Dependencies

Let \(p(z) = \sum _{k=0}^n c_k z^k\) be a complex polynomial of degree \(n\ge 1\). If \(p(a)\ne 0\), then every disk \(D\) around \(a\) contains an interior point \(b\) with \(|p(b)| {\lt} |p(a)|\)

TODO

The equation \(\binom n k = m ^l\) has no integer solutions with \(l \ge 2\) and \(4 \le k \le n - 4\).

\(e^r\) is irrational for every \(r\in \mathbb {Q}\setminus \{ 0\} \).

\(\pi ^2\) is irrational.

For every odd integer \(n \ge 3\), the number

is irrational.

The Borromean rings cannot be build from three perfect circles

The Borromean rings are nontrivial, and they are also not equivalent to Tait’s link No. 18

Let \(q = p^m\) be a prime power, \(n := 4q - 2\), and \(d := \binom {n}{2} = (2q - 1)(4q - 3)\). Then there is a set \(S \subseteq \{ +1, -1\} ^d\) of \(2^{n-2}\) points in \(\mathbb {R}^d\) such that every partition of \(S\), whose parts have smaller diameter than \(S\), has at least

parts. For \(q = 9\) this implies that the Borsuk conjecture is false in dimension \(d = 561\). Furthermore, \(f(d) {\gt} (1.2)\sqrt{d}\) holds for all large enough \(d\).

For every continuous map \(f : S^d \to \mathbb {R}^d\) from \(d\)-sphere to \(d\)-space, there are antipodal points \(x^*\), \(-x^*\) that are mapped to the same point \(f(x^*) = f(-x^*)\).

TODO

Every continuous function \(f : B^2 \longrightarrow B^2\) of an \(2\)-dimensional ball to itself has a fixed point (a point \(x \in B^2\) with \(f(x) = x\)).

If a short needle, of length \(\ell \), is dropped on paper that is ruled with equally spaced lines of distance \(d \geq \ell \), then the probability that the needle comes to lie in a position where it crosses one of the lines is exactly

If two \(3\)-dimensional convex polyhedra \(P\) and \(P'\) are combinatorially equivalent with corresponding pairs of adjacent congruent, then also the angels between corresponding pairs of adjacent facets are equal (and thus \(P\) is congruent to \(P'\)).

There are \(n^{n - 2}\) different labeled trees on \(n\) nodes.

There are \(n^{n - 2}\) different labeled trees on \(n\) nodes.

There are \(n^{n - 2}\) different labeled trees on \(n\) nodes.

There are \(n^{n - 2}\) different labeled trees on \(n\) nodes.

For primes \(p = 4m + 1\) the equation \(s^2 \equiv -1 (\mod p)\) has two solutions \(s \in \{ 1, 2, \dots , p - 1\} \), for \(p = 2\) there is one such solution, while for primes of the form \(p = 4m + 3\) there is no solution.

No number \(n = 4m + 3\) is a sum of two squares.

Every prime of the form \(p = 4m + 1\) is a sum of two squares, that is, it can be written as \(p = x^2 + y^2\) for some natural numbers \(x,y \in \mathbb {N}\).

Every prime of the form \(p = 4m + 1\) is a sum of two squares, that is, it can be written as \(p = x^2 + y^2\) for some natural numbers \(x,y \in \mathbb {N}\).

Every prime of the form \(p = 4m + 1\) is a sum of two squares, that is, it can be written as \(p = x^2 + y^2\) for some natural numbers \(x,y \in \mathbb {N}\).

In any configuration of \(n\) points in the plane, not all on a line, there is a line which contains exactly two of the points.

Let \(P\) be a set of \(n\ge 3\) points in the plane, not all on a line. Then the set \(\mathcal{L}\) of lines passing through at least two points contains at least \(n\) lines.

Let \(X\) be a set of \(n\ge 3\) elements, and let \(A_1, \dots , A_m\) be proper subsets of \(X\), such that every pair of elements of \(X\) is contained in precisely one set \(A_i\). Then \(m\ge n\) holds.

If \(K_n\) is decomposed into complete bipartite subgraphs \(H_1, \dots , H_m\), then \(m \ge n - 1\).

If a link consists of disjoint perfect circles that are pairwise not linked, then the link is trivial

For every \(d\ge 2\), there is a family of \(2^d\) pairwise touching \(d\)-simplices in \(\mathbb {R}^d\) together with a transversal line that hits the interior of every single on of them.

