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Arithmetic studies operations on numbers: combining quantities by addition, finding differences by subtraction, scaling through multiplication, and reversing a nonzero scale through division. Its usefulness goes beyond producing an answer. A calculation can express how resources are shared, how a measurement changes with scale, or why two different descriptions of a quantity agree.

The allowed numbers matter. Start with the natural numbers $0,1,2,\dots$. Counting collections gives $2+3=5$, but removing three from two calls for the integer $-1$. Sharing three units equally between two recipients gives the rational number $3/2$. These operations motivate larger number systems: integers supply additive inverses, and rational numbers supply division by nonzero integers. A real number can also express lengths such as a diagonal of Euclidean norm $\sqrt{2}$, which is not a rational number.

The beauty is that bookkeeping, geometry, and algebra can check one another. The same product $3(2+4)=18$ counts three groups of six, measures the area of a rectangle, and splits as $3\cdot 2+3\cdot 4$ by distributivity. A number line makes negative integers visible as positions; a fraction records an exact share even when its decimal representation does not terminate. An arithmetic rule becomes easier to remember when these interpretations fit together.

Form Champernowne constant by using concatenation to join the decimal representations of the positive integers $1,2,3,\dots$ after the decimal point. Champernowne proved in 1933 that this particular real number is a normal number in base ten. The construction is striking because its rule is completely explicit, while all finite words have the prescribed limiting digit block frequencies. ^[ref] Copeland and Erdős explicitly recall this result at the start of their own paper. ^[ref]

The first $100000$ digits of Champernowne constant have the following exact counts, in digit order $0,1,\dots ,9$:For the $99999$ adjacent pairs fully inside that prefix, the words $00,01,12,99$ occur $642,1753,2088,753$ times respectively. Thus, for example, $F_{100000}(12)=2088/99999$, which is about $0.02088$, rather than $0.01$. The prefix ends partway through the decimal representation of $22222$; the many leading ones in earlier positive integers help explain the visible imbalance.

These are finite counts, not a proof of the normal number theorem: a finite prefix can be followed by an all-zero tail. The theorem for Champernowne constant concerns the infinite limit, and establishes base-ten normality, rather than absolute normality. Nor do computations prove normality for familiar constants: the normality of Pi and Euler's number remains unproved, as discussed by Bailey and Borwein. ^[ref]

A prime number for which every cyclic rotation of its numerical digits is also a prime number. In decimal representation, $197$ is an example because $197,971,719$ are all prime numbers. A multi-digit circular prime can contain only $1,3,7,9$: any other digit can be rotated into the units position and force divisibility by $2$ or $5$.

One of the ten numerical digits $0,1,\dots ,9$ used in decimal representation.

A palindromic number in decimal representation, with no leading zeros.

For a reduced fraction $p/q$ with $0<p<q$, $q>1$, and $q,10$ coprime integers, its decimal representation repeats immediately after the decimal point. The shortest repeating block has lengthHere $\mathrm{ord}$ denotes multiplicative order, and $\phi$ denotes Euler's totient function. Changing the numerator preserves the decimal period as long as the fraction remains reduced.^[ref]

For example, long division of $1/7$ gives the remainders $1,3,2,6,4,5,1$, with emitted digits $1,4,2,8,5,7$. ConsequentlyEach remainder determines both the next digit and the next remainder; returning to one restarts the same six steps.

More generally, initialize long division with $r_0=p$. At each step, division with remainder givesso $r_j\equiv p10^j(\mathrm{mod}q)$. Since $p,q$ are coprime integers, the modular inverse of $p$ shows that $r_L=r_0$ exactly when $10^L\equiv 1(\mathrm{mod}q)$. To see why the digit block cannot repeat sooner than the remainders, a purely repeating decimal with a block of $L$ digits represents $D/(10^L-1)$ for an integer $D$. A reduced fraction $p/q$ with that value forces $q\mid 10^L-1$. Hence the shortest digit repeat and the multiplicative order coincide.

Knowing the decimal period predicts how many digits must be computed and stored: one block determines the whole repeating decimal. It also constructs the block directly. With $L={\mathrm{ord}}_q(10)$, the integeris the block, padded on the left with zeros to $L$ digits. For $1/11$, $L=2$ and $D=9$, so the block is $09$ and $1/11=0.\overline{09}$. Reduction matters: $7/21=1/3=0.\overline{3}$ has decimal period one, whereas ${\mathrm{ord}}_{21}(10)=6$.

