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

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- Jul 30th 2010, 02:20 PMDemandeurIntegers 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 PMVlasev
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 PMDemandeur
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?

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 PMVlasev
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 AMSoroban
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 AMundefined
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 AMAlso sprach Zarathustra
- Jul 31st 2010, 09:29 AMDemandeur
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 AMundefined
- Jul 31st 2010, 09:38 AMAlso sprach Zarathustra
I didn't see that... :( (Sorry Vlasev!)

- Jul 31st 2010, 09:39 AMDemandeur
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 AMAlso 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...] - Jul 31st 2010, 10:03 AMDemandeur
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 AMDemandeur
Found this in MathWorld:

Quote:

Originally 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. - Jul 31st 2010, 10:32 AMAlso sprach Zarathustra