1. I learned it in a class a while back (someone else, not me), but it involved showing it with sqrt(2). There's probably a better way though. That way seems ham handed.

2. What d'you mean? The proof is two lines. It's logically sort of sneaky but it's incredibly clean.

3. Originally Posted by Spark
Two line proof. Very, very sneaky proof too. The specific problem, I should say, is proving that there exists two irrationals a, b such that a^b is rational.
Not looking at e^pi*i are you?

4. i*pi isn't even real let alone rational

The hint is in the wording: Prove there exists. Not find it, just prove it exists. And as SS said, the simplest version of the proof involves sqrt(2).

5. Originally Posted by Spark
What d'you mean? The proof is two lines. It's logically sort of sneaky but it's incredibly clean.
Meh.

sqrt(2)^sqrt(2) = irrational. sqrt(2)^sqrt(2)^sqrt(2) = rational. So a = sqrt(2)^sqrt(2), b = sqrt(2).

*Saw it in class, just verified it online. I don't like it.

6. Yeah that's half the proof, because it relies on the knowledge that sqrt(2)^sqrt(2) is irrational which is actually quite tricky to prove. But that's no problem because, well, if you assume sqrt(2)^sqrt(2) irrational, then...

7. Well yea, but it's already been proven so just [cite] that ****. You don't have to do it .

8. Fancy explaining in words that would make sense to my Year 8s?

9. Well, the one I mentions relies on the fact that sqrt(2) is irrational. It has also been proven that sqrt(2)^sqrt(2) is also irrational.

However, sqrt(2)^sqrt(2)^sqrt^(2) simplifies to 2. So if a = sqrt(2)^sqrt(2), and b= sqrt(2), you have the scenario where if a and b are irrational, a^b can be rational. But as I said before, it's not a great proof. As spark said, it requires knowing sqrt(2)^sqrt(2) is irrational - which is not obvious and requires its own proof. As a student, you would just [cite] that and be done, but it's not really an elegant method.

I can't think of another way off the top of my head. Spark implied there's a more elegant way, so he might be able to show a better solution.

10. Wait. Duh.

I remembered from class. It doesn't matter.

You can prove it without proving sqrt(2)^sqrt(2).

You know sqrt(2) is irrational.
You know 2 is rational.

Let x be sqrt(2)^sqrt(2).

If x is rational, then the statement is true because a and b can both be sqrt(2).
If x is irrational, then the statement is true because a can be sqrt(2)^sqrt(2) and b can be sqrt(2) - which would equal 2.

Either way, the statement of a^b with a and b being irrational and the result being rational is true.

Non constructive proof FTL.

But it's nice. This is what Spark probably had in mind.

11. ^ That makes sense.

I can't, however, remember for the life of me how to compute irrational powers or even begin to get my head around why sqrt(2)*sqrt(2)*sqrt(2) = 2... Worth trying?

12. Originally Posted by silentstriker
Wait. Duh.

I remembered from class. It doesn't matter.

You can prove it without proving sqrt(2)^sqrt(2).

You know sqrt(2) is irrational.
You know 2 is rational.

Let x be sqrt(2)^sqrt(2).

If x is rational, then the statement is true because a and b can both be sqrt(2).
If x is irrational, then the statement is true because a can be sqrt(2)^sqrt(2) and b can be sqrt(2) - which would equal 2.

Either way, the statement of a^b with a and b being irrational and the result being rational is true.

Non constructive proof FTL.

But it's nice. This is what Spark probably had in mind.
(y)

Originally Posted by Neil Pickup
^ That makes sense.

I can't, however, remember for the life of me how to compute irrational powers or even begin to get my head around why sqrt(2)*sqrt(2)*sqrt(2) = 2... Worth trying?
Probably best to think of it as (sqrt(2)^sqrt(2))^sqrt(2) = 2.

13. Originally Posted by Neil Pickup
^ That makes sense.

I can't, however, remember for the life of me how to compute irrational powers or even begin to get my head around why sqrt(2)*sqrt(2)*sqrt(2) = 2... Worth trying?
Just rules of exponents. (a^b)^c = a^(b*c)

sqrt(2)*sqrt(2) = 2. So sqrt(2)^2 = 2.

14. It's 2am so I might be missing something, but I get ln(2/5) evaluating the attached problem but the correct answer apparently is (1/3)ln(5/2). What the hell am I missing?

EDIT: oh god, ignore me.

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