If the reduced denominator contains prime factors two or five, ten has no multiplicative order modulo that whole denominator. WriteThe decimal preperiod has exactly $k_0$ digits. If $m>1$, the remaining repeating decimal has decimal period ${\mathrm{ord}}_m(10)$. For instance, $28=2^2\cdot 7$ gives $1/28=0.03\overline{571428}$: two prefix digits, then six repeating digits. If $m=1$, the decimal representation is a terminating decimal with exactly $k_0$ fractional places. We use the representation ending in zeros, rather than its alternative ending in nines; for example $1/40=0.025$ terminates after three places.

A short derivation explains both cases. After $k$ digits, multiplying the reduced fraction by $10^k$ and discarding its integer part leaves a reduced fraction whose denominator is $q/\gcd (q,10^k)$. This denominator first becomes coprime to ten at $k=k_0$. It then equals $m$, giving an immediate repeat when $m>1$, or an integer when $m=1$. The same long division argument works for positional notation in any base $b\ge 2$: for a reduced denominator $q>1$ coprime to $b$, the shortest repeating block has length ${\mathrm{ord}}_q(b)$.

The shortest prefix before the repeating tail of a decimal representation. For a reduced fraction with denominator $2^u{5}^{v}m$, where $m>1$ is coprime to ten, its length is $\max (u,v)$; the tail length is given by decimal period from multiplicative order.

An integer is a whole number, positive, negative, or zero. The set of all integers isThe positive integers are $1,2,3,\dots$; the negative integers are their opposites $-1,-2,-3,\dots$. Zero is neither positive nor negative. The nonnegative integers include zero and all the positive integers. On a number line, consecutive integers are one unit apart, negative integers lie to the left of zero, and positive integers lie to its right. ^[ref]

The integers let counting describe changes in both directions. If a counter starts at $-3$ and increases by five, it reaches $2$: $-3+5=2$. Decreasing that result by four gives $2+(-4)=-2$. In addition on the number line, the sign of the added integer chooses the direction and its absolute value gives the distance. The additive inverse $-a$ reverses the change made by $a$.

The integers have closure under a binary operation for addition, subtraction and multiplication: whenever $a,b\in \mathbb{Z}$, the results $a+b$, $a-b$ and $ab$ are still integers. Together these operations give the integers their ring structure.^[ref] For instance, $-3+5=2$, $-3-5=-8$, and $(-3)\cdot 5=-15$. General division has no such closure: $1/2$ is a rational number but not an integer. Some divisions are exact, such as $(-6)/3=-2$; division by zero is undefined. Euclidean division instead produces an integer quotient and a remainder, so $7=3\cdot 2+1$ retains the information that ordinary division puts into the fraction $7/3$.

An integer is the value, whereas its written numeral is an encoding. The decimal representation $205_{10}$ and the binary representation $11001101_2$ name the same integer. Changing positional notation changes how that value is displayed and how calculations are organized; it does not change the value itself.

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]

A nonnegative integer whose positional notation reads the same forwards and backwards, with no leading zeros. This property depends on the base: $585$ is a palindromic number in both decimal representation and binary representation, since $585=(1001001001{)}_2$.

A deterministic finite automaton is a tuple $(Q,\Sigma ,\delta ,q_0,F)$ with finite state set $Q$, finite input alphabet $\Sigma$, transition function $\delta :Q\times \Sigma \to Q$, initial state $q_0$, and accepting-state set $F\subseteq Q$. It accepts a word when the extended transition function of a deterministic finite automaton carries $q_0$ to a state in $F$. A regular language is a language accepted by some deterministic finite automaton.

The pumping lemma for regular languages says that if a regular language is accepted by an automaton with $N$ states, every accepted word $w$ of length at least $N$ has a decomposition $w=xyz$ with $|xy|\le N$, $|y|>0$, and $x{y}^{i}z$ in the language for every $i\ge 0$. Indeed, among the $N+1$ states visited before and after the first $N$ symbols, two are equal by the pigeonhole principle. The intervening nonempty word $y$ labels a loop, which may be traversed any number of times without changing the final accepting state.

