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Truncation toward zero removes the fractional magnitude of a real number, giving the integer obtained by applying the floor function to its absolute value and restoring the original sign; a magnitude below one gives zero. More precisely,The result is zero for $-1<x<1$. For example, $\mathrm{trunc}(2.7)=2$ and $\mathrm{trunc}(-2.7)=-2$, whereas $\lfloor -2.7\rfloor =-3$. Python's math.trunc uses this towards-zero convention, so replacing it by math.floor changes calculations with negative noninteger inputs. ^[ref].

For every real number $x$,The discarded difference $x-\mathrm{trunc}(x)$ has the sign of $x$ when $x$ is noninteger. In contrast, the fractional part $x-\lfloor x\rfloor$ is always in $[0,1)$. At $x=-2.7$, these two differences are $-0.7$ and $0.3$. Thus specifying a signed fractional remainder requires specifying which integer-valued operation is used.

Unlike the floor function and ceiling function, truncation toward zero need not commute with adding an integer. At $x=-0.7$ and $j=1$, $\mathrm{trunc}(x+j)=\mathrm{trunc}(0.3)=0$, whereas $\mathrm{trunc}(x)+j=1$. This matters if negative coordinates are assigned to equal-width bins by converting a quotient to an integer: $x=-0.7$ belongs to the bin $[-1,0)$ under the floor function, but truncation gives zero. The appropriate operation follows from the declared bin intervals.

Truncation toward zero is a nondecreasing function with a discontinuity of a function at each nonzero integer. It has no discontinuity of a function at zero, because it is identically zero on $(-1,1)$. Its agreement with the floor function on the positive side and the ceiling function on the negative side explains the different closed endpoints in the diagram.

The two-factor tensor rank is the smallest number of pure tensors whose sum equals $t$, with zero allowed as the empty sum. For two finite-dimensional vector spaces over the same field, it equals the matrix rank of the tensor-product coefficient matrix. A sum of $r$ outer products has column space contained in the span of their $r$ first factors, so its matrix rank is at most $r$. Conversely, choose a basis $u_1,\dots ,u_r$ for the column space, express each column in that basis, and collect the coefficients into row vectors $w_k^T$. This yields $C={\sum }_{k=1}^r{u}_k{w}_k^T$, hence a sum of $r$ pure tensors.

The pure tensor $(2,1)\otimes (-1,3)$ therefore has two-factor tensor rank one, whereas $e_1\otimes f_1+e_2\otimes f_2$ has two-factor tensor rank two. Both have two factors: the number of factors and the minimum number of product summands are different counts. Over the real numbers, the singular value decomposition supplies a useful decomposition $C={\sum }_k{\sigma }_k{u}_k{v}_k^T$, with one nonzero term per matrix rank. The two-factor rank equality does not extend to three or more factors by merely choosing one matrix layout.

The ultrametric inequality, also called the strong triangle inequality, bounds a direct distance by the larger of the two distances through an intermediate point:A metric space satisfying this rule is an ultrametric space. For a Non-Archimedean absolute value, the corresponding inequality is $|a+b|\le \max \{|a|,|b|\}$, and the associated metric is $d(x,y)=|x-y|$. The ordinary triangle inequality follows because the maximum of two nonnegative real numbers is at most their sum.

For example, the 2-adic absolute value gives the three points $0,1,8$ distancesEach distance is bounded by the maximum of the other two. The two largest distances are equal: every triangle in an ultrametric space has this property. Otherwise its unique longest side would exceed the maximum of the other two, contradicting the ultrametric inequality. The drawn lengths of a triangle on paper need not represent these distances.

The ultrametric inequality gives a useful exact rule when $|a|\ne |b|$:To see why, suppose $|a|>|b|$. The upper bound is $|a+b|\le |a|$; applying the ultrametric inequality to $a=(a+b)-b$ forces the reverse bound. Equal sizes can instead cancel: $|1{|}_2=|1{|}_2=1$, but $|1+1{|}_2=1/2$. This makes a uniquely largest term decisive in a sum, a principle used in the Newton polygon and in arguments about roots of a polynomial.

