Assuming Schanuel's conjecture, one can show that e^e^e^79 is transcendental. Warut On Sat, Apr 28, 2012 at 5:07 AM, Allan Wechsler <acwacw@gmail.com> wrote:
Skewes' number, e^e^e^79, is a notoriously flamboyant upper bound in analytic number theory. As an upper bound, it has since been replaced by much more moderate results.
A <a href=" http://garden.irmacs.sfu.ca/?q=op/is_skewes_number_e_e_e_79_an_integer">conjecture</a> on the Open Problem Garden is that Skewes' number is not integral.
I am boggled. Of course we know that e^79 is nonintegral because e is not algebraic. (Is there an easier proof?)
But do we even know that e^e^79 is nonintegral? Do we know that e^e^10 is nonintegral? _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun