Vtyjkyuk

I found this on the lecture slides of my Discrete Mathematics module today. I think they quote the theorems mostly from the Susanna S.Epp Discrete Mathematics with Applications 4th edition.

Here's the proof:

1. We know from Theorem 4.7.1(Epp) that $sqrt2$ is irrational.


2. Consider $sqrt2^sqrt2$ : It is either rational or irrational.


3. 
   

   Case 1: It is rational:

   
   
   

   3.1 Let $p=q=sqrt2$ and we are done.

   
   


4. 
   

   Case 2: It is irrational:

   
   
   

   4.1 Then let p=$sqrt2^sqrt2$, and $q =sqrt2$

   
   
   

   4.2 p is irrational(by assumption), so is q (by Theorem 4.7.1(Epp))

   
   
   

   4.3 Consider $p^q = (sqrt2^sqrt2)^sqrt2$

   
   
   

   4.4 $=(sqrt2)^sqrt2 timessqrt2$, by the power law

   
   
   

   4.5 $=(sqrt2)^2=2$, by algebra

   
   
   

   4.6 Clearly $2$ is rational

   
   
   
   5. In either case, we have found the required p and q.
   
   

From what I understand from the proof, it's a clever construction of a number and splitting up into cases to prove that the given number is a rational number (one by clear observation and the other by some sort of contradiction). While the proof seems valid, though I somehow am forced to be convinced that the contradiction works, I wondered to myself if $sqrt2^sqrt2$ is actually rational since they constructed it.

My question would be if the example is indeed irrational, how is it possible an untrue constructed example could be used to verify this proof? If it's indeed rational, can someone tell me the $a/b$ representation of this number?

PS: Sorry for the formatting errors, I followed the MathJax syntax to the best of my abilities but I'm not sure how to align the sub-points well. Please help me edit the post, thanks.

The irrationality of $sqrt 2^sqrt 2$ (in fact, its transcendence) follows immediately from the Gelfond Schneider Theorem. This was the issue that motivated Hilbert's $7^th$ Problem.

The beauty of the argument here is its simplicity. It's a perfectly valid, non-constructive, argument. The claim is demonstrated though no explicit example is produced.

Here is a simple, constructive, way to settle the original issue: $$sqrt 3^log_34=2$$

Of course, $sqrt 3$ is irrational.

To see that $log_3 4$ is irrational work by contradiction. $$log_3 4=frac abimplies 4=3^frac abimplies 4^b=3^a$$ But if $a,bin mathbb N$ then this contradicts unique factorization.

The proof is non-constructive. We don't know whether $sqrt2^sqrt2$ is rational or irrational. The point of the proof is that in either case there will be two irrational numbers $p$ and $q$ such that $p^q$ is rational. In one case $p=sqrt2$ and in the other case $p=sqrt2^sqrt2$.