For log(2) : Sketch of an Irrationality Proof of the constant 2F1([[1, 1], [2], -1]) By Shalosh B. Ekhad Theorem: The constant of the title 2F1([[1, 1], [2], -1]) let's call it C, that is easily seen to be given by the following Alladi-Rob\ inson-type integral 1 / | 1 C = | ----- dx | 1 + x / 0 1/2 ln(3 + 2 2 ) + 1 is irrational, with an irrationality measure, 1 + ------------------- 1/2 -ln(3 - 2 2 ) - 1 that in decimals equals, 4.622100852 We need two lemmas Lemma ONE: , let A(n), B(n), be two sequences of rational numbers that satis\ fy the second-order recurrence (1 + n) X(n) 3 (2 n + 3) X(1 + n) ------------ - -------------------- + X(n + 2) = 0 n + 2 n + 2 Subject to the initial conditions A(0) = 0, A(1) = 2 B(0) = 1, B(1) = 3 A(n) Then, ----, approximates the constant of the title, 2F1([[1, 1], [2], -1]) B(n) let's call it C, that can be easily seen to be given by the Alladi-Robinson\ -type integral 1 / | 1 C = | ----- dx | 1 + x / 0 / 1 \n with an error that is OMEGA of, |-------------------------| , | 1/2 1/2 | \(3 + 2 2 ) (3 - 2 2 )/ that in floating point is, 0.0294372515228594143797353094836230571639374\ 954766231218798431441112102584547155337953495880683011271798338452290524\ 357009019673297911534482705420348007700230413280088699328869304005939861\ n 66615273142348 Proof, consider the Alladi-Robinson type-integral 1 /x (1 - x)\n / |---------| | \ 1 + x / F(n) = | ------------ dx | 1 + x / 0 Then , F(0) = B(0) - A(0), F(1) = B(1) - A(1) and F(n) also satisfies the above recurrence, thanks to the amazing multivar\ iable Almkvist-Zeilberger algorithm Hence, F(n) = B(n) - A(n) 1 By a simple bound of the integrand, F(n) is OMEGA of, -------------, / 1 \n |----------| | 1/2| \3 - 2 2 / and by the Poincare lemma, B(n) (and for that matter, A(n)) are OMEGA of, 1/2 n (3 + 2 2 ) 1 Dividing by B(n) gives that A(n)/B(n)-c is OMEGA of , -------------, QED. / 1/2\n |3 + 2 2 | |----------| | 1/2| \3 - 2 2 / we now claim that the sequence of RATIONAL numbers A(n),B(n), can be multipl\ ied by another sequence of rational numbers E(n) such that both A(n)E(n) and B(n)E(n) are integers Lemma TWO: Let E(n) be LCM(n) where LCM(1..m) is the least-common-multiple of the first m integers, and Then there exists a sequence G(n) of integers that is asymptotically negligi\ ble, i.e. the limit of log( G(n) )/n tends to 0 as n goes to infinity such that A1(n):=E(n)G(n)A(n), B1(n):=E(n)G(n)B(n) are BOTH integers For example, G(n) for n=, 2000, equals 1 whose log divided by n is 0. Indeed peanuts. Using standard asymptotics derived from the Prime Number Theorem it is readi\ ly seen that the limit of log(E(n))/n , and hence log(E(n)*G(n))/n, equals exactly, 1 that equals in decimals, 1. Just as a sanity check, the empircal numerical values of nu for E(n) from, 1980, to , 2001, are [0.9793251333, 0.9846469186, 0.9783256809, 0.9827790946, 0.9831580372, 0.9779669269, 0.9803232548, 0.9831346649, 0.9799322279, 0.9821453614, 0.9758077865, 0.9816063360, 0.9801360387, 0.9804189977, 0.9812363480, 0.9847484210, 0.9808837675, 0.9885073617, 0.9773804410, 0.9879983627, 0.9896853505] Multiplying F(n) by E(n)*G(n) we get E(n) F(n) G(n) = B1(n) - A1(n) and