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For a positive random variable $X$, let $D\in \{1,\dots ,9\}$ be its first nonzero base-ten digit. The Benford law is the discrete probability distribution with probability mass functionHere $X$ need not be an integer: $0.0042$, $4.2$ and $4200$ all have $D=4$. Multiplication by an integer power of ten changes the decimal point's position and leaves the first nonzero digit unchanged. Zero has no first nonzero digit, so exclude it. For negative inputs one may explicitly use their absolute values, still excluding zero; that convention must be stated rather than applying a logarithm to a negative number.

The Benford law favors small digits: $\mathbb{P}(D=1)={\log }_{10}2\approx 0.301029996$, whereas $\mathbb{P}(D=9)={\log }_{10}(10/9)\approx 0.04575749$. Thus the model assigns about $30.10\%$ to digit one and $4.58\%$ to digit nine. The masses are positive and their normalization is a finite telescoping series:

Why can digit one be so common? In a decimal significand interval $[1,10)$, the digit-one region $[1,2)$ occupies a larger fraction of the logarithmic scale than the digit-nine region $[9,10)$. The precise assumption is that the fractional part of the base-ten logarithm is distributed uniformly on $[0,1)$; Benford law from logarithmic uniformity turns those unequal interval lengths into probabilities. This makes a surprising digit pattern into a simple geometry of multiplicative size. The logarithmic explanation is also presented in Peter Sheridan Dodds, Benford's Law, essential story.

The Benford law is useful as a proposed model for leading digits when the mechanism supports logarithmic uniformity or an appropriate limiting argument, such as Benford frequencies for powers of an integer. Merely spanning many powers of ten does not establish that assumption. A discrepancy from the Benford law can prompt investigation of the model and data collection; by itself it does not establish fraud, and agreement does not establish authenticity.

Simon Newcomb's 1881 paper, Note on the Frequency of Use of the Different Digits in Natural Numbers, appeared in the American Journal of Mathematics 4, pp. 39–40. Its opening connects unequal leading-digit frequencies with the faster wear of early pages in logarithm tables. The publication metadata and original opening excerpt are available in Newcomb (1881), original publication record and ^[ref]. Frank Benford's The Law of Anomalous Numbers, Proceedings of the American Philosophical Society 78 (1938), pp. 551–572, likewise discusses worn logarithm tables on p. 551 and tabulates observations on p. 553. See Benford (1938), transcription of the original paper with page numbers and publication metadata. These observations motivated study of the digit pattern; they do not make the Benford law universal.

For a positive random variable $X$, write its base-ten logarithm as an integer part plus a fractional part:The floor function handles values below one as well: for $X=0.0042$, $K=-3$ and the decimal significand is $10^Z=4.2$. Since $10^Z\in [1,10)$, the first nonzero decimal digit is $D=\lfloor 10^Z\rfloor$. The monotonicity of the logarithm gives the exact event equivalenceIf $Z$ has the uniform distribution on $[0,1)$, the probability of this interval is its length. Consequently $\mathbb{P}(D=d)={\log }_{10}(d+1)-{\log }_{10}d$, proving the Benford law. This argument assumes uniformity of the fractional part of the logarithm, rather than uniformity of $X$ itself. See the logarithmic-coordinate explanation in Peter Sheridan Dodds, Benford's Law, essential story.

An exact constructed example is the log-uniform distribution on $[1,10)$, with probability density function $f_X(x)=1/(x\ln 10)$ there and zero elsewhere. Its total mass is ${\int }_1^{10}dx/(x\ln 10)=1$. The change of variables formula with $x=10^z$ gives $f_Z(z)=f_X(10^z)(\ln 10)10^z=1$ for $0\le z<1$. Equivalently, integrating directly givesFor example, the probability of first digit four is ${\log }_{10}(5/4)\approx 0.096910013$. By contrast, an ordinary uniform distribution for $X$ on $[1,10)$ has probability density function $1/9$ there, so $\mathbb{P}(D=d)={\int }_d^{d+1}dx/9=1/9$ for every digit. This is a concrete counterexample to any claim that the Benford law holds for every positive random variable.

More generally, a log-uniform distribution on $[A,B)$ with $0<A<B$ has uniform $Y={\log }_{10}X$ on $[{\log }_{10}A,{\log }_{10}B)$. If its logarithmic span ${\log }_{10}(B/A)=m$ is a positive integer, reduction to the fractional part wraps an interval of length $m$ around the unit interval exactly $m$ times, making $Z$ uniform even when $A$ is not a power of ten. Arbitrary partial-decade truncations generally do not have this property: for a log-uniform distribution on $[1,2)$, the first digit is always one.

