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Written in 1873, this proof uses the characterization of as the smallest positive number whose half is a zero of the cosine function and it actually proves that is irrational. [ 3 ] [ 4 ] As in many proofs of irrationality, it is a proof by contradiction .
Perhaps the numbers most easy to prove irrational are certain logarithms. Here is a proof by contradiction that log 2 3 is irrational (log 2 3 ≈ 1.58 > 0). Assume log 2 3 is rational.
A more recent proof by Wadim Zudilin is more reminiscent of Apéry's original proof, [6] and also has similarities to a fourth proof by Yuri Nesterenko. [7] These later proofs again derive a contradiction from the assumption that ζ ( 3 ) {\displaystyle \zeta (3)} is rational by constructing sequences that tend to zero but are bounded below by ...
For example, the square root of 2 is an irrational number, but it is not a transcendental number as it is a root of the polynomial equation x 2 − 2 = 0. The golden ratio (denoted or ) is another irrational number that is not transcendental, as it is a root of the polynomial equation x 2 − x − 1 = 0.
Rational numbers have irrationality exponent 1, while (as a consequence of Dirichlet's approximation theorem) every irrational number has irrationality exponent at least 2. On the other hand, an application of Borel-Cantelli lemma shows that almost all numbers, including all algebraic irrational numbers , have an irrationality exponent exactly ...
In the 18th and 19th centuries, there was much work on irrational and transcendental numbers. Lambert (1761) gave a flawed proof that π cannot be rational; Legendre (1794) completed the proof [11] and showed that π is not the square root of a rational number. [12]
Transcendental numbers therefore represent the typical case; even so, it may be extremely difficult to prove that a given number is transcendental (or even simply irrational). For this reason transcendence theory often works towards a more quantitative approach.
On the other hand, Euler proved that irrational numbers require an infinite sequence to express them as continued fractions. [1] Moreover, this sequence is eventually periodic (again, so that there are natural numbers N and p such that for every n ≥ N we have a n + p = a n ), if and only if x is a quadratic irrational .