Question

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

Answer 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

Answer 2

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.

Answer 3

http://www.homeschoolmath.net/teaching/proof_square_root_2_irrational.php

Answer 4

Just wondering, what math are you studying right now? Number theory?

Answer 5

seems like the contradiction method can be used just like the proof of sqrt(2) being irrational.

Answer 6

@zzr0ck3r Are you sure?

Answer 7

no, lol

Answer 8

Gelfond/Schneider thrm

Answer 9

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.

Answer 10

"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

Answer 11

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.

Answer 12

Sorry you went through the trouble of typing that up, @theEric .

Answer 13

Ah, whatever.

Answer 14

lol