Irrational number

In mathematics, an irrational number is any real number that is not a rational number, i.e., it is a number not of the form n/m, where n and m are integers. Almost all real numbers are irrational, in a sense which is defined more precisely below.

When the ratio of lengths of two line segments is irrational, the line segments are also described as being incommensurable, meaning they share no measure in common. A measure of a line segment I in this sense is a line segment J that "measures" I in the sense that some whole number of copies of J laid end-to-end occupy the same length as I.

History

The earliest known use of irrational numbers was in the Indian Sulba Sutras composed between 800-500 BC. It was known that the diagonal and side of a square are incommensurable with each other.

The first proof of the existence of irrational numbers is usually attributed to Hippasus of Metapontum, a Pythagorean who probably discovered them while identifying sides of the pentagram. However Pythagoras believed in the absoluteness of numbers, and could not accept the existence of irrational numbers. He could not disprove their existence through logic, but his beliefs would not accept the existence of irrational numbers and so, as legend had it, he had Hippasus drowned. Theodorus of Cyrene proved the irrationality of the surds of whole numbers up to 17, but stopped there probably because the algebra he used couldn't be applied to the square root of 17. It wasn't until Eudoxus developed a theory of irrational ratios that a strong mathematical foundation of irrational numbers was created. Euclid's Elements Book 10 is dedicated to classification of irrational magnitudes.

The sixteenth century saw the acceptance of negative, integral and fractional numbers. The seventeenth century saw decimal fractions with the modern notation quite generally used by mathematicians. The next hundred years saw imaginary numbers become a powerful tool in the hands of Abraham de Moivre, and especially of Leonhard Euler. For the nineteenth century it remained to complete the theory of complex numbers, to separate irrationals into algebraic and transcendental, to prove the existence of transcendental numbers, and to make a scientific study of a subject which had remained almost dormant since Euclid, the theory of irrationals. The year 1872 saw the publication of the theories of Karl Weierstrass (by his pupil Kossak), Heine (Crelle, 74), Georg Cantor (Annalen, 5), and Richard Dedekind. Méray had taken in 1869 the same point of departure as Heine, but the theory is generally referred to the year 1872. Weierstrass's method has been completely set forth by Pincherle (1880), and Dedekind's has received additional prominence through the author's later work (1888) and the recent endorsement by Paul Tannery (1894). Weierstrass, Cantor, and Heine base their theories on infinite series, while Dedekind founds his on the idea of a cut (Schnitt) in the system of real numbers, separating all rational numbers into two groups having certain characteristic properties. The subject has received later contributions at the hands of Weierstrass, Kronecker (Crelle, 101), and Méray.

Continued fractions, closely related to irrational numbers (and due to Cataldi, 1613), received attention at the hands of Euler, and at the opening of the nineteenth century were brought into prominence through the writings of Lagrange. Dirichlet also added to the general theory, as have numerous contributors to the applications of the subject.

Lambert proved (1761) that π cannot be rational, and that eⁿ is irrational if n is rational (unless n = 0). While Lambert's proof is often said to be incomplete, modern assessments support it as satisfactory, and in fact for its time unusually rigorous. Legendre (1794), after introducing the Bessel-Clifford function, provided a proof to show that π² is irrational, whence it follows immediately that π ; is irrational also. The existence of transcendental numbers was first established by Liouville (1844, 1851). Later, Georg Cantor (1873) proved their existence by a different method, that showed that every interval in the reals contains transcendental numbers. Charles Hermite (1873) first proved $e$ transcendental, and Ferdinand von Lindemann (1882), starting from Hermite's conclusions, showed the same for π. Lindemann's proof was much simplified by Weierstrass (1885), still further by David Hilbert (1893), and has finally been made elementary by Adolf Hurwitz and Paul Albert Gordan.

Example proofs

The square root of 2

One proof of the irrationality of the square root of 2 is the following reductio ad absurdum. The proposition is proved by assuming the contrary and showing that doing so leads to a contradiction (hence the proposition must be true).

1. Assume that ${\sqrt {2}}$ is a rational number. This would mean that there exist integers a and b such that a / b = ${\sqrt {2}}$.
2. Then ${\sqrt {2}}$ can be written as an irreducible fraction (the fraction is shortened as much as possible) a / b such that a and b are coprime integers and (a / b)² = 2.
3. It follows that a² / b² = 2 and a² = 2 b².
4. Therefore a² is even because it is equal to 2 b² which is also even.
5. It follows that a must be even (odd square numbers have odd square roots and even square numbers have even square roots).
6. Because a is even, there exists an integer k that fulfills: a = 2k.
7. We insert the last equation of (3) in (6): (2k)² = 2b² is equivalent to 4k² = 2b² is equivalent to 2k² = b².
8. Because 2k² is even it follows that b² is also even which means that b is even because only even numbers have even squares.
9. By (5) and (8) a and b are both even, which contradicts that a / b is irreducible as stated in (2).

