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Fix a prime number $p$. Write a nonzero rational number as $x=p^{k}a/b$, where $k\in \mathbb{Z}$ and the nonzero integers $a,b$ are not divisible by $p$. The exponent $k$ is uniquely determined. Its P-adic valuation is this signed exponent $v_p(x)=k$: equivalently, for any nonzero integer numerator $A$ and denominator $B$,Factors of $p$ in the numerator decrease the p-adic absolute value; factors in the denominator increase it. For $p=2$ the computations are:

This is an absolute value on a field measuring divisibility, rather than the magnitude on a real number number line. It satisfiesThe proof comes directly from P-adic valuations: exponents add in a product of numbers, and extracting a common power of $p$ from a sum gives $v_p(x+y)\ge \min \{v_p(x),v_p(y)\}$. If the two P-adic valuations differ, the remaining term with smaller exponent cannot cancel modulo $p$, so equality holds. Thus the p-adic absolute value is a Non-Archimedean absolute value. ^[ref]

The metric $d_p(x,y)=|x-y{|}_p$ makes two rational numbers close when their difference contains a high power of $p$. This packages successive modular congruences into a numerical error estimate, allowing convergence in a metric space and convergent series to encode exact arithmetic information.

For an ordered abelian group $\Gamma$, the Hahn series field $k((t^{\Gamma }))$ consists of formal sums with well-ordered support in $\Gamma$. The least exponent defines a valuation with value group $\Gamma$.

An Artinian ring satisfies the descending chain condition on ideals. A commutative ring is Artinian exactly when it is Noetherian and has Krull dimension zero.

The subring $R=K+XYK[X,Y]$ is not Noetherian. Its idealsform a strict ascending chain because multiplying by a nonconstant element of $R$ introduces at least a second power of $X$.

For a nonsquare integer $d$,The integers are Noetherian, the Hilbert basis theorem makes $\mathbb{Z}[X]$ Noetherian, and quotients preserve Noetherianity.

A polynomial ring over a Noetherian ring is Noetherian.

If a ring $R$ is finitely generated as an abelian group, then it is Noetherian. Every ideal is an additive subgroup of that finitely generated abelian group and is therefore generated by finitely many elements as an abelian group. The same elements generate it as an $R$-ideal because integer multiples are already scalar multiples by elements of $R$.

The ascending chain condition requires every chainof ideals to stabilize. A ring is Noetherian exactly when it satisfies this condition, equivalently when every ideal is finitely generated.

A Noetherian local ring is a Noetherian ring with a unique maximal ideal.

Two point masses $m_1,m_2$ separated by distance $r$ have potential energyThe corresponding attractive force is $-\nabla V$.

This is the Newtonian point-mass specialization of gravitational energy, with the reference value chosen at infinite separation.

If ${\nabla }^{2}f⪰mI$ and the Hessian is $M$-Lipschitz near a minimizer $x^{\ast }$, thenThus a sufficiently close initial point has an error exponent that doubles at each iteration.

Orthogonal projections of an isotropic Gaussian vector onto orthogonal subspaces are independent.

Any jointly normal random variables with zero covariance are independent.

For means ${\mu }_i$, standard deviations ${\sigma }_i$, and correlation $\rho$, the standardized quadratic form in the density is

Two subvectors of a jointly multivariate normal distribution are independent exactly when their cross-covariance matrix is zero.

If $X\sim N(\mu ,\Sigma )$ and $A$ and $c$ are deterministic, thenIn particular, $a^{T}X$ is normal with mean $a^T\mu$ and variance $a^T\Sigma a$.

For a positive-definite matrix $\Sigma$, the distribution $N_p(\mu ,\Sigma )$ has density

To verify a candidate order $r$, check $a^r\equiv 1$ and rule out $a^{r/q}\equiv 1$ for every prime divisor $q$ of $r$.

For a unit $a$ modulo $N$, the function $x\mapsto a^x\mathrm{mod}N$ is periodic on the integers.

For an $L$-digit decimal integer $n$, moving its leading digit to the end gives a cyclic rotation $c_1(n)$ satisfyingThus, when a modulus $d$ divides $10^L-1$, every rotation has the same divisibility by $d$ as $n$, up to multiplication by a power of ten.