Irrational number

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	Irrational number In mathematics, an irrational number is any real number that isn't a rational number, i.e., that cannot be written as a fraction a / b with a and b integers and b not zero. The irrational numbers are precisely those numbers whose decimal expansion never ends and never enters a periodic pattern. "Almost all" real numbers are irrational, in a sense which is defined more precisely below. There are two types of irrational numbers, the algebraic ones such as 2^1/2 (the square root of two) and 3^1/3 (the cube root of 3), and the transcendental numbers such as π and e.

Irrationality of certain logarithms

Perhaps the numbers most easily proved to be irrational are logarithms like log₂3. The argument by reductio ad absurdum is as follows:

Suppose log₂3 is rational. Then for some positive integers m and n, we have log₂3 = m/n.
Consequently 2^m/n = 3.
So 2^m = 3ⁿ.
But 2^m is even and 3ⁿ is odd, so that is impossible.

Irrationality of the square root of 2

The discovery of irrational number is usually attributed attributed to Pythagoras or one of his followers, who produced a (most likely geometrical) proof of the irrationality of the square root of 2.

One proof of this irrationality is the following reductio ad absurdum.

Assume that 2^1/2 is a rational number.
Then 2^1/2 can be written as an irreducible fraction a / b such that a and b are coprime integers and (a / b)² = 2.
It follows that a² / b² = 2 and a² = 2 b².
Therefore a² is even.
It follows that a must be even. (Odd numbers have odd squares.)
Therefore a² is divisible by 4.
So a² / 2 is even.
From (3) it follows that a² / 2 = b².
From (7) and (8) it follows that b² is even.
It follows that b must be even.
By (5) and (10) 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 2^1/2 is a rational number must be false.

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

Another reductio ad absurdum showing that √2 is irrational is less well-known and has sufficient charm that it is worth including here. It proceeds by observing that if √2=m/n then √2=(2n-m)/(m-n), so that 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 √2 is rational must be false. It is possible to construct from an isosceles right triangle whose leg and hypotenuse have respective lengths n and m, by a classic straightedge-and-compass construction, a smaller isosceles right triangle whose leg and hypotenuse have respective lengths m-n and 2n-m. That construction proves the irrationality of √2 by the kind of method that was employed by ancient Greek geometers.

Other irrational numbers

All transcendental numbers are irrational, and the article on transcendental numbers lists several examples. e^r is irrational if r ≠ 0 is rational; πⁿ is irrational for positive integers n.

Another way to construct irrational numbers is as zeros of polynomials: start with a polynomial equation

p(x) = a_n xⁿ + a_n-1 x^n-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 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 doesn't 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).

Numbers not known to be irrational

It isn't 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. It isn't known whether 2^e, π^e, π^√2 or the Euler-Mascheroni gamma constant γ are irrational.

The set of all irrational numbers

The set of all irrational numbers is uncountable (since the rationals are countable and the reals are uncountable). Using the absolute value to measure distances, the irrational numbers become a metric space which isn't 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.

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