Do there exist such that ?
If you do a bit of manipulation [LaTeX ERROR: Compile failed]
From this you get that you need positive integers a and b, such that
[LaTeX ERROR: Compile failed]
In essence you are asked if [LaTeX ERROR: Compile failed] is rational.
What theorems about properties of numbers do you have at your disposal? Given certain theorems this shouldn't be a difficult thing to answer, although it may be hard if you have to actually find and prove your answer.
It's almost as if you read my thought process. This was indeed what I was trying to determine - but having
failed to make any progress on it, I figured that it was equivalent to finding the integers a and b satisfying
the logarithmic relation in the first post, under the assumption that it might be easier to deal with it that
way.
What are the theorems? I know just the basic stuff - no more nor less advanced than what one usuallyWhat theorems about properties of numbers do you have at your disposal?
needs to show that . But if you have a 'non-elementary' answer (or a theorem that holds the key),
feel free to provide it - I will make the effort to understand (or apply) it.
If you have covered transcendental numbers before, you will know that if a number is transcendental, then it is irrational. Now, there is a theorem, the Gelfond-Schneider theorem, which characterizes numbers of the form [LaTeX ERROR: Compile failed] . It states that
If a and b are algebraic numbers with x ≠ 0, and if y is not a rational number, then any value of [LaTeX ERROR: Compile failed] a transcendental number.
Now, x = 2 is algebraic, since it is obviously the root of a polynomial with rational coefficients. Since y = [LaTeX ERROR: Compile failed] is irrational,then [LaTeX ERROR: Compile failed] is also irrational. Then by the G-S theorem, [LaTeX ERROR: Compile failed] must be transcendental. Hence it is irrational.
Now, unless this was the theorem you are supposed to use, there must be some other way to prove it, without invoking this theorem.
Hello, Demandeur!
This problem has an even more interesting form.
[But no, I haven't solved it yet.]
?
As vlasev pointed out: .
So we have: .
Note that: .
The equation becomes: .
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
Somewhere, there is a theorem about: .
But I can't recall it nor can I find the reference.
Does that sound familiar to anyone?
I don't know of that theorem, but according to Wikipedia, it is unknown whether and are irrational.
Irrational number - Wikipedia, the free encyclopedia (subheading: Open questions)
Perhaps the theorem you have in mind deals with algebraic irrationalities?
That's a very nice theorem, Vlasev, and a nice proof. It's really heavy theorem, though. I'm not doing this
for a class, so I won't need prove the theorem to use it (and I really doubt I would understand the proof).
It would be interesting if we could find a proof using contradiction or the other usual elementary techniques.
I don't think so... this theorem is a case of 7th Hilbert's problems - Wikipedia, the free encyclopedia,so there many mathematicians that thought about it... [but maybe all of them miss something...]
You misunderstood me, Nietzsche. I'm not trying to find a contradiction to this well established theorem (one
would have to be so vain to do that). I'm merely seeking an elementary way of proving that the square root
two raised as much is irrational. That is what I mean when I talk of the hope of deriving a contradiction and
elementary techniques, not Gelfond's theorem.
Found this in MathWorld:
So indeed my quest to find an elementary proof was in vain! It's probably what Nietzsche was trying toOriginally Posted by MathWorld
tell me as well - now that I realise it also says the same in the Wikipedia article whose link he's posted.