Irrational number

In mathematics, an irrational number is any real number that is not a rational number, i.e., one that cannot be written as a ratio of two integers, i.e., it is not of the form

{m over n-m}.

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

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Transcendental and algebraic irrationals

Almost all irrational numbers are transcendental and all transcendental numbers are irrational: the article on transcendental numbers lists several examples. er and πr are irrational if r ≠ 0 is rational; eπ is also irrational.

Related Topics:
Almost all - Transcendental number

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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

Related Topics:
Algebraic number - Polynomial

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

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where the coefficients ai are integers. Suppose you know that there exists some real number x with p(x) = 0 (for instance if n is odd and an 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 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 does not have any rational roots (the only candidates to check are ±1).

Related Topics:
Intermediate value theorem - Divisor

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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).

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The simplest irrationality proofs: certain 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:

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

Similar cases such as log102 can be treated similarly.

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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 recurring decimal, for example:

Related Topics:
Decimal - Long division - Recurring decimal

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:A=0.7,162,162,162,dots

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Since the length of the repitend is 3, multiply by 103:

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:1000A=7,16.2,162,162,dots

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and then subtract A from both sides:

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:999A=715.5,.

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Then

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:A=rac{715.5}{999}=rac{7155}{9990}=rac{135 imes 53}{135 imes 74}=rac{53}{74}.

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(The "135" above can be found quickly via Euclid's algorithm.)

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Numbers not known to be irrational

It is not known whether π + e and 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.

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It is not known whether 2e, πe, pi^sqrt{2} or the Euler-Mascheroni gamma constant γ are irrational.

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The set of all irrational numbers

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.

Related Topics:
Uncountable - Countable - Absolute value - Metric space - Complete - Homeomorphic - Sequence - Continued fraction - Baire category theorem

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Another irrational number

The Copeland-Erdős constant

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: 0.235711131719232931374143...

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obtained by concatenating the prime numbers is known to be irrational; in fact it is a normal number.

Related Topics:
Concatenating - Prime number - Normal number

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