# Integers in a logarithmic relation

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• Jul 30th 2010, 02:20 PM
Demandeur
Integers in a logarithmic relation
Do there exist $\displaystyle a,b\in\mathbb{Z}^+$ such that $\displaystyle \log(\frac{a}{b}) = \frac{1}{\sqrt{2}}\log{2}$?
• Jul 30th 2010, 02:32 PM
Vlasev
If you do a bit of manipulation $\displaystyle log(\frac{a}{b}) = \frac{1}{\sqrt{2}}log(2) = log(2^{\frac{1}{\sqrt{2}}})$

From this you get that you need positive integers a and b, such that

$\displaystyle \frac{a}{b} = 2^{\frac{1}{\sqrt{2}}}$

In essence you are asked if $\displaystyle 2^{\frac{1}{\sqrt{2}}}$ 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.
• Jul 30th 2010, 03:14 PM
Demandeur
Quote:

Originally Posted by Vlasev
$\displaystyle \frac{a}{b} = 2^{\frac{1}{\sqrt{2}}}$ In essence you are asked if $\displaystyle 2^{\frac{1}{\sqrt{2}}}$ is rational.

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.

Quote:

What theorems about properties of numbers do you have at your disposal?
What are the theorems? I know just the basic stuff - no more nor less advanced than what one usually
needs to show that $\displaystyle \sqrt{2}\notin\mathbb{Q}$. 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.
• Jul 30th 2010, 11:24 PM
Vlasev
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 $\displaystyle a^b$. 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 $\displaystyle x^y$ a transcendental number.

Now, x = 2 is algebraic, since it is obviously the root of a polynomial with rational coefficients. Since y = $\displaystyle \sqrt{2}$ is irrational,then $\displaystyle \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$ is also irrational. Then by the G-S theorem, $\displaystyle 2^{\frac{1}{\sqrt{2}}}$ 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.
• Jul 31st 2010, 07:23 AM
Soroban
Hello, Demandeur!

This problem has an even more interesting form.
[But no, I haven't solved it yet.]

Quote:

$\displaystyle \text{Do there exist }a,b\in\mathbb{Z}^+\text{ such that: }\;\log\left(\dfrac{a}{b}\right) \;=\; \frac{1}{\sqrt{2}}\log{2}$?

As vlasev pointed out: .$\displaystyle \log\left(\dfrac{a}{b}\right) \:=\:\log\left(2^{\frac{1}{\sqrt{2}}}\right)$

So we have: .$\displaystyle \dfrac{a}{b} \;=\;2^{\frac{1}{\sqrt{2}} }$

Note that: .$\displaystyle 2^{\frac{1}{\sqrt{2}}} \;=\;2^{\frac{\sqrt{2}}{2}} \;=\;2^{\left(\frac{1}{2}\cdot\sqrt{2}\right)} \;=\;\left(2^{\frac{1}{2}\right)^{\sqrt{2}} \;=\;\left(\sqrt{2}\right)^{\sqrt{2}}$

The equation becomes: .$\displaystyle \left(\sqrt{2}\right)^{\sqrt{2}} \;=\;\dfrac{a}{b}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Somewhere, there is a theorem about: .$\displaystyle \text{(irrational)}^{\text{irrational}}$

But I can't recall it nor can I find the reference.
Does that sound familiar to anyone?
• Jul 31st 2010, 07:54 AM
undefined
Quote:

Originally Posted by Soroban
Somewhere, there is a theorem about: .$\displaystyle \text{(irrational)}^{\text{irrational}}$

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 $\displaystyle \pi^e$ and $\displaystyle \pi^{\sqrt{2}}$ are irrational.

Irrational number - Wikipedia, the free encyclopedia (subheading: Open questions)

Perhaps the theorem you have in mind deals with algebraic irrationalities?
• Jul 31st 2010, 09:19 AM
Also sprach Zarathustra
• Jul 31st 2010, 09:29 AM
Demandeur
Quote:

Originally Posted by Soroban

Somewhere, there is a theorem about: $\displaystyle \text{(irrational)}^{\text{irrational}}$
But I can't recall it nor can I find the reference.
Does that sound familiar to anyone?

If $\displaystyle m,n\notin\mathbb{Q}$, it's possible that $\displaystyle m^n\in\mathbb{Q}$. (see here for proof - at the beginning of the page).
It's interesting the proof actually uses the concerned constant, without determining whether or not it's rational.
• Jul 31st 2010, 09:30 AM
undefined
Quote:

Originally Posted by Also sprach Zarathustra

I seem to be suffering from dementia... I read post #4 and almost immediately forgot having read it! (Speechless)
• Jul 31st 2010, 09:38 AM
Also sprach Zarathustra
I didn't see that... :( (Sorry Vlasev!)
• Jul 31st 2010, 09:39 AM
Demandeur
Quote:

Originally Posted by Vlasev
...

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.
• Jul 31st 2010, 09:45 AM
Also sprach Zarathustra
Quote:

Originally Posted by Demandeur
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...]
• Jul 31st 2010, 10:03 AM
Demandeur
Quote:

Originally Posted by Also sprach Zarathustra
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.
• Jul 31st 2010, 10:25 AM
Demandeur
Found this in MathWorld:
Quote:

Originally Posted by MathWorld
Both the Gelfand-Schneider constant $\displaystyle \sqrt{2}^{\sqrt{2}}$ and Gelfond's constant $\displaystyle e^{\pi}$ were singled out in the 7th of Hilbert's problems as examples of numbers whose transcendence was an open problem (Wells 1986, p. 45).

So indeed my quest to find an elementary proof was in vain! It's probably what Nietzsche was trying to
tell me as well - now that I realise it also says the same in the Wikipedia article whose link he's posted.
• Jul 31st 2010, 10:32 AM
Also sprach Zarathustra
Quote:

Originally Posted by Demandeur
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.

You misunderstood me, but I don't blame you, my English is very bad, sorry for that...

By the way, who it is in your picture avatar...?
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