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factor of 2. Roth's result is the best possible exponent. Charles Matthews 16:02, 2 Nov 2004 (UTC) My guess is that the Liouville's theorem mentioned in...

authors" that strengthened Roth's bound on the size of progression-free sets? It's spelled out in the linked main article, Roth's theorem on arithmetic progressions...

authors" that strengthened Roth's bound on the size of progression-free sets? It's spelled out in the linked main article, Roth's theorem on arithmetic progressions...

of Roth's theorem would be critical again to the understanding of what exponent is being referred to above. We don't yet have the Thue-Siegel-Roth theorem...

section on Mann's theorem refers to Theorem 1 and Theorem 1.1, but there are no numbered theorems in this article. Which theorems are they? Ntsimp 21:30...

mathematics, on top of head at elast the following ought to be inculded- : Roth's theorem on arithmetic progressions (after all there is Tao-Green is mentioned)...

on approximation of algebraic numbers is the best possible. In fact, Roth's theorem gives a much tighter bound, which leaves plenty of room for finding...

index theorem does not take on a special form for complex manifolds. If they are referring to the generalized Hirzebruch−Riemann−Roch theorem on complex...

anybody could help! Thanks in advance! Worth a mention, surely? Thue Siegel Roth (talk) 14:27, 8 February 2012 (UTC) 12 years later, I have added some information...

Waldschmidt (esp. in diophantine approximation, e.g. Thue-Siegel-Roth theorem, Gelfond-Schneider theorem, etc). I specifically wanted to know whether i was considered...

nanotubes. All the experiments were carried out in Roth's group by Meyer. My statement that Roth's group could lay claim to being the first to suspend...

algebraic number could be approximated by rationals" is the Thue–Siegel–Roth theorem and its proof proceeds by assuming the existence of an irrational algebraic...

non-zero measure. All numbers x with ω(x,1) > 1 are transcendental by Roth's theorem. Scott Tillinghast, Houston TX (talk) 23:59, 25 September 2010 (UTC)...

this is precisely the point of Roth’s paper which Fioravante Patrone en cites (see the quote on the first page of Roth’s paper). There is the matter of...

under one section, but that is a matter of taste. (Notice that, say, Roth's theorem is included in Hindry and Silverman's Diophantine Geometry.) At any...

Intense use is made in proves of transcendence of exponentials. In Roth's theorem's proof. The claim against Cramer's rule in many books is just done to...

Arithmetic mean, arithmetic progression, Dirichlet's theorem on arithmetic progressions, Roth's theorem on arithmetic progressions, arithmetic–geometric mean...

I've read that Fermat's Last Theorem is an easy consequence of this conjecture. If I get the chance I'll see if I can find out why and add it, unless...

terms equal to 1 (but it still compresses many others). There may be some theorem about this, but this is visible in the expansions of pi and e (and may...