For all \(d \ge 1\), we have \(f(d){\lt} 2^{d+1}\).

For every \(d\), one has the following chain of inequalities:

For every \(d\ge 2\), there is a set \(S\subset \{ 0, 1\} ^d\) of \(2\lfloor \frac{sqrt{6}}{9}(\frac{2}{\sqrt(3)})^d\rfloor \) points in \(\mathbb {R}^n\) (vertices of the unit \(d\)-cube) that determine only acute angels. In particular, in dimension \(d = 34\) there is a set of \(72 {\gt} 2*34 - 1\) points with only acute angels.

Let \(\mu \) be an ordinal number and denote by \(W_\mu \) the set of ordinal numbers smaller than \(\mu \). Then the following holds:

1. The elements of \(W_\mu \) are pairwise comparable.

2. If we order \(W_\mu \) according to their magnitude, then \(W_\mu \) is well-ordered and has ordinal number \(\mu \).

Any two ordinal numbers \(\mu \) and \(\nu \) satisfy precisely one of the relations \(\mu {\lt} \nu \), \(\mu = \nu \), or \(\mu {\gt} \nu \).

Every set of ordinal numbers (ordered according to magnitude) is well-ordered.

For every cardinal number \(\mathfrak {m}\), there is a definite next larger cardinal number.

Let the infinite set \(M\) have cardinality \(\mathfrak {m}\), and let \(M\) be well ordered according to the initial ordinal number \(\omega _{\mathfrak {m}}\). Then \(M\) has no last element.

Suppose \(\{ A_\alpha \} \) is a family of size \(\mathfrak {m}\) of countable sets \(A_\alpha \), where \(\mathfrak {m}\) is an infinite cardinal. Then the union \(\bigcup _\alpha A_\alpha \) has size at most \(\mathfrak {m}\).

The set of \(\mathbb {Q}\) of rational numbers is countable.

The set \(\mathbb {R}\) of real numbers is not countable

The set \(\mathbb {R}^2\) of all ordered pairs of real numbers (that is, the real plane) has the same size as \(\mathbb {R}\).

If each of two sets \(M\) and \(N\) can be mapped injectively into the other, then there is a bijection from \(M\) to \(N\), that is \(|M| = |N|\).

If \(c {\gt} \aleph _1\), then every family \(\{ f_\alpha \} \) satisfying \((P_0)\) is countable. If, on the other hand, \(c = \aleph _1\), then there exists some family \(\{ f_\alpha \} \) with property \(P_0\) which has size \(c\).

Suppose all roots fo the polynomial \(x^n + a_{n - 1}x^{n - 1} + \dots +a_0\) are real. Then the roots of are contained in the interval with the endpoints

Let \(f(x)\) be a real polynomial of degree \(n \ge 2\) with only real roots, such that \(f(x){\gt} 0\) for \( -1 {\lt} x {\lt} 1\) amd \(f(-1) = f(1) = 0\). Then

and equality holds in both cases only for \(n=2\).

Suppose \(G\) is a graph on \(n\) vertices without triangles. Then \(G\) has at most \(\frac{n^2}{4}\) edges, and equality holds only when \(n\) is even and \(G\) is the complete bipartite graph \(K_{n/2, n/2}\).

Suppose \(G\) is a graph on \(n\) vertices without triangles. Then \(G\) has at most \(\frac{n^2}{4}\) edges, and equality holds only when \(n\) is even and \(G\) is the complete bipartite graph \(K_{n/2, n/2}\).

Let \(\langle a, b \rangle \) be an inner product on a real vector space \(V\) (with the norm \(|a|^2 := \langle a, a \rangle \)). Then

holds for all vectors \(a, b \in V\), with equality if and only if \(a\) and \(b\) are linearly dependent.

Let \(a_1, \dots a_n\) be positive real numbers, then

with equality in both cases if and only if all \(a_i\)’s are equal.

Let \(a_1, \dots a_n\) be positive real numbers, then

with equality in both cases if and only if all \(a_i\)’s are equal.

Let \(a_1, \dots a_n\) be positive real numbers, then

with equality in both cases if and only if all \(a_i\)’s are equal.

Any line of the plane receives at most two different colors. The area of a rainbow triangle cannot be \(0\), and it cannot be \(\frac{1}{n}\) for odd \(n\).

For any blue point \(p_0 = (x_b, y_b)\), green point \((x_g, y_g)\), and red point \((x_r, y_r)\), the \(v\)-value of the determinant

is at least \(1\).

Every dissection of the unit square \(S = [0, 1]^2\) into finitely many triangles contains an odd number of rainbow triangles, and thus at least one.

Let \(p(x)\) be a real polynomial of degree \(n \geq 1\) with leading coefficient 1, and suppose that \(|p(x)| \leq 2\) for all \(x\) in the interval \([a, b]\). Then \(b - a \leq 4\).

If \(b\) is a multiple root of \(p'(x)\), then \(b\) is also a root of \(p(x)\).

We have \(p'(x)^2 \geq p(x)p''(x)\) for all \(x \in \mathbb {R}\).

Let \(f(z)\) be a complex polynomial of degree at least 1 and leading coefficient 1. Set \(C = \{ z \in \mathbb {C} : |f(z)| \leq 2 \} \) and let \(\mathcal{R}\) be the orthogonal projection of \(C\) onto the real axis. Then there are intervals \(I_1, \dots , I_t\) on the real line which together cover \(\mathcal{R}\) and satisfy

Let \(p(x)\) be a real polynomial of degree \(n \geq 1\) with leading coefficient 1, and all roots real. Then the set \(\mathcal{P} = \{ x \in \mathbb {R} : |p(x)| \leq 2\} \) can be covered by intervals of total length at most 4.

Let \(a_1, \dots , a_n\) be vectors in \(\mathbb {R}^d\), each of length at least 1, and let \(R_1, \dots , R_k\) be \(k\) open regions of \(\mathbb {R}^d\), where \(|x - y| {\lt} 2\) for any \(x, y\) that lie in the same region \(R_i\). Then the number of linear combinations \(\sum _{i=1}^{n} \epsilon _i a_i\), \(\epsilon _i \in \{ 1, -1\} \), that can lie in the union \(\bigcup _i R_i\) of the regions is at most the sum of the \(k\) largest binomial coefficients \(\binom {n}{j}\).

In particular, we get the bound \(\binom {\lfloor n/2 \rfloor }{n}\) for \(k = 1\).

for \(x\in \mathbb {R}\setminus \mathbb {Z}\).

The functions \(f\) and \(g\) are defined for all non-integral values and are continuous there.

Both \(f\) and \(g\) are periodic of period \(1\), that is \(f(x + 1) = f(x)\) and \(g(x + 1) = g(x)\) hold for all \(x\in \mathbb {R}\setminus \mathbb {Z}\).

Both \(f\) and \(g\) are odd functions, that is we have \(f(-x) = -f(x)\) and \(g(-x) = -g(x)\) for all \(x\in \mathbb {R}\setminus \mathbb {Z}\).

The two functions \(f\) and \(g\) satisfy the same functional equation: \(f(\frac{x}{2}) + f(\frac{x + 1}{2}) = 2f(x)\) and \(g(\frac{x}{2}) + g(\frac{x + 1}{2}) = gf(x)\).

By setting \(h(x) := 0\) for \(x \in \mathbb {Z}\), \(h\) becomes a continuous function on all of \(\mathbb {R}\) that shares the properties given in 26.2, 26.3, 26.4.

Consider the numbers \(1, 2, 3, \dots 2n\), and take away \(n + 1\) of them. Then there are two among these \(n + 1\) numbers which are relatively prime.

Suppose again \(A\subset \{ 1, 2, \dots , 2n\} \) with \(|A| = n + 1\). Then there are always two numbers in \(A\) such that one divides the other.

In any sequence \(a_1, a_2, \ldots , a_{mn+1}\) of \(mn+1\) distinct real numbers, there exists an increasing subsequence

of length \(m+1\), or a decreasing subsequence

of length \(n+1\), or both.

Suppose we are given \(n\) integers \(a_1, \dots , a_n\), which need not be distinct. Then there is always a set of consecutive numbers \(a_{k+1}, a_{k+2}, \dots , a_{\ell }\) whose sum \(\sum _{i=k+1}^{\ell } a_i\) is a multiple of \(n\).

If the graph \(G\) on \(n\) vertices contains no \(4\)-cycles, then

Let \(n \ge 2k\), and suppose we are given \(t\) distinct arcs \(A_1, \dots A_t\) of length \(k\), such that any two arcs have an edge in common. Then \(t \le k\).

The size of a largest antichain of an \(n\)-set is \(\binom {n}{\lfloor n/2\rfloor }\).

The largest size of an intersection \(k\)-family in an \(n\)-set is \(\binom {n - 1}{k - 1}\).

Let \(A_1, \dots A_n\) be a collection of subset of a finite set \(X\). Then there exists a system of distinct representatives if and only if the union of any \(m\) sets \(A_i\) contains at least \(m\) elements, for \(1 \le m \le n\).

Let \(\mathbb {Q} : \mathfrak {S}_n \longrightarrow \mathbb {R}\) be any probability distribution that defines a shuffling process \(\mathbb {Q}^*k\) with a strong uniform stopping rule whose stopping time is \(T\). Then for all \(k \geq 0\),

Let \(c \geq 0\) and \(k := \lceil n \log n + cn \rceil \). Then after performing \(k\) top-in-at-random shuffles on a deck of \(n\) cards, the variation distance from the uniform distribution satisfies

After performing \(k\) riffle shuffles on a deck of \(n\) cards, the variation distance from a uniform distribution satisfies

Let \(G = (V, E)\) be a finite weighted acyclic directed graph, \(A = \{ A_1, \dots , A_n\} \) and \(\mathcal{B} = \{ B_1, \dots , B_n\} \) two \(n\)-sets of vertices, and \(M\) the path matrix from \(A\) to \(\mathcal{B}\). Then

Let \(G = (V, E)\) be a finite weighted acyclic directed graph, \(A = \{ A_1, \dots , A_n\} \) and \(\mathcal{B} = \{ B_1, \dots , B_n\} \) two \(n\)-sets of vertices, and \(M\) the path matrix from \(A\) to \(\mathcal{B}\). Then

Every nonzero polynomial \(p(x) \in F[x_1, \dots , x_n]\) of degree \(d\) has at most \(dq^{n-1}\) roots in \(F^n\).

For every set \(E \subseteq F^n\) of size \(|E| {\lt} \binom {n+d}{d}\) there is a nonzero polynomial \(p(x) \in F[x_1, \dots , x_n]\) of degree at most \(d\) that vanishes on \(E\).

Any \((r \times n)\)-Latin rectangle, \(r {\lt} n\), can be extended to an \(((r+1) \times n)\)-Latin rectangle and hence can be completed to a Latin square.

Let \(P\) be a partial Latin square of order \(n\) with at most \(n - 1\) cells filled and at most \(\frac{n}{2}\) distinct elements, then \(P\) can be completed to a Latin square of order \(n\).

Let \(M = (m_{ij})\) be an \(n \times n\) matrix with entries in \(\{ 0, 1\} \), and let \(d_1, \dots , d_n\) be the row sums of \(M\), that is, \(d_i = \sum _{j=1}^{n} m_{ij}\). Then

The number \(L(n)\) of Latin squares of order \(n\) is bounded by

Let \(\vec{G} = (V, E)\) be a directed graph. A kernel \(K \subseteq V\) is a subset of the vertices such that

• \(K\) is independent in \(G\), and

• for every \(u \notin K\) there exists a vertex \(v \in K\) with an edge \(u \rightarrow v\).

A matching \(M\) of \(G = (X \cup Y, E)\) is called stable if the following condition holds: Whenever \(uv \in E \setminus M\), \(u \in X\), \(v \in Y\), then either \(uy \in M\) with \(y {\gt} v\) in \(N(u)\) or \(xv \in M\) with \(x {\gt} u\) in \(N(v)\), or both.

Let \(\vec{G} = (V, E)\) be a directed graph, and suppose that for each vertex \(v \in V\) we have a color set \(C(v)\) that is larger than the outdegree, \(|C(v)| \geq d^+(v) + 1\). If every induced subgraph of \(\vec{G}\) possesses a kernel, then there exists a list coloring of \(G\) with a color from \(C(v)\) for each \(v\).

A stable matching always exists.

We have \(\chi _\ell (S_n) = n\) for all \(n\).