this implies that | A1(n) | CONSTANT | -1 + ----- | <= ---------------- | B1(n) | (1 + delta) B1(n) 1/2 -ln(3 - 2 2 ) - 1 where , delta = ------------------- 1/2 ln(3 + 2 2 ) + 1 that in decimals is, , 0.2760828704 As you can see, delta is positive We leave it to the reader to fill-in the details. ---------------------------------- For GLC(1/3,1/3,1/5) Sketch of an Irrationality Proof of the constant 2F1([[1, 4/3], [8/3], -1/5]) By Shalosh B. Ekhad Theorem: The constant of the title 2F1([[1, 4/3], [8/3], -1/5]) let's call it C, that is easily seen to be given by the following Alladi-Rob\ inson-type integral 1 / 1/3 1/3 1 | x (1 - x) C = -------------- | --------------- dx Beta(4/3, 4/3) | 1 + x/5 / 0 is irrational, with an irrationality measure, 1/2 1/2 1/2 Pi 3 ln(55 + 10 6 5 ) + 3/4 ln(3) + 3/2 - ------- - ln(5) 12 1 + ---------------------------------------------------------- 1/2 1/2 1/2 Pi 3 -ln(55 - 10 6 5 ) - 3/4 ln(3) - 3/2 + ------- + ln(5) 12 that in decimals equals, 5.070283242 We need two lemmas Lemma ONE: , let A(n), B(n), be two sequences of rational numbers that satis\ fy the second-order recurrence (3 n + 7) (3 n + 4) X(n) (3 n + 7) (6 n + 11) X(n + 1) 25/3 ------------------------ - 55/3 ----------------------------- + X(n + 2) (8 + 3 n) (n + 2) (8 + 3 n) (n + 2) = 0 Subject to the initial conditions A(0) = 0, A(1) = 200/3 B(0) = 1, B(1) = 220/3 A(n) Then, ----, approximates the constant of the title, B(n) 2F1([[1, 4/3], [8/3], -1/5]) let's call it C, that can be easily seen to be given by the Alladi-Robinson\ -type integral 1 / 1/3 1/3 1 | x (1 - x) C = -------------- | --------------- dx Beta(4/3, 4/3) | 1 + x/5 / 0 / 1 \n with an error that is OMEGA of, |---------------------------------------| , | 1/2 1/2 1/2 1/2 | \(55 + 10 6 5 ) (55 - 10 6 5 )/ that in floating point is, 0.0020746977269100789332955676470610607923342\ 008873681401664421177029580660019971276222880878969246635201965851453874\ 126083165245163551408775264885415246857771097796382724489673712643397202\ n 602974725532500 Proof, consider the Alladi-Robinson type-integral 1 1/3 1/3 /x (1 - x)\n / x (1 - x) |---------| 1 | \ 1 + x/5 / F(n) = -------------- | ---------------------------- dx Beta(4/3, 4/3) | 1 + x/5 / 0 Then , F(0) = 1/5 B(0) - A(0), F(1) = 1/5 B(1) - A(1) and F(n) also satisfies the above recurrence, thanks to the amazing multivar\ iable Almkvist-Zeilberger algorithm Hence, F(n) = 1/5 B(n) - A(n) 1 By a simple bound of the integrand, F(n) is OMEGA of, --------------------, / 1 \n |-----------------| | 1/2 1/2| \55 - 10 6 5 / and by the Poincare lemma, B(n) (and for that matter, A(n)) are OMEGA of, 1/2 1/2 n (55 + 10 6 5 ) 1 Dividing by B(n) gives that A(n)/B(n)-c is OMEGA of , --------------------, / 1/2 1/2\n |55 + 10 6 5 | |-----------------| | 1/2 1/2| \55 - 10 6 5 / QED. we now claim that the sequence of RATIONAL numbers A(n),B(n), can be multipl\ ied by another sequence of rational numbers E(n) such that both A(n)E(n) and B(n)E(n) are integers Lemma TWO: Let E(n) be /3 n\ |---| \ 2 / 3 PP(0, 2/3, 0, 3, {2}, 3, n) ---------------------------------- n 5 PP(1/3, 2/3, 0, 3, {1}, 3, n) where for 0