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MathGroup Archive 2005

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Re: Re: Re: Types in Mathematica thread


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


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