Main Page | See live article | Alphabetical index In mathematics, an irrational number is any real number that is not a rational number, i.e., one 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. Some irrational numbers are algebraic numbers such as 21/2 (the square root of two) and 31/3 (the cube root of 3); others are transcendental numbers such as &pi and e. Table of contents 1 Irrationality of certain logarithms 2 Irrationality of the square root of 2 3 A different proof 4 Other irrational numbers 5 Irrational numbers and decimal expansions 6 Numbers not known to be irrational 7 The set of all irrational numbers 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 2m/n = 3. So 2m = 3n. But 2m is even (because at least one of its prime factors is two) and 3n is odd (because none of its prime factors is two (they're all three)) 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. The proposition is proved by assuming the opposite and showing that that is false, which in mathematics means that the proposition must be true. Assume that √2 is a rational number. Meaning that there exists an integer a and b so that a / b = √2. Then √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. It follows that a2 / b2 = 2 and a2 = 2 b2. Therefore a2 is even because it is equal to 2 b2 which is obviously even. It follows that a must be even. (Odd numbers have odd squares and even numbers have even squares.) Because a is even, there exists a k that fullfills: a = 2k. We insert the last equation of (3) in (6): 2b2 = (2k)2 is equivalent to 2b2 = 4k2 is equivalent to b2 = 2k2. Because 2k2 is even it follows that b2 is also even which means that b is even because only even numbers have even squares. 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 √2 is a rational number must be false. The opposite is proven. √2 is irrational. This proof can be generalized to show that any root of any natural number is either a natural number or irrational. A different proof 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. er 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) = an xn + an-1 xn−1 + ... + a1 x + a0 = 0 where the coefficients ai 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 an; 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 = (21/2 + 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). Irrational numbers and decimal expansions It is often erroneously assumed that mathematicians define "irrational number" in terms of decimal expansions, calling a number irrational if its decimal expansion neither repeats nor terminates. No mathematician takes that to be the definition, since the choice of base 10 would be arbitrary and since the standard definition is simpler and more well-motivated. Nonetheless it is true that a number is of the form n/m where n and m are integers, if and only if its decimal expansion repeats or terminates. When the long division algorithm that everyone learns in grammar school 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 repeating decimal, for example: Since the length of the repitend is 3, multiply by 103: and then subtract A from both sides: Then (The "135" above can be found quickly via the Euclid's algorithm.) Numbers not known to be irrational 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. It is not known whether 2e, πe, 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 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. This article is from Wikipedia. All text is available under the terms of the GNU Free Documentation License.