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Math Help - Quickie #13

  1. #1
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    Quickie #13

    This is a classic (very old).
    If you've seen it, please don't post the answer.


    An irrational number raised to an irrational power is always irrational.

    Prove it or provide a counterexample.

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  2. #2
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    Seen it. But I did not know it was a classic.
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  3. #3
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    Quote Originally Posted by Soroban View Post
    This is a classic (very old).
    If you've seen it, please don't post the answer.


    An irrational number raised to an irrational power is always irrational.

    Prove it or provide a counterexample.

    I'll give several counter examples;

    For the most part,

    [sqrt(x)^(sqrt(x))]^(sqrt(x)), where x is even, with the exception of when x is a square.

    EX:

    [sqrt(2)^(sqrt(2))]^(sqrt(2))

    [sqrt(2)^(sqrt(2))] = 2^(sqrt(2)/2)

    [2^(sqrt(2)/2)]^(sqrt(2)) = 2
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    Quote Originally Posted by AfterShock View Post
    I'll give several counter examples;

    For the most part,

    [sqrt(x)^(sqrt(x))]^(sqrt(x)), where x is even, with the exception of when x is a square.

    EX:

    [sqrt(2)^(sqrt(2))]^(sqrt(2))

    [sqrt(2)^(sqrt(2))] = 2^(sqrt(2)/2)

    [2^(sqrt(2)/2)]^(sqrt(2)) = 2
    Interesting, but you are assuming that \sqrt{2}^{\sqrt{2}} is irrational, else this isn't a counterexample. I have little doubt that it is, but this isn't a good counterexample without proving this fact.

    -Dan
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    Quote Originally Posted by topsquark View Post
    Interesting, but you are assuming that \sqrt{2}^{\sqrt{2}} is irrational, else this isn't a counterexample. I have little doubt that it is, but this isn't a good counterexample without proving this fact.

    -Dan
    Funny thing is you do not need to. The assumption is that irrational raised to irrational is always irrational. Thus, he is using this assumption. And arrives at a contradiction.
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    Funny thing is you do not need to. The assumption is that irrational raised to irrational is always irrational. Thus, he is using this assumption. And arrives at a contradiction.
    Aaaaah! I get it now.

    -Dan
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  7. #7
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    The "classic" solution goes like this:


    Theorem: . (\text{irrational})^{\text{irrational}} can be rational.


    Proof

    \text{Consider }a \:=\:\sqrt{2}^{\sqrt{2}}


    There are only two possibilities:

    . . (1) a is rational.

    . . (2) a is irrational.


    If (1) a is rational, the theroem is verified.


    If (2) a is irrational, consider: . a^{\sqrt{2}}
    . . an irrational number raised to an irrational power.

    Then we have: . a^{\sqrt{2}} \:=\:\left(\sqrt{2}^{\sqrt{2}}\right)^{\sqrt{2}} \:=\:\left(\sqrt{2}\right)^2\:=\:2 . . . a rational number.


    Either way, an irrational raised to an irrational power can be rational.


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

    I've always found this proof amusing.

    Note that we still don't know if \sqrt{2}^{\sqrt{2}} is rational or irrational
    . . but it doesn't matter . . .

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    I found a different counterexample.

    e is irrational.

    \ln 2 is irrational (but do not know how to show it).

    Then,
    e^{\ln 2}=2
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