Since we have found a contradiction, the assumption (1) that ${\sqrt {2}}$ is a rational number must be false; that is to say, ${\sqrt {2}}$ is irrational.

This proof can be generalized to show that any root of any natural number is either a natural number or irrational.

Another proof

Another reductio ad absurdum argument showing that ${\sqrt {2}}$ is irrational is less well-known:

• Assume that ${\sqrt {2}}$ is a rational number. This would mean that there exist integers m and n such that $m/n={\sqrt {2}}$.
• Then ${\sqrt {2}}={\frac {{\sqrt {2}}\cdot n({\sqrt {2}}-1)}{n({\sqrt {2}}-1)}}={\frac {2n-{\sqrt {2}}n}{{\sqrt {2}}n-n}}={\frac {2n-m}{m-n}}$.
• Since ${\sqrt {2}}>1$, it follows that $m>n\,$, and it can be shown that $m>2n-m\,$.

So a fraction in lowest terms is reduced to yet lower terms. That is a contradiction if n and m are positive integers, so the assumption that ${\sqrt {2}}$ is rational must be false.

Similarly, assume an isosceles right triangle whose leg and hypotenuse have respective integer lengths n and m. By the Pythagorean theorem, the ratio m/n equals ${\sqrt {2}}$. It is possible to construct by a classic compass and straightedge construction a smaller isosceles right triangle whose leg and hypotenuse have respective lengths m − n and 2n − m. That construction proves the irrationality of ${\sqrt {2}}$ by the kind of method that was employed by ancient Greek geometers.

The square root of 10

If √10 is a rational, say m/n, then m² = 10n². However, in decimal notation, every square ends in an even number of zeros. So then m² and 10n² in decimal must end in respectively an even and odd number of zeros, a contradiction.

More generally, in any radix r that is not itself a square, every square ends in an even numbers of zeros, whence √10_r in radix r is irrational, that is, √r is irrational. It follows that the only integers with rational square roots are squares. As a case in point, 2 is not a square, and 2 in binary is 10₂. (Note the convention of subscripting nondecimal numerals with their radix, to avoid ambiguity. As part of that convention the subscripts are understood to be in decimal, not being subscripted themselves.)

The golden ratio

When a line segment is divided into two disjoint subsegments in such a way that the ratio of the whole to the longer part equals the ratio of the longer part to the shorter part, then that ratio is the golden ratio, equal to

$\varphi ={1+{\sqrt {5}} \over 2}.$

Assume this is a rational number n/m in lowest terms. Take n to be the length of the whole and m the length of the longer part. Then n > m, and the length of the shorter part is n − m. Then we have

${n \over m}={\mathrm {whole} \over \mathrm {longer} \ \mathrm {part} }={\mathrm {longer} \ \mathrm {part} \over \mathrm {shorter} \ \mathrm {part} }={m \over n-m}.$

However, this puts a fraction already in lowest terms into lower terms—a contradiction. Therefore the initial assumption that φ is rational is false.

Logarithms

Perhaps the numbers most easily proved to be irrational are certain logarithms. Here is a proof by reductio ad absurdum that log₂3 is irrational:

• Assume log₂3 is rational. For some positive integers m and n, we have log₂3 = m/n.
• It follows that 2^m/n = 3.
• Raise each side to the n power, find 2^m = 3ⁿ.
• But 2 to any integer power greater than 0 is even (because at least one of its prime factors is 2) and 3 to any integer power greater than 0 is odd (because none of its prime factors is 2), so the original assumption is false.

Cases such as log₁₀2 can be treated similarly.

Transcendental and algebraic irrationals

Almost all irrational numbers are transcendental and all transcendental numbers are irrational: the article on transcendental numbers lists several examples. e^r and π^r are irrational if r ≠ 0 is rational; e^π is also irrational.