A sequence of functions $f_n:X\to Y$ converges uniformly to $f:X\to Y$, where $Y$ is a metric space with metric $d$, precisely whenFor real numbers, use the absolute value $d(u,v)=|u-v|$. Compare the universal quantification and existential quantification:The decisive change is whether $N$ can depend on $x$. Uniform convergence implies pointwise convergence.^[ref]

For functions taking values in real numbers on a nonempty domain of a function, the supremum error is $E_n={\sup }_{x\in X}|f_n(x)-f(x)|$, allowing $+\infty$. Then uniform convergence is equivalent to $E_n\to 0$. The supremum need not be a maximum: pointwise strict bounds below $\epsilon$ only give $E_n\le \epsilon$, so use tolerance $\epsilon /2$ when proving this equivalence. This single error bound controls every possible input, including one chosen after seeing the index.

For $f_n(x)=x^n$ and its pointwise limit on $[0,1]$, the error is zero at one but approaches one as $x\to 1$ from below. Consequentlyfor every positive integer $n$. This supremum is never attained: below one the error is $x^n<1$, and at one it is zero. Deleting the endpoint leaves the same supremum error on $[0,1)$. To disprove uniform convergence directly, choose $\epsilon =1/2$: for any proposed $N$, take $n=N$ and $x=2^{-1/N}<1$. The error equals $1/2$, violating the strict tolerance. Thus a moving input defeats every proposed common threshold.

On the smaller interval $[0,a]$, for a fixed $0\le a<1$, the pointwise limit is zero andFor $0<a<1$ and $0<\epsilon <1$, the exact first successful common threshold isThe logarithm and floor function turn an infinite approximation statement into a finite stopping rule. At $a=0.8$, $0.8^{20}=0.01152921504606846976$ fails tolerance $0.01$, while $0.8^{21}=0.009223372036854775808$ succeeds. At $a=1/2$ and $\epsilon =1/8$, the ratio is exactly three: $N=4$ is necessary because equality is insufficient. For $a=0$, every error is zero. The domain of a function determines whether one threshold can certify every input.

For another explicit stopping rule, let $h_n(x)=x/(n+1)$ on $[0,1]$. Its uniform limit is zero and its supremum error is $1/(n+1)$. For every $\epsilon >0$, $N=\max \{1,\lceil 1/\epsilon \rceil \}$ works, using the ceiling function. Unlike $x^n$ on $[0,1]$, it controls even the endpoint.

The practical beauty of uniform convergence is that one quantitative bound can survive further operations: the uniform limit theorem preserves continuity, and uniform convergence preserves integrals on a compact interval controls accumulated error. Differentiation needs additional hypotheses, such as uniform derivative convergence with an anchored value. These results distinguish an approximation that looks accurate at selected inputs from one that justifies replacing a whole function in a calculation.

For floating-point arithmetic with precision $p$ in radix $\beta$ and rounding to nearest, unit roundoff is the half-gap scale $u={\beta }^{1-p}/2$. If a nonzero exact real number $z$ lies in the normal finite range, its nearest rounded value satisfiesIndeed, the absolute distance to the nearest value is at most ${\Delta }_e/2$, and $|z|\ge {\beta }^e$ gives the bound. For a correctly rounded addition, subtraction, multiplication or division, apply this estimate to the exact result of that one operation on the stored operands. The bound does not include earlier roundoff error or automatically bound a whole algorithm. Results outside the finite range and tiny subnormal numbers require separate treatment. Rounding in a directed mode also changes the bound.^[ref]

For a bilinear map $B:V\times W\to U$, there is exactly one linear map $L:V{\otimes }_{F}W\to U$ satisfying $L(v\otimes w)=B(v,w)$. Equivalently, $B=L\circ \tau$, where $\tau (v,w)=v\otimes w$. The universal property of a tensor product concerns the vector space together with this distinguished bilinear map.^[ref]

Here is a complete calculation over the real numbers. Define the bilinear mapIts values on pairs of basis vectors determine $L$ on the four tensor-product basis elements:Consequently, for $t={\sum }_{i,j=1}^2{c}_{ij}{e}_i\otimes f_j$,For $v=(2,1)$ and $w=(-1,3)$, the direct calculation gives $B(v,w)=(-2+12+3,-1-6)=(13,-7)$. Using the expansion of $v\otimes w$ instead gives $L(v\otimes w)=-2(1,0)+6(2,-1)-(0,1)+3(1,0)=(13,-7)$. The linear map also handles a nonseparable tensor such as $t=e_1\otimes f_1+e_2\otimes f_2$: $L(t)=(2,0)$, even though no single pair has $v\otimes w=t$.