Under the uniform fractional-logarithm assumption, multiplication by any constant $c>0$ replaces $Z$ by $Z^{\prime }=\{Z+{\log }_{10}c\}$. This is a rotation modulo one, which preserves the uniform distribution and therefore preserves the Benford law; see scale invariance of decimal significands. For the constructed $[1,10)$ example, multiplying by $0.001$ gives a log-uniform distribution on $[0.001,0.01)$ with the same digit probabilities. The claim concerns the decimal significand law: there is no proper probability density function proportional to $1/x$ over all of $(0,\infty )$, since its integral diverges. The existing scale-invariant prior describes that unnormalizable measure, while a finite log-uniform distribution is normalizable.

A frequency function built from an affine variable and its fractional part. Irrational $\gamma$ and $\beta$ can give many exact additive relations because each pair sum has only two possible floor carries. For example $\beta =\sqrt{3}$, $\gamma =\sqrt{2}$ gives $\gg N^3$ additive quadruples on $[-N,N]$, but agreement with every ordinary affine function occurs on only $o(N)$ shifts. Integer floor points underlying affine agreement must be collinear.

For $x>0$, the decimal significand is the unique $S_{10}(x)\in [1,10)$ such that $x=10^k{S}_{10}(x)$ for an integer $k$. Explicitly, $k=\lfloor {\log }_{10}x\rfloor$ and $S_{10}(x)=10^{\{{\log }_{10}x\}}$, using the floor function and the fractional part. For $0.0042$, $4.2$ and $4200$, the decimal significand is $4.2$ and the first nonzero digit is $\lfloor S_{10}(x)\rfloor =4$. In the classical terminology of logarithm tables, the logarithmic mantissa is $\{{\log }_{10}x\}$, which differs from the decimal significand $10^{\{{\log }_{10}x\}}$.

For every irrational number $\alpha$ and positive integer $N$, there are integers $p,q$ with $1\le q\le N$ such thatThe proof applies the pigeonhole principle to the $N+1$ fractional parts of $0,\alpha ,\dots ,N\alpha$.

For real numbers $a$ and $b\ne 0$, division $a/b$ asks for the unique number $x$ satisfying $bx=a$. Equivalently, $a/b=a{b}^{-1}$ uses the multiplicative inverse of $b$. This description separates the role of the denominator from the result: the quotient undoes a specified multiplication.

For example, $3/2$ is the number whose double is three, so it is exactly $1.5$. A second interpretation asks how many portions of size $2$ fit into an amount $3$; the answer is one and a half portions. Equal sharing and measuring with a specified portion size lead to the same quotient, although the units assigned to the result can differ.

In the natural numbers, exact division need not be possible. Euclidean division instead keeps a remainder: $17=5\cdot 3+2$, with $0\le 2<5$. Thus seventeen objects form three complete groups of five with two left over. As a rational number, the same quotient is $17/5=3+2/5=3.4$. The remainder and the fractional part describe the same unfinished group in different ways.

Division by zero cannot retain the unique-solution rule. For $a\ne 0$, the equation $0x=a$ has no solution; for $a=0$, every real number $x$ solves $0x=0$, so there is no unique quotient. This explains both failures through multiplication, rather than assigning $a/0$ a value that would contradict ordinary arithmetic. ^[ref]

The floor function and ceiling function turn a real number into directed integer bounds. They answer two different questions: which integer is the largest one that does not exceed $x$, and which is the smallest one that is not less than $x$?Both outputs are integers. These are exact mathematical operations; they need not choose the closest integer, and they do not refer to roundoff error in floating-point arithmetic.

For example,The fractional part is $\{x\}=x-\lfloor x\rfloor$. The negative example matters: discarding the digits after the decimal point would give $-2$, which is truncation toward zero, rather than the floor function's value $-3$. At an integer, the floor function and ceiling function agree and the fractional part vanishes.

The usefulness is a bridge from continuous quantities to discrete decisions. An integer count in a closed interval uses both bounds. A ceiling count for fixed-size batches counts how many containers or pages are needed; a floor function counts how many complete groups fit. In Euclidean division, the same bounds specify the quotient, while the fractional part records the scaled remainder. A small amount of notation removes many cases that would otherwise have to be stated separately.

The reflection identityfollows by reflecting the ordered integers across zero: the largest one below $-x$ becomes the smallest one above $x$. For every integer $m$, translation gives $\lfloor x+m\rfloor =\lfloor x\rfloor +m$ and $\lceil x+m\rceil =\lceil x\rceil +m$. These formulas turn upper bounds into lower bounds and move examples between neighboring intervals.