Another way to construct irrational numbers is as irrational algebraic numbers, i.e. as zeros of polynomials with integer coefficients: start with a polynomial equation

p(x) = a_n xⁿ + a_n-1 xⁿ⁻¹ + ... + a₁ x + a₀ = 0

where the coefficients a_i are integers. Suppose you know that there exists some real number x with p(x) = 0 (for instance if n is odd and a_n is non-zero, then because of the intermediate value theorem). The only possible rational roots of this polynomial equation are of the form r/s where r is a divisor of a₀ and s is a divisor of a_n; there are only finitely many such candidates which you can all check by hand. If neither of them is a root of p, then x must be irrational. For example, this technique can be used to show that x = (2^1/2 + 1)^1/3 is irrational: we have (x³ − 1)² = 2 and hence x⁶ − 2x³ − 1 = 0, and this latter polynomial does not have any rational roots (the only candidates to check are ±1).

Because the algebraic numbers form a field, many irrational numbers can be constructed by combining transcendental and algebraic numbers. For example 3π+2, π + √2 and e√3 are irrational (and even transcendental).

Decimal expansions

The decimal expansion of an irrational number never repeats or terminates, unlike a rational number.

To show this, suppose we divide integers n by m (where m is nonzero). When long division is applied to the division of n by m, only m remainders are possible. If 0 appears as a remainder, the decimal expansion terminates. If 0 never occurs, then the algorithm can run at most m − 1 steps without using any remainder more than once. After that, a remainder must recur, and then the decimal expansion repeats!

Conversely, suppose we are faced with a recurring decimal, we can prove that it is a fraction of two integers. For example:

$A=0.7\,162\,162\,162\,\dots$

Here the length of the repitend is 3. We multiply by 10³:

$1000A=7\,16.2\,162\,162\,\dots$

Note that since we multiplied by 10 to the power of the length of the repeating part, we shifted the digits to the left of the decimal point by that exactly many positions. Therefore, the tail end of 1000A matches the tail end of A exactly. Here, both 1000A and A have repeating 162 at the end.

Therefore, when we subtract A from both sides, the tail end of 1000A cancels out of the tail end of A:

$999A=715.5\,.$

Then

$A={\frac {715.5}{999}}={\frac {7155}{9990}}={\frac {135\times 53}{135\times 74}}={\frac {53}{74}},$

which is a quotient of integers and therefore a rational number.

Open questions

It is not known whether π + e or π − e are irrational or not. In fact, there is no pair of non-zero integers m and n for which it is known whether mπ + ne is irrational or not. Moreover, It is not known whether the set {π, e} is algebraically independent over Q.

It is not known whether 2^e, π^e, π^√2, Catalan's constant, or the Euler-Mascheroni gamma constant γ are irrational.

The set of all irrationals

The set of all irrational numbers is uncountable (since the rationals are countable and the reals are uncountable). The set of algebraic irrationals, that is, the non-transcendental irrationals, is countable. Using the absolute value to measure distances, the irrational numbers become a metric space which is not complete. However, this metric space is homeomorphic to the complete metric space of all sequences of positive integers; the homeomorphism is given by the infinite continued fraction expansion. This shows that the Baire category theorem applies to the space of irrational numbers. Whereas the set of all reals with its usual topology is connected, this Baire space, topologized in the same way as the reals, namely with the order topology, is totally disconnected: there is no path from any irrational to any other along the irrational line.

If removing the rationals from the continuum (the reals) totally disconnects the space, one might imagine that having two copies of every rational, ordered so that one is less than the other, would connect it even better than with one copy. But two copies makes the continuum just as totally disconnected as no copies, though not homeomorphic to Baire space but instead to Cantor space (provided we also include as endpoints ±∞). The nature of the total disconnection in both cases is that at every rational, both Baire space and Cantor space partition as the disjoint union of two clopen sets, one on each side of the selected rational. The difference is that whereas the clopen sets of Baire space have no least or greatest element, the selected rational being missing, those of Cantor space have both a least and greatest element, the selected rational showing up in both intervals. The reason both intervals are clopen is that for Baire space both are obviously open but the complement of an open set is closed, so both are closed; for Cantor space both are obviously closed but again the complement of a closed set is open. In contrast, when we partition the continuum at any rational as a disjoint union of two intervals, the selected rational itself must belong to one interval or the other and so one interval is open at that point while the other is closed. The open interval thereby created is not closed, and its complement is not open, the essential difference between the continuum and either Baire space or Cantor space.

See also

• Dedekind cut
• Proof that e is irrational

External links

• Weisstein, Eric W. "Irrational Number". MathWorld.
• Square root of 2 is irrational

References

• Adrien-Marie Legendre, Éléments de Géometrie, Note IV, (1802), Paris
• Rolf Wallisser, "On Lambert's proof of the irrationality of π", in Algebraic Number Theory and Diophantine Analysis, Franz Halter-Koch and Robert F. Tichy, (2000), Walter de Gruyer