In base two, the powers of two have representations $10^n$, so their language is the regular expression $10^{\ast }$ and is regular.

Suppose instead that their base-ten representations formed a regular language. Apply the pumping lemma to a sufficiently long decimal power of two $xyz$, and put $d=|y|>0$. Pumping gives decimal integers $a_i$ represented by $x{y}^{i}z$, all powers of two. A direct place-value calculation shows thatfor a constant $C$ independent of $i$. The lengths, and hence the exponents in $a_i=2^{n_i}$, tend to infinity. Dividing by $a_i$ givesFor large $i$, the right side has absolute value less than one while the left side is an integer, so it must vanish. This would make a power of two equal to $10^d=2^d{5}^d$, impossible for $d>0$ by unique prime factorization. Thus the decimal powers of two are not a regular language.

In positional notation, a numeral digit contributes its value multiplied by a weight determined by its position. Fix an integer radix $b\ge 2$. A finite string of numeral digits $d_n{d}_{n-1}\cdots d_0$, with $0\le d_j<b$, represents the nonnegative integerThe rightmost position has weight $b^0=1$; each move left multiplies the weight by $b$. This weight is the place value. For a nonzero integer, the canonical string has $d_n\ne 0$. Zero is written with the single numeral digit $0$. Leading zeros are permitted in a fixed-width display but omitted in the canonical form: $00205_{10}=205_{10}$. A negative integer is written with a separate minus sign and the positional notation for its absolute value; the minus sign is not a numeral digit.^[ref]

For example,The zero is essential inside this string: $25_{10}=25$, not $205$. It records that the tens position contributes nothing while keeping the hundreds and units in their proper positions. The same integer has binary representation

Repeated Euclidean division constructs the representation exactly. Set $N_0=N$. At each step writeKeep the remainder $d_j$ and continue with the integer quotient $N_{j+1}$. A positive quotient strictly decreases, so eventually it is zero. Substituting the successive equations back into the first gives $N=d_0+b{d}_1+b^2{d}_2+\cdots$: the numeral digits are obtained from right to left. The remainder modulo $b$ uniquely fixes $d_0$, and then the quotient uniquely fixes the remaining string by the same reasoning. This proves existence and uniqueness once leading zeros are excluded. The needed unique quotient and remainder are precisely the conclusion of the Euclidean division theorem.^[ref]

This finite expansion concerns nonnegative integers. Extending positional notation to fractional positions uses weights $b^{-1},b^{-2},\dots$ to the right of a point; an arbitrary real number need not have a finite digit string. In particular, the decimal representation $0.1$ represents $1/10$, while the base-two string $0.1_2$ represents $1/2$. The radix is part of the interpretation.

The benefit is compact arithmetic: $k$ positions can encode every nonnegative integer below $b^k$. To see the upper endpoint, fill all positions with $b-1$; the finite geometric series gives $(b-1)(1+b+\cdots +b^{k-1})=b^k-1$. Carrying in addition exchanges $b$ units of one place value for one unit of the next. Thus a short string stores a large count and also supplies an algorithm for computing with it.

A real number is simply normal in base $K$ if each individual digit in its positional notation has limiting digit block frequency $1/K$. This is the length-one part of the definition of a normal number. It places no equal-frequency requirement on longer words.

For example, $x=0.010101{\dots }_2=1/3$ is a simply normal number in base two: each binary digit has frequency $1/2$. But $00$ and $11$ never occur, and $01$ and $10$ each have frequency $1/2$. Thus $x$ is not a normal number in base two. Even a sequence that looks more complicated, such as the deterministic Thue–Morse sequence, must be tested against the actual digit block frequency definition; visual complexity is no substitute for it.

The decimal representation $y=0.01234567890123456789\dots$ makes the same point in base ten. Every digit has frequency $1/10$, yet the only adjacent pairs areEach has limiting digit block frequency $1/10$; the other ninety pairs have frequency zero. A normal number in base ten requires $1/100$ for every one of the hundred pairs. This rational number is a simply normal number in base ten, but it is not a normal number in that base.

A decimal representation with only finitely many nonzero fractional digits. A reduced fraction has this property exactly when its denominator has no prime factors other than two and five. The zero tail is distinguished from the genuine repeating block measured by decimal period from multiplicative order.