For a historical example of the notation in use, Kenneth E. Iverson's 1962 conference paper A Programming Language lists the floor function, ceiling function and residue operation alongside the ordinary arithmetic operations. It states the two defining inequalities above, then extends operations component by component to vectors and matrices. This shows their role as tools for describing algorithms, rather than merely conventions for writing decimal numbers. ^[ref]

The fractional part of a real number isIt is the amount left after removing the lower enclosing integer, even for negative inputs. Thus $\{-2.7\}=-2.7-(-3)=0.3$. It is not the signed string of digits obtained by truncation toward zero.

Translation by an integer leaves the fractional part unchanged: $\{x+m\}=\{x\}$. It is therefore a periodic function with period one, increasing from zero toward one on each unit interval, then returning to zero at the next integer. In Euclidean division $N=kq+r$ with $0\le r<k$, $\{N/k\}=r/k$; for $102/20$, this is $2/20=0.1$.

The notation is useful for following repeated motion modulo an integer. In normality and equidistribution under integer multiplication, the values $\{K^{n}x\}$ shift the digits of a positional notation; the fractional part discards the leading integer portion without losing the remaining digits. Its braces must be distinguished by context from the braces denoting a set.

For an all-integer dictionary row $x_B+{\sum }_j{a}_j{x}_j=b$ with nonnegative variables, this cut follows because $x_B+{\sum }_j\lfloor a_j\rfloor x_j$ is an integer and ${\sum }_j\{a_j\}{x}_j\ge 0$. Consequently that integer is at most $\lfloor b\rfloor$. Negative coefficients use the usual fractional part $u-\lfloor u\rfloor$, so $\{-1/5\}=4/5$. Normalize a resulting inequality sensibly before introducing a new integer-valued slack variable.

A real number $x$ is a normal number in base $K$ exactly when its iterates under the integer multiplication map on the circle, $T_K(x)=\{Kx\}$, form an equidistributed sequence. Braces denote the fractional part. Multiplication by $K$ shifts the positional notation one digit to the left:A length-$r$ word with base-$K$ integer value $a$ specifies the half-open interval $[a/K^r,(a+1)/K^r)$, whose length is $K^{-r}$. Visiting that interval is exactly encountering that word at the next digit position. These digit intervals form arbitrarily fine grids, so their correct limiting visiting frequencies imply equidistribution on every interval. This converts a question about digits into one about a dynamical system. Bailey and Crandall develop this connection using equidistribution of the fractional parts of $K^{n}x$. ^[ref]

In base ten the word $12$ corresponds to $[0.12,0.13)$. For Champernowne constant, counting the visits $T_{10}^n(x)$ for $0\le n\le 99998$ gives precisely the $2088$ occurrences of $12$ in its first $100000$ digits. For $x=1/3$, the base-two integer multiplication map on the circle instead alternates between $1/3$ and $2/3$; it never visits $[0,1/4)$, the interval for $00$. This two-point orbit is not an equidistributed sequence.

A real number is normal in an integer base $K\ge 2$ when every finite word of digits occurs in its positional notation with the same limiting digit block frequency as every other word of the same length. Write its fractional part as $x=0.d_1{d}_2\dots$ in base $K$, using the expansion that is not eventually all $K-1$ digits when two expansions are possible. For every length $r\ge 1$ and every word $w\in \{0,\dots ,K-1{\}}^r$, the requirement isOccurrences overlap: the word $11$ occurs twice in $111$. There are $K^r$ possible words of length $r$, so $K^{-r}$ is the equal share for each. In a decimal representation, this means frequency $1/10$ for each digit, $1/100$ for each pair, and $1/1000$ for each triple, continuing through every finite length. The definition agrees with that used by Copeland and Erdős in their work on explicit normal numbers. ^[ref]

The appeal is that a precise, deterministic real number can have the same limiting digit block frequencies as repeated draws from a uniform distribution on a finite set, with independence between draws. Champernowne constant is an explicit example. Normality is useful for studying digit distribution, but balanced digits alone establish only a simply normal number, and a finite sample cannot establish any of these infinite limits.

Émile Borel introduced normal numbers and established their prevalence in his 1909 paper applying probability to number theory; the stronger all-base statement is discussed under absolutely normal number. ^[ref]

Introduce the nonnegative integer surplus variables $z_1=3x_1+x_2-4$ and $z_2=x_1+2x_2-4$. Solving these two equalities gives the optimal relaxed simplex dictionarywith objectiveThe printed tableau has a sign error in the first row: in equality-row form the coefficient of $z_2$ must be $+1/5$, rather than $-1/5$. The defining constraints and the printed objective row both confirm this correction. The second row, from which the required cut is obtained, is unaffected.

