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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 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.