Re: Re: Re: Types in Mathematica thread

*To*: mathgroup at smc.vnet.net*Subject*: [mg62935] Re: [mg62927] Re: [mg62891] Re: Types in Mathematica thread*From*: Andrzej Kozlowski <akoz at mimuw.edu.pl>*Date*: Fri, 9 Dec 2005 05:10:14 -0500 (EST)*References*: <dmp9na$hi2$1@smc.vnet.net> <roadnYOk3NcFDw7eRVn-jg@speakeasy.net> <200512050837.DAA08425@smc.vnet.net> <200512051840.NAA21063@smc.vnet.net> <200512060503.AAA02736@smc.vnet.net> <dn3jsl$8s0$1@smc.vnet.net> <5_ydnSmM8KqB-gjenZ2dnUVZ_v6dnZ2d@speakeasy.net> <dn5npi$nef$1@smc.vnet.net> <200512080504.AAA11638@smc.vnet.net> <200512080836.DAA18262@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

On 8 Dec 2005, at 17:36, Andrzej Kozlowski wrote: > But if you really have an idea how to deduce that Pi is > transcendental from the fact that E is then perhaps you might wish to > prove that E+Pi is transcendental because somehow nobody has so far > been able to do it. However; it is amusing that it is trivial to prove that either Pi+E or Pi*E is transcendental. (This fact is mentioned on MathWorld, but without a proof, and as I just thought of how to do it I could not resist posting it, although I am sure many people have seen it). Consider the polynomial equation eq = (x - Pi)*(x - E) == 0; It's roots are Pi and E. If we expand it we get (Collect[#1, x] & ) /@ ExpandAll[eq] x^2 + (-E - Pi)*x + E*Pi == 0 so if both E+P and E*P are algebraic so are E and P; a contradiction. So at least one of them is transcendental. This is of course unknown. So once I realised it was so easy to prove I tried: Assuming[Element[Pi+E,Algebraics],FullSimplify[Element [Pi*E,Algebraics]]] but the answer was disappointing :-( Andrzej Kozlowski

**References**:**Re: Types in Mathematica thread***From:*Kristen W Carlson <carlsonkw@gmail.com>

**Re: Re: Types in Mathematica thread***From:*Andrzej Kozlowski <akoz@mimuw.edu.pl>

**Re: Re: Re: Types in Mathematica thread***From:*Kristen W Carlson <carlsonkw@Gmail.com>

**Re: Types in Mathematica thread***From:*"Steven T. Hatton" <hattons@globalsymmetry.com>

**Re: Re: Types in Mathematica thread***From:*Andrzej Kozlowski <akoz@mimuw.edu.pl>