The cutting-plane method solves the linear programming relaxation and, if the optimum is fractional, adds an inequality valid for every feasible integer point but violated by that optimum. For an all-integer tableau rowwrite each coefficient as its integer part plus its fractional part. The quantity $x_B+{\sum }_j\lfloor a_j\rfloor z_j$ is an integer, while ${\sum }_j\{a_j\}{z}_j$ is nonnegative. Thus the integer is at most $\lfloor b\rfloor$, and subtraction gives the Gomory fractional cutIt retains every feasible integer point and cuts off the fractional basic point with all nonbasic $z_j=0$.

Apply this to the row $x_2+\frac{1}{5}{z}_1-\frac{3}{5}{z}_2=\frac{8}{5}$. Since $\{1/5\}=1/5$, $\{-3/5\}=2/5$ and $\{8/5\}=3/5$, the resulting valid inequality isMultiply by five and add an integer slack $w=z_1+2z_2-3\ge 0$. Its row is $w-z_1-2z_2=-3$. At the old relaxed optimum, $w=-3$ is negative, while the minimization objective coefficients $2/5,9/5$ remain nonnegative. This is the starting configuration for the dual simplex algorithm: retain dual feasibility and pivot to repair the negative basic value.

The cut row permits either $z_1$ or $z_2$ to enter. The dual ratio test givesThus $z_1$ enters and $w$ leaves. Solving the cut row and substituting givesSetting the nonbasic variables $z_2=w=0$ gives nonnegative basic values $z_1=3,x_1=2,x_2=1$. The new objective coefficients are nonnegative, so this is optimal for the cut relaxation. Its decision variables are already integer and satisfy the original inequalities. ThereforeIndeed the positive coefficients of both nonbasic variables make this relaxed optimizer unique. No additional cut is needed.

Introduce nonnegative slack variables $s_1,\dots ,s_4$ and write $z$ for the objective value. The initial simplex dictionary, recovered from the original PDF rather than the corrupted converted table, isFor this minimization convention, nonnegative objective coefficients give dual feasibility. The dual simplex algorithm repairs a negative basic value while retaining those nonnegative reduced costs. If the leaving row is $x_B=b+{\sum }_j{a}_j{x}_j$, $b<0$, an entering variable must have $a_j>0$; choose the smallest ratio of its objective coefficient to $a_j$.

Choose $s_1$ to leave. Eligible columns are $x_1,x_3$, with ratios $2/1=2$ and $18/6=3$. Thus $x_1$ enters. Solving its row and substituting givesNow only $s_3$ has a negative basic value. Its only eligible entering column is $x_3$, so pivot there, with $s_3$ leaving. The resulting dictionary isSetting the nonbasic variables $s_1,x_2,s_3$ to zero gives a basic feasible solution. The final objective row also proves $z\ge 23$ for every feasible point, because its nonbasic variables are nonnegative. Hence the unique relaxed optimum is

For the integer program, all four original slack variables are integers whenever the decision variables are integers. The fractional row can be writtenThe Gomory fractional cut keeps the fractional parts of these coefficients: $\{-1\}=\{-2\}=0$ and $\{-1/2\}=1/2$. It therefore gives $\frac{1}{2}{s}_3\ge \frac{1}{2}$, or $s_3\ge 1$. Its validity can also be seen directly: the displayed row makes $s_3$ an odd nonnegative integer. In original variables this cut is $2x_1+10x_3\le 18$.

Introduce the new integer-valued slack $u=s_3-1\ge 0$, with initial row $u=-1+s_3$. It is primal infeasible but retains dual feasibility, so one more dual-simplex pivot has $u$ leave and $s_3$ enter. Substitute $s_3=1+u$ to obtainAll basic values at $s_1=x_2=u=0$ are nonnegative integers, including $s_3=1$. The objective row proves a lower bound of $26$ on every point satisfying the valid cut, hence on every integer-feasible point. It is attained atThe strictly positive nonbasic objective coefficients also show uniqueness. One Gomory cut and its reoptimization are sufficient.