If a graph \(G = (V, E)\) on \(n\) vertices has no \(p\)-clique, \(p \geq 2\), then

If a graph \(G = (V, E)\) on \(n\) vertices has no \(p\)-clique, \(p \geq 2\), then

If a graph \(G = (V, E)\) on \(n\) vertices has no \(p\)-clique, \(p \geq 2\), then

If a graph \(G = (V, E)\) on \(n\) vertices has no \(p\)-clique, \(p \geq 2\), then

If a graph \(G = (V, E)\) on \(n\) vertices has no \(p\)-clique, \(p \geq 2\), then

Whenever \(T = \{ v^{(1)}, \dots , v^{(m)}\} \) is an orthonormal representation of \(G\) with constant \(\sigma _T\), then

Every family of at most \(2^{d-1}\) \(d\)-sets is \(2\)-colorable, that is, \(m(d) {\gt} 2^{d-1}\).

Every family of at most \(2^{d-1}\) \(d\)-sets is \(2\)-colorable, that is, \(m(d) {\gt} 2^{d-1}\).

For every \(k \geq 2\), there exists a graph \(G\) with chromatic number \(\chi (G) {\gt} k\) and girth \(\gamma (G) {\gt} k\).

Let \(G\) be a simple graph with \(n\) vertices and \(m\) edges, where \(m \ge 4n\). Then

If \(A\) is a real symmetric \(n \times n\) matrix that is not diagonal, that is \(\operatorname {Od}(A) {\gt} 0\), then there exists \(U \in O(n)\) such that \(\operatorname {Od}(U^TAU){\lt}\operatorname {Od}(A)\).

Let \(p(x)\) be a real polynomial of degree \(n \geq 1\) with leading coefficient 1. Then

If a system of homogeneous linear equations with integer coefficients has a positive real solution, then it also has a positive integer solution.

For every real symmetric matrix \(A\) there is a real orthogonal matrix \(Q\) such that \(Q^{T}AQ\) is diagonal.

Suppose that we are given two finite sets \(R\) and \(C\) and a subset \(S \subseteq R \times C\). Whenever \((p, q) \in S\), then we say \(p\) and \(q\) are incident. If \(r_p\) denotes the number of elements that are incident to \(p \in R\), and \(c_q\) denotes the number of elements that are incident to \(q \in C\), then

\(e\) is irrational

\(e ^2\) is irrational

\(e ^4\) is irrational

Let \(G\) be any simple plane graph with \(n{\gt}2\) vertices. Then \(G\) has at most \(3*n - 6\) edges.

Let \(G\) be any simple plane graph with \(n{\gt}2\) vertices. Then \(G\) has a vertex of degree at most \(5\).

Let \(G\) be any simple plane graph with \(n{\gt}2\) vertices. If the edges of \(G\) are two-colored, then there is a vertex of \(G\) with at most two color-changes in the cyclic order of the edges around the vertex.

For \(a \not\equiv 0 (\mod p)\),

If \(G\) is a connected plane graph with \(n\) vertices, \(e\) edges and \(f\) faces, then

TODO

TODO

TODO

There exists an \(n \times n\) matrix with entries \(\pm 1\) whose determinant is greater than \(\sqrt{n!}\).

TODO

TODO

For \(a \not\equiv 0 \mod p\),

All planar graphs \(G\) can be \(5\)-colored:

Collecting the proofs from the chapter

Suppose that \(G\) is a finite graph in which any two vertices have precisely one common neighbor. Then there is a vertex which is adjacent to all other vertices.

Every nonconstant polynomial with complex coefficients has at least one root in the field of complex numbers.

There is an arrangement of \(2k + d\) points on \(S^d\) such that every open hemisphere contains at least \(k\) of these points.

TODO

If \(p(x)\in \mathbb {R}_+[x_1, \dots , x_n]\) is a \(H\)-stable and homogeneous of degree \(n\), then either \(p'\cong 0\), or \(p'\) is \(H\)-stable and homogeneous of degree \(n - 1\). In either case

Let \(G\) be a finite simple graph with vertex set \(V\) and edge set \(E\) and let \(d(v)\) denote the degree of a vertex \(v\). Then

TODO

Let \(K \subseteq F^n\) be a Kakeya set. Then

We have

TODO

TODO

TODO

TODO

If the \(d\)-sphere \(S^d\) is covered by \(d + 1\) sets,

such that each of the first \(d\) sets \(U_1, \dots , U_d\) is either open or closed, then one of the \(d + 1\) sets contains a pair of antipodal points \(x^*, -x^*\).

Given any finite configuration of “black” and “white” points in the plane, not all on one line, there is always a “monochromatic” line: a line that contains at least two points of one color and none of the other.

It is not possible to dissect a square into an odd number of triangles of equal algebra area.

