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badreferences
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Looking for reference: the proof where \(2^{\sqrt2}\) was demonstrated to be ir/rational.
 one year ago
 one year ago
badreferences Group Title
Looking for reference: the proof where \(2^{\sqrt2}\) was demonstrated to be ir/rational.
 one year ago
 one year ago

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asnaseer Group TitleBest ResponseYou've already chosen the best response.1
this is the best reference I could find: http://answers.yahoo.com/question/index?qid=20100909093754AA7A3o3 hope its not a "bad" reference @badreferences :D
 one year ago

badreferences Group TitleBest ResponseYou've already chosen the best response.0
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.
 one year ago

theEric Group TitleBest ResponseYou've already chosen the best response.0
http://www.homeschoolmath.net/teaching/proof_square_root_2_irrational.php
 one year ago

Herp_Derp Group TitleBest ResponseYou've already chosen the best response.0
Just wondering, what math are you studying right now? Number theory?
 one year ago

zzr0ck3r Group TitleBest ResponseYou've already chosen the best response.0
seems like the contradiction method can be used just like the proof of sqrt(2) being irrational.
 one year ago

Herp_Derp Group TitleBest ResponseYou've already chosen the best response.0
@zzr0ck3r Are you sure?
 one year ago

zzr0ck3r Group TitleBest ResponseYou've already chosen the best response.0
Gelfond/Schneider thrm
 one year ago

theEric Group TitleBest ResponseYou've already chosen the best response.0
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 GelfondSchneider theorem, though.
 one year ago

theEric Group TitleBest ResponseYou've already chosen the best response.0
"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
 one year ago

badreferences Group TitleBest ResponseYou've already chosen the best response.0
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.
 one year ago

badreferences Group TitleBest ResponseYou've already chosen the best response.0
Sorry you went through the trouble of typing that up, @theEric .
 one year ago

badreferences Group TitleBest ResponseYou've already chosen the best response.0
Ah, whatever.
 one year ago
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