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A finite simple graph that is a connected graph and a planar graph, with $n\ge 3$ vertices and $m$ edges, satisfies $m\le 3n-6$. Choose a plane graph: each facial boundary walk has length at least three, so $3f\le 2m$. Substitute $f=2-n+m$ from the Euler formula for a connected planar graph.^[ref]

The complete graph $K_4$ attains equality: $6=3\cdot 4-6$. The bound also holds for a disconnected finite simple graph that is a planar graph with $n\ge 3$: place its connected components of a graph in disjoint regions and join them with new bridges in a graph, then apply the connected bound.

For an integer matrix $A$ and prime $p$, the commuting-matrix binomial theorem and divisibility of $\left(\genfrac{}{}{0ex}{}{p}{j}\right)$ for $0<j<p$ give

For every odd prime,

Euler's criterion at $a=-1$ gives

This is a ball for the Euclidean metric.

The closed Euclidean ball of radius $R$ about $a\in {\mathbb{R}}^n$ is $\{x:\Vert x-a{\Vert }_2\le R\}$; replacing $\le$ by $<$ gives the open ball.

A proper Euclidean motion has the form $E(x)=Ax+b$ with $A\in SO(3)$. It preserves lengths, dot products, cross products, orientation, curvature, and torsion.

This is the orientation-preserving subgroup of the three-dimensional Euclidean group.

In $d$ dimensions, the conformal Killing equation is

Attouch--Wets convergence of closed sets is locally uniform convergence of their distance functions on bounded subsets. Decreasing closed sets converge to their nonempty intersection in this topology.

Spectral pollution is convergence of finite-section eigenvalues to points outside the spectrum of the limiting operator. For bounded operators, such pollution lies in the essential numerical range.

For a measure-preserving transformation $T$ and $f\in L^1(\mu )$, the ergodic averagesconverge almost everywhere to $\mathbb{E}[f\mid \mathcal{I}]$, the conditional expectation on the invariant sigma-algebra. The limit has the same integral as $f$.

A measurable transformation $T:X\to X$ preserves a measure $\mu$ when $\mu (T^{-1}A)=\mu (A)$ for every measurable set $A$.

If a fixed graph $H$ has $v$ vertices, $e$ edges, and $N_H(n)$ unlabelled copies in $K_n$, then the number $X_H$ of its copies in $G(n,p)$ satisfiesThis follows by writing $X_H$ as a sum of indicator random variables.

For every fixed $\epsilon >0$,whereas the probability tends to zero when $p\le (1-\epsilon )\log n/n$. Above the threshold a union bound excludes every component of size at most $n/2$; below it isolated vertices remain with high probability.

Let $N$ be the number of isolated vertices in $G(n,p)$. Indicator variables giveand, because a specified pair is simultaneously isolated precisely when the $2n-3$ incident edges are absent,

Topological equivalence does not determine uniform convergence. For the usual and arctangent metrics on $\mathbb{R}$, maps that change the value $n$ to $2n$ at a single domain point have arctangent sup-error tending to zero while their usual sup-error diverges.

On $\mathbb{R}$, the usual metric and $|\arctan x-\arctan y|$ induce the same topology, but their ratio along $(0,n)$ tends to zero. Thus no positive global lower comparison constant exists.

An adiabatic equation of state relates pressure and volume along a process with no heat transfer. It commonly has the form $p{V}^{\gamma }=\text{constant}$.

An isotropic ideal gas whose particles obey the ultrarelativistic dispersion $\epsilon =pc$ satisfies

Let $X=(X_1,\dots ,X_n)$ be a discrete random vector, and let $\mathcal{S}$ be a collection of subsets of $[n]$ in which every index occurs at least $r$ times. ThenApply the chain rule for information entropy to each projection $X_S$ and use conditioning reduces entropy to compare every term with the corresponding conditional entropy in the full chain rule.

Every nonconstant entire function takes every complex value with at most one exception. Equivalently, its image is either $\mathbb{C}$ or $\mathbb{C}\setminus \{a\}$ for one complex number $a$. The hypothesis of being nonconstant matters: a constant function has only one value. The Little Picard theorem guarantees existence of preimages, without requiring infinitely many preimages for each attained value. ^[ref]

For a nonconstant polynomial $p$, the fundamental theorem of algebra applied to $p(z)-w$ supplies a preimage for every $w\in \mathbb{C}$. The complex exponential function has precisely one omitted value, zero, as the explicit formula for its preimages shows. In contrast, the complex sine function has no omitted value. To solve $\sin z=w$, set $u=e^{iz}$ and rearrange to the quadratic equationIts roots have product $-1$, so they are nonzero. Every such root has a preimage under the complex exponential function, giving a solution for $z$. Adding $2\pi k$ supplies further solutions by the periodicity of the complex sine.

The force of the Little Picard theorem is that it applies to every nonconstant entire function, without an explicit formula. The Liouville theorem already gives a dense image of a nonconstant entire function, but a dense subset could still leave out many individual points. The Little Picard theorem limits these missing points to one. In particular, establishing two omitted values proves that an entire function is constant.

For a nonconstant meromorphic function on the complex plane, the corresponding bound is two omitted values in the Riemann sphere $\mathbb{C}\cup \{\infty \}$, where poles take the value infinity. ^[ref] For example,omits zero and one, and has poles at $2\pi ik$ for $k\in \mathbb{Z}$. Every other finite value $w$ is attained by solving $e^z=1-1/w$. For an entire function, infinity is already an omitted value in the Riemann sphere, leaving at most one additional finite omitted value.

Proof outline: use the version of Montel's theorem for two fixed finite omitted values. ^[ref] It makes the rescaled holomorphic functions $g_n(\zeta )=f(a+n\zeta )$ a normal family on the unit disc. The fixed value $g_n(0)=f(a)$ gives pointwise anchoring of a holomorphic normal family, hence a subsequence with a finite holomorphic function as its locally uniform convergence limit. The Cauchy integral formula bounds its derivatives at zero, whereas $g_n^{\prime }(0)=n{f}^{\prime }(a)$. Consequently $f^{\prime }(a)=0$; arbitrary $a$ makes $f$ constant.