e^e

Pafnuty Tschebyscheff

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Mar 13, 2010, 4:59:47 PM3/13/10

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Is e^e rational?
Can anyone give me any references to this problem?
Thanks in advance.

bill

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Mar 13, 2010, 7:20:48 PM3/13/10

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e^e = 1 + e + e^2/2! + e^3/3! ad infinitum.

Does this help?

regards, Bill J

Tony

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Mar 14, 2010, 8:54:35 AM3/14/10

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No.

Bart Goddard

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Mar 14, 2010, 9:17:37 AM3/14/10

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Tony <temp...@freemail.hu> wrote in news:fa015042-af9c-4f4d-9271-
e05f5a...@t20g2000yqe.googlegroups.com:

An old "West Coast Number Theory" problem, (before 1980,
I think) asks for the existence of a "Humdrum Number"
which was defined as an integer N such that e^(e^N) is
an integer. As far as I know, this remains unresolved.

B.

--
Cheerfully resisting change since 1959.

Pafnuty Tschebyscheff

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Mar 14, 2010, 1:23:35 PM3/14/10

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On Sun, 14 Mar 2010, Bart Goddard wrote:

> An old "West Coast Number Theory" problem, (before 1980,
> I think) asks for the existence of a "Humdrum Number"
> which was defined as an integer N such that e^(e^N) is
> an integer. As far as I know, this remains unresolved.
>

Can you give me any references to this problem?
I was not able to find anything on "Humdrum Number"...

Transfer Principle

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Mar 14, 2010, 6:34:39 PM3/14/10

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> No.

Today's Pi Day. So what's this thread about that _other_
transcendental number doing here?

Even though numbers like e^e and pi^pi are most likely
transcendental, I'd personally find it interesting if
either were rational (or even algebraic). That's because
I'm interested in the operation of tetration (since e^e
and pi^pi are just e^^2 and pi^^2). If e^e (respectively
pi^pi) turned out to be rational, it would mean that e
(respectively pi) would be the super square root (i.e.,
root of tetration) of a rational number.

And of course, if e^e and pi^pi are transcendental, then
we can keep trying with e^e^e and pi^pi^pi (i.e., e^^3
and pi^^3) for super cube roots, and so on.

Phil Carmody

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Mar 24, 2010, 7:01:21 PM3/24/10

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Bart Goddard <godd...@netscape.net> writes:

> Tony <temp...@freemail.hu> wrote:
>> On Mar 14, 1:20 am, bill <b92...@yahoo.com> wrote:
>>> On Mar 13, 1:59 pm, Pafnuty Tschebyscheff <th...@SDF-EU.ORG> wrote:
>>>
>>> > Is e^e rational?
>>> > Can anyone give me any references to this problem?
>>> > Thanks in advance.
>>>
>>> e^e = 1 + e + e^2/2! + e^3/3! ad infinitum.
>>>
>>> Does this help?
>>

>> No.
>
> An old "West Coast Number Theory" problem, (before 1980,
> I think) asks for the existence of a "Humdrum Number"
> which was defined as an integer N such that e^(e^N) is
> an integer. As far as I know, this remains unresolved.

E r \in |Q : e^e^r \in |Q ?

Phil
--
I find the easiest thing to do is to k/f myself and just troll away
-- David Melville on r.a.s.f1

Gerry Myerson

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Mar 24, 2010, 9:43:23 PM3/24/10

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In article <87y6hht...@kilospaz.fatphil.org>,
Phil Carmody <thefatphi...@yahoo.co.uk> wrote:

> Bart Goddard <godd...@netscape.net> writes:
> > Tony <temp...@freemail.hu> wrote:
> >> On Mar 14, 1:20 am, bill <b92...@yahoo.com> wrote:
> >>> On Mar 13, 1:59 pm, Pafnuty Tschebyscheff <th...@SDF-EU.ORG> wrote:
> >>>
> >>> > Is e^e rational?
> >>> > Can anyone give me any references to this problem?
> >>> > Thanks in advance.
> >>>
> >>> e^e = 1 + e + e^2/2! + e^3/3! ad infinitum.
> >>>
> >>> Does this help?
> >>
> >> No.
> >
> > An old "West Coast Number Theory" problem, (before 1980,
> > I think) asks for the existence of a "Humdrum Number"
> > which was defined as an integer N such that e^(e^N) is
> > an integer. As far as I know, this remains unresolved.
>
> E r \in |Q : e^e^r \in |Q ?

My hunch is it's no easier (nor harder) for rationals than
for integers.

--
Gerry Myerson (ge...@maths.mq.edi.ai) (i -> u for email)