prove it if you can

[vaishakh]

This is an interesting question but not that easy to be proved. The statement is that an irrational number again with an irrational power can be rational. It is very easy to point out at the numbers in natural logarithm table developed to the base 'e' which is irrational and all the values except log1 are irrational. But, can you prove this using simple algebra rather than an example. What I need is a concrete proof, which one can write even without the logarithm table. The proof should be also convincing. Statements such as 'the summation of this series is irrational since each term adds a new digit to the actual value' will not at all be considered since they have no mathematical certainty.

[José Carlos Santos]

Let _s_ be the square root of 2. I don't know whether s^s is rational or
not. If it is, then you're done. Otherwise, consider

(s^s)^s = s^{s^2} = s^2 = 2.

So, if s^s is irrational, you have the example that you want.

Best regards,

Jose Carlos Santos

[Michael Jørgensen]

"vaishakh" <vendva...@yaho.co.in> wrote in message
news:8891659.11309307279...@nitrogen.mathforum.org...

Assume sqrt(2) is irrational.

Let a = sqrt(2) ^ sqrt(2)

Let b = a ^ sqrt(2)

Now, upon simplification we find that b = sqrt(2) ^ ( sqrt(2) * sqrt(2) ) =
sqrt(2) ^ 2 = 2. Therefore, b is rational.

Now we have two cases, depending on whether a is rational or irrational.

- a is rational. Then you immediately have what you asked for: a is a
rational number that is an irrational number raised to an irrational power.

- a is irrational. Then you have instead that *b* is rational number that is
an irrational number raised to an irrational power.

QED

-Michael.

[roh5337]

"vaishakh" <vendva...@yaho.co.in> wrote in message
news:8891659.11309307279...@nitrogen.mathforum.org...

Three proofs are presented on:
http://en.wikipedia.org/w/index.php?title=Nonconstructive_proof&oldid=24852721

[Joseph Fagan]

"José Carlos Santos" <jcsa...@fc.up.pt> wrote in message
news:3sro10F...@individual.net...

Beautiful

[Proginoskes]

José Carlos Santos wrote:
> [...]

> Let _s_ be the square root of 2. I don't know whether s^s is rational or
> not. If it is, then you're done. Otherwise, consider
>
> (s^s)^s = s^{s^2} = s^2 = 2.
>
> So, if s^s is irrational, you have the example that you want.

For the record, s^s is not only irrational, but transcendental. This is
a consequence of Lindemann's Theorem.

--- Christopher Heckman

[Robert Low]

How so? All I know about Lindemann's Theorem is what I
googled for twenty seconds or so ago, and I can't see
how it implies that. The theorem tells us that if
a is algebraic, then exp(a) is transcendental. But
s^s is exp(ln(s)s), and surely ln(s)s isn't algebraic.

[Gerry Myerson]

In article <3sspfmF...@individual.net>,
Robert Low <mtx...@coventry.ac.uk> wrote:

I think it needs Gelfond-Schneider, which came after Lindemann.
If a is algebraic and not 0 or 1, and b is algebraic and irrational,
then a^b is transcendental.

http://mathworld.wolfram.com/GelfondsTheorem.html

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

[Proginoskes]

Yes. I got those two mixed up.

--- Christopher Heckman

[Robert Low]

Gerry Myerson wrote:
> Robert Low <mtx...@coventry.ac.uk> wrote:
>>Proginoskes wrote:

[s is sqrt(2)]

>>>For the record, s^s is not only irrational, but transcendental. This is
>>>a consequence of Lindemann's Theorem.

>> ...

> I think it needs Gelfond-Schneider, which came after Lindemann.
> If a is algebraic and not 0 or 1, and b is algebraic and irrational,
> then a^b is transcendental.
> http://mathworld.wolfram.com/GelfondsTheorem.html

Ah, thanks. I could have sworn that when I first saw that
proof about s^s and (s^s)^s I was told that whether s^s was
irrational or not was open, and that was much more recently
than 1934. I guess that it's just not as well-known as
it should be.