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Thread: The Maths Thread

  1. #586
    The Wheel is Forever silentstriker's Avatar
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    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.
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  2. #587
    Global Moderator Spark's Avatar
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    What d'you mean? The proof is two lines. It's logically sort of sneaky but it's incredibly clean.
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  4. #589
    Cricket Web Staff Member / Global Moderator Neil Pickup's Avatar
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    Quote Originally Posted by Spark View Post
    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?
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  5. #590
    Global Moderator Spark's Avatar
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    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).
    Last edited by Spark; 28-12-2012 at 06:25 AM.

  6. #591
    The Wheel is Forever silentstriker's Avatar
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    Quote Originally Posted by Spark View Post
    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.

  7. #592
    Global Moderator Spark's Avatar
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    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...

  8. #593
    The Wheel is Forever silentstriker's Avatar
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    Well yea, but it's already been proven so just [cite] that ****. You don't have to do it .

  9. #594
    Cricket Web Staff Member / Global Moderator Neil Pickup's Avatar
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    Fancy explaining in words that would make sense to my Year 8s?

  10. #595
    The Wheel is Forever silentstriker's Avatar
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    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.

  11. #596
    The Wheel is Forever silentstriker's Avatar
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    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.

  12. #597
    Cricket Web Staff Member / Global Moderator Neil Pickup's Avatar
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    ^ 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?

  13. #598
    Global Moderator Spark's Avatar
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    Quote Originally Posted by silentstriker View Post
    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)

    Quote Originally Posted by Neil Pickup View Post
    ^ 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.

  14. #599
    The Wheel is Forever silentstriker's Avatar
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    Quote Originally Posted by Neil Pickup View Post
    ^ 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.
    Last edited by silentstriker; 28-12-2012 at 10:30 PM.

  15. #600
    Global Moderator nightprowler10's Avatar
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    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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    Last edited by nightprowler10; 20-01-2013 at 01:46 AM.
    RIP Craigos

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