Write $e(t)=\exp (2\pi it)$ and use $\Vert t\Vert$ for distance to the nearest integer. The Dirichlet approximation theorem, in its pigeonhole form, says that for every real $\alpha$ and integer $Q\ge 1$ there are coprime integers $a,q$ with $1\le q\le Q$ and $|\alpha -a/q|\le 1/(qQ)$. Indeed put the $Q+1$ fractional parts of $0,\alpha ,\dots ,Q\alpha$ into $Q$ equal intervals; two lie in the same interval. Their difference gives $|q\alpha -a|\le 1/Q$, and reducing the fraction preserves the claimed bound. This includes rational $\alpha$.

For $t\ge s$, the finite geometric series has length $L=t-s+1$. When $\theta \notin \mathbb{Z}$,The last inequality uses the concavity of sine on $[0,\pi /2]$. The triangle inequality also gives $L$. At integer $\theta$ the sum has absolute value $L$, interpreting the reciprocal as infinity. Thus the precise exponential geometric sum bound isThe interval must be nonempty; for $t<s$ its sum is zero and its length should be defined as $\max (0,t-s+1)$, rather than using a potentially negative bound.

For the separated points, assume $Q\ge 2$ and $\delta >0$. An arc of length $2r\le 1$ contains at most $1+2r/\delta$ of the points, by ordering them along the arc. Points with $\Vert {\theta }_i\Vert \le 1/Q$ contribute $O(Q+{\delta }^{-1})$. In the dyadic annulus $2^j/Q<\Vert {\theta }_i\Vert \le 2^{j+1}/Q$, each summand is $O(Q{2}^{-j})$ and there are at most $1+2^{j+2}/(Q\delta )$ points. There are $O(\log (2Q))$ nonempty annuli, giving the stronger separated reciprocal-distance sumIf there is only one point the bound is immediate; otherwise $\delta \le 1/2$. The printed $\log Q$ requires a lower bound away from $Q=1$; $\log (2Q)$ avoids that endpoint defect.

A useful version of the quadratic Weyl inequality is the following, with $(a,q)=1$, $1\le q\le Q$ and $|\alpha -a/q|\le 1/(qQ)$:For $N\ge 2$, expand the square of this quadratic exponential sum by $h=n-m$:The exponential geometric sum bound therefore gives $|S{|}^2\ll N+{\sum }_{h\le N}\min (N+1,\Vert 2\alpha h{\Vert }^{-1})$. For $q\ge 8$, partition these $h$ into blocks of length at most $\lfloor q/4\rfloor$. Two different points in one block have difference $d$ with $0<|d|<q/4$. The reduced denominator of $2a/q$ is at least $q/2$, so $\Vert 2ad/q\Vert \ge 1/q$, whileThus the rotations $2\alpha h$ are separated by at least $1/(2q)$ in each block. The separated reciprocal-distance sum bounds each block by $O(N+q\log (2N))$. There are $O(N/q+1)$ blocks, and henceFor $q<8$ or $N<2$, the trivial bound supplies the same assertion with an absolute constant. The same proof applies to ${\sum }_{n=1}^{N}e(\alpha n^2)$, whose difference sums also have interval endpoints.

We finally prove a polynomial rate of quantitative quadratic recurrence, without using the optional finite-group assumption. For $0<\epsilon \le 1/2$, use the periodic triangular bump $b(t)=(1-\Vert t\Vert /\epsilon {)}_{+}$. It is ${\epsilon }^{-1}$ times the periodic convolution of two interval indicators, and therefore has the nonnegative Fourier coefficientsIts Fourier series is absolutely convergent, ${\sum }_r\hat{b}(r)=b(0)=1$, and the coefficient tail is ${\sum }_{|r|>R}\hat{b}(r)\ll 1/(\epsilon R)$. Suppose that $\Vert \alpha n^2\Vert \ge \epsilon$ for every $1\le n\le N$. Then ${\sum }_{n}b(\alpha n^2)=0$. Choose $R=\lceil C{\epsilon }^{-2}\rceil$ with a sufficiently large absolute $C$, so that the contribution of frequencies $|r|>R$ has absolute value at most $N\epsilon /2$. The zero frequency contributes $N\epsilon$. Since all coefficients are nonnegative and their sum is one, some $1\le r\le R$ must satisfyApply the Dirichlet approximation theorem to $r\alpha$ with $Q=N$, obtaining a reduced $a/q$, $q\le N$. The quadratic Weyl inequality then impliesSet $\epsilon =N^{-1/8}$. For sufficiently large $N$, the $N\log (2N)$ term is absorbed on the left, giving $q\ll {\epsilon }^{-2}\log (2N)$. The integer $n=rq$ consequently satisfiesMoreover the approximating integer is $arq$, andThis contradicts the assumed avoidance. Increasing the absolute constant handles the finitely many smaller $N$, uniformly for every real $\alpha$. Thus