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badreferences

  • 3 years ago

Looking for reference: the proof where \(2^{\sqrt2}\) was demonstrated to be ir/rational.

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  1. asnaseer
    • 3 years ago
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    this is the best reference I could find: http://answers.yahoo.com/question/index?qid=20100909093754AA7A3o3 hope its not a "bad" reference @badreferences :D

  2. badreferences
    • 3 years ago
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    My name will never be left alone. Also, IIRC, the proof was published sometime before Hilbert's death. An old paper that's probably open access.

  3. theEric
    • 3 years ago
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    http://www.homeschoolmath.net/teaching/proof_square_root_2_irrational.php

  4. Herp_Derp
    • 3 years ago
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    Just wondering, what math are you studying right now? Number theory?

  5. zzr0ck3r
    • 3 years ago
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    seems like the contradiction method can be used just like the proof of sqrt(2) being irrational.

  6. Herp_Derp
    • 3 years ago
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    @zzr0ck3r Are you sure?

  7. zzr0ck3r
    • 3 years ago
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    no, lol

  8. zzr0ck3r
    • 3 years ago
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    Gelfond/Schneider thrm

  9. theEric
    • 3 years ago
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    The link I posted above, and here http://www.homeschoolmath.net/teaching/proof_square_root_2_irrational.php uses proof by contradiction. It's not incredibly tough, either! It just takes a little while to understand. I don't know of the Gelfond-Schneider theorem, though.

  10. theEric
    • 3 years ago
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    "Let's suppose √2 were a rational number. Then we can write it √2 = a/b where a, b are whole numbers, b not zero. We additionally make it so that this a/b is simplified to the lowest terms, since that can obviously be done with any fraction. It follows that 2 = a2/b2, or a2 = 2 * b2. So the square of a is an even number since it is two times something. From this we can know that a itself is also an even number. Why? Because it can't be odd; if a itself was odd, then a * a would be odd too. Odd number times odd number is always odd. Check if you don't believe that! Okay, if a itself is an even number, then a is 2 times some other whole number, or a = 2k where k is this other number. We don't need to know exactly what k is; it won't matter. Soon is coming the contradiction: If we substitute a = 2k into the original equation 2 = a2/b2, this is what we get: 2 = (2k)2/b2 2 = 4k2/b2 2*b2 = 4k2 b2 = 2k2. This means b2 is even, from which follows again that b itself is an even number!!! WHY is that a contradiction? Because we started the whole process saying that a/b is simplified to the lowest terms, and now it turns out that a and b would both be even. So √2 cannot be rational." - http://www.homeschoolmath.net/teaching/proof_square_root_2_irrational.php

  11. badreferences
    • 3 years ago
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    No, no, no, no! Not \(\sqrt2\). It's \(2^\sqrt2\). The former I could pick up in a textbook. The latter's proof is hidden in some famous publication somewhere.

  12. badreferences
    • 3 years ago
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    Sorry you went through the trouble of typing that up, @theEric .

  13. badreferences
    • 3 years ago
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    Ah, whatever.

  14. zzr0ck3r
    • 3 years ago
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    lol

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