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A more general proof shows that the mth root of an integer N is irrational, unless N is the mth power of an integer n. [7] That is, it is impossible to express the mth root of an integer N as the ratio a ⁄ b of two integers a and b, that share no common prime factor, except in cases in which b = 1.
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
The extraction of decimal-fraction approximations to square roots by various methods has used the square root of 7 as an example or exercise in textbooks, for hundreds of years. Different numbers of digits after the decimal point are shown: 5 in 1773 [4] and 1852, [5] 3 in 1835, [6] 6 in 1808, [7] and 7 in 1797. [8]
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 ...
The fact that any rational number has a unique representation as an irreducible fraction is utilized in various proofs of the irrationality of the square root of 2 and of other irrational numbers. For example, one proof notes that if could be represented as a ratio of integers, then it would have in particular the fully reduced representation ...
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.
Likewise, tan 3 π / 16 , tan 7 π / 16 , tan 11 π / 16 , and tan 15 π / 16 satisfy the irreducible polynomial x 4 − 4x 3 − 6x 2 + 4x + 1 = 0, and so are conjugate algebraic integers. This is the equivalent of angles which, when measured in degrees, have rational numbers. [2] Some but not all irrational ...
The condition that ξ is irrational cannot be omitted. Moreover the constant 5 {\displaystyle {\sqrt {5}}} is the best possible; if we replace 5 {\displaystyle {\sqrt {5}}} by any number A > 5 {\displaystyle A>{\sqrt {5}}} and we let ξ = ( 1 + 5 ) / 2 {\displaystyle \xi =(1+{\sqrt {5}})/2} (the golden ratio ) then there exist only finitely ...