A short proof uses a tensor-product basis: assign $L(e_i\otimes f_j)=B(e_i,f_j)$ and extend by linearity. Expanding both inputs verifies $L(v\otimes w)=B(v,w)$ by bilinearity; the assigned values on a basis force uniqueness. More generally, every multilinear map $F:V_1\times \cdots \times V_r\to U$ factors uniquely through a linear map from $V_1\otimes \cdots \otimes V_r$. This lets formulas involving several inputs define linear maps without choosing coordinates; tensor contraction uses the evaluation pairing between a vector and a covector.

Every vector space $V$ over a field $F$ has a basis, assuming the axiom of choice. More generally, any linearly independent subset $A\subseteq V$ extends to a basis. This is an algebraic Hamel basis: each representation uses a finite linear combination, even when the basis is infinite.^[ref]

Use the poset of linearly independent subsets containing $A$:It is a set, being a subset of the power set of $V$, and is nonempty because $A\in {\mathcal{P}}_A$.

A short proof outline supplies both essential steps:

Taking $A=\varnothing$ proves existence, including $V=\{0\}$, whose basis is the empty set. The chain in a partial order need not be indexed by the natural numbers: only the finitely many members needed for a particular linear relation are compared. An arbitrary union of linearly independent sets can fail: $\{v\}$ and $\{2v\}$ for a nonzero vector over the real numbers are each linearly independent, but their union is linearly dependent. The nesting of a chain in a partial order is what prevents this failure.

For a parabola described by the function graph $y=a(x-h{)}^2+k$, with nonzero $a$ in the real numbers, its vertex of a parabola is the point $(h,k)$. This point gives its minimum or maximum value of $y$ in these coordinates.

Vieta's formulas connect two descriptions of a polynomial: its coefficients and its roots of a polynomial. They let us read the sum and product of the roots of a polynomial without finding the individual roots of a polynomial, and turn a proposed factorization of polynomials into checkable coefficient identities. The key is that coefficients encode elementary symmetric polynomials, which are unchanged by a permutation of the roots of a polynomial.

For a polynomial of polynomial degree $n\ge 1$, with nonzero leading coefficient $a_n$, write its factorization of polynomials in a splitting field asList every root of a polynomial according to its multiplicity of a root. ThenIn particular, ${\sum }_i{r}_i=-a_{n-1}/a_n$ and ${\prod }_i{r}_i=(-1{)}^n{a}_0/a_n$. With coefficients in the real numbers or complex numbers, the fundamental theorem of algebra ensures that a list of $n$ roots of a polynomial in the complex numbers exists when multiplicity of a root is counted.^[ref]

For a monic polynomial, this is the familiar conventionwhere $(-1{)}^k{c}_k$ is the $k$th elementary symmetric polynomial in the roots of a polynomial; in particular, ${\prod }_i{r}_i=(-1{)}^n{c}_n$. The ratio by $a_n$ in the general statement matters: multiplying every coefficient by the same nonzero scalar preserves both the roots of a polynomial and these ratios.

A short derivation explains every alternating sign. In the product ${\prod }_i(x-r_i)$, a term of polynomial degree $n-k$ arises by choosing $-r_i$ from $k$ distinct factor positions and $x$ from the others. Summing the resulting products gives $(-1{)}^k{e}_k$. Values at different positions may coincide: the selection is of positions, not distinct root values. Matching coefficients gives the formula. This is the distributive property organized by the number of chosen root factors.

A weakly linear partial function on the real numbers has a domain closed under rational scalar multiplication and addition, and it preserves addition on that domain. A nonempty domain is a vector space over the rational numbers, and additivity implies rational homogeneity of an additive function by integer multiplication and division. The empty function also satisfies the defining conditions vacuously. This terminology does not assert homogeneity under arbitrary real scalars.

For a nonempty bounded array of real numbers, each column infimum is at most every row supremum, because ${\inf }_k{x}_{k,n}\le x_{m,n}\le {\sup }_j{x}_{m,j}$ for all $m,n$. Taking the supremum over columns and then the infimum over rows proves the displayed inequality. Equality need not hold: for $x_{m,n}=1_{\{m=n\}}$ on positive integer indices, every column infimum is zero and every row supremum is one. This elementary bound is the order-theoretic starting point for stronger minimax theorems with additional hypotheses.

