Irrational Number

An irrational number is a real number that cannot be written as $p/q$, where $p$ and $q$ are integers and $q\ne0$. Irrational numbers therefore constitute precisely the real numbers that are not rational. Their set is written $\mathbb{R}\setminus\mathbb{Q}$. Unlike rational numbers, their decimal expansions neither terminate nor eventually repeat. They represent exact quantities, not merely imprecise numerical values. (math.mit.edu)

Decimal representation

In decimal positional notation, a real number is rational if and only if its expansion terminates or eventually repeats a fixed block of digits. For example, $1/8=0.125$, while $1/6=0.1666\ldots$. An irrational expansion continues indefinitely without eventual periodicity. This does not mean that individual digits or finite blocks never recur; only permanent repetition of a fixed block is excluded. (math.mit.edu)

A repeating decimal can be converted into a fraction by shifting its repeating portion and subtracting. If $x=0.272727\ldots$, then $100x-x=27$, so $x=27/99$. Conversely, division of integers has only finitely many possible remainders, so it must terminate or eventually repeat. These arguments establish the decimal criterion; examining a finite sequence of digits alone cannot establish irrationality. (math.mit.edu)

The familiar ambiguity $0.999\ldots=1$ concerns rational numbers: a terminating expansion also has an alternative ending in repeating nines. It does not provide an irrational representation of a rational number. (math.mit.edu)

Examples and a fundamental proof

A central example is $\sqrt2$. In geometry, the Pythagorean theorem gives this as the diagonal length of a square whose side has length one. The diagonal and side are incommensurable: no common positive length measures both an integer number of times. Its decimal expansion begins $1.414213562\ldots$. (nrich.maths.org)

The standard proof uses contradiction. Suppose $\sqrt2=p/q$, with positive integers $p,q$ having no common factor. Squaring gives

Thus $p^2$, and hence $p$, is even. Writing $p=2k$ yields $q^2=2k^2$, so $q$ is also even. This contradicts the assumption that the fraction was reduced. Therefore $\sqrt2$ is irrational. (math.mit.edu)

More generally, the square root of a positive integer is irrational unless that integer is a perfect square. Another important example is π, the ratio of a circle’s circumference to its diameter. Its irrationality requires a different argument from the elementary proof for square roots. (nrich.maths.org)

Arithmetic properties

Irrationality is not preserved by every operation of arithmetic. Two irrational numbers can have a rational sum or product:

Consequently, the irrational numbers do not form a field under the usual operations. By contrast, both the rational numbers and the real numbers do. (im.kendallhunt.com)

Adding a rational number to an irrational number always produces an irrational number. Multiplication by a nonzero rational number also preserves irrationality. For example, if $rx$ were rational with $r\ne0$ rational, then $x=(rx)/r$ would be rational, contradicting its assumed irrationality. The restriction that the multiplier be nonzero is essential. (dpmms.cam.ac.uk)

Algebraic and transcendental irrationals

An algebraic number is a root of a nonzero polynomial with rational coefficients, equivalently with integer coefficients. The number $\sqrt2$ is algebraic because it satisfies $x^2-2=0$. An irrational real number that satisfies no such polynomial equation is called a transcendental number. Irrational numbers therefore divide into algebraic irrationals and transcendental reals. (math.mit.edu)

These classifications answer different questions. Irrationality excludes representation as a ratio of integers; transcendence excludes every polynomial relation of the specified kind. In particular, irrationality does not imply transcendence, as $\sqrt2$ demonstrates. (math.mit.edu)

Size and distribution

In set theory, the rational numbers form a countable set, whereas the real numbers are uncountable. Since a union of two countable sets remains countable, the irrational numbers must be uncountable. There is no sequence that lists every irrational real number. This distinction concerns the size of infinite sets rather than the numerical magnitude of their elements. (math.mit.edu)

Both rational and irrational numbers form dense subsets of the real line: every nonempty open interval contains numbers of both kinds. Density therefore does not imply countability, and neither kind occupies a separate continuous segment of the line. (math.mit.edu)

Under Lebesgue measure, the rationals have measure zero. Consequently, within any interval, the irrational numbers have the full measure of that interval. This is the precise measure-theoretic sense in which almost every real number is irrational, despite the rationals being dense. (ocw.mit.edu)

Approximation and foundations

An irrational number can be approximated arbitrarily closely by rational numbers without equaling any of them. Continued fractions provide particularly useful approximations; for $\sqrt2$, successive examples include $3/2$, $7/5$, $17/12$, and $41/29$. A finite approximation remains rational, while the limit can be irrational. (nrich.maths.org)

Rigorous nineteenth-century foundations incorporated irrational numbers into the real-number system. Richard Dedekind’s construction, published in 1872, represents real numbers through cuts in the ordered rational numbers. Another construction uses equivalence classes of rational Cauchy sequences. These approaches give arithmetic and ordering a precise foundation without requiring a pre-existing geometric interpretation of irrational lengths. (mathshistory.st-andrews.ac.uk)