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

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

badreferences
 2 years ago
Best ResponseYou've already chosen the best response.0My 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.

theEric
 2 years ago
Best ResponseYou've already chosen the best response.0http://www.homeschoolmath.net/teaching/proof_square_root_2_irrational.php

Herp_Derp
 2 years ago
Best ResponseYou've already chosen the best response.0Just wondering, what math are you studying right now? Number theory?

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

Herp_Derp
 2 years ago
Best ResponseYou've already chosen the best response.0@zzr0ck3r Are you sure?

zzr0ck3r
 2 years ago
Best ResponseYou've already chosen the best response.0Gelfond/Schneider thrm

theEric
 2 years ago
Best ResponseYou've already chosen the best response.0The 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.

theEric
 2 years ago
Best 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

badreferences
 2 years ago
Best ResponseYou've already chosen the best response.0No, 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.

badreferences
 2 years ago
Best ResponseYou've already chosen the best response.0Sorry you went through the trouble of typing that up, @theEric .
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