The multiplicative group of a finite field is cyclic

For any museum with \(n\) walls, \(\lfloor \frac{n}{3} \rfloor \) guards suffice.

If \(P\) and \(Q\) are equidecomposable, then one can place a positive number of pearls (that is, assign positive integers) to all the segments of the decompositions \(P = P_1 \cup \cdots \cup P_n\) and \(Q = Q_1 \cup \cdots \cup Q_n\) in such a way that each edge of a piece \(P_k\) receives the same number of pearls as the corresponding edge of \(Q_k\).

Every elementary triangle \(\Delta = \operatorname {conv}\{ p_0, p_1, p_2\} \subset \mathbb {R}^2\) has area \(A(\Delta )=12\)

The area of any (not necessarily convex) polygon \(Q\subset \mathbb {R}^2\) with integral vertices is given by

where \(n_{int}\) and \(n_{bd}\) are the numbers of integral points in the interior respectively on the boundary of \(Q\).

If \(n\) objects are placed in \(r\) boxes, where \(r {\lt} n\), then at least one of the boxes contains more than one object.

TODO

TODO

TODO

If \(n\ge 3\) points in the plane do not lie on one single line, then they determine at least \(n - 1\) different slopes, where equality is possible only if \(n\) is odd and \(n \ge 5\).

Any partial Latin square of order \(n\) with at most \(n - 1\) filled cells can be completed to a Latin square of the same order.

Suppose that some “big” triangle with vertices \(V_1, V_2, V_3\) is triangulated (that is, decomposed into a finite number of “small” triangles that fit together edge-by-edge). Assume that the vertices in the triangulation get “colors” from the set \(\{ 1, 2, 3\} \) such that \(V_i\) receives the color \(i\) (for each \(i\)), and only the colors \(i\) and \(j\) are used for vertices along the edge from \(V_i\) to \(V_j\) (for \(i \neq j\)), while the interior vertices are colored arbitrarily with 1, 2, or 3. Then in the triangulation there must be a small “tricolored” triangle, which has all three different vertex colors.

A natural number \(n\) can be represented as a sum of two squares if and only if every prime factor of the form \(p = 4m + 3\) appears with an even exponent in the prime decomposition of \(n\).

For all natural \(n, k\) such that \(n \ge 2k\), at least one of the numbers \(n, n - 1, \dots , n - k - 1\) has a prime divisor \(p\) greater than \(k\), or, equivalently the binomial coefficient \(\binom {n}{k}\) always has a prime factor \(p {\gt} k\).

Given any set of \(n \ge 3\) points in the plane, not all on one line, there is always a line that contains exactly two of the points.

For any positive natural number, there is a prime which is greater than it, but no more than twice as large.

For all \(n \in \mathbb {N}\)

For all \(n \in \mathbb {N}\)

A finite set \(\{ p_1, \dots , p_r\} \) cannot be the collection of all prime numbers.

The set of primes \(\mathbb {P}\) is infinite.

The prime counting function is unbounded

There are infinitely many primes. (Six + infinitely many proofs)

If the sequence \(S = (s_1, s_2, s_3, \dots )\) is almost injective and of subexponential growth, then the set \(\mathbb {P}_S\) of primes that divide some member of \(S\) is infinite.

Any two Fermat numbers \(F_n := 2^{2^n} + 1\) are relatively prime.

The series \(\sum _{p\in \mathbb {P}}\frac1p\) diverges.

There is no largest prime.

TODO

Whenever a rectangle is tiled by rectangles all of which have at least one side of integer length, then the tiled rectangle has at least one side of integer length.

Whenever a rectangle is tiled by rectangles all of which have at least one side of integer length, then the tiled rectangle has at least one side of integer length.

Whenever a rectangle is tiled by rectangles all of which have at least one side of integer length, then the tiled rectangle has at least one side of integer length.

Collecting the proofs from the chapter...

A proper subring \(R\subset K\) is a valuation ring with respect to some valuation \(v\) into some ordered group \(G\) if and only if \(K = R \cup R^{-1}\).

The field of real numbers \(\mathbb {R}\) has a non-Archimedean valuation to an ordered abelian group

such that \(v(\frac{1}{2}) {\gt} 1\).

Let \(M = (m_{ij})\) be a doubly stochastic \(n \times n\) matrix. Then

and equality holds if and only if \(m_{ij} = \frac{1}{n}\)

Every finite division ring is commutative