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In 1840, Liouville published a proof of the fact that e 2 is irrational [10] followed by a proof that e 2 is not a root of a second-degree polynomial with rational coefficients. [11] This last fact implies that e 4 is irrational. His proofs are similar to Fourier's proof of the irrationality of e.
In 1948, Erdős showed that the constant E is an irrational number. [3] Later, Borwein provided an alternative proof. [4] Despite its irrationality, the binary representation of the Erdős–Borwein constant may be calculated efficiently. [5] [6]
In other words, the n th digit of this number is 1 only if n is one of 1! = 1, 2! = 2, 3! = 6, 4! = 24, etc. Liouville showed that this number belongs to a class of transcendental numbers that can be more closely approximated by rational numbers than can any irrational algebraic number, and this class of numbers is called the Liouville numbers ...
Since e is an irrational number (see proof that e is irrational), it cannot be represented as the quotient of two integers, but it can be represented as a continued fraction. Using calculus, e may also be represented as an infinite series, infinite product, or other types of limit of a sequence.
Here is a proof by contradiction that log 2 3 is irrational (log 2 3 ≈ 1.58 > 0). Assume log 2 3 is rational. For some positive integers m and n , we have
Copeland and Erdős's proof that their constant is normal relies only on the fact that is strictly increasing and = + (), where is the n th prime number. More generally, if is any strictly increasing sequence of natural numbers such that = + and is any natural number greater than or equal to 2, then the constant obtained by concatenating "0."
There’s a completely rational reason that The Irrational won’t be on tonight: There are no more episodes ready to air. Last week’s hour of the NBC procedural, titled “The Real Deal ...
This same constant can be used to prove that "an irrational elevated to an irrational power may be rational", even without first proving its transcendence. The proof proceeds as follows: either 2 2 {\displaystyle {\sqrt {2}}^{\sqrt {2}}} is a rational which proves the theorem, or it is irrational (as it turns out to be) and then