - quem mihi, quem tibi finem di dederint, Leuconoe, nec Babylonios temptaris numeros. Ut melius, quidquid erit, pati, seu pluris hiemes seu tribuit Iuppiter ultimam, quae nunc oppositis debilitat pumicibus mare Tyrrheneum. Sapias, vina liques, et...

For ##n = 1## the theorem is true, ## x=0## is the solution. Provided that ##n \gt 1## let ##G(x)## be the function defined for the ##\left[0, \frac {n-1} n \right]## interval $$G(x) = f\left(x+ \frac 1 n\right) - f(x).$$ If ##G(0) = 0## then the theorem is proved, ## x=0## is the solution. If...

I noticed my own mistake. Correctly: $$ |S(t)| = \left| \frac 1 {g(t) -\lambda} s(t) \right|$$ function is greater than ## i |s(t)|.## Tomorrow I will consider the above solution.

Question 4 Self-adjoint linear operator's eigenvalues are real $$ \lambda (x,x) = (\lambda x, x) = (Ax,x) = (x, Ax) = (x,\lambda x,) = \bar \lambda (x,x) $$ However, in the Euclidean space ## (x,x) =0 ## follows ##x = 0##, so ## \lambda = \bar \lambda##. The eigenvectors belonging to...