The weighted arithmetic mean combines values according to how much each contributes to the quantity being summarized. For a finite list of real numbers $x_i$ and nonnegative weights $w_i$ with $W={\sum }_i{w}_i>0$, defineZero weights ignore an entry; at least one weight must be positive. Multiplying every weight by the same positive number leaves the result unchanged. Writing $p_i=w_i/W$ makes the weighted mean a convex combination, since $p_i\ge 0$ and ${\sum }_i{p}_i=1$. It therefore lies between the smallest and largest input values. ^[ref]

For example, give the values $90$ and $60$ weights $2$ and $8$:The larger weight pulls the result closer to $60$. The balancing identityexplains a geometric interpretation: if the weights were masses placed at positions $x_i$, the weighted mean would be their center of mass. Here $8(66-60)=2(90-66)=48$, so the two sides balance even though their distances from the mean differ.

Weights need an interpretation. They can represent numbers of observations, masses, or probabilities in a finite probability distribution. These meanings lead to the same formula, while answering different questions. For instance, combining means of groups uses group sizes as weights to recover the mean over individual observations. For a function $f$ and nonnegative weight function $w$, the analogous integral is $\int wf/\int w$, provided $wf$ is integrable and the total weight $\int w$ is positive and finite.

For positive real numbers $x_i$ and nonnegative weights with ${\sum }_i{w}_i=1$, the weighted geometric mean isIt takes a weighted mean of logarithms. A zero weight excludes that input; equal weights give the ordinary geometric mean. If $x_i$ occurs $m_i$ times in a list of total length $M$, its correct weight is $w_i=m_i/M$. For the list $(12,3,3,3)$,and replacing all four entries by $G_w$ preserves the product $324$.

There is also an optimization interpretation. A logarithmic transformation turns multiplicative discrepancies into additive ones. For a positive representative $c$, the weighted mean minimizes squared deviations identity givesThus $c=G_w$ uniquely minimizes weighted quadratic loss in logarithmic coordinates. The weighted arithmetic mean minimizes squared discrepancies in the original coordinates. These two objectives explain why different choices of what constitutes an error select different summaries.

The strictly concave function $\log x$ and the reversed Jensen's inequality give $G_w\le A_w$, where $A_w={\sum }_i{w}_i{x}_i$ is the weighted arithmetic mean. Apply the same reasoning to the multiplicative inverses to get $H_w\le G_w$ for the weighted harmonic mean. Thus $H_w\le G_w\le A_w$; equality holds exactly when all positive-weight inputs agree. For this list, $H_w=48/13$, $G_w=3\sqrt{2}$, and $A_w=21/4$. The weights specify which inputs are represented and must stay the same across the comparison.

For positive real numbers $x_i$ and nonnegative weights with ${\sum }_i{w}_i=1$, the weighted harmonic mean isEqual weights give the harmonic mean, while a zero weight excludes that input. For travel at positive constant speeds $v_i$ over positive leg lengths $d_i$, the weights are fractions of total distance:For $30\mathrm{km}$ at $30km/h$ and $90\mathrm{km}$ at $60km/h$, the weights are $1/4,3/4$, elapsed time is $5/2$ hours, and the average speed is $48km/h$. Weighting the speeds by their time fractions instead gives the same result as a weighted arithmetic mean: $(2/5)30+(3/5)60=48$. The two weight systems describe the same journey in different coordinates and must not be interchanged. ^[ref].

For a finite list of real numbers $x_i$ and nonnegative weights of positive total $W$, the weighted mean uniquely minimizes the weighted sum of squared deviationsExpanding around ${\bar{x}}_w$ and using ${\sum }_i{w}_i(x_i-{\bar{x}}_w)=0$ givesEvery displacement from the weighted mean increases the objective. This is the simplest one-parameter weighted least squares problem: represent all entries by one common value.

For the weighted values $90$ and $60$ with weights $2$ and $8$,Thus $S(66)=1440$, while $S(75)=2250$. The minimum is defined by this squared-error criterion; a different loss can select a different summary, such as a median under unweighted absolute deviations.